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Digitized by the Internet Archive 
 
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THE ELEMENTS 
 
 OF 
 
 PLANE TRIGONOMETRY. 
 
THE ELEMENTS 
 
 OF 
 
 PLANE TRIGONOMETRY 
 
 
 BY 
 
 ^i--^ 
 
 R 
 
 LEVETT, 
 
 SECOND MASTER 
 
 AND 
 
 M.A. 
 
 C. 
 
 DAVISON, 
 
 M.A. 
 
 MATHEMATICAL MASTER 
 
 KING Edward's high school, Birmingham 
 
 MACMILLAN AND CO. 
 
 AND NEW YORK 
 
 1892 
 All 7'ighfs reserved 
 
LA 
 
 ^^. 
 
 /X. 
 
PREFACE. 
 
 In this treatise on the Elements of Plane Trigonometry 
 the subject is divided into three parts, dealing respectively 
 with arithmetical, real algebraical, and complex, quantity. 
 Such an arrangement appears to be a natural one, and 
 has the advantage of introducing the new names and 
 formulae that belong to the subject before the student 
 encounters the difficulty of the application of signs to 
 denote the sense and direction of lines. Part I. is 
 further simplified by the postponement of the treatment 
 of the circular measurement of angles. For many 
 practical purposes, e.g. in surveying and in the applica- 
 tion of trigonometry to elementary mechanics, the short 
 introduction to the subject comprised in Part I. will be 
 found useful. 
 
 The immediate substitution of the real for the tabular 
 logarithm of the trigonometrical ratios was recommended 
 by Professor De Morgan, and seems likely to be generally 
 adopted as the simpler method in teaching and the more 
 expeditious and more accurate one in working. 
 
 In Part II. the theory of the Circular and Hyperbolic 
 Functions, for real variables, is presented in some 
 detail, the analogy between the circular and hyperbolic 
 
 797949 
 
vi PREFACE. 
 
 functions is exhibited by similarity of method of treat- 
 ment, and an essay has been made to lead the student to 
 deal with infinite series with due caution. The chapter 
 on the Solution of Triangles and the applications to 
 Surveying has been made as practical as possible, in 
 order to add to the interest the student will find in this 
 part of the subject. The chapter on Factors is a de- 
 velopement of the consequences of the elegant theorem 
 given by Professor Adams in the Transactions of the 
 Carnhndge Philosophical Society. 
 
 No apology is needed for the introduction of geometrical 
 methods in Part TIL The methods are essentially general, 
 and the student who learns to think of complex numbers 
 as lines will gain a clearness of conception and a means 
 of testing results he can acquire in no other way. 
 
 Abundant examples for exercise have been collected 
 from University and other examination papers. The 
 student is advised to work through the shorter sets in 
 order to gain skill and readiness in using formulae, and 
 to select from the longer sets such examples as may 
 appear interesting or useful. The sets marked A and B 
 are alternative and may be used when the same portion 
 of the subject is read in class in consecutive terms. 
 Some of the longer sets of examples have been divided 
 into sections, arranged in order of difficulty. 
 
 The repetition of matter the student will have read 
 elsewhere has, so far as possible, been avoided. In 
 Chapters I.-X. a knowledge of Euclid and a few 
 well-known additional theorems in Geometry and of 
 elementary Algebra, including simple properties of 
 logarithms, is assumed ; Chapters XL-XXII. will probably 
 be read in connection with Geometrical Conies and the 
 
PREFACE. vii 
 
 more advanced Algebra given in such treatises as those 
 of Ur. Todhunter, Mr. Charles Smith, or Messrs. Hall 
 and Knight. 
 
 A text-book of Plane Trigonometry, intended for use 
 in schools, can, from its nature, contain little original 
 matter. The present work differs mainly from those 
 most generally read in the extent to which the treatment 
 adopted by Professor De Morgan has been followed. The 
 influence of De Morgan's writings will be seen throughout 
 the book, and, in particular, in the use of the negative 
 hypotenuse in defining the ratios (necessary if the proofs 
 of some of the fundamental theorems are to be general), 
 in the more definite meaning assigned to the notation for 
 inverse functions, the manner in which the addition 
 formulae are extended to any number of variables, the 
 geometrical treatment of the hyperbolic functions and of 
 complex numbers, and in the two-fold generalisation of a 
 logarithm to a given base. 
 
 Our acknowledgements are due to Professor Chrystal 
 for the aid we have derived from his masterly and ex- 
 haustive work on Algebraical Analysis. In the classifi- 
 cation and terminology of convergent and divergent 
 series, in the use made of the continuity of a series up 
 to the limit of convergence, and in the geometrical form 
 given to the proof of the Binomial Theorem we have 
 followed his treatment. The chapters dealing with 
 imaginary quantities and infinite series must, of 
 necessity, contain much that is due directly or indirectly 
 to Cauchy's Analyse Algehrique. Arts. 183, 239 are 
 derived from recent numbers of Mathesis ; Arts. 210, 211 
 from Schlomilch's Homdhnch der algehraischen Analysis. 
 The short chapters on the direct and inverse exponential 
 
viii PREFACE. 
 
 functions follow the lines laid down in the early para- 
 graphs of Professor Cay ley's Article on Functions in the 
 Encyclopaedia Britannica. 
 
 We are under great obligations to Mr. R. Tucker, of 
 University College School, for the very valuable assist- 
 ance he has given while the work has passed through 
 the press, and for his kindness in testing the results of 
 examples ; and our best thanks are also offered to Mr. 
 E. M. Langley, of the Bedford Modern School, and to our 
 past and present colleagues, Mr. C. H. P. Mayo, Mr. W. H. 
 Wagstaff and Mr. F. O. Lane for the kindly interest they 
 have taken in the book, and the pains they have given to 
 make it as free from error as possible. Our thanks are 
 due to the publishers for permission to make use of a 
 portion of the map of the Mer de Glace of Chamouni 
 given in the Life and Letters of Professor Forbes. 
 
 R LEVETT. 
 
 C. DAVISON. 
 King Edward's School, 
 
 Birmingham, January, 1892. 
 
CONTENTS. 
 
 PART L— ARITHMETICAL QUANTITY, 
 
 CHAPTER I. 
 Measurement of Angles. 
 
 ARTS. PAGE 
 
 1. Sexagesimal Measure, , . 1 
 
 Examples I. a, b, . . . . . . . . . 2 
 
 CHAPTER II. 
 
 Trigonometrical Ratios of Acdte Angles. 
 
 2-5. Definitions, 4 
 
 6-14. Relations between the Ratios, ...... 5 
 
 15-20. Ratios of 0°, 18°, 30°, 45°, 60°, 90°, ..... 10 
 
 Viv^ Voce Examples, • . . 13 
 
 21, 22. Variations in the Ratios, . . . . . . .14 
 
 23. Solution of Equations, 15 
 
 Examples II. A, B, .16 
 
 Examples III. , 19 
 
 CHAPTER III. 
 
 Trigonometrical Ratios of Compound Angles. 
 
 24-29. Ratios of a ±/3, 21 
 
 30-34. Ratios of 2a, 3a, etc., 25 
 
 35. Products expressed as the Sum or Difference of Ratios, . 29 
 Viva Voce Examples, .29 
 
 36. Sums or DiflFerences expressed as Products of Ratios, . . 30 
 Viva Voce Examples, . 33 
 
 ix 
 
X CONTENTS. 
 
 AKT8. PAOB 
 
 37. Transformations and Solution of Equations, .... 34 
 
 Examples IV. a, B, 35 
 
 Examples V., 42 
 
 CHAPTER IV. 
 
 Use of Mathematical Tables. 
 
 38-43. Logarithms of Numbers, 44 
 
 44-47. Logarithms of Trigonometrical Ratios, .... 48 
 Examples VI. a, b, 61 
 
 CHAPTER V. 
 
 )LUTioN OF Right- Angled Tbllngles and Practical 
 Applications. 
 
 48-57. Solution of Right-Angled Triangles, .... 54 
 
 Examples VII. a, b, 59 
 
 58-63. Heights and Distances, 60 
 
 64. Dip of the Horizon, 62 
 
 65. Dip of a Stratum, 63 
 
 Examples VIII. a, b, 63 
 
 Miscellaneous Examples I., 70 
 
 PART II.— REAL ALGEBRAICAL QUANTITY. 
 
 CHAPTER VL 
 
 CracuLAR Measure of Angles. 
 
 66-76. Definitions and Fundamental Propositions, ... 76 
 
 77-79. Change of Units of Angular Measurement, ... 83 
 
 Viva Voce Examples, 85 
 
 Examples IX. a, b, 86 
 
 Examples X., 88 
 
 CHAPTER VII. 
 
 General Definitigns of the Circular Functions. Formulae 
 INVOLVING One Variable Angle. 
 
 § 1. Definitions. 
 
 80-82. Sense of Lines and Angles, 91 
 
 83. Projection of Point and Line, 92 
 
 84. Extended Definitions of the Ciicular Functions, ... 93 
 
CONTENTS. xi 
 § 2. Fundamental Properties of the Circular Functions. 
 
 ARTS. PAGE 
 
 85. Circular Functions one-valued, 94 
 
 86, 87. Signs of the Functions, 95 
 
 88-90. Periodicity and Continuity of the Functions. Formulae, 97 
 
 Viv4 Voce Examples, 101 
 
 § 3. Reduction of Functions of nZ^O. 
 
 2 
 
 91,92. Even and Odd Functions, 102 
 
 93-97. Functions of n'^±d expressed as Functions of ^, . . 103 
 
 Viva Voce Examples, 107 
 
 98. Geometrical Proofs, . . . ^ 108 
 
 § 4. Inverse Functions. 
 
 99. Definitions of COS" ^ a, Cos" ^a, etc., 109 
 
 100-106. Expressions for all Angles which have a given Circular 
 
 Function, . .111 
 
 Viva Voce Examples, 116 
 
 § 5. Curves of the Circular Functions. 
 
 107, 108. Curves, 117 
 
 Examples XI. A, b, 122 
 
 CHAPTER VIII. 
 Circular Functions of Two or More Variable Angles. 
 
 109-112. General Proofs of the Addition Formulae, . . .124 
 113-115. cos a and sin a in terms of cos 2a or sin 2a, . . .129 
 
 Viv4 Voce Examples, 133 
 
 116. Ex. 1, 2. If J +5 + (7= 180°, then 2 sin ^= 411 cos :|, 
 
 Scos2^+2ncos^ = l, 133 
 
 Ex.4. tan"^a; + tan"^v = ?i7r + tan"^ ^^ , where 7i = or +1, 134 
 
 i-xy 
 
 Ex. 5. Euler's and Machin's Values of - , 
 
 4 
 
 Ex.8. Solution of a cos ^ + 6 sin ^ = c. 
 
 Examples XII. A, B, 
 
 117. Illustrative Examples, .... 
 
 Examples XIII. , . . . . 
 
 135 
 
 137 
 139 
 
 148 
 150 
 
xii CONTENTS. 
 
 CHAPTER IX. 
 
 Relations between the Elements of a Triangle. Solution 
 OF Triangles and Practical Api'lications. 
 
 § 1. Relations between the Elements of a Triangle. 
 
 ARTS. PA6> 
 
 118-121. a = 6co8(7+ccos5, 186 
 
 a/sm A = 6/sin B = r/sin 0. 
 a2 = 62 + c2-2&ccos^. 
 
 122-125. cos^=^«"(^^ sin ^=^£1^15, sm A =2Slbc, 167 
 
 126. tan^Z^=J^cot^-, 170 
 
 2 b + e 2 
 
 127. Illustrative Examples, 171 
 
 Ex. 4. Relation between the six lines joining four points in 
 
 a plane 172 
 
 Examples XIV. a, b, 172 
 
 Examples XV., 175 
 
 l__ 128-140. § 2. Solution of Triangles. 179 
 
 Examples XVI. A, b, 188 
 
 Examples XVII., 189 
 
 / § 3. Practical Applications. 
 
 141-147. Use of Chain and Theodolite, 191 
 
 148-151. Survey of Mer de Glace by Prof. Forbes, . . .199 
 
 152-154. Measurement of Heights, 204 
 
 155. Dip of a Stratum, 207 
 
 Examples XVni., 208 
 
 CHAPTER X. 
 
 Applications to the Geometry of Triangles, Polygons 
 AND Circles. 
 
 156. Enunciations of Geometrical Theorems, 
 
 157. *S' = ^bc sin A = sfsis - a)(8 - 6)(.s - c), 
 
 158. Q = ^^/{{s - a){s - b){s - c)(« -d)- abed cos-w} 
 / 159, 160. Area of Polygons, .... 
 ^ 161. JR = al2 8UiA=abcl4JS, .... 
 
 219 
 220 
 221 
 222 
 223 
 
CONTENTS, xiii 
 
 ARTS. PA.GE 
 
 162. r = >S'/s = 4i?sin^sin:|sin^, 224 
 
 163. ?-i = *9/(,9-a) = 4i?sin^cos:|cos^, 225 
 
 Ji "Ji z 
 
 164. SO"^ = R'^{1- 8 COS A cos Bco^C), .226 
 
 SP = B^~-2Rr. 
 
 OP ^2r^-4R^-cos A COS B COB C. 
 
 165. Feuerbach's Theorem, ........ 228 
 
 166. 167. Area of Circle, 228 
 
 168. Illustrative Examples, 229 
 
 Examples XIX., 231 
 
 Miscellaneous Examples II., . 247 
 
 CHAPTER XI. 
 
 Hyperbolic Functions. 
 
 169, 170. Definitions, 257 
 
 171-173. Elementary Relations between the Hyperbolic Functions, 260 
 
 174-176. Geometrical Properties of the Rectangular Hyperbola, . 262 
 
 177-179. Addition Formulae, 265 
 
 180. Gudermannian Function, 267 
 
 181. Curves of the Hyperbolic Fvmctions, 268 
 
 Examples XX., . . , 271 
 
 CHAPTER XII. 
 
 Inequalities and Limits. 
 § 1. Inequalities. 
 
 182-184. sin d>e- 6^6, cos ^ < 1 - 6^2 + 6^/24, tan ^ > ^ + d^/-i, . 274 
 
 185. sinh X > a: > tanh x% cosh a; > 1 + a;-/2, 277 
 
 186. Ex. 1. Jsin^>^> Jvers^, 277 
 
 Examples XXI. , 279 
 
 §2. limits. 
 
 187. 188. Fundamental Propositions, 281 
 
 189-191. i.4^"'^'^\ i.^. fcos-Y%tc., 283 
 
 a:=0\ X I n=a)\ ul 
 
 Examples XXII., .286 
 
xiv CONTENTS. 
 
 CHAPTER XIII. 
 
 Series. 
 arts. page 
 
 192-199. § 1. The Addition Formulae Extended. 288 
 
 § 2. Series of Powers of a Cosine or Sine. 
 200-204. 2cosn^ = (2cos^)"-'i(2co8^)"-2 + ^(n-3)i(2co8^)«-''-... 
 
 + (-l)'-*V-r-l)^_i(2co8^)"-2'-+...,etc., . 296 
 r 
 
 § 3. Summation of Series. 
 205, 206. '^ S cos(a + rj8) = co8(a + (7i-l)^}8in^/8in^, . . 306 
 
 r=0 *• ^-' ^ I ^ 
 
 207. Table of Diflference-forms, 309 
 
 208. Illustrative Examples, 310 
 
 209-214 § 4. Convergency and Continuity of Series. 312 
 
 § 5. Infinite Series for Cosines and Sines. 
 
 215-217. cosa; = S(-lf'J'*, cosh a: = 2^ 321 
 
 [2r I2r 
 
 sin r = S( - 1 Y^ — -, sinh x = S-^- - , . 
 ^ ' 2r+l \2r+l 
 
 '218-221. cosha; = ^(e* + e-*), sinha: = i(ef-e-'), . . . .326 
 Examples XXm., ........ 329 
 
 CHAPTER XIV. 
 Factors. 
 § 1. Fundamental Theorem on Trigonometrical Factors. 
 222-228. If v„ = 2 cos nx, 2 cosh nx, or a;" + — , then will 
 
 v„-2coana = ~ll^(vi-2cos(a + r—\\ . . .339 
 
 §2. Products for cos w^, =^, coshmt, ~^' . 
 wsin^ wsinhw 
 
 229-232. cos nd = cos"^ II 1 1+ tan ^/tan — ^ ir j ; etc. , . .348 
 
CONTENTS. XV 
 
 § 3. Infinite Products for the Cosines and Sines of x. 
 
 ARTS, PAGE 
 
 where l>i^^>l ^^^;etc., . . . .354 
 
 (r-- l)7r-^ 
 
 Examples XXIV., .363 
 
 CHAPTER XV. 
 Approximations. 
 237-241. § 1. Approximations and Errors, 369 
 
 Examples XXV., 376 
 
 242-249. §2. Theory of Proportional Parts, 380 
 
 Miscellaneous Examples III. , 388 
 
 PART III. -COMPLEX QUANTITY. 
 
 CHAPTER XVI. 
 
 Complex Numbers. 
 
 250, 251, Representation of Numbers by Straight Lines, . . 398 
 
 252-258. Addition and Multiplication of Complex Numbers, . 401 
 
 259. Conjugate Complex Numbers, 405 
 
 260. Powers and Roots, 405 
 
 261. Resolution of Complex Numbers. Demoivre's Theorem, . 409 
 
 262. 263. Some Applications of Complex Numbers, . . . 412 
 Examples XXVI., 419 
 
 CHAPTER XVII. 
 
 Series of Complex Numbers. 
 
 264-270. Convergency and Continuity of Series of Complex Numbers, 424 
 
 271,272. Summation of Series, 431 
 
 Examples XXVII., 434 
 
 CHAPTER XVIII. 
 
 273-277. The Binomial Theorem, 437 
 
 Examples XXVIII., 441 
 
xvi CONTENTS. 
 
 CHAPTER XIX. 
 
 The Exponential Series, 
 
 artb. page 
 
 278, 279. exp(a;)xexp(y) = exp(a: + y), 444 
 
 280. exp(arO = cos a: + » sin x, 445 
 
 281,282. exp(a; + yi) = exp(a;)(co8 y + e sin y), .... 446 
 
 Examples XXIX 447 
 
 CHAPTER XX. 
 
 Logarithms of Complex Numbers. 
 
 283-285. Log(r, ^) = log r + (^ + 27i7r)i, 449 
 
 286. When a: is real, ((a))* = exp{a:Loga), 451 
 
 287-290. Logarithmic Series, .452 
 
 291, 292. Gregory's Series. Numerical Value of tt, . . . 456 
 
 293. Some Trigonometrical Series, 457 
 
 Examples XXX., 463 
 
 CHAPTER XXI. 
 
 295-300. Complex Indices, 467 
 
 Examples XXXI. , 472 
 
 CHAPTER XXII. 
 
 CiRCCJLAR AND HYPERBOLIC FUNCTIONS OF COMPLEX NUMBERS. 
 
 301-305. Definition and Fundamental Properties of Circular and 
 
 Hyperbolic Functions of a Complex Variable, , . 474 
 
 306. Formulae of Interchange of Circular and Hyperbolic Functions, 479 
 
 307-309. Inverse Functions, 481 
 
 Examples XXXII., 486 
 
 Miscellaneous Examples IV. , , . . . . . 489 
 
 Mathematical Tables, ....... 498 
 
 Answers to Examples, 501 
 
PART I. 
 AKITHMETICAL QUANTITY. 
 
 ''And to fuch as delight in matter feruifable for the State J hope 
 this Introduction fhal not he umvelcome : meaning as I fee the fame 
 gratefully accepted, hereafter to impart the reft, leaving at this time 
 farther to tvade in the large Sea 0/ Algebra d: numbers Cofsical."— 
 Stratioticos. 
 
 CHAPTER I. 
 MEASUEEMENT OF ANGLES. 
 
 1. In selecting a unit of angular measurement for 
 practical purposes, it is necessary that the unit should 
 be : (1) constant, (2) easily obtained, and (3) of such a 
 magnitude that the angles most frequently measured may 
 be expressed by integers that are as a rule not very 
 great. 
 
 A right angle satisfies the first two of these conditions, 
 and, by sub-division, the third also. It has therefore 
 been adopted as the primary unit in the only system now 
 in use for the practical measurement of angles. 
 
 In this system, a right angle is divided into 90 equal 
 parts called degrees, a degree into 60 equal parts called 
 minutes, and a minute into sixty equal parts called 
 seconds. An angle containing 47 degrees, 39 minutes, 
 17 seconds is written 47° 39' 17". 
 
 (g A 
 
2 l^EA&UREMENT OF ANGLES. 
 
 Examples I a. 
 
 1. Reduce 57° 14' 46" to seconds, and 121475" to degrees 
 
 etc. 
 
 2. Express 69° 47' 42" and 58° 12' 18'' as decimals of a 
 
 right angle. 
 
 3. Find the number of degrees in the angle of a regular 
 
 octagon. 
 
 4. Find the number of sides in the regular polygon each 
 
 angle of which contains lo7|°. 
 
 5. The angles of a triangle are in arithmetical progression, 
 
 and the greatest angle is double of the least. Find 
 the number of degrees in each angle. 
 
 6. The angles of a triangle are such that the first contains 
 
 a certain number of degrees, the second 10 times as 
 many minutes, and the third 120 times as many 
 seconds. Find the angles. 
 
 7. The numerical measures of the angles of a quadrilateral 
 
 when referred to units containing 1°, 2°, 3°, 4° 
 respectively, are in arithmetical progression, and 
 the difference between the second and fourth is 
 equal to a right angle. Find the angles. 
 
 Examples I. b. 
 
 1. Reduce 35° 18' 47" to seconds, and 210501" to degrees 
 
 etc. 
 
 2. Express 8° 15' 81" and 85° 3' 2" as decimals of a right 
 
 angle. 
 
 3. Find the number of degrees in the angle of a regular 
 
 quindecagon. 
 
 4. Find the number of sides in the regular polygon each 
 
 angle of which contains 162°. 
 
MEASUREMENT OF ANGLES. 3 
 
 5. An isosceles triangle has each of the angles at the base 
 
 double of the third angle. Find the number of 
 degrees in each angle. 
 
 6. The angles of a quadrilateral are in arithmetical pro- 
 
 gression, and the difference between the greatest 
 and least is a right angle. Find the number of 
 degrees in each angle. 
 
 7. One regular polygon contains twice as many sides as 
 
 another, and an angle of the first is double an 
 angle of the second. Find the number of sides in 
 each polygon. 
 
CHAPTER II. 
 
 TRIGONOMETRICAL RATIOS OF AN ACUTE ANGLE. 
 
 2. Let POM be an acute angle, and let it be denoted 
 ^ by a. From P, any point in either 
 
 of the bounding lines, draw PM 
 perpendicular to the other. 
 
 The following are called the trig- 
 
 ^ j^^— ooiometrical ratios of the angle a : 
 
 Base Oif /hypotenuse OP is the cosine of a. 
 Perpendicular Pi//hypotenuse OP is the sine of a. 
 Perpendicular PM /h&se DM is the tangent of a. 
 H3^potenuse OP/base OM is the secant of a. 
 Hypotenuse OP/perpendicular PM is the cosecant of a. 
 Base Oif /perpendicular PM is the cotangent of a. 
 These ratios are written as follows : cos a, sin a, tan a, 
 sec a, cosec a and cot a. 
 
 3. Powers of trigonometrical ratios may be denoted in 
 the usual way, as (cos a)^ (tan a)^ etc ; but positive inte- 
 gral powers are generally written thus : cos^a, sec^a, etc. 
 
 4. Inverse Notation. — A notation similar in form is 
 used to denote angles having a given cosine, etc. The 
 angle whose cosine is J is written cos~^|, the angle whose 
 tangent is 3 is written tan-^S, etc. It should be borne 
 
TRIGONOMETRICAL RATIOS. ' 5 
 
 in mind that these expressions are entirely different from 
 the first negative powers of the ratios, which are written 
 in the usual way, as (cos a)"^ (tan a)"\ etc. 
 
 5. The cosine of an angle depends on the angle only. 
 Let POM be the given angle, and let it be denoted by 
 
 a. Let P, P' be any „ 
 
 points on one bounding r^ 
 
 line of the angle, and P" p^ 
 
 any point on the other. 
 
 From P, P\ F' draw 
 
 perpendiculars Pif , P'M\ q 
 
 P"M" to the other bounding line. 
 
 Now, the angle POM is common to the three triangles 
 POM, PVM\ P'VM") and the right angles PMO, 
 FM'O, P"M"0 are equal. 
 
 .-. the triangles POM, FOM\ F'OM" are similar 
 (Eucl. VI. 4). 
 
 OMIOP = OMjOP' = OM"IOP", 
 i.e. the cosine of the angle a is the same wherever the 
 point P be taken on either bounding line. 
 
 .•. the cosine of an angle depends on the angle only. 
 
 Cor. — Similarly, it may be shewn that the other trigon- 
 ometrical ratios of an angle depend on the angle only. 
 
 Relations between the Trigonometrical Ratios of an 
 Acute Angle. 
 
 6. To shew that the cosine and secant of an angle are 
 reci'procal, and likewise the sine and cosecant, and the 
 tangent and cotangent. 
 
 Let POM (see figure of art. 2) be the given angle, and 
 let it be denoted by a. Then 
 
TRIGONOMETRICAL RATIOS 
 
 OM OP , 
 cosa.seca = ^.^ = l, 
 
 cosa = l/seca and seca = l/cosa \ 
 
 PM OP 
 Also, sin a . cosec OL—TTp ' -pTf = 1, 
 
 sin a = 1/cosec a and cosec a = 1/sin a . . . . 
 
 . , , . PM OM ^ 
 
 And, tana.cota=^p^-p^=l, 
 
 tan a = 1/cot a and cot a — 1/tan a. 
 
 (A)' 
 
 7. If the sum of two angles is equal to a right angle, 
 each angle is called the complement of the other. 
 
 8. To shew that the cosine of an angle is equal to the 
 sine of its complement, the tangent of an angle to the 
 cotangent of its complement, and the secant of an angle 
 to the cosecant of its complement 
 
 Let POM (see fig. of art. 2) be the given angle, and 
 let it be denoted by a. The angle 0PM is the comple- 
 ment of a, since the angle OMP is a right angle. Now, 
 
 cos a = OMjOP = sin 0PM = sin(90° - a),\ 
 sin a=^PM/OP= cos 0PM = cos(90°-a), 
 t3ina = PM/0M= cot 0PM = cot(90°-a), 
 sec a = OP/OM= cosec 0PM = cosec(90° - a), 
 cosec a =OP/Pif= sec OPif= sec(90°-a), 
 cot a =Oilf/Pif= tan OPif= tan(90°-a),> 
 
 9. To prove that tan a = » and cot a = 
 
 '■ cos a sm a 
 
 Let POM (see fig. of art. 2) be the given angle a. Then 
 
 * Formulae that should be remembered are denoted by capital letters 
 at the end of the lines in which they occur. 
 
 •(B). 
 
OF AN ACUTE ANGLE. 
 
 sin a _PM OM_PM_ 
 cosa~ OP ' OP~OM~^''''' 
 cos a _0M PM_OM_ 
 sin a OP ' OP~PM~^'^^''' 
 
 ...(C). 
 
 10. To prove that cos^a4-sin2a = l, l-\-tein^a = sec^a, 
 and cot^a + 1 = cosec^a. 
 
 Let POM (see figure of art. 2) be the given angle a. 
 Since OMP is a right angle, 
 
 03P + PM' = 0P\ (End I. 47.) 
 
 Dividing both sides by OP^, we have 
 0M\ PM^_ 
 
 cos^a + sin^a = 1 (D) . 
 
 Again, dividing both sides of the first equation by Oilf ^ 
 we have 
 
 PM''_ OP^ 
 
 1 + tan^a = sec^a, (D). 
 
 Lastly, dividing both sides of the same equation by 
 PM\ we have 
 
 01/2 
 
 + 1 
 
 OP^ 
 
 PM^ ' PM^ 
 
 cot^a+l =cosec%, (D). 
 
 V 11. Example.— Prove that 
 
 2(cos'5a + sin^a) - 3(cos*a + sin^a) +1=0. 
 2(cos<'a + sin'^a) - 3(cos*a + sin''a) + 1 
 
 = 2(cos-a + sin2a)(cos^a ~ cos^a sin^a + sin^a) 
 
 — 3{(cos2a + sin^a)^ - 2 cos^a sin^a} + 1 
 = 2(cos*a - cos^a sin^a + sin-^a) - 3( 1 - 2 cos^a sin^a) + 1 
 = 2{(cos% + sin^a)^ - 2 cos^a sin^a - cos^a sin^a} -3(1-2 cos-a sin^a) + 1 
 = 2(1-3 cos^a sin^a) - 3(1 - 2 cos2a sin^a) + 1 
 = 0. 
 
8 
 
 TRIGONOMETRICAL RATIOS 
 
 12. The cosine of an angle being given, to express the 
 other trigonometrical ratios in terms of it. 
 
 Let a denote the angle whose cosine is given. 
 
 (1) Algebraical method. — By art. 10, we have 
 
 cos^a + sin^a^l, 
 
 sin a = V(l — cos^a). 
 Again, tan a = sin a/cos a (art. 9), 
 
 = ;^(1— cos2a)/cosa. 
 Also, sec a = 1/cos a, 
 
 cosec a = l//v/(l — cos^a), 
 cot a = cos al^{\ — cos^a). 
 
 (2) Geometrical method. — Let POM be the angle a, 
 and let its cosine be denoted by c. 
 Regarding the hypotenuse OP as 
 the unit of length, the base OM 
 contains c of these units, and there- 
 fore the perpendicular PM contains 
 ^(l-c2) units (Eucl. I. 47). 
 
 .-. sin a = PM/OP = V(l - o2)/l = ^(l _ cos^a), 
 
 tan a = PMJOM = s/{\- c^)/c = J(l - cos^aVcos a, 
 sec a = OP/OM = l/c = 1/cos a, 
 
 cosec a = OP/PM = 1/^(1 - c^) = 1/^^(1 - cos^a), 
 cot a = OM/PM= cj JO - c2) = cos aj J{1- cos^a). 
 Cor. — Similarly, the trigonometrical ratios may be ex- 
 pressed in terms of the sine, secant, or cosecant of the angle. 
 
 13. Example. — If cos a =f, find the other trigonometrical ratios 
 of a. . P Let POM be the angle a. Regarding the 
 
 hypotenuse OP as containing 5 units of 
 length, the base OM contains 4 such units, 
 and therefore the perpendicular PM con- 
 tains 3. (Eucl. I. 47.) 
 .'. sina = f, tana = f, seca = f, cosec a = |, 
 and cot a =|. 
 
OF AN ACUTE ANGLE. 9 
 
 14. The tangent of an angle being given, to express the 
 other trigonometrical ratios in terms of it 
 Let a denote the angle whose tangent is given. 
 
 (1) Algebraical method. — By art. 10, we have 
 
 sec^a = 1 + tan^a, 
 cos a = 1/sec a = 1/^(1 +tan2c(). 
 Also, sin a/cos a — tan a, 
 
 sin a = tan a . cos a 
 
 = tan aU{ 1 + tan'^a) ; 
 and sec a = ^/O- + tan^a), 
 
 cosec a = jj{ 1 + tan^a)/ tan a, 
 cot a — 1/tan a. 
 
 (2) Geometrical method. — Let POM be the angle a, 
 and let its tangent be denoted by t. 
 Regarding the base OM as the unit 
 of length, the perpendicular PM 
 contains t of these units, and there- 
 fore the hypotenuse OP contains 
 s/il+f^) units (Eucl. I 47). 
 
 .-. cos a=OM/ OP = l/J{l-ht^) = l/^(i+ta.n^a), 
 sin a = PM/OP = tlJ{l + f") = tan «/ V(l + tan^a), 
 sec a = OP 1 031 = J(l+ 1^)/! = J(l + tan^a), 
 cosec a = OPIPM= ^(1 + t^)lt = ^(1 + tan2a)/tan a, 
 cot a = 1/tan a. 
 
 Cor. — Similarly, if the cotangent of an angle be given, 
 the other trigonometrical ratios may be expressed in 
 terms of it. 
 
10 
 
 TRIGONOMETRICAL RATIOS 
 
 Trigonometrical Ratios of Particular Acute Angles. 
 15. To find the trigonometrical ratios of angles of G0° 
 and 30^ 
 
 Let ABC be an equilateral triangle. Draw BD per- 
 j3 pendicular to AG. Then BD bisects 
 
 both the angle ABC and the base AG 
 (Eucl. I 26). 
 
 Then, angle 5^D = 60°, 
 and angle ABD = ^0\ 
 Now, BD^=AB^-AD^ 
 
 ' =AB^-iAB^=iAB\ 
 
 BD^^.AB. 
 
 COS 60° = sin 30° = AB\AB=\AB\AB = \, 
 sin 60°= cos^O° = BDIAB=^^ABIAB = '4 
 
 tan 60°= cot30° = J5i)/^i)=^^5/J^j5 = V3, 
 
 sec60° = cosec30° = 2, 
 
 2 
 
 cosec 60° 
 cot 60° 
 
 sec 30° = 
 
 tan 30' 
 
 x/3' 
 1 
 
 73' 
 
 The values of these ratios may be remembered by 
 the aid of the accompanying figure. 
 
 16. To find the trigonometrical ratios of an angle of 4)5°. 
 B Let ABG be an isosceles triangle, right- 
 angled at (7, so that each of the angles A 
 and B is 45°. 
 Now, AB'' = AG''^BG'' = ^AG'' = 2BG\ 
 
 AG=BG=^.AB. 
 
 v2 
 
OF AN ACUTE ANGLE. 
 
 11 
 
 cos 45° = sin 45° = ^(7/^5 
 
 ^^ABIAB. 
 
 1 
 
 tan45° = cot45° = 50/Aa=l, 
 sec 45° = cosec 45° = ^2. 
 
 The values of these ratios may be remembered 
 by aid of the accompanying figure. 
 
 17. Definition of a Limit. — If two quantities, A and 
 B, be so connected that, when any change is made in B, 
 a corresponding change is consequently made in A, the 
 limit of A for a given value of B is that value towards 
 which A (from and after a certain value) continually 
 approaches, and from which it can be made to differ as 
 little as we please by making B approach near enough to 
 its given value. 
 
 For example, let APB be a circular arc, OA and OB 
 radii perpendicular to one an- 
 other, and OP any other radius. " 
 Draw PM perpendicular to OA. 
 Then the lengths of PM and 
 OM depend upon the magni- 
 tude of the angle AOP. As 
 this angle diminishes, the length 
 of PM (from and after the value 
 
 OB) continually diminishes, and 6 M A 
 
 may be made to diflfer from zero as little as we please by 
 making the angle AOP small enough. Thus, the limit 
 of PM, when the angle AOP vanishes, is zero; or, the 
 length of PM is ultimately zero. In like manner, the 
 limit of OM, when the angle AOP vanishes, is OA. And, 
 when the angle A OP is a right angle, the limit of PM is 
 OB or OA, and the limit of OM is zero. 
 
12 
 
 TRIGONOMETRICAL RATIOS 
 
 18. To find the tHgonometrical ratios of an angle of 0°. 
 
 Let AOP be a very small angle, APsl circular arc with 
 centre 0. Draw PM perpendicular 
 to OA. 
 O M Then, by the preceding article, 
 
 when the angle AOP vanishes, the limit of OM is OA, 
 and the limit of PM is zero. 
 
 Hence, when the angle AOP vanishes, the limit of 
 OMjOP is unity, i.e. the limit of cos ^ OP is unity. 
 
 This is usually written, for brevity, cosO° = l. 
 
 Similarly, sin 0" = 0, tan 0° = and sec 0° = 1. 
 
 Again, when the angle AOP vanishes, the limit oi PM 
 is zero. Hence, the limit of OP/PM is infinitely great, 
 
 cosec 0° = 00 , and, similarly, cot 0° = x . 
 
 19. To find the trigonometrical ratios of an angle 
 of 90°. 
 
 Let AOP be an angle very nearly equal to 90°, APB 
 a quadrant, centre 0. Draw PM 
 perpendicular to OA. 
 
 Then, by art. 17, when the 
 angle AOP is a right angle, the 
 limit of OM is zero, and the limit 
 ofPif is 05 or OP. 
 
 Hence, when the angle AOP 
 is a right angle, the limit of 
 OM/OP is zero, i.e. the limit of 
 cos J. OP is zero. 
 
 This is usually written, for brevity, cos 90° = 0. 
 
 Similarly, sin 90° = 1, tan 90° = oo , sec 90°= oo , 
 
 cosec 90° = 1 and cot 90° = 0. 
 
 O M 
 
OF AN ACUTE ANGLE. 
 
 13 
 
 20. To find the sine of 18°. 
 
 Let ABC be a triangle having each of the angles at 
 the base BG double of the third angle A, D 
 a point in AB such that the rect. AB . BD 
 is equal to the square on AD \ then AD is 
 equal to BG (Eucl. IV. 10). '> " . 0.\j " 
 
 Draw AE bisecting the angle BAG, and 
 therefore bisecting the base BG at right 
 angles (Eucl. I. 4). 
 
 The angles A, B, C are in the propor- B 
 tion of 1 : 2 : 2, and therefore the angle BAG=\ of 2 
 right angles = 86°; therefore angle 5^^=18°. 
 
 Let AB = a, AD = x; 80 thsit BE =x/2. 
 
 Then, a{a — x) = x\ 
 
 x^ + ax — a^ = 0, 
 
 X- 2 - 2 • 
 
 The + sign must be taken in this expression, for the — 
 sign would give a value numerically greater than a. 
 
 sin 18 =-i-^=— ^^ 
 
 AB '. 
 
 ■^a 
 
 V5-1 
 
 Viva Voce Examples. 
 Express in degrees the following angles : 
 
 1. cos-ii 7. tan-il. 
 
 2. tan-V^- 8. sin-^O. 
 
 3. cosec"^l. 9. cos"^0. 
 
 4. cot-V3. 10. sec-V2. 
 
 5. sec~^l. 
 
 6. sec-^x 
 
 11. cosec"^ 
 
 V3- 
 
12. 
 
 tan'^oo. 
 
 13. 
 
 cot-n. 
 
 14. 
 15. 
 
 • 1 1 
 
 16. 
 
 sin-ij. 
 
 17. 
 
 tan-iQ.' 
 
 18. 
 
 19. 
 
 cosec-^2. 
 
 20. 
 
 cos-^1. 
 
 21. 
 
 ^-.f. 
 
 14 TRIGONOMETRICAL RATIOS 
 
 22. sec -12. 
 
 23. cosec-^oo. 
 
 24. cot"ioo. 
 
 25. sin'U. 
 
 26. tan-i-io- 
 
 27. cosec'V^. 
 
 2 
 
 28. sec-i-7K- 
 
 29. cos-i^o- 
 
 30. cot- 10. 
 
 Variations in the Trigonometrical Ratios when the 
 Angle changes. 
 
 21. To find the limits between which the trigonometrical 
 ratios of an acute angle lie. 
 
 Let POM (see figure of art. 2) be an acute angle ; from 
 any point P, in either bounding line, draw PM perpen- 
 dicular to the other. 
 
 Since the angle PMO is a right angle, it is not less 
 than either of the angles 0PM or POM, 
 
 .-. the side OP is never less than either of the sides OM 
 or PM, 
 
 .'. OMjOP and PM/OP are never greater than unity, 
 while OP/OM and OP/PM are never less than unity, 
 i.e. the cosine and sine of an angle are never greater 
 than unity, and the secant and cosecant are never less 
 than unity. 
 
 Again, when the angle POM is zero, the limit of OM 
 
OF AN ACUTE ANGLE. 16 
 
 is OP, that of PM is zero ; and, when the angle POM is 
 a right angle, the limit of OM is zero, and that of PM is 
 OP. Hence, the values of the cosine and sine of an acute 
 angle lie between and 1, those of the tangent and 
 cotangent lie between and oo , and those of the secant 
 and cosecant between 1 and oo . 
 
 22. To trace the changes in the trigonometrical ratios 
 of an angle as the angle increases from 0° to 90°. 
 
 Let AOB be a quadrant (see the figure of art. 17), OA 
 and OB its bounding radii ; and let the radius OP re- 
 volve from the position OA to the position OB, so that 
 the angle AOP increases from 0° to 90°. Draw PM 
 perpendicular to OA. 
 
 As the angle AOP increases from 0° to 90°, OM 
 diminishes from OA to zero, and PM increases from zero 
 to OB or OA. Hence, as the angle AOP increases from 
 0° to 90°, 
 
 cos A OP decreases from 1 to 0, 
 sin ^ OP increases from to 1, 
 tan A OP increases from to oo, 
 sec -4 OP increases from 1 to oo, 
 cosec -4 OP decreases from oo to 1, 
 cot -4 OP decreases from oo to 0. 
 
 23. Example 1. — Solve the equation * 
 
 6cos2(9+17sin(9=13. 
 Since cos2^=l-sin2^, 
 
 6-6sin2^+17sin^=13,' 
 6sin2(9-17sia(9 + 7 = 0, 
 (2sin6'-l)(3sin(9-7)=--0, 
 
 sin^ = ^or|. 
 
 *In Part I. the phrase " Solve the equation" must be understood to 
 mean ' ' Find the angle or angles between 0° and 90° inclusive, which 
 satisfy the equation." 
 
IC TRIGONOMETRICAL RATIOS 
 
 Of these roots, only the first is admissible, since the sine of an 
 angle cannot be greater than unity. Now, the sine of 30° is ^, 
 ^=30''. 
 
 Example 2. — Solve the equation 
 
 6 tan ^ + 5 cot ^=11. 
 Since cot 6= 1/tan 6, 
 
 6tan^+-A-.=ll, 
 tan^ 
 
 6tan2^-lltan^+5 = 0, 
 
 (tan^-l)(6tan^-5)=0, 
 
 tan ^=1 or f . 
 
 Both of these roots are admissible, since the tangent of an angle 
 
 can have any value between and c» ; 
 
 ^=45° or tan~^f. 
 
 Example 3. — To eliminate 6 between the equations 
 acos ^ + 6sin ^=c and 6cos ^-asin^=o?. 
 Squaring both sides of each equation, we have 
 
 a''^cos-^+ 2a6 cos 6 sin 6-\-hHin-$=c'^, 
 \ h\os''6 - 2ab cos ^ sin ^ + a^sin"^ = d\ 
 
 Vdding a-(cos-^ + sin^^) + b-{cos^e + sin^^) ^(r+d', 
 
 Examples II. a. 
 I Prove the following identities - 
 
 1. (cos a + sin a)^ = 1 + 2 cos a sin a. 
 
 tan^a + l^cot^a + l 
 
 .« o cot^a 
 
 ^4. tan a + cot a = sec a cosec a. 
 
 5. sin^a sec^^S + tan^^ cos^a = sin^a + tan-^. 
 
 6. { ;y/(sec a + tan a) + ^/(sec a — tan a) }^ = 2( 1 + sec a). 
 
 7. cos^^a + sin^a = 1-3 sin^a + 3 sin^a. 
 „ cosec a — sec a _ cot a — tan a 
 
 cot a + tan a cosec a + sec a 
 
OF AN ACUTE ANGLE. 17 
 
 9. tan^a sec^a + cot-acosec^a = sec^acosec^a-Ssec^acosec^a. 
 
 ^_ l+sin« — cosa , l+sina + cosa _ 
 
 1 0. r--^ , h , ,— ^ = 2 cosec a. 
 
 1 + sin a + cos a 1 + sin a — cos a 
 
 11. If cos a = ^, iind sin a, tan a and cosec a. 
 
 12. If sin a = fy, find cot a, sec a and cosec a. 
 
 13. If tan a = Yy find cos a, cot a and sec a. 
 
 *yi3 
 
 14. If sec a = "—- , find cos a, sin a and tan a. 
 
 2 
 
 15. If cot a = —7^, find sin a, sec a and cosec a. 
 
 16. If tan a = -?, — Thy find cos a and sin a. 
 
 17. Find cos 18°. 
 Prove that : 
 
 18. cosec 60°cot 30° = sec245°. 
 
 19. tan260°- 2 tan245° = cot^SO"- 2 sin^SO"- f cosec245°. 
 
 20. cos 30°sin 30° + cos 45°sin 45° + cos 60°sin 60° 
 
 = sin 30° + sin 60°. 
 
 21. sin 90° + cos260° = (2 sin 18° + sin 30°)^. 
 Find the value of: 
 
 22. tan 60°cos 80° - cos 0°tan 45° + 4 sin 18°. 
 
 23. cosec "^x + cos -^0 — sec "^2. 
 
 24. 2cos-il+cot-i0-3sec-ii 
 
 25. 2 cosec -12 -cos -1-^ + 3 cos-i^-sin-^l. 
 
 26. tan-^Go — cot'^— T^ + sin-^^ — cot'^l. 
 
 Solve the equations : 
 
 27. tan = cot a 
 
 28. cos20 + sin0 = l. 
 
 29. V3(tan0 + cot0) = 4. 
 
18 TRIGONOMETIUCAL RATIOS 
 
 30. sec20=^3taue + l. 
 
 31. 2cos3e + sm2a-l=0. 
 
 32. If cos = tan ft prove that sin = 2 sin 18°. 
 Eliminate between the following equations : 
 
 33. cot^ = a, sec = 6. 
 
 34. a sec — c tan Q = dy 6 sec 0+ cZ tan Q = c. 
 
 35. a tan20 + & tan + c = 0, acoi'^O + 6'cot + c' = 0. 
 
 Examples II. b. 
 
 Prove the following identities : 
 
 1. cos^a tan^a + sin^a cot2a = 1. 
 
 2. sec^a + cosec^a = sec^a cosec^a. 
 
 3. l+TH = seca. 
 
 1+seca 
 
 J. cos g + cos ^ sing + sin ;g 
 sin g — sin (3 cos g — cos/3 ~ * 
 
 5. (cot0 + 2)(2cot0+l) = 2cosec2e + 5cota 
 
 6. sin2g(l + n cot^g) + cos2g(l + n tan^a) 
 
 = sin2g(7i + cot^g) + co^\{n + tan^g \ 
 • 7. sec^g — tan^g = 1 + 3 tan^g sec^g. 
 
 8. (4 cos^g — 1 )2tan^g + (3 — 4 cos^g)^ = sec^g. 
 
 9. (tan^a -f tan2/3)cos2g cos"^^ = cos^g + cos2;5-2 cos^a cos^/?. 
 
 10. (1 + secg + tan g)(l + cosecg4-cotg) 
 
 = 2(1 + tan g + cot g + sec g + cosec g). 
 
 11. If cos g = W, find sin g, tan g and cot g. 
 
 12. If sin g = y^^, find cos g, cot g and sec g. 
 
 13. If tan g = if, find cos g, sin g and cosec g. 
 
 14. If sec g = ||-, find sin g, tan g and cosec g. 
 
 15. If cosec g = f¥» ^^^^ ^^^ «> *^^ « and cot g. 
 
 2771/ 
 
 16. If cosg = 3-- — 9, find sing and tang. 
 
 1+m^ 
 
 17. Find tan 18°. 
 
OF AN ACUTE ANGLE. 19 
 
 Prove that 
 18. tan230° + 3 sin245° = sec245° - J cot^GO^. 
 l + cot60° ^/ l + cos30Y 
 l-cotG0°""Vl-cos30V* 
 
 20. sin 90°cot 30° - cot 45'tan 60° = cosec245'^ - 8 siii230°. 
 
 21. 2 cos218° - sec245° = cos 72° - siii245°. 
 Find the value of: 
 
 22. (2 cos 0°sin 30°tan 45° 
 
 +COS 30°sin 45°tan 60°-cosec 30''cos245°)2 
 
 23. sin-iO + 2 sec-loo -3tan-V3. 
 
 24. 4tan-i0+3sec-V2-2cosec-i-^. 
 
 25. cos-ii + sin-i-y^-cosec-il+tan-U-2cot-V3. 
 
 26. 2cosec-V2 + sin-i^-3sec-n-5tan-ii 
 
 . Solve the equations : 
 
 27. 2 sin = tan a 
 
 28. 2cos20 + llsin0-7 = O. 
 
 29. 3tan20-7sece+5 = O. 
 
 30. cot20(2 cosec - 3) + 3(cosec - 1) = 0. 
 
 31. sec cosec — cot = 2. 
 
 32. tan 0+ sec = 2. 
 
 Eliminate between the following equations : 
 
 33. sec = (X, cosec = 6. 
 
 34. ^ cosec +g' cot = r, s cosec — r cot = g. 
 
 35. m cos20 + -71 cos = p, m'sec20 + Ti'sec =_p'. 
 
 Examples III 
 
 1. Simplify ( — ^— ^ -\ ^ r-^ ) x cos^a sin^a. 
 
 •^ Vsec^a — cos^a cosec^a — sin^a/ 
 
20 TRIGONOMETRICAL RATIOS. 
 
 " ^ Vcos a + tan'^a sin a cos a cot^a + sin a/ 
 sec a cosec a — 1 
 cosec a — sec a 
 
 3. Express sec^0 in terms of tan Q. 
 
 7— a; 
 
 4. If cos a= ^ :r> and a? be positive, show that x can- 
 
 not be less than 2. 
 
 2 4- cc 
 
 5. If cosec a = ^ «> show that x cannot be greater than 5. 
 
 G. If ^ = ^2, and ^^ = ^3, find a and ^. 
 sinj8 ^ tan^ ^ ' '^ 
 
 7. If sin a = m sin ^8 and cos a = ti cos /3, find tan a and 
 
 tan ;8. 
 
 Vtan ^/ Vtana/ • Vtan^/' 
 / 1 Y ^ / cos e y / sin^ Y 
 Vsin cf)) Vsin a/ Vsin /3/ 
 9. If a tan a = 6 tan /3, and a^^^ = a^ — 6-, show that 
 (1 - aj2sin2^)(i _ x^co^^a) = l-x\ 
 
 10. Eliminate Q between 
 
 cos — sin = a and tan Q=c sec^0. 
 
 11. Eliminate between 
 
 cosec ^ — sin = a, sec — cos = 6. 
 
 12. Find the least value of a^sec^^ + ^^cos^^, where a and h 
 
 are constant quantities. 
 
 <9^ 
 
CHAPTER III. 
 
 TEIGONOMETEICAL EATIOS OF COMPOUND ANGLES. 
 
 24. The principal object of the present chapter is to 
 express the trigonometrical ratios of the sum or difference 
 of two or more angles in terms of the ratios of the com- 
 ponent angles, and those of the multiples or sub-multiples 
 of a given angle in terms of the ratios of that angle. In 
 the latter part of the chapter, we shall shew how these 
 relations may be used for effecting the transformation of 
 trigonometrical expressions. 
 
 Throughout the chapter every angle, both component 
 and compound, is supposed to be acute ; but it should be 
 remarked that this limitation only applies to the first 
 two propositions. These propositions are shewn to be 
 true for all real values of the angles in Chapter VIII. 
 
 . 25. To exijress the cosine and sine of the sum of tiuo 
 angles in terms of the cosines and sines of the angles 
 themselves. 
 
 Let the given angles be denoted by a and /3. Draw- 
 the angle AOB equal to a and BOG equal to ^, so that 
 AOG is equal to the compound angle a + j3. 
 
 In OC, one of the bounding lines of the compound 
 angle, take any point P. Draw PM and PK perpen- 
 
 21 
 
22 
 
 TRIGONOMETRICAL RATIOS 
 
 dicular to OA and OB, KL perpendicular to OA arid 
 KR to PM. Then, in the triangle PRK, the angle KPR 
 is equal to a, for 
 angle KPR = complement of PKR 
 = angle RKO 
 = angle AOB. 
 Now, 
 
 , , ^, OM OL--ML 
 
 OP 
 
 OL-RK 
 
 OP 
 PK 
 OP 
 
 _qL^ OK RK 
 ~OK'OP PK 
 
 = cos a cos ^ — sin a sin /3 
 
 KL+RP 
 
 .(A) 
 
 ., . , ^ r.. PM MR+RP 
 Also, 8in(a + i8) = ^p= ^ — 
 
 OP 
 
 ^KL OK RP PK 
 OK' OP"^ PK' OP 
 = sin a cos /3 + cos a sin /3., 
 
 (B) 
 
 26. To express the cosine and sine of the difference of 
 two angles in terms of the cosines and sines of the angles 
 themselves. 
 
 Let the given angles be denoted by a and ^. Draw 
 the angle AOB equal to a, and 
 BOD equal to ft so that AOD is 
 equal to ' the compound angle 
 a-/3. 
 
 In OD, one of the bounding 
 lines of the compound angle, 
 take any point Q. Draw QiV 
 and QK perpendicular to OA 
 and OB, KL perpendicular to OA and KS to NQ pro- 
 
Also, sin(a — /3) 
 
 OF COMPOUND ANGLES. 23 
 
 duced. Then, in the triangle QSK, the angle KQS is 
 
 equal to a, for 
 
 angle iiTQ/Sf = complement of QKS = bx\^q SKB 
 
 = angle AOB. 
 
 , ^, ON OL + LN OL+KS 
 Now, cos(a-/5) = -^^=— ^g— =— ^g— 
 
 _0L OK KS QK 
 ~0K' OQ^QK'OQ 
 = cosaCOs/3 + sin a sin/3 (C) 
 
 QN_ JNS-QS _ LK-Q8 
 
 0Q~ OQ OQ 
 _LK OKQS QK 
 ~OK'OQ QK'OQ 
 = sin a COS ^ - cos a sin fi (D) 
 
 27. To express the tangent of the sum and difference 
 of two angles in terms of the tangents of the angles 
 themselves. 
 
 Let the given angles be denoted by a and ^. 
 
 (1) Algebraical proof 
 
 ( I Q\ _ ^^"(<^ + /^) _ ^^^ g cos ^ + cos a sin /3 
 tan(a + ^)- ^^^^^ _^ ^^ - ^^^ ^ ^os ^ - sin a sin /3' 
 
 Dividing the numerator and denominator of this frac- 
 tion by cos a cos /3, we have 
 
 sin g sin ^ 
 ^ , , ^. cos a cos/3 tan a + tan 8 .j.. 
 
 ^°(" + ^^ = ^-iWihr^ = l-tanatan/3 (^> 
 
 COS a cos /3 
 
 . . ^ . r,. sin(a-/3) sinacos^-cosasin^ 
 
 Again, tan(a--/3)= — 7 ^ = 15-, — — o 
 
 ^ ' ^ ^^ cos(a — j5) cosaCOs/5 + smasm/5 
 
 _ tan a — tan /3 /-px 
 
 "r+tana'tajT/S ^ 
 
24 TRIGONOMETRIC A L RA TIOS 
 
 (2) Geometrical Proof. — (See figure of art. 25.) 
 , . ^ox P^^ MR+RP 
 
 LK,RP LK RP 
 OL^OL OL^OL 
 
 RK ._^PP 
 OL RP'OL 
 
 Now, the triangles KPR, KOL are similar, for the 
 angles KPR and KOL are equal, and the angles KRP 
 and KLO are right angles. 
 
 RP_PK_. ^ 
 
 OL~OK~^^^f^' 
 
 ^ '^^ 1 — tan a tan /5 
 Similarly, making use of the figure in Art. 26, we may 
 
 show that t8Lu(a — 8) = ^ . J^ r — ^. 
 
 ^ '^^ 1+tanatanp 
 
 Govs. — Similarly, it may be shewn, both algebraically 
 
 and geometrically, that 
 
 ., , Q. cotacotiS — 1 
 C0t(a + /3) = — 7 — , ^. o ' 
 ^ ^^ cota + cot/3 
 
 J x/ m cot a cot -5+ 1 
 
 and cot(a-/3) = — -^ — ^^^ — • 
 
 ^ '^^ cot p — cot a 
 
 28. The formulae of the preceding articles may be used 
 to obtain the trigonometrical ratios of angles which are 
 the sums or diflferences of angles whose ratios are known. 
 
 For example, 
 
 cos 75° = cos(45° + 30°) = cos 45°cos 80° - sin 45°sin 30° 
 
 _J JS 1 i_V-^-i 
 
 J2' 2 J~2'2~ 2\/2 ' 
 
OF COMPOUND ANGLES. 25 
 
 sin 75° = sin (45° + 30°) = sin 45°cos 30° + cos 45°sin 30° 
 
 = _L x/? , J 1_ V3 + 1 
 
 V2' 2 "^V2'2~ 2V2 ' 
 
 , ^j,o , //i-o , oAox tan 45° + tan 30° 
 tan 75 = tan(4a + 30 ) = r — r — — ^r — ^7^ 
 ^ ^ 1 — tan4o tan 30 
 
 ^+73 V3+t 4 + 2^/3 
 
 Similarly, or by art. 8, it may be shewn that 
 cosl5°=^^±i, sin 15°-^^2^,andtanl5° = 2-V3. 
 
 29. The product of the sines of the sum and difference 
 of two angles is equal to the difference of the squares of 
 the sines of the component angles. 
 
 Let the angles be denoted by a and /5. 
 Then, sin (a + /5)sin(a - ^) 
 
 = (sin a cos |8 + cos a sin /3)(sin a cos ^ — cos a sin /3) 
 
 = sin^a cos^yg — cos^a sin^/^ 
 
 = sin^a — sin^^ (G) 
 
 Cor. 1. — sin(a + |8)sin (a — /3) = cos^/3 — cos^a. 
 Cor. 2. — Similarly it may be shewn that 
 
 cos(a+/8)cos(a — P) = cos^a — sin^/^ = cos"/3 — sin^a. 
 
 30. To express the cosine of 2a in terms of the cosine 
 and sine of a. 
 
 (1) Algebraical proof 
 
 cos 2a = cos(a + a) = cos a cos a — sin a sin a 
 
 = cos^a — sin^a (H) 
 
 Putting sin^a = 1 — cos^a in this equation, we obtain an 
 expression for cos 2a in terms of cos a, namely, 
 
 cos 2a = cos^a — (1 — cos^a) = 2 cos^a — 1 (I) 
 
26 
 
 TRIGONOMETRICAL RATIOS 
 
 Again, putting cos^a = 1 — sin^a in (H), we obtain an 
 expression for cos 2a in terms of sin a, namely, 
 
 cos 2a = 1 — sin-a — sin^a = 1 — 2 sin^a (J) 
 
 (2) Geometrical proof. — Let AG he the diameter of a 
 circle, centre 0. Draw the angle GAP equal to a : then 
 
 the angle- GOP is equal to 2a. Draw PM perpendicular 
 to AG, and join PG. Then APG is a right angle, and 
 angle (7Pif= complement of angle PGA = angle PAG= a. 
 
 T., „ OM 2.0M {2.0G-2.MG) 
 Now, cos2a = ^ = 2-^ = ^^jj^ 
 
 _ AM-MG _AM AP^MG PG^ 
 ~ AG ~AP'AG PG'AG 
 = cos^a — sin^a. 
 The formulae (I) and (J) may be obtained in a similar 
 manner. 
 
 Govs. — From equations (I) and (J) we have 
 
 cos2a = Kl+cos2a) (K) 
 
 and sin2a = J(l — cos2a) ; (L) 
 
 equations which give the cosine and sine of an angle in 
 terms of the cosine of double the given angle. 
 
OF COMPOUND ANGLES. 27 
 
 31. To express the sine of 2a in terms of the cosine and 
 sine of a. 
 
 (1) Algebraical jpr oof 
 
 sin 2a = sinfa + a) = sin a cos a + cos a sin a 
 
 = 2 cos a sin a (M) 
 
 (2) Geometrical jproof — (See figure of Art. 30.) 
 
 • 9 _PM_2.PM_2PM AP 
 ^'"^ ''~0P~2.0P~ AP 'AG 
 = 2 sin a cos a. 
 
 32. To express the tangent of 2a in terms of the tan- 
 gent of a. 
 
 (1) Algebraical proof 
 
 tan 2a = tan (a + a) 
 
 tan a + tan a 
 
 1 — tan a . tan a 
 2 tana 
 
 1 — tan^a *'" 
 
 (2) Geometrical proof — (See figure of art. 30.) 
 PM 2PM 2PM 
 
 m 
 
 tan 2a = 
 
 OM 20M~AM-MG 
 2PM 
 AM 2 tan a 
 
 . MG PM 1-tanV 
 
 pm'am 
 
 Gor. — 2 cot 2a = cot a — tan a, for 
 
 o .o 2(l-tan2a) 1 
 
 2cot2a = -^T7T -=7 tana 
 
 Z tan a tan a 
 
 = cot a — tan a. 
 
28 TRIGONOMETRICAL RATIOS 
 
 33. To express the tAgonometrical ratios of 3a in 
 tervis of those of a. 
 
 cos 3a = cos(2a + a) = cos 2a cos a — sin 2a sin a 
 = (2 cos^a — l)cosa — 2 sin a cos a . sin a 
 = 2 cos^a — cos a — 2 cos a(l— cos^a) 
 = 4 cos^a — 3 cos a (O) 
 
 sin 3a = sin(2a + a) = sin 2a cos a + cos 2a sin a 
 = 2 sin a cos a . cos a + (1 — 2 sin2a)sin a 
 = 2 sin a(l — sin^a) + sin a — 2 sin^a 
 = 3 sin a — 4 sin^a (P) 
 
 tan 3a = tan(2a + a) 
 
 _ tan 2a + tan a 
 ~1— tan 2a . tan a 
 2 tan a 
 
 1 — tan^a 
 
 + tana 
 
 ^ 2tana . 
 1 — ^ — 77 — o- . tan a 
 1 — tan'^a 
 
 _ 2 tan a + tan a — tan^a 
 
 1 — tan^a — 2 tan^a 
 _ 3 tan a — tan^a 
 
 l-3tan2a * 
 
 34. The cosine and sine of any multiple of a may thus 
 be expressed in terms of the cosine and sine of a. The 
 general formulae for obtaining them are 
 
 cos(n + l)a = 2 cos na cos a — cos{n — l)a, 
 sm(n + 1 )a = 2 sin na cos a — ^m{n — l)a, 
 formulae which may be easily proved by means of t])e 
 results obtained in Arts. 25, 26. Putting 7i = 3, 4, 5, etc., 
 successively in the formulae, we find the cosines and 
 sines of 4a, ba, 6a, etc. 
 
OF COMPOUND ANGLES. 29 
 
 Trigonometrical Transformations. 
 
 35. We have already proved that ' ::i 
 
 cos(a + /3) = cos a cos P — sin a sin ^, 
 cos(a — j8) = cos a cos /3 + sin a sin ^, . ^ 
 
 sin(a + /3) = sin a cos ^ + cos a sin ^, 
 sin(a — ^) — sin a cos /3 — cos a sin /3. 
 By addition and subtraction of the first and second of 
 these formulae, and also of the third and fourth, we obtain, 
 2 cos a cos /3 = cos(a + /3) + cos(a — P); 
 2 sin asin/5 = cos(a — /3) — cos(a + ^),l /qx 
 
 2 sin acos P = sin(a + /3) + sin(a — ^), 
 2 cos asin/3 =sin(a4-/3) — sin(a — /3).- 
 By means of these formulae, we can express twice the 
 product of two cosines, or of two sines, or of a sine and 
 cosine, as the sum or difference of two cosines or sines. 
 
 It should be noticed that the third formula is used 
 when we have the sine of the greater angle, and the 
 fourth when we have the cosine of the greater angle ; 
 also, that, in the second formula, the cosine of the greater 
 angle, a + ^, is subtracted from the cosine of the lesser, 
 a — /3: for the cosine of an angle diminishes as the angle 
 increases. 
 
 YiVA Voce Examples. 
 
 Transform the following expressions into the sums or 
 differences of two cosines or sines : 
 
 1. 2 cos 4a cos 2a. 7. 2 sin 7a sin 4a. 
 
 2. 2 cos 5a cos a. 8. 2 sin 10a sin 3a. 
 
 3. 2 cos 10a cos 7a. 9. 2 sin 4a sin 3a. 
 
 4. 2 cos a cos 3a. 10. 2 sin 5a sin 5a. 
 
 5. 2 cos 7a cos 6a. 11. 2 sin 3a sin 18a. 
 
 6. 2 cos 3a cos 13a. 12. 2 sin a sin 12a. 
 
80 
 
 TRIGONOMETRICAL RATIOS 
 
 13. 2 sin 6a cos 4a. 
 
 14. 2 sin 5a cos a. 
 
 15. 2 sin 11a cos 9a. 
 
 16. 2 sin 3a cos 2a. 
 
 17. 2 sin 7a cos 4a. 
 
 18. 2 sin 10a cos a. 
 
 19. 2 cos Hasina. 
 
 20. 2 cos 14a sin 3a. 
 
 21. 2 cos 10a sin 5a. 
 
 22. 2 cos 2a sin a. 
 
 23. 2 cos a sin a. 
 
 24. 2 cos 7a sin 2a. 
 
 25. 2 cos a sin 15a. 
 
 26. J sin 3a cos 8a. 
 
 27. cos 3a . cos 7a. 
 
 28. sin 3a. sin 11a. 
 
 29. 2sin(2a + 3/3)cos(a-i8). 
 
 30. cos(2a+;8)cosa. 
 
 31. cos(a-)8)cos(2a + 4^). 
 
 32. sin3asin(2/3-3a). 
 
 33. cos(5a-2/3)sin(a-4^). 
 
 34. sin(8a-3/3)sin(5/3-3a). 
 
 35. sin(45° + a)sin(45° — a). 
 
 36. cos(30° + 2a)sin(30°-a). 
 
 36. The four formulae (Q) of the preceding article serve 
 also for expressing the sum or difference of two cosines 
 or sines as the product of sines or cosines. But they 
 may be put into a more convenient form by writing 
 a+/3 = o- and a — /8 = <5, 
 
 so that 
 
 + S 
 
 and /? = ' 
 
 The formulae thus become 
 COSO-+COS 8 
 
 cos ^ — coso-: 
 
 sin (T + sin 6 
 
 2 cos — y- cos -^ 
 2 sin --r— sin —^ 
 
 2 sin ^7^- cos 
 
 sin 0- — sm d = 2 cos — 7^— sin 
 
 2 
 
 .(R) 
 
 2 "" 2 
 They may be thus expressed : 
 The sum of the cosines of two angles is equal to twice 
 the product of the cosine of half their sum by the 
 cosine of half their difference. 
 
OF COMPOUND ANGLES. 
 
 31 
 
 The difference of the cosines of two angles is equal to 
 twice the product of the sine of half their sum by 
 the sine of half their inverted difference. 
 
 The sum of the sines of two angles is equal to twice the 
 product of the sine of half their sum by the cosine 
 of half their difference. 
 
 The difference of the sines of two angles is equal to twice 
 the product of the cosine of half their sum by the 
 sine of half their direct difference. 
 
 Geometrical 'proof . — Draw the angle AOG equal to o-, 
 and the angle AOD equal to S, 
 so that the angle GOD is equal 
 to o- — ^. Bisect the angle GOD 
 by OB, making the angles BOD, 
 GOD each equal to ^{cr — S), 
 and, consequently, the angle 
 AOB equal to ^ + i(a--^), i.e. to 
 
 In OB take any point K, and o 
 through K draw PKQ perpendicular to OB, meeting OG 
 OD in P, Q. Draw KL, PM, QN perpendicular to OA, 
 and through K draw RKS perpendicular to PM and NQ 
 produced. Then, by elementary geometry, we have 
 
 OP=OQ, PM+ QF= 2KL, PM- QN= 2PE, 
 
 0M+0N=20L, and ON-OM=2ML = 2RK. 
 
 Also, in the triangle PRK, we have 
 
 angle ZPJS = complement of angle PKR 
 
 angle EKO = 
 
 cr + (5 
 
82 TRIGONOMETRICAL RATIOS 
 
 „ ^ , OM ON OM+ON 
 
 Hence, cos ^+^^^^ = 'qp+-qq = — Qp — 
 
 201 _qL OK 
 ~ OP ~ OK OP 
 
 = 2 cos — ^— cos — o— • 
 
 ON-OM 2RK 
 
 cos 6-C03(r = ^p =-Qp- 
 
 RK PK 
 ~ PK OP 
 
 = 2 Sin ^-sin-g-. . 
 
 . , . , PM+QN 2KL 
 sin o-+sin S= QjT— = ^TT 
 
 _^ OK 
 ~ OK' OP 
 
 = 2 sin ^— cos 
 
 sin a — sin S = 
 
 2 2 
 
 PM-QN_2PR 
 OP ~ OP 
 PR PZ 
 "^PZ" OP 
 
 „ cr + S . (T — S 
 
 = 2 cos ^ sm 
 
 2 2 
 
 Oo7's. — The cosine of an angle being equal to the sine 
 of its complement, we have 
 
 coso-+sin ^ = 2 cosf 45°+^-s~) cos f 45° — 
 
 2 
 + S' 
 
 cos 0- — sin (5 = 2 sin (45° +^-2—) sin f45° — ^^^ 
 
OF COMPOUND ANGLES. 
 
 83 
 
 Viva Voce Examples. 
 
 Transform the following expressions into the products 
 of sines and cosines : — 
 
 1. cos 5a + cos a. 
 
 2. cos 7a + cos 3a. 
 
 3. cos 10a + cos 2a. 
 
 4. cos 1 5a + cos 3a. 
 
 5. cos 9a + cos 8a. 
 
 6. cos 2a + cos a. 
 
 7. cos a — cos 3a. 
 
 8. cos3a — cos9a. 
 
 9. cos 5a — cos 6a. 
 
 10. cos 8a — cos 11a. 
 
 11. cos 5a — cos 10a. 
 
 12. cos 6a — cos 16a. 
 
 13. 
 
 sin 6a + sin 4a. 
 
 14. 
 
 sin 4a + sin2a. 
 
 15. 
 
 sin 9a + sin a. 
 
 16. 
 
 sinl3a + sinlla. 
 
 17. 
 
 sin 1 2a + sin 7a. 
 
 18. 
 
 sin 8a + sin 5a. 
 
 19. 
 
 sin 9a — sin 3a. 
 
 20. 
 
 sin 11a — sin a. 
 
 21. 
 
 sin 10a — sin 2a. 
 
 22. 
 
 sin 5a — sin 4a. 
 
 23. 
 
 sin a- sin |. 
 
 24. 
 
 sin 3a — sin a. 
 
 25. 
 
 sin 3a — sin 5a. 
 
 26. 
 
 sin 7a + sin 13a. 
 
 27. 
 
 cos 3a + cos 9a. 
 
 28. 
 29. 
 30. 
 
 31. 
 32. 
 33. 
 
 cos 9a — cos 2a. 
 
 sin 11a — sin 15a. 
 
 cos 1 la + cos 2a. 
 
 • a .3a 
 sin ^ — sin— . 
 
 5a a 
 
 cos y- cos 2- 
 
 . K .5a 
 sin 7a — sin -^. 
 
 34. 
 
 35. 
 36. 
 
 37. 
 38. 
 39. 
 40. 
 41. 
 
 42. 
 
 43. 
 44. 
 45. 
 
 sin a + sin 
 
 3a 
 
 cos 9a + cos 10a. 
 
 sin 11a — sin 7a. 
 
 . 7a . 11a 
 
 o ^a 
 
 cos 8a — cos -^. 
 
 . 5a , . 3a 
 sm-^ + sin-. 
 
 3a 
 cos-j — cos 
 4 
 
 sin. -sin 
 
 .a .a 
 sm^-sin^. 
 
 11a 
 4 ' 
 15a 
 
 sin(2a+3^)+sin(2a+/3). 
 cos(3a-j8) + cos(a + 5/3). 
 cos(2a-/3)-cos3/3. 
 
34 TRIOONOMETRWAL RATIOS 
 
 46. 8in(3a+8/3)-sin(7a-3/3). 49. sin 63° -- sin 27°. 
 
 47. cos 12° -cos 48°. 50. cos 60° + cos 20". 
 
 48. sin 75° + sin 15°. 
 
 51. cos{a + (r-J)/3}-cos{a + (r+J)^}. 
 
 52. sin{a + (r + i)^}-sin{a + (r-J)^}. 
 
 37. The following examples are given to shew the use 
 of the formulae proved in this chapter : — 
 Example 1. — Prove that 
 
 tana=-«Hl2^=lz^^, and tan^a-^ "^^'^a 
 
 1 + cos 2a sin 2a 1 + cos 2a 
 
 sina_ 2sinacosa _ sin 2a 
 cos a 2 cos^a 1 + cos 2a' 
 2 sin^a 1 - cos 2a 
 
 2 sin 2a 1 - cos 2a 1 — cos 2a 
 
 tana= 
 
 Also, tana=^ . . « 
 
 2 sin a cos a sm 2a 
 
 Hence, by multiplication, 
 
 tan^a . . 
 
 1 + cos 2a sin 2a 1 + cos 2a 
 
 Example 2. — Prove that 
 
 sin a + 2 sin 3a + sin 5a _ , o„ 
 
 cos a + 2 cos 3a + cos 5a 
 
 sin a + 2 sin 3a + sin 5a _ (sin 5a + sin a) + 2 sin 3a 
 
 cos a 4- 2 cos 3a + cos 5a (cos 5a + cos a) + 2 cos 3a 
 
 _ 2 sin 3a cos 2a + 2 sin 3a 
 
 2 cos 3a cos 2a + 2 cos 3a 
 
 _ 2 sin 3a(cos 2a + 1 ) 
 
 2cos3a(cos2a+l) 
 
 = tan 3a. 
 
 Example 3. — Prove that 
 
 3 sin 3a ,. 3 cos 3a sin 4a 
 cos^a . — - — + siira . ^ — = — - — 
 3 3 4 
 
 «"^^ + sin^a . 2^^ 
 3 3 
 
 = ^[cos*a(3 sin a - 4 sin^a) + sin'a(4 cos^a - 3 cos a)] 
 = ^(3 cos^a sin a - 3 sin^a cos a) 
 = cos a sin a(cos^a — sin'^a) 
 =^ sin 2a. cos 2a 
 =1 sin 4a. 
 
OF COMPOUND ANGLES. 85 
 
 Example 4. — If tana=| and tail ^8 = ^, find tan(2a-/3). 
 
 l-tan^a 1-^ ^' 
 tan(2a- ^)^ tan 2a-tan/? _i-i ^ ^ 
 ^ '^^ l+tan2atan/? l+l-i 
 
 Example 5. — Express 4 cos a cos /? cos y as the sum of four 
 cosines. 
 4 cos a cos /8 cosy = 2 cos a . 2cos /3cosy = 2cosa{cos(/8 + y) + cos(/8-y)} 
 
 = 2 cos a cos(/3 + y) + 2 cos a cos(^ - y ) 
 
 = cos(a + /? + y) +cos(^ + y - a) + cos(y + a - /S) + cos(a + /S - y). 
 
 Example 6. — Solve the equation 
 
 2sin^sin3^=sin22(9. 
 2sin6'sin36'-sin22l9=0, 
 cos 2(9 - cos 4^ - sin22(9 = 0, 
 cos W - (2 cos22^ - 1 ) - ( 1 - cos22^) - 0, 
 cos2^-cos22(9=0, 
 cos 2(9(1 -cos 2(9) = 0, 
 
 cos 2^=0 or 1, 
 2^=90° or 0°, 
 (9=45° or 0°. 
 
 Examples IV. a. 
 
 Find the values of : 
 
 1. cos(a+j8) and sin(a — /?), if sma = xT ^^^ mi\^ = -^^. 
 
 2. cos(a — /8) and sin(a + /3), if sin a = f and cos/3 = ^|. 
 
 3. cos(a + /3) and sin(a + /5), if tana = ^f and cot|8 = ff. 
 
 4. tan(a — /5), if tan a = a_/^ and tan /5 = xt^q' 
 
 5. cot(a + 18), if cot a = 7 and cot |8 = f . 
 
 6. Shew that tan 75° + cot 75' = 4. 
 
 7. The value of 
 
 cos(7i + l)a cos(7i — l)a4-sin(7i + l)a sin(?i — l)a 
 is independent of n. 
 
 8. sinacos(/3+y) — sin /5cos(a + y) = sin(a — /9)cosy. 
 
36 TRIGONOMETRICAL RATIOS 
 
 A L ^ I o sin(/5±a) 
 9. cota±cot/8 = -^-^^^-^ — 5. 
 ' sin a sin p 
 
 iA / _L /o\ sec a sec /3 
 
 10. sec(a±/3) = 7TrT r^-^- 
 
 '^^ 1 + tan a tan ^ 
 
 11. cos^a - cos a cos(60° + a) + sin2(80° - a) = f . 
 
 12. Find cos(a+/3+y) in terms of the cosines and sines 
 
 of a, /5, y; and hence shew that, if a+/3+y = 90°, 
 tan |8 tan y + tan y tan a + tan a tan ^ = 1 . 
 
 13. Find tan(a + j8 + y) in terms of the tangents of a, /3, 
 
 y ; and hence shew that, if a+/3+y = 90°, 
 tan |8 tan y + tan y tan a + tan a tan ^ = 1. 
 
 14. cos^a + cos2/3 — 2 cos a cos /3 cos(a + /3) = sin2(a + ^8). 
 
 1 5. tan^a — tan^/S = sin(a + /5)sin(a — ^)sec^a sec^/?. 
 
 16. sin44"cos74° = l-sinn4°. 
 
 Find the values of : 
 
 17. cos 2a, when (1) cos a = f, (2) sin a = i, (3) cos a = f . 
 
 18. sin 2a, when (1) cosa = -^f, (2) tana = ^V 
 
 19. tan 2a, when (1) tan a = A, (2) cot a = 4. 
 
 20. cos a, sin a, and tan a, when cos 2a = f i. 
 
 21. tan a, when (1) tan 2a = ^8, (2) cos | = i- 
 
 22. tan(a + 2/3), when tan a = ^3 and tan /3 = 2 - ^3. 
 
 /3 / 3 — 1 
 
 23. cos 3a, when cos a = '^- ; sin 3a, when sin a= - ^ /o > 
 
 and tan 3a, when tan a = i- 
 
 24. Find the cosine, sine and tangent of 22^°, and shew 
 
 that 2 cos 11° 15'= V {2+ V(2+ V2)}. 
 
 25. sin 18° and sin 54° are roots of the equation 
 
 4»2_2^5i»+l=0. 
 
 o^ n 1 " tan^a 
 
 26. cos2a = =-rT — 2- 
 
 l + tan^a 
 
 27. 2 cosec 2a = sec a cosec a. 
 
28. sec 2a — 
 
 OF COMPOUND ANGLES. 37 
 
 cot a + tan a 
 
 cot a — tan a 
 
 29. tan(45° + a)=sec2a + tan2a. 
 
 30. Simplify Z^^^ + ^^^y. 
 
 ^ *" l + cosa + cos2a 
 
 Q^ c<- i-r 2(l + tanatan2a) 
 
 31. Simplify y, , . 7 — ^. 
 
 ^ ^ 2 + tan a tan 2a 
 
 32. tan(45°-|) + tan(45°+|) = 2seca. 
 
 33. sin2(22r + 1) - sin2(22i° - 1) = -1 sin a. 
 
 l-tan2(45°-|) 
 
 34. Simplify ^-• 
 
 l + tan2U5°-|j 
 
 35. Simplify cos(36°+a)cos(36°-a)+cos(54°+a)cos(54°-a). 
 
 36. cos^a + sin^a cos 2(3 = cos^^ + sin^^ cos 2a. 
 
 37. {cos2a + cos2^ + 2cos(a + |8)}2 
 
 + {sin2a + sin 2/3 + 2 sin(a + iS)}2 = 16 cos*^^. 
 
 38. tan 3a tan 2a tan a = tan 3a — tan 2a — tan a. 
 
 39. cos4a = 8cos*a — 8cos2a + l. 
 
 40. Find the value of 4 tan-i^ 
 
 41. Express cos Qa in terms of cos a. 
 
 42. sin 3a = 4 sin(60° - a)sin a sin(60° + a). 
 
 43. tan 3a = tan(60° - a)tan a tan(60° + a). 
 
 44. ^i^«^ = tan4^. 
 cosa + cos/5 2 
 
 sin a — Sin |8 2 2 
 
 46. cos(4a + /5) = 2 cos a cos(3a + /5) - cos(2a + /5). 
 
 47. sin 1 1 a sin a + sin 7a sin 3a = sin 8a sin 4a. 
 
 48. sin a(cos 2a + cos 4a + cos 6a) = sin 3a cos 4a. 
 
38 TRlGONOMEriUOAL RATIOS 
 
 49. Simplify '^'°"+""f"+^'"f ^. 
 
 cos a + cos 2a + cos 3a 
 
 50. Simplify si" «-«'" 4«+8i n 7«-sin 10a 
 
 COS a — COS 4a + COS 7a — COS 1 Oa 
 
 51. sin 10°+ sin 50° = sin 70°. 
 
 52. cos 55° + sin 25° = sin 85°. 
 
 53. sina+sin(72°+a)+sin(36°- a) = sin(72°- a)+sin(36°+a). 
 
 54. Prove geometrically that cot(a+j8)= ^ ^ , To • 
 
 55. Also that tana = =— ; =r-- 
 
 1 + cos 2a 
 
 Solve the equations : 
 
 56. sin(e+a)-sin(0-a) = ^2sina. 
 
 5/. -^sm0+^cose = ^2- 
 
 58. 2cos20-2sine-l = O. 
 
 59. (1 + V3)tan20(l~tan0) = 2tana 
 
 60. sin 7^ + sin 30 = cos 2a 
 
 61. sin 20 + 1= cos + 2 sin a 
 
 62. cos 60 + cos 40 + cos 20 + 1 = 2 cos 20 cos 0. 
 
 63. cos + ^3 sin = ^3 cos - sin = 2 cos(0 + <p). 
 Eliminate between the following equations : 
 
 64. cos = a, cos 20 = b. 
 
 65. cos + sin = a, cos 20 = 6. 
 
 66. acos0 + 6cos20 = c, asin0 + 6sin 26 = d. 
 
 67. a cos + 6 sin = J(^^^^)> tan 20 = c. 
 
 Examples IV. b. 
 
 Find the values of : 
 
 1. cos(a + ^) and cos(a — /3), if cos a = f f and cos /9 = f f . 
 
 2. sin(a + /8) and am(^ — a), if sin a = H ^^^ cos ^ = |f . 
 
OF COMPOUND ANGLES. 39 
 
 12 
 ~5~- 
 
 3. cos(a + ^) and sin(a - ^), if tan a = V- and cot /S ^ 
 
 4. tan(a + P) and tan(a — /3), if tan a = J and tan ^ = J. 
 
 5. cot(a - B). if sin a = tVt and cos (3 = f f. 
 
 6. Shew that 4 sin 75° sin 15° = 1. 
 
 ^ ^. yn tan('yi + l)a — tsin(n — l)a 
 
 7. bimplity i_|_tan(7i+l)atan('^-l)a' 
 
 8. cos(a + /3)sin ^ — cos(a + y)sin y 
 
 = sin(a + /5)cos ft — sin(a + y)cos y. 
 
 f^ i. _L.j. /o sin(a±/8) 
 
 9. tana±tan/3= ^ '^i- 
 
 '^ cos a cos /5 
 
 , . ^. cosec a cosec /3 
 
 10. cosec(a±/5) = -^^-^^^^-. 
 
 11. sin2a-sinasin(60° + a) + cos2(30°-a) = f. 
 
 12. Find sin(a + /3 + y) in terms of the cosines and sines of 
 
 a, A y. 
 
 1 3. Find tan(a + /5 + y), when tan a = J, tan /5 = f , tan y = | . 
 
 14. cos^a + cos2/3 — 2 cos a cos ft cos(a — ft) = sm\a — ft). 
 
 1 5. sin(a + /5)sin(a — ft) = (sin a + sin /3)(sin a — sin ft), 
 
 16. sin55°cos25° = f — sin25°. 
 Find the values of: 
 
 17. cos 2a, when (1) cosa = H» (2) sin a= iV» (3) sin a = 4- 
 
 18. sin 2a, when (1) sin a = f, (2) tan a = ^^!^. 
 
 19. tan 2a, when (1) tana = f, (2) sin a= — ^ — r- 
 
 20. cos a, sin a and tau a, when cos 2a = 2^. 
 
 21. tan a, when sec 2a = 3; and tan ^, when sin a = ff. 
 
 22. tan(2a + iS), when tana = l and tan^ = J. 
 
 23. cos 3a, when cos a= \^',^ ; sin 3a, when sina = J; 
 
 and tan 3a, when tana = -y-. 
 
40 TRIGONOMETRICAL RATIOS 
 
 Shew that : 
 
 24. seen 5° = 4 tan 1 5°, and tan 7° 30' = ^6 + v^2- 2- ^3. 
 
 25. 4sml8°cos36° = l. 
 2 tan a 
 
 26. sin 2a 
 
 27. sec 2a = 
 
 1 + tanV 
 sec^a 
 
 2 — sec^a' 
 
 28. 2 cosec 2a = tan a + cot a. 
 
 29. c»««+8in« = sec2a+tau2a. 
 COS a — sin a 
 
 30. cot§— cota = coseca. 
 
 31. l±sina = 2sin2('45^±«Y 
 
 32. cot(45" + a) + cot(45° - a) = 2 sec 2a. 
 
 33. cot<45° + ^) = |^^^^^|^^«. 
 
 \ 2/ 2 cosec 2a 4- sec a 
 
 34. 224|5;±2_) = sec2a-tan2a. 
 cos(4o —a) 
 
 „^ cos 2a sin 2a _ 1 
 
 cos a + sin a cos a — sin a ~~ v^2 cos(a + 45°)' 
 36. sin^a — cos^a cos 2/3 = sin^^ — cos^/S cos 2a. 
 
 2 
 
 .37. seca-^^2_^^^2 + 2cos4a)r 
 
 38. From the expressions for cos(a + /3 + y), sin(a + iS + y) 
 
 and tan(a + i5 + y), deduce the values of cos 3a, 
 sin 3a and tan 3a. 
 
 39. 4 cos^a — 4 sin^a = 4 cos 2a — sin 2a sin 4a. 
 
 40. Express tan 4a in terms of tan a. 
 
 41. Express sin 5a in terms of sin a. 
 
 42. sin 3(a - 15°) = 4 cos(a - 45°)cos(a + 15°)sin(a - 15°). 
 
 43. tan 3a = tan a cot(30° - a)cot(30° + a). 
 
OF COMPOUND ANGLES. 41 
 
 COS p — COS a '1 
 
 45. 52i^±£^=cot^cot^«. 
 
 COS a — cos/3 2 2 
 
 46. sec(45° + a)sec(45° — a) = 2sec2a. 
 
 47. sin 2a sin 5a + sin 3a sin 10a = sin 5a sin 8a. 
 
 48. sin 2a(cos 3a + cos 7a + cos 1 la) = sin 6a cos 7a. 
 
 An o- Ti* sin a + 2 sin 3a + sin 5a 
 
 49. Simplify -.— - — , ^ . „ — ;— r— ^_-. 
 
 ^ "^ sin 3a + 2 sm 5a + sin 7a 
 
 -A o- vr cosa — cos 3a + cos 5a — cos 7a 
 
 oO. bimplity ^ — ; -. ^ =— . 
 
 '^ "^ cos 2a + cos 4a — cos 6a — 1 
 
 51. cos 25° — sin 5° = cos 35°. 
 
 52. sin 33° + cos 63° = cos 3°. 
 
 53. 4sin20°sin40°sin80° = sin60°. 
 
 54. Prove geometrically that cot(a — /3) = ^ — ^— ^ . 
 
 ^ ^ V A-/ cot/3-cota 
 
 55. Also that 2 cot 2a = cot a — tana. 
 Solve the equations : 
 
 56. isin0+^cos^ = -^. 
 
 57. tan(45° + ^) = 4tan(45°-e). 
 
 58. sill 20 cos = sin 0. 
 
 59. coslie + cos50 = cos3a 
 
 60. cos 0- cos 90 = ^2 sin 4a 
 
 61. sin 20 + sin = cos 20 + cos a 
 
 62. sin 30 + sin 50-sin 7a- sin = 2^2 sin 20 sin 6. 
 
 63. 4 cos(0 + 60°) -J2 = jQ-4! cos(0 + 30°). 
 
 Eliminate between the following equations : 
 
 64. sin0 — cos = (X, sin 20 = 6. 
 
 65. cos20 + cos0 = a, sin20 + sin0 = 6. 
 
 66. atan(0 + a) = c, 6tan(0 + /3) = d 
 
 67. cos 30 + sin 30 = a, cos 0- sin = 6. 
 
42 TRIGONOMETRICAL RATIOS 
 
 Examples V. 
 
 1. Prove geometrically, by means of Eucl. VI. 8, that 
 
 cos 75°=^^^ and cos 15° = ^5^^. 
 
 2. If tan Q = b/a, then a cos 20 +b sin 26 = a. 
 
 o. Express 4 sin a cos ^ cos y as the sum of four sines. 
 
 4. Find the maximum value of cos a + sin a. 
 
 5. Shew that cos 6 cos(a --0) is a maximum when 6 = Ja, 
 
 and that its value is then cos^^. 
 
 6. If a sin(e +a) = b am(d + fi), then 
 
 ^g^asiDa-6sin^ 
 6 COS p — a cos a 
 
 7. Shew that acos(0+a)+6cos(0+iS) may be written 
 
 in the form ccos(0 + y), and find c and y in terms 
 of a, b, a and /3. 
 
 8. V3 + tan 40° + tan 80° = ^3 tan 40°tan 80°. 
 
 9. sin^a + sin*2a = f(l — cos a cos 3a) — J sin 3a sin 5a. 
 
 10. If tan a = tm% then 2 tan 2a = tan 2/3 sin 2/3. 
 
 11. If cos(a — y)cos /3 = cos(a — /3+y), then tan a, tan /3, 
 
 tan y are in harmonical progression. 
 
 12. If cos(a-^)sin(y-^) = cos(a + /3)sin(y+^), then 
 
 cot S = cot a cot ^ cot y. 
 
 13. If tan I = ^^7^^ , then sin = ^. 
 
 14. If tan /3 = ^£_«.^, then tan(a - iS) = (1 - ^)tan a. 
 
 1 5. If cot ^(tan^a - 1) = 2 tan /3, then 
 
 sin(a + |5)sec a = cos a cosec(a — /3). 
 
 16. Shew that {2«+icosec 2'^'ri^f 
 
 = (2^cosec 2^a)2 + (2^sec 2^a)\ 
 
OF COMPOUND ANGLES. 43 
 
 irr Ti* COS a — COS i8 COS y ,, 
 
 17. If cos a = — —.—>,-. -^^ \ then 
 
 sin p sm y 
 
 1 — sec^a — sec2/3 — sec^y + 2 sec a sec /3 sec y = 0. 
 
 18. If tan0+tan0 = secO, find the relation between 6 
 
 and (p. . 
 
 19. If — a, 0, + a be three angles whose cosines are in 
 
 harmonical progression, then cos = ^2 cos J 
 
 20. If the angle ^J5(7 Ls divided by the straight line 5Q 
 
 into two angles, a and /?, and the circle ^P(7, 
 which touches AB and 5(7, is cut by BQ, in P, Q : 
 shew that sin a sin ^8 varies as the square of the 
 intercepted chord PQ of the circle. 
 
CHAPTER IV. 
 
 USE OF MATHEMATICAL TABLES. 
 
 38. The principal tables required in trigonometrical 
 calculations are: (1) a table of the logarithms of all 
 numbers from 1 to 100,000; (2) a table of the trigono- 
 metrical ratios of angles for every minute from 0° to 90° ; 
 and (3) a table of the logarithms of these ratios. 
 
 In the tables at the end of this book are given all 
 the logarithms of numbers, trigonometrical ratios and 
 logarithms of trigonometrical ratios that will be required 
 for the working of the examples given in the text. The 
 logarithms of numbers are printed in nearly the same 
 form as in books of Mathematical Tables. The reader 
 is, however, strongly recommended to obtain a collection 
 of tables, for example, Chambers' Mathematical Tables, 
 and to work exclusively with them. 
 
 (For the fundamental properties of logarithms and 
 their proofs, see C. Smith's Elementary Algebra, second 
 edition, or any other treatise on Algebra.) 
 
 Logarithms of Numbers. 
 
 39. The table of logarithms of numbers given at the 
 
 end of this book contains eleven columns. In the first 
 
 column, are the first four digits of the number, the fifth 
 
 44 
 
USE OF MATHEMATICAL TABLES. 45 
 
 and last digit being given at the head of each of the 
 remaining ten columns. In these columns, the mantissse 
 of the logarithms of the numbers are given to seven 
 places of decimals, the first three being given in the 
 second column only. The characteristic of the logarithm 
 is omitted, since it is easily determined by inspection of 
 the number. 
 
 For example, consider a number whose first four digits 
 are 2296. These digits are given in the first column. 
 In a line with them, and in the second column (headed 
 0), are the figures 3609719 ; these forming the mantissa 
 of the logarithm of 22960. In the same line, and in the 
 third column (headed 1), are the figures 9908, which form 
 the last four figures of the mantissa of the logarithm 
 of 22961, the three figures 360 being omitted, so that the 
 complete mantissa is 3609908. Again, in the same line, 
 and in the fourth column (headed 2), are the figures 
 0097, the bar above the figures * indicating that the last 
 of the three figures 360 is to be increased by 1 ; so that 
 the mantissa of the logarithm of 22962 is 3610097. 
 Thus, log 22960 = 4-3609719, 
 
 log 229610 =5-3609908, 
 
 log 2-2962 =0-3610097, 
 
 log 0-22963 =1-3610286, 
 
 log 0-00022964 = 4-3610475. 
 
 40. In the examples just given, the logarithms can be 
 found at once from the tables ; but, frequently, this is not 
 the case, the number whose logarithm is required lying 
 between two numbers in the tables. 
 
 The logarithm of 229613, for example, must lie be- 
 
 * Sometimes the bar is placed over the first figure only, thus : 0097. 
 
46 USE OF MA THEM A TICAL TABLES. 
 
 tween the logarithms of 229610 and 229620, i.e., between 
 5-3609908 and 5-3610097. To determine log 229613, we 
 make use of the Principle of Proportional Parts, which 
 may be thus stated : 
 
 If there be three numbers such that the difference 
 between any two of them is small in comparison with 
 either, then the difference between any two of the 
 numbers is proportional to the difference between their 
 logarithms. 
 
 This theorem, it should be mentioned, is only approxi- 
 mately true. The limits between which it is applicable 
 are investigated in a later chapter. 
 
 Applying this principle to find log 229613, we have 
 the difference between 229610 and 229620 is to the 
 difference between 229610 and 229613 as the difference 
 between log 229610 and log 229620 is to the difference 
 between log 229610 and log 229613. The difference be- 
 tween log 229610 and log 229620 is '0000189: let the 
 difference between log 229610 and log 229613 be denoted 
 by 5; then 10 :3 :: -0000189 : 5, 
 
 from which we find ^ = '0000057 
 to seven places of decimals. 
 
 log 229613 = log 229610+^ 
 
 = 5-6309908 + 00000057 
 = 5-6309965. 
 
 In practice, this example should be written out as 
 in the first of the following article. 
 
 41. Example 1.— To find log 229613. 
 
 log 229620 = 5-3610097, 
 log 229613=5-3609908 + 8, 
 log 229610=5-3609908, 
 
USE OF MA THEM A TWA L TA BLES. 47 
 
 10: 3:: -0000189: 8, 
 8= -0000057. 
 log 229613= 5-3609908 
 +0-000()057 
 = 5-3609965. 
 
 Example 2.— To find log 0-0356124 
 
 log 0-035613 =2-5516086, 
 log 0-0356124 = 2-5515964 + 8, 
 log 035612 =2-5515964, 
 10 : 4 :: 0-0000122 : 8, 
 
 8 = 0-0000049, 
 log 0-0356124= 2-5515964 
 + 0-0000049 
 = 2-5516013. 
 
 42. In a similar manner, we find a number whose 
 logarithm is known from the two numbers (given in 
 the table) whose logarithms have mantissse respectively 
 just greater and less than that of the given logarithm. 
 
 Example 3. — To find the number whose logarithm is 3*5970721. 
 3-5970806 = log 0-0039544, 
 3-5970721 = log (0-0039543 + 8), 
 3-5970696 = log 0*0039543, 
 110 : 25 :: 22 : 5 :: 00000001 : 8, 
 8 = 0-000^005-7-22 = 0-00000002, nearly, 
 
 3-5970721 = log 0*00395432, nearly. 
 
 43. Example 4.— To find the 61st root of 329 x 3079-i-425130 to 
 six places of decimals. 
 
 The logarithm of the required root 
 
 = 6\(log 329 + log 3079 -log 451230) 
 2-5171959 
 
 -^1 + 3-. 
 
 4884097-5-6543980/ 
 ^1 / 6-0056056 \ 
 ^V- 5-6543980/ 
 = ^^ of 0-351 2076 = 0-0057575. 
 
48 USE OF MA THEM A TIGAL TA BLES. 
 
 Now, 0-0057809 = log 1-0134, 
 
 0-0057575 = log( 1-0 133 + 8), 
 0-0057380 = log 1-0133, 
 429: 195:: -0001 : 8, 
 8=0-000045, 
 the required root = 1 '01 3345. 
 
 Logarithms of Trigonometrical Ratios. 
 
 44. It is unnecessary to give examples of finding by- 
 means of tables the trigonometrical ratios of angles; 
 the method being, in every respect but one, the same as 
 that for finding the logarithms of the same ratios. It 
 differs only on account of the form in which the ratios 
 and their logarithms are printed in the tables. 
 
 The cosines and sines of all acute angles, the tangents 
 of angles less than 45°, and the cotangents of angles 
 greater than 45°, being less than unity, the logarithms of 
 all such ratios have negative characteristics. To avoid 
 printing such characteristics, it is usual to add 10 to the 
 logarithm, forming the tabular logarithm of the ratio, 
 written L cos a, etc. The letter L is used as an abbrevi- 
 ation of the words ' tabular logarithm ' ; thus, 
 Xcos a = log cos a + 10. 
 In working with the tables, however, the tabular loga- 
 rithm should not be used : the 10 should be subtracted 
 from it whilst copying the logarithm. For example, we 
 should write at once, 
 
 logcos 57° 10' = 1-7341572, 
 
 log sin 1° =2-2418553, 
 
 W tan 49° 13' = 00641 556. 
 
USE OF MA TEEM A TICAL TABLES. 49 
 
 45. In finding the logarithms of trigonometrical ratios 
 which are not given exactly in the tables, it is necessary 
 to distinguish the cases in which the ratio increases or 
 decreases as the angle increases. The sine, secant, and 
 tangent of an angle increase as the angle increases, while 
 the cosine, cosecant, and cotangent decrease as the angle 
 increases (Art. 22), Taking, first, an example of the 
 former, let us find the logarithm of sin 16° 23' 27". 
 
 From the table, we find log sin 16° 23' = 1-4503452, and 
 log sin 16° 24' = 1-4507747. The required logarithm must 
 therefore lie between these values. To obtain it, we 
 make use of the Principle of Proportional Parts, as 
 enunciated above (Art. 40), with the requisite verbal 
 changes. 
 
 The limits between which the principle is applicable 
 in this case are investigated in chapter xv. : for the 
 present, it may be stated that the result will not be 
 accurate to the seventh place of decimals if the angles 
 differ by less than a few degrees from 0° or 90°. 
 
 Now, the difference between 16° 23' and 16° 24' is 60"; 
 and that between 16° 23' and 16° 23' 27" is 27". Again, 
 the difference between log sin 16° 23' and log sin 16° 24' is 
 0-0004295. Let the difference between log sin 16° 23' and 
 log sin 16° 23' 27" be denoted by S. Then 
 
 60 : 27:: 00004295 :^, 
 
 from which we find S = 00001933. 
 
 log sin 16° 23' 27" = log sin 16° 23'+^ 
 
 = T'4503452 + 00001933 
 
 = 1-4505385. 
 
 In practice, we should work this example as follows : 
 
 D 
 
51) USE OF MATHEMATICAL TABLES. 
 
 Example 5.— To find log sin 16° 23'_27". 
 
 log sin 16° 24' =1-4507747, 
 log sin 1 6° 23' 27" == 1 '4503452 + 8, 
 logsinl6°23' =14503452, 
 60 : 27 : : 0-0004295 : 8. 
 •0004295 
 
 9 
 
 20 ) -0038655 
 •0001933 
 log sin 16° 23' 27"= 1-4503452 
 + 0-0001933 
 = T-4505385 
 
 46. Similarly, we find an angle the logarithm of whose 
 sine is given, from the two angles (given in the table) the 
 logarithm of whose sines are respectively just greater and 
 just less than the given logarithm. 
 
 Example 6. — To find the angle the logarithm of whose sine is 
 i -9658931. 
 
 1 -9659285 = log sin 67° 36', 
 1-9658931 =log sin 67° 35' S", 
 1 -9658764 = log sin 67° 35', 
 521:167::60:S. 
 
 167 
 60 
 521 ) 10020 ( 19", nearly. 
 521 
 4810 
 4689 
 121 
 r9658931 =log sin 67° 35' 19". 
 
 47. Example 7.— To find log cos 53° 14' 51". 
 log cos 53° 14' =1-7771060, 
 log cos 53° 14'51"=T-7771000-S, 
 log cos 53° 1 5' =1-7769369, 
 
USE OF MATHEMATICAL TABLES. 51 
 
 .'. 
 
 -60 :-51 :: -0001691: 8. 
 
 
 •0001691 
 51 
 
 
 1691 
 
 8455 
 
 
 60 ) -0086241 
 
 
 •0001437 = 8. 
 
 • • 
 
 log cos 53° 14' 51"= 1-7771060 
 
 - 0-0001437 
 
 = 1-7769623. 
 
 Example 8.- 
 1-9447435. 
 
 —To find the angle the logarithm of whose cosine is 
 1-9447862= log cos 23° 17', 
 
 
 1-9447435 = log cos 23° 17' 8", 
 
 
 1-9447182 = log cos 23° 18', 
 
 ,*. 
 
 680: 427:: -60 -.-8. 
 
 
 427 
 3 
 
 34 ) 1281 ( 38", nearly. 
 102 
 
 261 
 1-9447435 =log cos 23° 17' 38", nearly. 
 
 Examples VLa. 
 
 Find the logarithms of the following numbers : 
 
 1. 249317. 4. 9152-06. 
 
 2. 0-751204. 5. 539-005. 
 
 3. 39172500. 
 
 Find the numbers whose logarithms are : 
 
 6. 5-2471903. 9^ 2-3058769. 
 
 7. 8-7192855. 10. 21572693. 
 
 8. 4-2145028. 
 
52 VSE OF MATHEMATICAL TABLES. 
 
 Find the values of : 
 
 _, 72x5301 , „ , CA ' ^ 
 
 6 19x70 025' P ^^^^ decimals. 
 
 12. 4/(3914-26 X 130-72), to 4 places of decimals. 
 
 ,^ (1-0012)2 X (-059)^ 
 
 13. ^ '7 a.qi9q\4 — > ^° ' places oi decimals. 
 
 Find the logarithms of: 
 
 14. sin 17° 13' 28". 17. sin 54° 15' 2". 
 
 15. tan 19° 57' 12''. 18. tan 17° 41' 39". 
 
 16. cos 78° 24' 50". 19. cos 11° 3' 12". 
 
 Find the values of a, when : 
 
 20. log sin a = 1-5127043. 23. log sin a = 1-8318514. 
 
 21. log tan a = 1-4587148. 24. log tan a = 0-7974903. 
 
 22. log cos a = 1-9849418. 25. logcos a = r9447285. 
 
 Find: 
 
 26. sin 12° 3' 12". 28. cot 52° 41' 21". 
 
 27. cos 71° 14' 39". 
 
 Find the values of a, when : 
 29. sin a = -37 14051. 30. cos a = "2. 
 
 Examples VI. b. 
 Find the logarithms of the following numbers : 
 
 1. 632503. 4. 0-000531058. 
 
 2. 00513517. 5. 7912-02. 
 
 3. 73-1459. 
 
 Find the numbers whose logarithms are : 
 
 6. 0-2175004. 9. 5-3971142. 
 
 7. 1-3721153. 10. 4-2456173. 
 
 8. 1-3190725. 
 
USE OF MATHEMATICAL TABLES. 53 
 
 Find the values of: 
 
 .. 035 X 1-2502 , ^ , rA ' ^ 
 
 11. .,-, -, o — ,, ^r,m > to 9 places or decimals. 
 
 12. a/ L^^3 ,^ J, to 5 places of decimals. 
 
 13. (0001528x517)7, to 6 places of decimals. 
 Find the logarithms of: 
 
 14. sin 52° 15' 17''. 17. sin 5° 51' 22". 
 
 15. tan 74° 11' 9". 18. tan 14° 19' 38". 
 
 16. cos 32° 30' 14". 19. cos 12° 43' 18". 
 
 Find the values of a, when : 
 
 20. log sin a = 1-9703652. 23. log sina = ro012301. 
 
 21. log tan a = 0-5175023. 24. log tan a = 1-8115891. 
 
 22. log cos a = 1-6838142. 25. log cos a = 1-5354010. 
 
 Find : 
 
 26. sin 19° 12' 37". 28. cosec 24° 14' 9". 
 
 27. cos 43° 15' 5". 
 
 Find the values of a, when : 
 29. sin a = -5502471. 30. cos a = "8026140. 
 
CHAPTER V. 
 
 SOLUTION OF EIGHT- ANGLED TEIANGLES AND 
 PEACTICAL APPLICATIONS. 
 
 Solution of Right-Angled Triangles. 
 
 48. In every triangle there are six elements, the 
 three sides and the three angles. If any three of these 
 elements, except the three angles, be given, they are, as a 
 general rule, sufficient to enable the other three to be 
 determined. The process by means of which the remain- 
 ing elements are found is called the solution of the 
 triangle. In the present chapter, we confine ourselves to 
 the solution of right-angled triangles. 
 
 It is obvious that, if two angles of a triangle be known, 
 the third may be determined from the fact that the three 
 angles are together equal to two right angles. And, in 
 a right-angled triangle, if one acute angle be known, the 
 second may be found in the same way. 
 
 We have thus four cases to consider, in which we are 
 given (1) the hypotenuse and one acute angle; (2) one 
 side and one acute angle; (3) the hypotenuse and one 
 side; and (4) the two sides. 
 
 49. If ABC be any triangle, the length of the sides 
 
 BG, GA, and AB, opposite respectively to the angles J., 
 
 j5, and 0, will be denoted by the letters a, 6, and c. 
 
 54 
 
En^Fish Miles 
 
RIGHT-ANGLED TRIANGLES. 55 
 
 50. Case I. — The hypotenuse and one acute angle being 
 given, to find the two sides and the other angle. 
 
 Let C be the right angle, and let the hypotenuse c and 
 the angle A be given. ^ 
 
 We have 4+^ = ^0", 
 
 B = 90°-A. 
 Also, a/c = sin A and b/c = sin B, 
 
 a = c sin A and 6 = c sin i?, a 
 .-. log a = log c+ log sin J. and log6 = logc + logsin5. 
 51. Example l.~If c = 3701 and ^ = 41° 13' 24", to find a, b ami B. 
 B=m°- 41° 13' 24" =48° 46' 36". 
 log sin 41° 14' =1-8189692, 
 log sin 41° 13' 24" = 1-8188250 + 5, 
 log sin 41° 13' =1-8188260, 
 60'': 24":: -0001442: 8, 
 8= -0000577, 
 log sin 41° 13' 24"= 1-8188827. 
 
 log a=log 3701 + log sin 41° 13' 24" 
 = 3-5683191 
 + 1-8188827 
 = 3-3872018. 
 Now, 3-3872118 = log 2439-0, 
 
 3-38720 1 8 = log(2438-9 + S), 
 3-3871940 = log 2438-.9, 
 178:78 :: -1 : 8, 
 
 8=0-04, nearly, 
 « = 243^-94. 
 Similarly, we find log sin 48° 46' 36" =1-8763025, log 6 = 3-4446216, 
 and therefore 6 = 2783-69. 
 
 i?=48°46'36",] 
 a = 2438 -944, ^ 
 6=2783-69. J 
 
 52. Case II. — One side and one acute angle being 
 given, to find the hypotenuse, the other side and the 
 other angle. 
 
56 RIGHT-ANGLED TRIANGLES 
 
 Let G be the right angle, and let the side a and the 
 angle B be given. 
 
 Then, ^ = 90' -5. 
 
 Also, cja = sec B and hja = tan B, 
 
 c = a sec jB and 6 = 66 tan 5, 
 log c = log a H- log sec B and log 6 = log a + log tan B. 
 
 53. Example 2.— If a = 374 and B=2V 34', to find c, 6 and A. 
 
 ^ = 90°-2r34'=68°26'. 
 Now, c = a sec i5, 
 
 log c = log 374 + log sec 21° 34' 
 = 2-5728716 
 + 0315215 
 = 2-6043931. 
 Now, 2-6043989 = log 402-16, 
 
 2-6043931 = log(402-15 + 8), 
 2-6043881= log 402-15, 
 
 108 : 50:: -01 : 8, 
 8 =-005, nearly, 
 c = 402-155. 
 Again, h = a tan 5, 
 
 log 6 = log 374 + log tan 2 r 34' 
 = 2-5728716 
 + 1-5968776 
 = 2-1697492. 
 Now, 2-1697626 = log 147-83, 
 
 2-1697492 = log(147-82 + 8), • 
 2-1697332 = log 147-82, 
 
 294 : 160:: -01 : 8, 
 8 =-005, nearly, 
 6=147-825. 
 A = i 
 c- 
 
 54. Case III. — The hypotenuse and one side being 
 given, tojind the other side and the tivo acute angles. 
 
I 
 
 AND PRACTICAL APPLICATIONS. 57 
 
 Let C be the right angle, and the hypotenuse c and the 
 iide a be given. 
 ' Now, c2 = a2 + &^ 
 
 62^c2-a2 = (c + aXc-«), 
 log6 = Hlog(c+«) + log(c-a)}. 
 Also, H\nB = hlc, 
 
 log sin B = log 6 — log c. 
 
 55. Example 3. — If c = 569, and a =435, to find b, A, and B. 
 
 b = J{c + aXc - «) = Vl004 X 1 34, 
 log6=Klog 1004 + log 134) 
 ^i/ 3-001 7337 \ 
 ^V + 2-1271048/ 
 = 1(5-1288385) 
 = 2-5644193. 
 Now, 2-5644293 = log 366-80, 
 
 2-5644193 = log(366-79 + 8), 
 2-5644175 = log 366-79, 
 
 118: 18:: -01: 6, 
 8 = -0015, nearly, 
 6 = 366-7915. 
 Again, sin B=blc, 
 
 log sin ^ = log 6 - log 569 
 
 = 2-5644193 
 
 -2-7551123 
 
 = 1.8093070. 
 
 Now, 1 -80941 89 = log sin 40° 9', 
 
 1 -8093070= log sin 40° 8' 8", 
 1-809269 1 = log sin 40° 8', 
 
 1498: 379:: 60": 8, 
 8=15", nearly, 
 
 5=40° 8' 15", and therefore yl = 49° 51' 45", 
 6 = 366-7915 
 .4 = 49° 51' 45" 
 5=40° 8' 15" 
 
58 iii(!iiT-Ai\(Uj':ii rniAMUJCS 
 
 66. Case IV. — The. Uvo sidcH co7itainin<j the right 
 aiujta heiiKj <jlven, to find the hypotenuse and the two 
 acute angles. 
 
 Lot C bo tho right angle, and let the sidoH a and h bo 
 given. 
 
 Then, tan A = a/6, 
 
 log tan A = log a — log h. 
 From this equation A can be fouud and B from the 
 equation B=iW — A. 
 
 Also, cja = cosec A , 
 
 c=acosecil, 
 log c = log a + log cosec A . 
 
 57. Example 4.— If a -371 and 6-504, to find 0, /I, and B. 
 
 tan yl -a/6-371/504, 
 
 log tan i( - log 371 - log 604 
 
 - 25693739 
 -2-7024305 
 
 - 1-8669434. 
 Now, T'8070937 - log tan 36" 22', 
 
 1 •8(J60434 - log tan 30° 2 1' 6", 
 i-8668291-logtan36°21', 
 2646: 1143:: 60": 8", 
 8-26", nearly, 
 il-36'21'20". 
 5-53' 38' 34". 
 Again, 0— aaec/?, 
 
 log c - log 37 1 + log Bee 53° 38' 34". 
 Now, log Hec 53" 39' -0-2271532, 
 
 log HOC n:r 38' 34" -0-226981 6 + 8, 
 log sec 53° 38' -0*2269815, 
 
 60": 34":: -0001717: 8, 
 8 --0000973, 
 log sec 63' 38' 34" -0-2270788. 
 
AJ^D VRAGTICAL APPLICATIONS, 50 
 
 logo- 2*5693730 
 +0-2270788 
 - 2-7964627. 
 N o\y, 2-7064564- log 625-88, 
 
 2-7964527 - log(625 -82 + S), 
 2-7064404-log 625-82, 
 70 ! 33 :: -01 : 8, 
 fi-«-005, ueftrly, 
 0-625-825. 
 0-625- 
 il-36' 
 J5-53" 
 
 15-825 ] 
 j»2r26" Y 
 r 38' 84" J 
 
 Examples VII. a. 
 
 (-Yoee.— -Angles shouUl bo fouiul to tho nearest second, 
 sides to six signilicai^t iignvOvS.) 
 
 Solve the following triangles, having given 0^= 00 \ an»l ; 
 
 1. 0=172,^1 = 85'' 27'. 7. a=821, 6=1002. 
 
 2. = 3041, B = 24M 9'. 8. a = 792, 6 = 562. 
 
 X a = 004, B = 41° 17'. 9. a = 249, A = 02*^ 15' 42". 
 
 1. 6 =40128, ^ = 64* 8'. 10. o =204137, ^, = 91423. 
 
 ». o=712,a=408. 11. c =898, 2^=15^4' 8". 
 
 (I. =3052, 6 = 796. 12. tt = 401, 6=602. 
 
 Examples VII. b. 
 (See note at the head of Exaiuph^s \\\ \.) 
 Solve the following triangles, having' K^^ ^'^ ^' = ^^^^ \ ^^^ • 
 
 1. o-4018,il-7ri6r. 7. a =518, 6 = 927. 
 
 2. = 5702, li = 59** 48'. 8. a = 8012, 6 = 5148. 
 
 'X a=48, ^=10' 12'. 9. a=700, B^Sir 14' 16". 
 
 k 6-619, 5=25'* 48'. 10. a-69, 6=78. 
 
 >. =6941, a=1026. 11. o =621, 6=507. 
 
 (). =57102, 6=31040. 12. c =7205, £-31" 14'66\ 
 
60 RIGHT-ANGLED TRIANGLES 
 
 Practical Applications. 
 
 58. Several simple problems in the measurement of 
 heights and distances may be solved as applications of 
 the preceding propositions. The principal instruments 
 required for the purpose, and the method of using them, 
 are described in Chapter IX. They are the chain, for 
 measuring distances, and the theodolite, for measuring 
 angles in a vertical or horizontal plane. 
 
 59. Definitions. — The angle of elevation (occasionally 
 called the elevation or the altitude) of any point above 
 tKe horizontal plane through the observer's eye is the 
 inclination to the plane of the straight line joining the 
 point and the observer's eye. If the point be below this 
 plane, the angle is then called the angle of depression 
 (or the depression) of the point. 
 
 60. To find the height of a toiver standing on a 
 horizontal plane, the foot of the tower being accessible. 
 
 Let BC represent the tower. 
 From the foot, C, measure the 
 length of any line, GA, called a base- 
 line, in the horizontal plane on 
 which the tower stands; and at the 
 other end, A, of this line measure 
 the angle of elevation, BA C, of the top of the tower. 
 Now, BCIAG=t8inBAG, 
 
 BG= AG tarn BAG, 
 i.e., the height of the tower above the observer's eye is 
 equal to the length of the base-line multiplied by the 
 tangent of the angle of elevation of the top of the tower 
 at the other end of the base-line. 
 
AND PRACTICAL APPLICATIONS. 
 
 61 
 
 61. To find the height and distance of a tower stand- 
 ing on a horizontal plane, the foot of the toiuer being 
 inaccessible. ^ 
 
 Let CD represent the tower. 
 
 In the horizontal plane on 
 which the tower stands, measure 
 the length of any base-line, AB, 
 directed towards the foot of the 
 tower, D ; and, at each end of ^ 
 the base-line, measure the angles of elevation, CAD and 
 GBD, of the top of the tower. 
 
 Now, AD = CD cot A and BD = CD cot B, 
 AB = AD - BD=^CD(cot A- cot B) 
 
 — cnf^^^ ^ — ^^^ ^^ 
 
 = CD. 
 
 Vsin A sin BJ 
 sin B cos A — cos B sin A 
 
 sin A sin B 
 
 62. 
 
 = CD ^^^(^-^) 
 ' sin A sin B 
 
 CD = AB sin A sin B/smiB- A). 
 To find the breadth of a river, or the distance of an 
 
 inaccessible object on a horizontal plane. 
 
 Let a base-line, AB, be 
 measured close to one bank 
 of the river, the ends, A 
 and B, being marked by 
 prominent objects, such as 
 
 trees. Selecting some ob- ^ J-> B 
 
 ject, C, close to the other bank of the river and visible 
 from both A and B, measure the angles CAB and CBA. 
 Draw CD per])endicular to AB. 
 
62 
 
 RIGHT-ANGLED TRIANGLES 
 
 Now, AD = CD coi A and BD = CD cot B, 
 AB=AD+BD = GD(cot ^ + cot B) 
 
 * sin A sin B' 
 GD=ABBmA^\nBl^m{A+B). 
 
 63. Example 1. — If the base-line, AB^ be 127 yards, and the 
 angles CAB and CBA be 41° 27' and 35° 14', to find the breadth of 
 the river to the nearest yard. 
 
 log CD=\og AB + log sin A + log sin B - log sin(^ + B) 
 
 =log 127 + log sin 41° 27' + log sin 35° 14' - log sin 76° 41' 
 = 2-1038037 
 
 + 1-8208358 
 
 +1-7611063-1-9881628 
 = 1-6857458 
 
 -19881628 
 = 1-6975830 
 = log 49-84, nearly, 
 the breadth of the river is about 50 yards. 
 
 64. Dip of the Horizon. — The dip of the horizon at 
 any point above the earth's surface is the angle of depres- 
 sion, at that point, of a point on 
 the horizon. 
 
 Let the circle in the annexed 
 figure represent a section of the 
 earth (supposed spherical) through 
 the centre (7, and a point above 
 the earth's surface. Draw OB 
 perpendicular to OC, OA touching 
 the circle, and Al) perpendicular 
 to 00, Then, the angle BOA is 
 the dip of the horizon at 0. 
 Let c denote the radius of the earth, and h the height 
 of above its suiface, so that OG=c-\-h. 
 
AND PRACTICAL APPLICATIONS. 
 
 63 
 
 Now, angle 50^ = complement of angle AOG 
 = angle AGO, 
 and Qo^AGO = ACIOG=cl{c+h). 
 
 the dip of the horizon at = cos~^ y-. 
 
 65. The following example illustrates a further applica- 
 l^ion of the methods explained in this chapter : — 
 
 Example 2. — The dip of a stratum is 8 degrees to the east. 
 Find its apparent dip in the direction degrees south of east. 
 
 Let A be any point on the surface of the stratum and AB a 
 vertical line of any length through A. 
 From B, draw horizontal lines, BC and 
 BD^ in directions east and 6 degrees 
 south of east respectively, meeting the 
 surface of the stratum in the points C 
 and D. 
 
 Then, angle ACB=8, angle CBD^O, 
 and angle ADB (denoted by a) is the 
 apparent dip of the stratum in the given 
 direction. Also, the angle BCD is a 
 right angle, since CD is a horizontal 
 line on the surface of the stratum. 
 
 Now, BD=BCsece 
 
 =AB cot 8 sec 6. 
 tan a = ABjBD = AB/AB cot S sec 9 
 = tan 8 cos 6. 
 
 ^o-f ^ 
 
 t 
 
 Examples VIII. a. 
 
 (Note. — In the following examples, unless otherwise 
 stated, the ground is supposed to be level, and the height 
 of the observer's eye above the ground is not to be taken 
 
J: 
 
 64. RIGHT-ANGLED TRIANGLES 
 
 into account. See also the note at the head of Examples 
 VILA.) 
 
 1. The angle of elevation of the top of a vertical cliff at 
 
 a point 660 feet from its foot is 34° 16', find the 
 height of the cliff. 
 
 2. Find the length (to the nearest inch) of the shadow 
 of a vertical flagstaff, 53 feet high, the altitude 
 of the sun being 22°. 
 
 3. At a spot 67 ft. distant from the base of a chimney, 
 
 the angle of elevation of its top is 23° 10'; find 
 the height of the chimney to the nearest inch, the 
 observer's eye being 5 ft. 6 ins. above the ground. 
 
 4. The angle of elevation of the cairn of a mountain, 
 
 2314 feet high, from a point on the sea-level 
 is 11° 14'; find, to within a hundredth of an inch, 
 the distance between the two points as represented 
 on a map drawn on a scale of 6 ins. to the mile. 
 
 5. A ship sailing due east at the rate of 10 miles an 
 
 hour is observed from a lighthouse to bear due 
 north at a certain moment, and, half an hour later, 
 to bear N. 32° 20' E. Find the distance between 
 the ship and the lighthouse in both positions. 
 
 6. A statue is placed on the top of a column. At a 
 
 point on the ground, 1 24 feet distant from the base 
 of the column, the angles of elevation of the tops 
 of the statue and column are 45° 27' and 43° 12', 
 respectively ; find the heights of the statue and 
 column to the nearest inch. 
 
 7. Find the height of a chimney, when it is found that 
 
 walking towards it 100 feet along a straight line 
 through its base, changes the angle of elevation 
 from 24° 40' to 46° 10'. 
 
AND PRACTICAL APPLICATIONS. 65 
 
 8. From the top of a hill two consecutive milestones 
 
 are seen on a horizontal road running directly 
 Mk from its base. Their angles of depression are 
 
 m. found to be 14° 3' and 3° m'. Find the height 
 
 W of the hill to the nearest foot. 
 
 9. Three consecutive milestones on a road running east 
 
 and west are seen from a hill due south of the 
 first milestone. The second milestone bears N. 
 24° 30' E. Find the distance of the hill from the 
 road, and the bearing from the hill of the third 
 milestone. 
 
 10. At two points 1200 yds. apart on a road running 
 
 east and west, the bearings of a spire are observed 
 to be N. 57° E. and N. 43° W. ; find the distance 
 of the spire from the road. 
 
 11. The centre of the base of a tower, surmounted b}?- a 
 
 spire, lies directly between two stations 200 feet 
 apart, and the angles of elevation of the top of the 
 spire at these two points are 43° 17' and 31° 23'. 
 Find the height of the top of the spire. 
 
 12. A mountain, 4360 feet high, rises from a narrow 
 
 peninsula. At two points on the sea-level, on 
 opposite sides of the mountain, the angles of 
 elevation of its summit are 10° 50' and 13° 12'. 
 Find the distance between these points, supposing 
 the straight line joining them to pass vertically 
 below the top of the mountain. 
 
 13. The dip of a stratum is 38° in the direction K 47° E. 
 
 Find its apparent dip in the direction E. 5° S. 
 
 14. Find the dip of the horizon at the top of a mountain 
 
 15000 feet high, the radius of the earth being 
 4000 miles. 
 
66 RIQHT-ANQLED TRIANGLES 
 
 15. An observatory and distant chimney are 3520 feet 
 
 apart. The angle of depression at the observatory 
 of the top of the chimney was found at a given 
 moment to be 5° 17' 2", and some years later to be 
 5° 19' 23". Find, to within a hundredth of an inch, 
 the amount of subsidence of the ground on which 
 the chimney is built, supposing the observatory to 
 be stationary. 
 
 16. A base-line, I feet long, is drawn from a point on a 
 
 horizontal plane in a direction at right angles to 
 the line joining that point to the base of a tower 
 standing on the plane. The angle of elevation of 
 the top of the tower from the two ends of the base- 
 line are 30° and 18°. Find the height of the tower. 
 
 17. The angle of elevation of a column, as viewed from a 
 
 station due north of it, being a, and, as viewed 
 from a station due east of the former and at a dis- 
 tance c from it, being /3 ; shew that the height of 
 
 ,1 1 . c sin a sin /3 
 
 the column is •- =-. 
 
 {sin(a + /3)sin(a-^)}* 
 
 18. At the top of a castle, which stood on a hill near the 
 
 sea-shore, the angle of depression of a ship at 
 anchor was observed to be 5° 10'; and, at the 
 bottom of the castle, it was observed to be 4° 25'. 
 If the castle be 60 feet high, find the horizontal 
 distance of the vessel, and the height of the hill 
 above the sea-level. 
 
 19. A flagstaff, a feet high, stands on a plane inclined 
 
 towards the south at an angle of a degrees. If the 
 altitude of the sun at noon be ^ degrees, find the 
 length of the shadow then cast by the flagstaff on 
 the plane. 
 
AND PRACTICAL APPLICATIONS. 67 
 
 Find the length of the shadow, if a = 50 feet, 
 a = 12°, /5=37°. 
 20. Three consecutive milestones are situated on a road 
 inclined at an uniform angle to the horizon, the 
 road running east and west. From a station due 
 south of the lowest milestone, and in the same 
 horizontal plane with it, the angles of elevation of 
 the second and third are found to be a and ^, 
 respectively. Find the inclination of the road and 
 its distance from the station. 
 
 Shew that tan/3 must lie between tana and 
 2 tan a. 
 
 Examples VIII. b. 
 
 (See note at the head of Examples YIII. A.) 
 
 1. At a point 300 feet distant from the base of a monu- 
 
 ment, the angle of elevation of the top is 18° 42'; 
 find the height of the monument. 
 
 2. The shadow cast on the ground by a statue on the 
 
 top of a column is 14 "2 2 feet long, the altitude of 
 the sun being 41° 10' ; find the height of the statue, 
 and also that of the column if the shadow of the 
 head be 56*14 feet from the foot of the column. 
 
 8. The angle of elevation of the top of a cathedral tower 
 is 18° 30' at a point on the ground 320 feet from 
 its base ; find the height of the tower. 
 
 4. The distance on a map drawn to the scale of 6 miles 
 to the inch, of the points representing the summit 
 of a mountain, 2412 feet high, and a point 134 feet 
 above the sea-level, is 32*54 inches ; find the angle 
 of elevation at this point of the mountain summit. 
 
68 RIGHT-ANGLED TRIANGLES 
 
 5. A balloon passes vertically over two points, A and B, 
 
 12 miles apart. When vertically over A, its angle 
 y of elevation at J? is 12° 40' ; and, when vertically 
 
 over B, its angle of elevation at ^ is 14° 12'. Find 
 the inclination of its path to the horizon. 
 
 6. At the top of a castle, built on the edge of a vertical 
 
 cliff, the angle of depression of a ship anchored at 
 a distance of one mile from the foot of the cliff is 
 4° 52' ; at the bottom of the castle, it is 4° 2'. Find 
 the heights of the castle and the cliff. 
 
 7. In order to ascertain the height of a tower, a base- 
 
 line, 93 feet long, was measured in a direct line 
 through its base, and the angles of elevation at the 
 two ends of the line were found to be 33° 20' and 
 55° 54'; find the height of the tower. 
 
 8. From the top of a hill, the angle of depression of two 
 
 consecutive milestones on a road running directly 
 towards the hill are 14° 3' and 3° 56'; find the 
 height of the hill. 
 
 9. Find the distance between two objects lying in a 
 
 horizontal straight line from the foot of a tower, 
 84 feet high, when their angles of depression at 
 the top of the tower are 57° 30' and 25° 15'. 
 
 10. To determine the breadth of a river, a base-line, 
 
 516 feet long, is measured close to one bank, and 
 it is found that the direction of this base-line is 
 N.W. and S.E. From the two ends of the base- 
 line, the bearings of a tree on the opposite bank 
 are observed to be E. 7° S. and N. 10° W. Find 
 the breadth of the river. 
 
 11. ^ and B are two stations, 2 miles apart, A being due 
 
 north of B. At A^ the altitude of a cloud is 32° 
 
AND PRACTICAL APPLICATIONS. 69 
 
 to the S. ; and, at B, it is 41° to the N. Find the 
 height of the cloud. 
 
 12. On a map, drawn to the scale of 25 inches to a mile, 
 
 the distance between a church spire and a road 
 running N. and S. is represented as 9*5 inches. 
 Find the actual distance between two points on 
 the road at which the spire bears N. 14° E. and 
 N. 35° E. 
 
 13. The dip of a stratum is 18° N. Find its apparent 
 
 dip in the direction W. 10° N. 
 
 14. Find the dip of the horizon at the top of a mountain, 
 
 17500 feet high, the radius of the earth being 
 4000 miles. 
 
 15. The summit of a spire is vertically over the centre 
 
 of a horizontal square enclosure, whose side is 
 a feet long ; the height of the spire is h feet above 
 the level of the square. If the shadow of the spire 
 just reaches the corner of the square when the 
 sun has an altitude 6, shew that h,^2 = a tan 0. 
 Find h, having given a = 1000 feet, = 27° 29' 48''. 
 
 16. The angle of elevation of a steeple at a place due 
 
 south of it is 45°, and at another place due west 
 of the former it is 15° ; shew that the height of 
 the steeple is |(x(3i — 3"*), a being the distance 
 between the places. 
 
 17. A balloon at starting is a miles north of an observer, 
 
 and, after travelling due east for a given interval, 
 
 (is seen by him in a direction degrees E. of N. 
 and at an altitude of a degrees. Find the height 
 of the balloon at this instant. 
 A building on a square base, A BCD, has two of its 
 sides, AB and CD, parallel to the bank of a river. 
 
70 RIGHT-ANGLED TRIANGLES 
 
 An observer, standing on the river's bank in the 
 same straight line with DA, finds that the side 
 AB subtends at his eye an angle of 45°. Having 
 walked a yards along the bank, he finds that the 
 side DA subtends an angle sin"^J. Prove that 
 the length of each side of the building is .aj^^ 
 yards. 
 
 19. If the dip of the horizon at the summit of a moun- 
 
 tain, 3 miles high, be 2° 13' 27", find the diameter 
 of the earth. 
 
 20. A statue, 10 feet high, stands on the top of a column, 
 
 and, at a point in the horizontal plane through 
 the base of the column, the angles of elevation of 
 the top of the statue and the top of the column 
 are 32° 20' and 30°, respectively. Find the height 
 of the column. 
 
 Miscellaneous Examples. — I. 
 
 a. 
 
 1. Reduce 95741'' to degrees, and find the number of 
 
 degrees in the angle of a regulg-r dodecagon. 
 
 2. cot^a — cos^a = cot^a cos^a. 
 
 8. Solve the equation cosec20 = l + cot0. 
 
 4. If cos a = ff and tan ^8 = H, find the value of cos(a — /8) 
 
 and sin (a + /3). 
 
 5. (cos a + sin a)* + (cos a — sin a)^ = 3 — cos 4a. 
 
 6. cos 60° + cos 72° = cos 36°. 
 
 7. If the shadow of a column 120 feet high be 164 feet 
 
 long, find the altitude of the sun. 
 
 /3. 
 1. Each angle of a regular polygon contains 174°; find 
 the number of its sides. 
 
AND PRACTICAL APPLICATIONS. 71 
 
 2. sin2a(4 cos^a - 1)^ + cos^aC^ cos^a - 3)2 = 1. 
 
 3. If cot a = 2 - JZ, then sec a = jQ-\- jj2. 
 
 '4. The tangents of 60°, 45°, and 15° are in arithmetical 
 progression. 
 
 K cu irx. I /. X sin(a-/3) + sina + sin(a + /3)_sin a 
 
 5. Shew that: (i.) -7—7 ^^f- — . ; — ^^ — ro\~- • 
 
 ^ ' sin(y-/3) + siny + sin(y + /3) sin y 
 
 (ii.) cos 40° + cos 80° = cos 20°. 
 
 6. Solve the triangle in which c = 14157, 6 = 10253, 
 
 (7=90°. 
 
 7. A tower stands on level ground. At two points in 
 
 the same straight line through the foot of the 
 tower and 44 yards apart, the angles of elevation 
 of the top are 26°14' and 63°47'. Find the height 
 of the tower. 
 
 y- 
 
 1. The angles of a triangle are in arithmetical progression, 
 and the difference between the greatest and least 
 is equal to half the angle of a regular pentagon ; 
 find the number of degrees in each angle. 
 
 2- { ^(cosec a + cot a) - ^(cosec a - cot a) } ^ = 2 (cosec a - 1 ). 
 
 8. Solve the equation 8 cos + 4 sec = 7. 
 
 4. Shew that tan 36° = V(5- 2^5). 
 
 5. Express cos 7a in terms of cos a. 
 
 6. Simplify si" 3a + 2 sin 5a + sin 7a ^„j ^^^^ ^^^^ 
 
 sin 5a + 2 sm 7a + sm 9a 
 
 sin2(22i° + |)-sin2(22i°-|) = ^sina. 
 
 7. When the altitude of the sun is 32°40', the cross of the 
 
 top of a church-spire casts a shadow 18 feet long 
 on level ground. Find the length of the cross to 
 the nearest inch. 
 
72 RIGHT-ANGLED TRIANGLES 
 
 S. 
 
 ^ cos a , cos^ _ cos a , cos ,8 
 
 sina + cos^ sin /3 — cos a sin a — cos ^8 sin^+cosa* 
 
 2. The sine of a certain angle is K^+^g) ^ : find the 
 
 values of its other trigonometrical ratios. 
 
 3. If tan a = J and tan /3 = J, find the values of tan(2a + 1^) 
 
 and tan(2a — /3). 
 
 4. (1 + cot a + cosec a)(l + cot a — cosec a) = cot ^ — tan J 
 
 5. Shew that: (i.) sin(GO° + a) - sin(60° - a) = sin a. 
 
 ,. . . cos 2a — cos 4a cos a — cos 3a sin a 
 
 sin 4a — sin 'la sin 3a — sin a cos 2a cos 3a" 
 
 6. Solve the triangle in which a = 1025, 6 = 2531, = 90". 
 
 7. A lighthouse, 65 feet high, is built on the top of a cliff*. 
 
 The angles of depression of a ship from the top 
 and bottom of the lighthouse are 14°30' and 12°46'. 
 Find the height of the cliff" and the distance of the 
 ship from it. 
 
 e. 
 
 1. If the measures of the angles of a triangle, referred to 
 
 1°, 100', and 10000" as units, be in the proportion 
 of 2, 1, 3, find the angles. 
 
 2. cosec^a — cot^a = 3cosec2acot2a + l. 
 
 3. Solve the equation cos Q sin 0+cos = cos2^+sin 0. 
 
 4. Find the values of cos a and sin a, when cos 2a = -^. 
 
 5. 2-%in « - 2- %in 2|ri = 2«sin |.(sin ^^ . 
 
 6. Eliminate between tan 20 = a and cot = 6. 
 
 7. A cloud just grazing the top of a mountain 2310 feet 
 
 high, is seen at an altitude of 30° 15' by a man at 
 the sea-level. It is driven along by the wind at 
 
AND PRACTICAL APPLICATIONS. 73 
 
 the same height and directly away from him, and, 
 twenty minutes later, he finds its altitude to be 
 12°S0'. Find the velocity of the wind in feet per 
 second. 
 
 1. If tan a = ^ \ find the other trigonometrical 
 
 ratios of a. 
 
 2. Eliminate between 
 
 tan + cot = a and c cos = 6 — c sin 6. 
 
 3. cotra — cot(r+l)a = 
 
 sm a 
 
 sin7"asin(r+l)a 
 
 l-6tan2| + tan*| 
 4. cos 2a = 
 
 (l + K)^ 
 
 5. Shew that: (i^ ^^^^^4^^±^^^ = 4cos2acos4a. 
 
 tan oa — tan da 
 
 (ii.) tan 70° + tan 20° = 2 sec 50°. 
 
 6. Solve the triangle in which c = 1000, A = 19°36', C= 90°. 
 
 7. A man, observing at noon a cloud due south of him, 
 
 finds its angle of elevation to be 33°10', and that 
 of the sun 54°20. At the same time he notes the 
 position of its shadow, and afterwards ascertains 
 its distance to be 1020 feet. Find the height of 
 the cloud. 
 
 V' 
 1. If A be the right angle of an isosceles right-angled 
 triangle ABC, and if BL, BM be drawn to make 
 equal angles with BG and to meet ^(7 in X, M; then 
 AL.GM+AM.GL = AB.LM. 
 
74 RIGHT-ANGLED TRIANGLES 
 
 2. (3 - taii2a)cot a = (cot^a - 3)tan 3a. 
 
 3. Find the value of cos 3a when cosa = f, and of sin 3a 
 
 when sin a = J. 
 
 4. sin(a + /3)cos/3-sin(a + y)cosy = sin(|8-y)cos(a + )Q+y). 
 
 5. Solve the equatioDS : (i.) cos 90+ cos 30 = cos 60. 
 
 (ii.) tan 20 = 3 tan a 
 
 6. Solve the triangle in which a = 7l, ^ = 9°54', 0=90°. 
 
 7. The dip of a stratum is 24°30' towards the N.E. ; find 
 
 its apparent dip in the direction E. 1 5° S. 
 
 0. 
 (Examples on Chapter II.) 
 - - sin^a „ /. tan^aX 
 
 2. cot% + cot2a = cosec*a — cosec^a. Also verify this for- 
 
 mula when a = 30°. 
 
 3. If cos a= j^ and cos/5 = ff, find the value of 
 
 tan a sin ^ — sin a tan ^. 
 
 4. (sin 30° + 2 cos 45° - 1)^ = 3 sin260° - cosec 45°. 
 
 5. Solve the equation 5(1 — sin 0) = 008^0(5 — 2 sin 0). 
 
 6. Eliminate between 
 
 sec^0 + cosec^0 = a, and tan = 6 sec^0. 
 
 7. Find the least value of cos^O+sec^O. 
 
 K. 
 
 (Examples on Chapter III.) 
 
 1. If sin a = -5^ and sing= ^ ^, , find the value of 
 
 tan(a+/3). 
 
 2. ia,n(a + r/3)-tan(a + r^/3)= — , , —^'1 / _l /3^ • 
 
 ^ '^ '^ cos(a+r-l|8)cos(a + r/3) 
 
 3. Express as a single term 2 cos a cos 2a + sin a sin 2a. 
 
AND PRACTICAL APPLICATIONS. 75 
 
 4. Shew that : 
 
 (i.) sin(0 + ^) - sin = cos S sin s(l - tan 6 tan |\ 
 
 (ii.) tan 50° + tan 40° = 2 sec 10°. 
 
 5. Solve the equation 
 
 cos 20+ cos 40 -cos 80 -cos 100 = 2^2 cos sin 30. 
 
 6. If cos 80 + cos 40 = a and sin 80 + sin 40 = 6, then 
 
 2cos20 = V(»H6'). 
 
 7. Two parallel chords of a circle, lying on the same side 
 
 of the centre, subtend respectively 72° and 144° at 
 the centre. Shew that the distance between the 
 chords is half the radius of the circle. 
 
 X. 
 
 (Examples on Chapter III.) 
 
 1. tan 5a — tan 3a — tan 2a = tan ha tan 8a tan 2a. 
 
 2. Find the value of tan a, when tan 2a = x/l5. 
 
 3. Simplify - cos2a + sin2« 
 
 2 cos a + sm a — 2(cos'^a + sm^a) 
 
 4. Shew that : 
 
 /. X sin a + 2 sin 3a + sin 5a_4sin a — 3coseca 
 cosa — 2 cos3a + cos5a 4cosa — 3seca" 
 (ii.) l+cosl8° + cos36° + cos54° = 4cos9°cosl8°cos27°. 
 
 5. Solve the equation tan(0 + 3O') = (7 + 4v^3)tan(0-3O°). 
 
 6. Eliminate between 
 
 acos0 + 6sin0 = cand2a6cos20-(a2-62)sm20 = d2 
 
 7. Points A, B, G, D are taken on the circumference of a 
 
 circle, so that the arcs AB, BG, and GD subtend 
 respectively at the centre angles of 108°, 60°, and 
 36°. Shew that 
 
 AB=BG+GR 
 
PART II. 
 REAL ALGEBRAICAL QUANTITY. 
 
 CHAPTER VI. 
 CIECULAR MEASUEE OF ANGLES. 
 
 66. For practical purposes, as we have seen (art. 1), it 
 is essential that the unit of angular measurement should 
 be constant, easily obtained and of a convenient magni- 
 tude for measuring the angles most frequently in use. 
 
 For theoretical purposes, it is essential that the unit 
 should be constant, and that it should be so chosen that 
 the expressions and formulae in which it is employed 
 should by means of it be reduced to the simplest attain- 
 able form. 
 
 67. Definitions. — The length of the circumference of a 
 circle is the limit of the length of the perimeter of a 
 regular inscribed polygon when the number of sides in 
 the polygon is indefinitely increased. 
 
 The length of the arc of a circle is the limit of the 
 length of a broken-line which consists of a series of 
 
CIRCULAR MEASURE OF ANGLES. 77 
 
 consecutive equal chords of the arc, when the number 
 of these chords is indefinitely increased. 
 Two assumptions are made in these definitions, namely : 
 
 (1) That the length of the perimeter of the inscribed polygon 
 (or broken-line) tends to a limit ; 
 
 (2) That this limit is the same for all inscribed polygons (or 
 broken-lines) when the number of sides is indefinitely increased. 
 For example, we may first inscribe a square in the circle, then, by 
 bisecting the arcs, we get regular figures of 8, 16, 32, ...sides ; or 
 beginning with an equilateral triangle, we may suppose regular 
 figures of 6, 12, 24, ... sides, or of 9, 27, 81, ... sides, to be inscribed, 
 or we may proceed by any other law. The assumption is that the 
 perimeter of the polygon tends to the same limit in all cases. 
 
 (For proofs of these j)ropositions, see Kouch^ et De Comberousse, 
 Geometrie Elementaire, art. 290.) 
 
 68. If two circular arcs stand upon the same straight 
 line, the length of the exterior arc is greater than that of 
 the interior. 
 
 Let AG DEB and AFOHB be two convex broken-lines 
 
 A "B 
 
 standing upon the same straight line AB. Produce AF, 
 FG, GH to meet the outer line in a, h, c. 
 Then, AG-\-Ga> AF+Fa, 
 
 Fa+aD-\-Db>FG+Gb, 
 Gh + hE+EoGH+Hc, 
 and Hc+cB>HB. 
 
 :. by addition, the length of the broken-line AGDEB is 
 greater than the length of the broken-line AFGHB. 
 
78 CIRCULAR MEASURE OF ANGLES. 
 
 Now, suppose AG, CD, BE, EB and AF, FG, GH, HB 
 to be series of consecutive equal chords of two circular 
 arcs standing upon the straight line AB; then, the length 
 of the exterior broken-line is greater than the length of 
 the interior broken-line. This is true whatever be the 
 number of chords in the two arcs, and therefore when 
 the number in each is infinite, in which case, the lengths 
 of the broken-lines become the lengths of the correspond- 
 ing arcs. 
 
 Hence, the length of the exterior circular arc is greater 
 than the length of the interior. 
 
 Gov. — In the same way, if BAG be a circular arc 
 standing on a straight line BG 
 and if jBT, GT be the tangents 
 at B and (7, it may be shewn 
 that the length of the sum of 
 the tangents BT, TG is greater 
 than that of the arc BAG^ and the length of the arc BAG 
 greater than that of the chord BG. 
 
 69. The circumference of a circle varies as its radius. 
 Let 0, 0' be the centres of two circles of different radii, 
 and let regular polygons ABGD ... , A'BG'B' . . . , be in- 
 
 scribed in them, both having the same number {n) of 
 sides. Join OA, OB, ..., 0'A\ 0'B\ .... 
 
CIRCULAR MEASURE OF ANGLES. 79 
 
 Then, in the triangles OAB, O'A'B', the angles AOB, 
 A'O'E are equal, each being the same fraction of four 
 right angles. 
 
 Again, the sum of the angles OAB, OBA is equal to 
 the sum of the angles O'A'B', O'B'A' (Eucl. I. 32) ; and 
 the angles OAB, OBA are equal to one another, and also 
 the angles O'A'B', O'B'A' (Eucl. L 5); 
 .-. angle 0^5=angle O'A'B'd^ndi angle 05^= angle 0'FA\ 
 :. the triangles OAB, O'A'F are similar (Eucl. VI. 4). 
 AB:A'E=OA:0'A\ 
 n.AB'.n.A'E=OA'.0'A\ 
 .'. perim. of polygon A BCD. . . : perim. of polygon A'B'G'D' 
 = radius OA : radius O'A'. 
 This is true whatever be th^ number of sides in the 
 two polygons ; and it is therefore true when the number 
 of sides in both is infinitely great, in which case the 
 length of the perimeter of either polygon is the length of 
 the circumference of the circle in which it is inscribed, 
 circumf. of circle ABG : circumf of circle A'B'G' 
 = radius OA : radius 0'A\ 
 i.e. the circumference of a circle varies as its radius. 
 
 70. It follows, from this proposition, that the ratio of 
 the circumference of a circle to its diameter is constant. 
 This ratio is denoted by the letter tt. 
 
 Hence, if r be the length of the radius of a circle, the 
 length of the circumference is 27rr. 
 
 It has been shewn that the number tt is incommensur- 
 able with unity.* Various methods have been employed 
 
 * For the proof of this proposition, see Chrystal's Algebra, chap. 
 
80 CIRCULAR MEASURE OF ANGLES. 
 
 for obtaining its approximate value, several of them 
 being given in later chapters. By one of these methods 
 its value has been found to more than 700 places of 
 decimals. To the first 20 places it is 
 
 3-14159265358979323846. 
 Convenient approximate values of tt are -^ and W\. 
 In the first of these, the error is about ^-nnr* ^.nd, in the 
 second, less than TirgTr ooo o? ^^ ^^ ^^'^^ value. 
 
 71. Definition. — The unit of circular measure is the 
 angle at the centre of a circle which stands on an arc 
 equal in length to the radius. 
 
 This unit is called a radian. 
 
 72. The radian is a constant angle. 
 
 Let be the centre of a circle of any radius (see figure 
 of next article), and let ^-S be an arc equal in length to 
 the radius. Join OA, OB. 
 
 Then, angle A OB : 4 right angles 
 
 :: arc AB : circumference of circle (Eucl. YI 33). 
 
 ::r : lirr 
 
 ::l:2x, 
 
 angle A OB x 27r = 4 right angles. 
 Hence, the angle AOB is constant, whatever be the 
 radius of the circle. 
 
 Cor. 1. — Since the angle AOB is one radian, it follows 
 from the last equation that 
 
 4 right angles = 27r radians. 
 2 „ =7r „ 
 
 1 ,, =7r/2 „ 
 
 Cor. 2— One radian = 18073-14159265... =57° 17' 45'', 
 nearly. 
 
CIRCULAR MEASURE OF ANGLES. 
 
 81 
 
 73. The number of radians in any angle at the centre 
 of a circle is equal to the length of the arc on which the 
 angle stands divided by the radius of the circle. 
 
 Let AOC be the angle, being the centre of a circle 
 of any radius ; and let AB be an arc equal in length to 
 the radius, so that the angle AOB is one radian. 
 
 Then, angle AOC : angle AOB 
 f. =arc ^0 : arc AB (Eucl. VI. 33) 
 
 = arc AG : radius. 
 
 .-. angle AOC 
 
 arc AG , . ^ „ 
 
 = — T-. — X angle A OB. 
 radius ° 
 
 .•. the number of radians in the 
 
 angle AOG= arc A (7/radius. 
 
 74. If be the number of radians in an acute angle, 
 is greater than sin 0, but less than tan 0. 
 
 Let AOB be the given angle, AB the arc of a circle 
 with centre and any radius. 
 
 Draw BN perpendicular to OA 
 and produce it to G so that NO is 
 equal to BN. Then OG is equal 
 to OB (Eucl. I. 4), and therefore q. 
 the circle of which AB is an arc 
 passes through G. Also, CT is 
 equal to BT, and GT touches the 
 circle (Eucl. I. 4, IIL 16). 
 
 Now, sum of tangents BT, GT > slyc BAG > chord BG 
 (art. 68. Cor.). 
 
 BT> SiYcAB >B]Sr. 
 BT/OB > arc AB/OA > BN/OB. 
 tan0> e >smO, 
 
 that is, is intermediate between sin and tan 0. 
 
 F 
 
82 CIRCULAR MEASURE OF ANGLES. 
 
 75. If 6 he the number of radians in an acute angle, 
 tJie limiting values of sin 0/0 and t&n 6/0, when is 
 indefinitely diminished, are each unity. 
 
 By the last article, we have 
 
 tan > > sin 0, 
 
 tan O/sin > O/sin 0>1, 
 
 sec > O/sin 0>1. 
 
 Now, when is indefinitely diminished, the limit of 
 
 sec is unity. 
 
 .*. the limit of O/sin 0, and therefore also of sin 0/0, 
 
 when is indefinitely diminished, is unity. 
 
 . . tan sin 1 
 
 Again, -^^- = — ^r— . -, 
 
 ^ cosO 
 
 and, when is indefinitely diminished, the limits of both 
 
 sin 0/0 and 1/cos are unity, 
 
 the limit of tan 0/0 is unity. 
 
 Hence, sin 0, and tan vanish in a ratio of equality. 
 
 76. If he the number of radians in an acute angle, 
 
 0^ . . 
 
 cos is greater than 1 — q-, ctnd sinO is greater than 
 
 '-T 
 
 
 We have 
 
 cose=l-2sin2|, 
 
 and 
 
 . 0^0 
 «^^2"^2' 
 
 hence, 
 
 cos 0> 1-2.^, 
 
 i.e. 
 
 >'-f- 
 
 Again, 
 
 6 
 sin = 2 cos ^ ^^^9' 
 
 and 
 
 .0,0 
 smg^tan^cosg, 
 
CIRCULAR MEASURE OF ANGLES. 83 
 
 hence, sin 6 = 2 tan ^ cos^^ 
 
 %.e. 
 
 = 2tau|(l-sin^|) 
 
 >-f{'-(Dl. 
 
 /^ 6' 
 
 >^-4- 
 
 (art. 74) 
 
 Change of Units. 
 
 77. In this section, and in the corresponding examples, 
 reference will be made to another system of angular 
 measurement no longer in use, the centesimal system, 
 in which the right angle is the fundamental unit'. A 
 right angle is divided into 100 equal parts called grades, 
 a grade into 100 equal parts called minutes, and a minute 
 into 100 equal parts called seconds. An angle of 89 
 grades, 71 minutes, 47 seconds, is written 89^ 71' 47". 
 
 It will be noticed that any angle whose centesimal 
 measure is known can be at once expressed as a decimal 
 of a right angle. Thus, 89^ 71' 47'' is -897147 of a right 
 angle ; and 7^ 45' 3" is '074503 of a right angle. 
 
 So, also, -157423 of a right angle is equal to 15^ 74' 23", 
 and '000503 of a right angle is equal to 5' 3". 
 
 78. If D degrees, G grades and radians be the 
 sexagesimal, centesimal and circular measures of the 
 same angle, then D/180 = G/200 = O/tt. 
 
 1 degree = 1/180 of two right angles, 
 .0 degrees = D/180 „ 
 Also, 1 grade = 1/200 
 
 G^ grades = (?/200 -„ 
 
84 CIRCULAR MEASURE OF ANGLES. 
 
 And, 1 radian = I/tt of two right angles, 
 
 radians = O/tt „ „ 
 
 But the given angle is the same fraction of two right 
 angles in whatever measure it is expressed, 
 D/18O = G^/2OO = 0/7r. 
 
 79. Example 1. — To find the centesimal measure of 23° 18' 36". 
 23° 18' 36" = 23° 18' -6 
 =23°-31 
 
 = -259 of a right angle 
 = 25^90\ 
 Example 2.— To find the sexagesimal measure of 13^ 38^ 12^ 
 
 •133812 
 
 13^ 38^ 12^'= -133812 of a right angle 90 
 
 = 12° -04308 12-04308 
 
 = 12°2'-5848 ^^ 
 
 = 12°2'35"-088. ^^^1o 
 
 35-088 
 Example 3.— To find the circular measure of 12° 16'. 
 Let 6 radians be the circular measure of 12° 16', i.e., of 736'. 
 
 Now, ^'—TEK — ^ of two right angles 
 
 60 
 
 36 
 
 60 
 
 18-6 
 
 90 
 
 23-31 
 
 
 •259 
 
 12° 16'= 
 
 180x60 "" ' ~°"~' 
 
 736 
 
 180 X 60 " 
 and ^ radians =^/7r „ „ 
 
 n_ 7367r _467r 
 180x60 676" 
 Example 4. — Find the number of seconds in the angle sub- 
 tended at the centre of a circle, whose radius is one mile, by an 
 arc 5^ inches long (tt being taken equal to 3^). 
 
 The number of radians in the angle — — 1^= 
 
 63360 11520 
 Let 8 be the number of seconds in the same angle, 
 
 *^^''' 180x60x6o"Tl520"^''' 
 
 / ^_ 180x60x60 _ 1575 
 
 1152077 88 
 
 - =1711 seconds. 
 
CIRCULAR MEASURE OF ANGLES. 85 
 
 Viva Voce Examples. 
 
 Find the number of degrees in the angles whose 
 measures in radians are : 
 
 1. 7r/4. 
 
 
 9. tt/IO. 
 
 17. 47r/3. 
 
 2. x/3. 
 
 
 10. 7r/12. 
 
 18. 57r/12. 
 
 3. S7r/2. 
 
 
 11. 7r/15. 
 
 19. 77r/6. 
 
 4. 87r/4. 
 
 
 12. 7r/18. 
 
 20. Il7r/18. 
 
 5. 7r/8. 
 
 
 13. 7r/36. 
 
 21. 47r/9. 
 
 6. 7r/6. 
 
 
 14. 57r/6. 
 
 22. 27r/5. 
 
 7. 37r/8. 
 
 
 15. 57r/3. 
 
 23. 137r/9. 
 
 8. 57r/4. 
 
 
 16. 27r/3. 
 
 24. 77r/12. 
 
 Express 
 
 the following: ansfles in 
 
 radians : 
 
 25. r. 
 
 
 30. 270°. 
 
 35. 120°. 
 
 26. 30^. 
 
 
 31. 225°. 
 
 36. 11° 15'. 
 
 27. 15°. 
 
 
 82. 22° 30'. 
 
 37. 7° 30'. 
 
 28. 60°. 
 
 
 33. 330°. 
 
 38. 0° 30'. 
 
 29. 135°. 
 
 
 34. 210°. 
 
 39. 67° 30'. 
 
 Express 
 
 the 
 
 following angles 
 
 as decimals of 
 
 angle : 
 
 
 
 
 40. 39fi'14^24^ 43. 91^8^3". 46. 4^ 2^ r\ 
 
 41. 485^2^31". 44. 4^' 15' 87". 47. 3' 2'\ 
 
 42. 19^ 34' 5". 45. 15^ 3' 2". 48. 17". 
 
 Express the following decimals of a right angle in 
 centesimal measure : 
 
 49. -375984. 52. -051403. 55. -00003. 
 
 50. -58441. 53. 70001. 56. -91005. 
 
 51. -793602. 54. -006017. 57. '385. 
 
86 CIRCULAR MEASURE OF ANGLES. 
 
 Examples IX. a. 
 
 1. A railway train is travelling on a circular arc of half 
 
 a mile radius at the rate of 20 miles an hour; 
 through what angle does it turn in 10 seconds ? 
 
 2. Find the length of an arc which subtends an angle 
 
 of 50° at the centre of a circle of radius 8 feet 
 (7r = 31416). 
 
 3. The length of a degree on the earth's surface in the 
 
 neighbourhood of the equator is G9'07 miles ; find 
 the radius of the earth, to the nearest mile. 
 
 4. On a circle, 10 feet in radius, it was found that an 
 
 angle of 22° 30' at the centre was subtended by an 
 arc 3 feet 11 J inches in length ; hence find the 
 value of TT to three places of decimals. 
 
 5. Prove geometrically that a radian is less than 60°. 
 
 6. Find approximately the distance of a tower whose 
 
 height is 51 feet, and which subtends at the eye 
 an angle of h^\' (tt = 3|). 
 
 7. By considering an angle of 15°, shew that tt is greater 
 
 than 310 and less than 322 (art. 74). 
 
 8. Find the limit of ^inn^jn, when n is indefinitely 
 
 diminished. 
 
 9. Find the limit of -^ir^tan-, when n is indefinitely 
 
 increased. 
 
 10. Find the sine of 1° to 5 places of decimals, by means 
 
 of art. 76 (7r = 314159). 
 
 11. If TYh and fjL be the numbers of sexagesimal and 
 
 centesimal minutes in the same angle, find the 
 relation between m and />t. 
 
 12. Find the sexagesimal measures of the following angles : 
 
 (1) 12^ 47^ 93". (2) 0^ 1^ 2". (3) 87^ 0^ 13". 
 
CIRCULAR MEASURE OF ANGLES. 87 
 
 13. Find the centesimal measures of the following angles: 
 
 (1) 29°14'15"'. (2) 75° 47' 22". (8) 43° 19' 20''. 
 
 14. Find the number of radians in the following angles : 
 
 (1) 35° 1' 12". (2) 47^ 0^ 30". 
 
 Examples IX. b. 
 
 1. Find the circular measure of the angle subtended at 
 
 the centre of a circle, whose radius is one mile, by 
 an arc 22 inches long. 
 
 2. If the radius of a circle be 4000 miles, find the 
 
 length of an arc which subtends at the centre an 
 
 angle of ^ radians. 
 
 3. A rail way- train, travelling due north, begins to move 
 
 on a line in the form of a circular arc. After 
 travelling 5 J miles, its direction of motion is N.W. 
 Find the radius of the circle (7r = 3^). 
 
 4. If an arc of 6*283 inches subtends an angle of 30° at 
 
 the centre of a circle one foot in radius, find the 
 value of TT to two places of decimals. 
 
 5. Prove that tt is greater than 3 and less than 2^3, 
 
 by considering the lengths of the perimeters of 
 regular hexagons described in and about a given 
 circle. 
 
 6. Find approximately the diameter of the sun, which 
 
 subtends at the earth an angle of 32' 1", its dis- 
 tance from the earth being 91 million miles 
 
 7. By considering an angle of 18°, shew that tt is greater 
 
 than 309 (art. 74). 
 
 8. Find the limit of tan7i°/7i, when n is indefinitely 
 
 diminished. 
 
88 CIRCULAR MEASURE OF ANCLES. 
 
 9. Find the limit of Jnr^in— , when n is indefinitely 
 increased. 
 
 10. Find the sine of 3° to 3 places of decimals, by means 
 
 of art. 76 (7r = 3-14159). 
 
 11. If s and (T be the numbers of sexagesimal and 
 
 centesimal seconds in the same angle, find the 
 relation between s and o-. 
 
 12. Find the sexagesimal measure of the following angles: 
 
 (1) 53^ 81^ 7". (2) 5^ 12^ 9". (3) 0^ 0^ 3". 
 
 13. Find the centesimal measure of the following angles: 
 
 (1) 5° 3' 14". (2) Sr 13' 17". (3) 0° 0' 5". 
 
 14. Find the number of radians in the following angles : 
 
 (1) 87° 37' 15". (2) 1^ 25' 4". 
 
 Examples X. 
 
 1. What must be the unit angle, if the sum of the 
 
 measures of a degree and grade be 3 ? 
 
 2. What unit is such that the number of units in a 
 
 radian is equal to the circular measure of a 
 grade ? 
 
 3. Divide the angle 77° into two parts, so that the 
 
 number of sexagesimal minutes in one part may 
 equal the number of centesimal minutes in the 
 other part. 
 
 4. The number of degrees in an angle is n times the 
 
 number of minutes in its complement; find the 
 number of radians in the angle. 
 
 5. The angles of a triangle are in arithmetical pro- 
 
 gression, and the greatest contains as many 
 degrees as the smallest contains grades ; find the 
 angles in degrees. 
 
CIRCULAR MEASURE OF ANGLES. 89 
 
 6. Divide a right angle into two parts so that the 
 
 number of grades in their difference may be 
 equal to the number of degrees in the whole 
 angle. 
 
 7. The difference between two angles which contain the 
 
 same number of degrees and grades, respectively, 
 
 is -^ radians ; find the angles. 
 
 8. One angle of a triangle is 45°, and another is IJ 
 
 radians. Find the third, both in degrees and 
 radians (7r = 8f). 
 
 9. The angles of a quadrilateral are in arithmetical pro- 
 
 gression, and the difference between the greatest 
 and least is a right angle. Find the number of 
 degrees, and also the number of radians, in each 
 angle. 
 
 10. If the circumference of a circle be divided into 
 
 five parts in arithmetical progression, the greatest 
 part being six times the least, express in 
 radians the angle which each subtends at the 
 centre. 
 
 11. The angles of a triangle are in arithmetical pro- 
 
 gression, and the ratio of the number of radians 
 in the least to the number of degrees in the mean 
 is 1 : 120. Find what multiple of a right angle 
 is the greatest angle. 
 
 12. The perimeter of a certain sector of a circle is equal 
 
 to the length of the arc of a semicircle having the 
 same radius. Express the angle of the sector in 
 degrees, etc. (tt = 8^). 
 
 13. Prove geometrically that cos 0>1-^. 
 
90 CIRCULAR MEASURE OF ANGLES. 
 
 14. The angles of a triangle, when referred, in ascending 
 order of magnitude, to three units in geometrical 
 progression, are represented by numbers in arith- 
 metical progression. The mean angle is equal to 
 the sum of the first two units or to half the sura 
 of the last two, and the greatest angle is four 
 times the mean unit. Determine the angles. 
 
CHAPTER VII. 
 
 GENEKAL DEFINITIONS OF THE CTECULAR 
 
 FUNCTIONS. FORMULA INVOLVING ONE VARIABLE 
 
 ANGLE. 
 
 § L Definitions. 
 
 80. Negative Lines. — A straight line may be generated 
 by the motion of a point. The point may move in either 
 of two opposite directions or senses. 
 
 The sense of a line is denoted by the order of the 
 letters. Thus, " the line AB " means " the line generated 
 by a point moving from A to B!' 
 
 If one of the two senses in which a line is generated 
 be regarded as positive, the opposite sense is said to be 
 negative. Thus AB and BA are equal in magnitude but 
 opposite in sense, and, if AB be regarded as positive, BA 
 is equal to AB in magnitude and is negative. Hence, 
 AB+BA=0, or AB=-BA. 
 
 A C B B A O 
 
 If A, B, C be points in any order in the same line, then 
 AB = AC-\-GB in all cases. 
 
 81. Negative Angles.— An angle may be generated by 
 
 the rotation of a line about a fixed point. The line may 
 
 rotate in either of two opposite dii'ections or senses. 
 
 91 
 
92 DEFINITIONS OF THE 
 
 The sense of an angle is denoted by the order of the 
 letters. Thus, "the angle AOB" means "the angle gen- 
 erated by a line rotating about from OA to OB." 
 
 If one of the two senses in which an angle is generated 
 be regarded as positive, the opposite sense is said to be 
 negative. Thus, the angles AOB and BOA are equal in 
 magnitude but opposite in sense, and, if AOB be regarded 
 as positive, BOA is equal to AOB in magnitude and is 
 negative. Hence, lAOB-^-lBOA = 0, or lAOB= -lBOA. 
 
 If OA, OB, OC he lines drawn from in any order, 
 then lAOB==lAOG-\-lCOB in all cases. 
 
 82. The positive sense of a line is arbitrary, and must 
 therefore be defined explicitly or implicitly in each case 
 considered. 
 
 The positive sense of an angle is likewise arbitrary, 
 but, for the purposes of Plane Trigonometry, it is sufficient 
 in all cases to fix as the negative sense that in which the 
 hands of a watch placed face upwards on the plane rotate. 
 
 Hence, in all cases of rotation, positive and counter- 
 clockwise are equivalent terms, and so also negative and 
 clockwise are equivalent. 
 
 83. Def. — The foot of the perpendicular from a point 
 on a line is called the 'projection of the point on the line. 
 
 If the point be on the line, the point and its projection 
 coincide. 
 
CIRCULAR FUiVCTIOJVS. 
 
 93 
 
 Def. — The intercept between the projections of the 
 ends of a line on another line is called the projection of 
 the first line on the second. 
 
 The projection of a line is to be considered with regard 
 to magnitude and sense. 
 
 The sum of the projections of the parts of a continuous 
 
 B 
 
 b d 
 
 broken-line is equal to the projection of the line joining 
 its extremities. Thus, 
 
 ah-\-hc-\-cd = ad. 
 Hence, the sum of the projections of the sides of a 
 closed polygon is zero. 
 
 84. Defs. — Let a line rotate about from OX through 
 any positive or negative angle a to the position OA ; let 
 
 X' 
 
 if> 
 
 o 
 
 N 
 
 M 
 
 X' 
 
 N 
 
 / 
 T 
 
 M 
 
 X 
 
 OF be a line making an angle |^ in the positive sense 
 with 0X\ and letO^, OX, OF be the positive senses of 
 
94 FUNDAMENTAL PROPERTIES OF THE 
 
 the lines OA, OX, OY. Let a length OP, of any mag- 
 nitude and of either sense, be measured along OA ; and 
 let OM, ON be the projections of OP on OX, OY, 
 respectively. 
 
 The ratio OM : OP is called the coaiTie of the angle a, 
 ON : OP the sine of a, OiV^ : OM the tow^eTi^ of a, OP : Oif 
 the secant of a, OP : ON the cosecant of a, and Oilf : OJV 
 the cotangent of a. 
 
 These ratios are called the Circular Functions of the 
 angle a. 
 
 Two other ratios are occasionally used, and are defined as follows : 
 If the length OP be equal in magnitude to OX, and positive in 
 
 sense, and if OY=OX, the ratio MX : OP is called the versine of a, 
 
 and NY : OP the coverdne of a. 
 
 The abbreviations of these ratios are vers a and covers a. 
 
 § 2. Fundamental Properties of the Circular 
 Functions. 
 
 85. Each circular function of a given angle has one 
 value only. 
 
 Consider the cosine of a given angle XOA or a, and 
 suppose that OP, Op are lengths of any magnitude and 
 either sense measured along OA. We shall shew that 
 OM : 0P= Om-.Op, where OM and Om are the projections 
 of OP and Op on OX. 
 
 The angles POM, pOm are equal, and the right angles 
 PMOy pmO are equal, therefore the triangles POM, pOm, 
 
CIRCULAR FUNCTIONS. 
 
 95 
 
 X" 
 
 m M X 
 
 F' 
 
 are similar (Eucl. YI. 4), and therefore, regarding the 
 magnitude only of the lines, 
 
 0M:0P==0m:0p. 
 
 Again, if P and ^ are on 
 the same side of 0, OP and 
 Ojp are of the same sense, and 
 so also are OM and Om ; 
 while, if P and p are on op- 
 posite sides of 0, OP and Op 
 are of opposite sense, and so 
 also are OM and Om. Hence, 
 regarding the sense as well as the magnitude of the lines, 
 
 0M:0P = 0m:0p, 
 i.e., the cosine of a has one definite value independent of 
 the magnitude or sense of the length OP. 
 
 In like manner, it may be shewn that sin a, tan a, sec a, 
 etc., are one- valued functions of a. 
 
 86. Signs of the Circular Functions. — If an angle lies 
 between and ^, it is said to be in the first right angle ; 
 if between J and tt, in the second right angle ; and so on. 
 
 TT 
 
 If an angle lies between and — ^, it is said to be in the 
 first negative right angle ; if between — ^ and — tt, in the 
 
 second negative right angle ; and so on. 
 
 Referring to the diagrams of art. 84, and supposing 
 OP to be measured in the positive sense in all cases, we 
 see that : 
 
 When a is in the first right angle, OM and ON are 
 positive, and therefore all the ratios are positive ; 
 
 When a is in the second right angle, OM is negative 
 
96 FUNDAMENTAL PROPERTIES OF THE 
 
 and ON positive, and therefore the sine and cosecant are 
 positive, and the other ratios negative ; 
 
 When a is in the third right angle, OM and ON are 
 both negative, and therefore the tangent and cotangent 
 are positive, and the other ratios negative ; 
 
 When a is in the fourth right angle, OM is positive 
 and ON negative, and therefore the cosine and secant are 
 positive, and the other ratios negative. 
 Thus, the succession of signs is as follows : 
 cosine and secant, + — — + 
 
 sine and cosecant, + + — — 
 
 tangent and cotangent, + — + — 
 
 87. Values of the Functions of 0, ^, tt, etc.— Let the 
 angle a (see fig. of art. 84) have the series of values 
 
 0» 9> '^f IT' ^"^^ •••' ^^®^ ^^ ^^^ ^^® series of values 
 OP, 0, -OP, 0, OP, ..., and ON the series 0, OP, 0, 
 — OP, 0, .... Hence, we see that 
 
 cosO = l, cosj=0, cos7r=-l, cos~=0, cos27r=l,... 
 
 sinO=0, sin- = l, sm7r=0, sin— =-1, sin27r=0,... 
 
 tan 0=0, tan ^= GO, tan7r=0, tan— =00, tan27r=0, ... 
 
 secO = l, sec^=oo, sec7r=-l, sec — = qo, sec27r=l, ... 
 2t 2 
 
 cosecO = oo, cosec^=l, cosec7r=Qo, cosec— =-1, cosec27r=oo, ... 
 Z 2 
 
 cotO=oo, cot^=0, cot7r=Q0, cot-J = 0, cot27r=oo, ... 
 
 Example. — Trace the changes in the sign and magnitude of 
 sin 0, as increases from to 27r : 
 
 As d increases from to ^, sin Q is positive, and increases from 
 Otol: 
 
CIRCULAR FUNCTIONS. 97 
 
 As ^ increases from - to tt, sin 9 is positive, and decreases 
 from 1 to ; 
 
 As increases from tt to -^, sin 6 is negative, and increases 
 
 numerically from to - 1 ; 
 As 9 increases from ~ to Stt, sin 9 is negative, and decreases 
 
 numerically from - 1 to 0. 
 
 88. Periods of the Circular Functions.— If an angle 
 d pass in order through a series of values from to 27r, 
 the cosine of passes through a series of values ranging 
 from +1 to — 1, and returning from — 1 to + 1 ; if 
 pass in order through a second series of values from 27r 
 to 47r, the values of cos Q recur in the same order as in 
 the first series ; and this series of values of the cosine is 
 continually repeated after each complete revolution of 
 the generating line of the angle 0. Thus, if a, a + 'lnir 
 be values of 6 differing by an}' multiple of 27r, then 
 cos a = cos (a + Stitt). From this property of the recur- 
 rence of its values, the cosine is called a periodic function 
 of the angle, and the increment Stt of the angle after 
 which the values of the cosine recur is called the period 
 of the cosine. 
 
 In like manner, sin 0, sec 6 and cosec are periodic 
 functions of of period 2x. 
 
 In the case of the tangent or cotangent, the values 
 recur after an increment tt of the angle, hence tan 6 and 
 cot 6 are periodic functions of of period tt. 
 
 In any given period, a given value of cos 6, sin 0, 
 sec or cosec corresponds to two values of 0, while 
 a given value of tan or cot 6 occurs once only in each 
 period. 
 
98 
 
 FUNDAMENTAL PROPERTIES OF THE 
 
 89. Continuity of the Circular Functions.— Def. If a 
 circular function of an angle be such that, as the angle 
 increases from one given value to another, an infinitely 
 small change in tlie function corresponds to an infinitely 
 small change in the angle, then the function is said to be 
 continuous between those two given values of the angle. 
 For example, if 6 and O' be two values of an angle, 
 and if we can shew that cos r^ cos $' is infinitely small 
 when '-' 6' is infinitely small, then cos will be a 
 continuous function of 6. 
 
 If and 6' be in the same right angle, we need only 
 consider the case in which and S' are in the first right 
 angle ; for, if and 0' be in any other right angle, the 
 only difference is in the sign of the function and the 
 order in which it passes through its diflferent numerical 
 values. If 6 and 6' be in consecutive right angles, as 
 might be tlie case when is less than, but very nearly 
 equal to, a multiple of Jtt, it will be seen that the same 
 reasoning holds true.* 
 
 Let the line OP rotate through a right angle from the 
 position OA, in the positive sense, 
 to the position OB ; and, from any 
 intermediate position OP, through 
 an infinitely small angle, in the 
 same sense, to the position OP'. 
 Let 0, 0' denote the angles AOP, 
 AOP\ Draw PM, P'M perpen- 
 dicular to OA, and PL perpendic- 
 ular to P'M'. 
 
 * In this case, however, in considering the continuity of the tangent 
 and cotangent, the formula for sin {&' - 6) is supposed true when 0' is 
 greater than 5" ; this is proved in the next Chapter (art. 1 10). 
 
CIRCULAR FUNCTIONS, 99 
 
 Cosine and Sine. — We have, from the figure, 
 OM^OM' MM LP 
 
 cos — COS 0' = 
 
 OP ~OP~OP 
 
 PP' SiYcPP' . ^a' a 
 
 <OP^-^P'''-''^^-^' 
 
 Since 6' — is infinitely small, it follows that cos 
 — cos & is infinitely small ; and this is true for all values 
 of the angle between = and = Jtt. 
 
 Hence, the cosine is continuous for all values of the 
 angle 6. 
 
 Similarly, the sine is continuous for all values of the 
 angle Q. 
 
 Secant and Cosecant — Again, 
 
 a' a nv( ^ 1 ^ OP. MM 
 
 sec — sec = 0P\ —yt, — 
 
 UM OMJ ~ OM' OM 
 
 OP . arc PP' . OP^ arcPP' 
 
 '^ OM'.OM ''^'^'^OM .OM' OP 
 
 OP^ 
 
 Since O' — O is infinitely small, sec 0' — sec is so also, 
 provided OM' . OM is not infinitely small. If, however, 
 be very nearly equal to Itt, we can make OM' . OM 
 comparable in magnitude with OP^(0' — 6), or less than 
 this, if we please, by making the difference between 6 
 and Jtt small enough. In this case, it does not follow that 
 sec 0' — sec 6 is infinitely small. We know, indeed, that 
 it is not, for, as passes from a value a little less than 
 Jx up to Itt, sec changes from a quantity that is finite 
 to one that is infinitely great. 
 
 Hence, the secant is continuous for all values of the 
 angle 0, except those which differ by an infinitely small 
 
100 FUNDAMENTAL PROPERTIES OF THE 
 
 quantity, either in defect or excess, from an odd number 
 of right angles. 
 
 Similarly, the cosecant is continuous for all values of 
 the angle ^, except those which differ by an infinitely 
 small quantity, either in excess or defect, from zero or 
 any even number of right angles. 
 
 Tangent and Cotangent — Lastly, 
 
 , ^ , ^ sin 0" sin 
 
 tan Q —tan Q = yr, ^ 
 
 cos Q cos Q 
 
 _sin ^cos — cos Q' sin 0_sin {B' — 0) 
 
 ~ cos & COS ~ cos 6' cos 
 
 .-. tan0'-tan0< -^ A^-0). 
 
 cos V COS d 
 
 Since O' — O is infinitely small, tan 0' — tan ^ is also 
 infinitely small, provided cos & cos is not infinitely 
 small, i.e. provided is not very nearly equal to Jtt. 
 
 Hence, as in the case of the secant, the tangent is 
 continuous for all values of the angle 0, except those 
 which differ by an infinitely small quantity, either in 
 defect or excess, from an odd number of right angles. 
 
 Similarly, the cotangent is continuous for all values of 
 the angle 0, except those which difter by an infinitely 
 small quantity, either in excess or defect, from zero or 
 any even number of right angles. 
 
 90. Fundamental Formulae.— The following formulae 
 may be obtained immediately from the definitions of 
 art. 84, as in arts. 8, 9 : ^ :: 
 
 f cos sec = 1, sin Q cosec = 1, tan cot = 1 ; 
 
 . ^ sin , r. cos d 
 
 ^ tan0 = ^, cotO = ^— ^. r 
 
 cos Q sm 6 
 
 We have, also, by Eucl. I. 34 and 47, 
 
 OM''+ON^=OF^; 
 
CIRCULAR FUNCTIONS. 101 
 
 whence, dividing by OP'^, OM'^, and ON"^, we get, iu 
 
 succession, (^ cos^^4-sin20=l, 
 
 ? l + tan''^0 = sec2e, 
 '> cot20+l=cosec2a " 
 
 Of these eight formulae, five only are independent ; the 
 
 fifth, seventh, and eighth may easily be obtained from 
 
 the other five. 
 
 "We have, also, from the definitions of Art. 84, 
 
 n MX OX-OM , OM . n 
 
 -"''^^^ OP^—OF- = ^- 0P = ^-'''^^ 
 
 -, n NY OF- ON , ON , . n 
 
 and covers ^=-^-=—^^^ =l-_p = l-sm^. 
 
 Viva Voce Examples. 
 
 What are the signs of: 
 1. sin 120°. 
 
 10. 
 11. 
 12. 
 13. 
 
 cosjf. 
 
 15. 
 
 cot^. 
 4 
 
 . Sir 
 sm -p. 
 4 
 
 16. 
 
 27r 
 sec-—. 
 
 a 
 
 tan^=-. 
 
 17. 
 
 cos-^. 
 
 . ox 
 sm~. 
 
 18. 
 
 tan-^. 
 4 
 
 2. cos 150°. 
 
 3. tan 210°. 
 
 4. cot 300°. 
 
 5. sec 260°. 
 
 6. cosec 250°. 
 
 7. cos 320°. 
 
 8. tan 140°. 
 
 9. sin 245°. .. «^„^« ^"^ 
 
 14. cosec -J-. 
 4 
 
 19. Trace the changes in sign and magnitude of cos 6 as 
 
 increases from to 2x. 
 
 20. Trace the changes in sign and magnitude of tan as 
 
 increases from ~ tt to + tt. 
 
102 
 
 FUi\DAMENTAL VROPELiTIES OF THE 
 
 e 
 
 § 3. Reduction of Functions of 
 
 91. Even and Odd Functions.— If, in any integral 
 rational expression involving even powers only of a 
 
 x^ x^ 
 variable x, such as I — ^-f ^, we change x into —x, the 
 
 expression is unaltered in magnitude and sign; if the 
 
 OC/ X 
 
 expression involves odd powers only, such as a;- rr + j^y 
 
 its magnitude is unaltered but its sign reversed by the 
 reversal of the sign of the variable. Hence, by a simple 
 extension of the meaning of the words even and odd, we 
 have the following definition : 
 
 Def. — l{f{ — x) be equal to f(x) for all values of x, then 
 f(x) is said to be an even function of x; and if /( — a?) be 
 equal to -f(x) for all values of x, then f{x) is said to be an 
 odd function of x. 
 
 92. To prove that cos and sec are even functions of 
 0, and that sin 0, cosec 0, tarn 0, and cot are odd func- 
 tions of 0. 
 
 Let a line rotate about from the position OX, through 
 an angle 6 to the position OF ; 
 also, let a line rotate from OX 
 through an angle equal to 
 in magnitude and of contrary 
 X sense, to the position Op ; then 
 OP and Op have the same pro- 
 jection OM on XX\ while their 
 projections on YT, namely, ON" 
 and On, are in all cases equal 
 
 
 
 r 
 
 / 
 
 0^ 
 
 ir^^^^ 
 
 
 / 
 
 o 
 
 P 
 
 \^_ 
 
 [_^^ 
 
 y 
 
 in magnitude and of contrary sense. 
 
Hence, cos =-rrT, = -it- = ^^K ~ ^)' 
 
 OJr Up 
 
 . . /, ON —On . / m 
 
 and sm0=^^= ^^— -=-sin(-0), 
 
 OP Oj9 
 
 Similarly, sec = sec( — 0), cosec 6= — cosec( — 0), 
 tan0=-tan(-0), cot 0= -cot(-O). 
 
 93. To _proi;e that cos\Q-{-'^] = —sin 0. 
 Let a line rotate about 0, from the position OX, through 
 
 Y 
 
 an angle 0, to the position OP ; let it further rotate from 
 the position OP, through an angle ^ in the positive sense, 
 to the position OQ ; then angle X0P = 6 and angle 
 
 Leti M be the projection of P on OF, and N that of Q 
 on XX\ 
 
 Then, in the triangles POM, QON, the angles POM and 
 QON are equal (since POQ and ifOiV are right angles), 
 the right angles PMO and QNO are equal, and OP = OQ; 
 therefore, the triangles are geometrically equal in every 
 respect, and, therefore, OM=ON in magnitude. 
 
104 
 
 FUNDAMENTAL PROPERTIES OF THE 
 
 Again, since OQ is always a right angle in the positive 
 sense in advance of OP, therefore, when P is on the same 
 side of XX' as F, Q is on the opposite side of YY' to X ; 
 and, therefore, when OM is positive, ON is negative. In 
 like manner, when OM is negative, ON is positive. 
 Hence, in magnitude and sense, ON— — OM, 
 
 ■ m^_qM 
 
 OQ OP' 
 
 {e+|)=-sina 
 
 cos 
 
 94. To prove thit dniO + 1-) = (^os 0. 
 
 Let a line rotate about 0, from the position OX, 
 
 Y' Y' 
 
 through an angle Q, to the position OP ; let it further 
 rotate from the position OP, through an angle ^ in the 
 positive sense, to the position OQ, ; then, angle XOP = 6 
 and angle XOQ = 0-\--^- 
 
 Let M be the projection of P on XX\ N that of Q 
 on YT. 
 
 Then, in the triangles POM, QON, the angles P031, 
 QON are equal, since POQ and AWN are right angles, 
 
CIRCULAR FUNCTIONS. 105 
 
 the right angles PMO, QNO are equal, and OP = OQ; 
 therefore the triangles are geometrically equal in e very- 
 respect, and therefore Oili = OiV in magnitude. 
 
 Again, since OQ is always a right angle in the positive 
 sense in advance of OP, therefore, when P is on the 
 same side of YY' as X, Q is on the same side of XX' 
 as F; and therefore, when OM is positive, OiV is also 
 positive. In like manner, when OM is negative, ON is 
 also negative. Hence, in magnitude and sense, 0N= OM. 
 
 ON_OM 
 
 OQ~OP' 
 
 sin( 0+« ) = cos0. 
 95. Since, cosf + |^J= —sin Q and sinf + |^j = cos 0, 
 
 ( + ^ ) = 7 r = : — ^ = - cosec 0, 
 
 V -1) ^^^n^ -sm0 
 
 sec 
 
 cos 
 
 A , fn^A '^'"V^"^27 cose ,. 
 
 COS 
 
 So, also, cosecf + ^j =sec 0, 
 
 and cot f + 1^ j = — tan 0. 
 
 96. By arts. 93-95, we can express any circular 
 function of an angle in terms of a function of an angle 
 less or greater than the original angle by a right angle, 
 and, therefore, by repeated operations, we may diminish 
 or increase the angle by any multiple of a right angle ; 
 
106 FCyDAMENTAL PROPERTIES OF THE 
 
 by art. 92 we may change the sign of the angle ; hence, 
 
 we can express the circular functions of an angle n^ ±6, 
 where n is an integer, in terms of a function of 0. 
 Thus, we have 
 
 sinf ^ — j = cos( — 0) = cos 0, 
 
 sm{7r — 0) = cos(~ — 0]=: — sin( — 0) = sin 0, 
 
 and so on. 
 
 97. From the above propositious, we see that, when 
 
 n is even, any circular function of (n^±0] is equal in 
 
 magnitude to the same function of 0; and that, when 
 71 is odd, the sine, tangent and secant of the one angle 
 are respectively equal in magnitude to the cosine, 
 cotangent and cosecant of the other. 
 
 The relation between the signs of the functions of the 
 two angles can be readily seen by inspection of the 
 figure in which ^ is a positive acute angle. 
 
 Example 1.— Simplify cos(180° - a). 
 
 Here, the number of right angles is even^ therefore we retain the 
 cosine; in the simplest case i80°-a is in the second right angle, 
 and therefore cos(180° - a) is negative ; 
 
 cos(180° - a)= - cos a. 
 
 Example 2. —Simplify sin(270° - a). 
 
 Here, the number of right angles is odd. and the sine is negative 
 in the third right angle, 
 
 sin(270° - a) = - cos a. 
 
 Example 3.— Simplify tan 210". 
 
 Rejecting two right angles, and observing that the tangent is 
 positive in the third right angle, we get 
 
 tan2I0° = tan30°=-L, 
 
 v3 
 
CIRCULAR FUNCTIONS. I07 
 
 Viva Voce Examples. 
 Simplify : 
 
 1. sin (180° - a). 14. cos (270° + a). 
 
 2. tan (180° -a). 15. sin (270° + a), 
 a sec (180° -a). _ . /x^ 
 
 4. cot (180° -a). ^^' ^^H2+« 
 
 5. cosec (180° — a) 
 
 6. sin(7r + a). 
 
 7. tan(27r-a). 
 
 8. tan(7r + a). 
 
 9. sec (27r-a). ^^ . /Stt 
 
 1" 
 
 17. tan(|^ + a). 
 
 18. cot(^-a). 
 
 in «^. / I \ 19. sm (-^ + a 
 
 JO. cos (-TT + a). \ 2 
 
 11. sin(90° + a). 4'ur^o. _. /Stt \ 
 
 12. cos(90° + a). -^. 20. sec^-^-aj. 
 
 13. tan (270° -a). 
 Find the values of: 
 
 21. cos 150°. :V 
 
 22. sin 150°. '^^- ^^^ T' 
 
 23. tan 120°. . 97r 
 
 24. cot 135°. '^^- ^'" To* 
 
 25. sec 135°. . IItt 
 
 26. sin 210°. '^^- ""^^ 10 ■ 
 
 27. cos 300°. .,^ 77r 
 
 28. cot 330°. 37. cos ^. 
 
 29. sec 240°. qq . ^tt 
 
 30. cosec 225°. ^^' *^^" 4 * 
 
 31. tan ^. 39. cot ^. 
 
 4 o 
 
 32. sin ?J. 40. sec ^^ 
 
 33. cos , 
 o 
 
 3 3 
 
 57r 
 
108 
 
 FUNDAMENTAL PROFEHTIES OF THE 
 
 98. All the formuloe connecting the functions of 
 71+0 with those of 6 may be proved directly from a 
 diagram for all values of by the method of arts. 92-94. 
 
 Example 1. — To prove that sin(180°- a) = sina. 
 Let a line rotate about from the position OX, through an 
 Y angle a, to the position OP ; also, 
 
 let a line rotate from OX through 
 180° in the positive sense to the 
 position OX', and then let it further 
 rotate from OX', through an angle 
 
 X'\ ZP^ 1^' <^f magnitude a and of contrary 
 
 sense to a, to the position OP' 
 so that angle XOP = a and angle 
 XOP' = 180" -a] then OP and OP' 
 have the same projection ON on 
 y YT, 
 
 sin(180° - a) = sin a. 
 
 Example 2. — To prove that cos(a+18()°;= -cos a. 
 Let a line rotate about from the position OX, through an 
 angle a, to the position OP ; also, let it further rotate from the 
 
 
 \ 
 
 \^ 
 
 ^\/ 
 
 P'\ 
 
 A^ > 
 
 position OP, through an angle of 180° in the positive sense, to the 
 position OQ, so that angle XOP=a and angle XOQ = a+ 180°. 
 Let M be the projection of P on XX'. 
 
CIRCULAR FUNCTIONS. 
 
 109 
 
 Regarding P as a point in OQ, we have 
 
 cos(a+180°)=y-^, where OP is negative, 
 and, regarding P as a point in OP, we have 
 
 cos a=77pj where OP is positive, 
 
 cos(a + 1 80°) = - cos a. 
 
 Example 3. — To prove that cot(270°-a)=tan a. 
 
 Let a line rotate about from the position OX, through an 
 angle a, to the position OP; also, y 
 
 let a line rotate from OX, through 
 an angle 270° in the positive sense, 
 to the position OY', and then let it 
 further rotate, through an angle 
 
 equal in magnitude to a, but of x'\ ^ ^ ^,X 
 
 contrary sense, to the position Op. 
 
 Let M, m be the projections of 
 P, p on XX', N, n the projections 
 on YY', then 0?n = OiV'and On=OM 
 in magnitude. 
 
 Also, since angle Y'Op= -angle XOP, we see that p crosses the 
 line YY' when P crosses XX' ; hence Om= - ON. 
 
 Similarly, On=-OM, 
 
 .^f/o^A° N Om -ON ON , 
 cot(270 -«)-^-ZCT=^^=tana. 
 
 X 
 
 "'^ 
 
 "^^ 
 
 P 
 
 M 
 
 ^ 
 
 \ 
 
 \ 
 
 / 
 
 '« 1 
 
 \ 
 
 / 
 
 A^^^^X 
 
 / 
 
 P 
 
 ^-~- 
 
 
 § 4. Inverse Functions. 
 
 99. Defs. — If cos = a, the angle 6 is called an inverse- 
 cosine of a. Thus, if cos0 = J, any one of the angles 
 TT Stt Vtt tt' Stt 77r 
 
 3' 3 ' 3 
 
 3' 
 
 . . ., is an inverse-cosine of J. 
 
110 J'UNDAMENTAL PROPERTIES OF THE 
 
 The symbol Cos"^a is used as an abbreviation of the 
 words " inverse-cosine of «," or " any angle whose cosine 
 is a," the capital letter being used here, and in other 
 similar cases, to indicate that the symbol is many- 
 valued. 
 
 Similar definitions may be given of the inverse-sine, 
 inverse-tangent, etc., and a similar notation will be em- 
 ployed ; thus, Tan-^1 denotes any one of the group of 
 
 , TT Stt Qtt Stt Ttt IItt 
 
 angles -r, -j-, -j-, .... — r-, — j-, r-, .... 
 
 o 4'4'4' ' 4' 4 4' 
 
 It is convenient to define some one of the group of 
 
 angles denoted by an inverse circular function as the 
 
 piHncipal value of that inverse function. We select as 
 
 the principal value the numerically least angle, taking 
 
 the positive one when there are two of equal numerical 
 
 value. Tt is evident from the definitions of art. 84, or by 
 
 considering the curves of the functions (see art. 107), that 
 
 the principal values of Cos"^a, and Sec"% lie between 
 
 and TT, while those of Sin " ^a, Tan ~ ^a, Cosec " ^a, and Cot " ^a 
 
 lie between — ^ and -^. 
 
 The symbol cos-^a is used as an abbreviation of the 
 words " principal value of the angle whose cosine is a," 
 the small initial letter being used to indicate that the 
 symbol is one-valued. Similarly, the principal values of 
 Sin~%, Tan"^a, etc., are represented by sin~%, tan "%, etc. 
 
 Thus, tan-il=^, cos-i(- J) = ^|^, sin-i(-i)= -|. 
 co8ec"^^2 = Y, tan-^(-f-x)= , tan-^( — oo ) = — ^, 
 
 TT 
 
 cot-K + 0) = |, cot-V-0)=-|. 
 
CIRCULAR FUNCTIONS. 
 
 Ill 
 
 100. To find an expreasion for all angles which have a 
 given cosine. 
 
 Let a be the given cosine. 
 
 Describe a circle with centre 0, and radius OX equal to 
 unity. On OX take a length 
 OM equal in magnitude and 
 sense to the given quantity a. 
 Through M draw the chord 
 PP' at right angles to OX, and 
 join OP, OF. 
 
 Then all angles bounded by 
 OX and OP, and all angles 
 bounded by OX and OP', and 
 no other angles, have their cosines equal to a. 
 
 If a be one of the angles bounded by OX and OP, then 
 since any other of the angles bounded by these lines 
 differs from a by a multiple of four right angles, we see 
 that all the angles bounded by OX and OP are included 
 
 in the formula 2ti7r + a, (1) 
 
 where n is zero or any integer positive or negative. 
 
 Again, since the angles XOP, XOP' are geometrically 
 equal and of conti*ary sense, it follows that one of the 
 angles bounded by OX and OP' is —a, and consequently 
 all the angles bounded by OX and OP' are included in 
 
 the formula 2nir — a, (2) 
 
 where 71 is zero or any integer positive or negative. 
 
 The formulse (1) and (2) are together equivalent to the 
 single formula ^nir ±a (o) 
 
 Got. 1. — If a — cos"^a, the theorem becomes 
 Cos " % = 2')i7r ± cos " ^a. 
 
 Cor. 2. — If a=\,P and P' coincide with X, and we get 
 Cos-n = 27i7r. 
 
112 
 
 FUNDAMENTAL PROPERTIES OF THE 
 
 2n7r±-^ or Cos-i0 = m'7r+2. 
 
 If a= — 1, P and P' coincide with X\ the other ex- 
 tremity of the diameter through X, and we get 
 Cos-i(-l)=(27i + l)7r. 
 If a = 0, P and P' coincide with Fand Y\ and we get 
 
 Cos -10: 
 
 Cor. 3. — In like manner, if a be one of the angles which 
 have a given secant a, we can prove, by taking 0M= 1/a, 
 that all the angles with the given secant are included in 
 the formula Snx ± a. 
 
 101. Example.— Solve the equation 
 
 sec ^- 2 cos ^=1. 
 Multiply both sides of the equation by cos ^, 
 then l-2cos2^=cos^. 
 
 2cos2^+cos^-l=0, 
 (2cos6>-lXcos6'+l)=0, 
 
 cos 6=^ or -1, 
 =27117 ±l7r or (2n+l)7r. 
 
 102. To find an expression for all angles which have 
 a given sine. 
 
 Let a be the given sine. 
 
 Describe a circle with centre 0, and radius OX equal 
 to unity. Draw YY' the dia- 
 meter at jight angles to XX\ 
 On F take a length OJN' equal 
 in magnitude and sense to the 
 X given quantity a. Through N, 
 draw the chord PQ parallel to 
 OX, and join OP, OQ. 
 
 Then, all angles bounded by 
 OX and OP, and all angles 
 bounded by OX and OQ, and no other angles, have their 
 sine equal to a. 
 
CIRCULAR FUNCTIONS. 113 
 
 If a be one of the angles bounded by OX and OP, then, 
 since any other of the angles bounded by these lines 
 differs from a by a multiple of four right angles, we see 
 that all the angles bounded by OX and OP are included 
 
 in the formula 2m7r + a, (1) 
 
 when m is zero or any integer positive or negative. 
 
 Again, since the angles XOP, X'OQ are geometrically 
 equal and of contrary sense, it follows that one of the 
 angles bounded by OX and OQ is tt — a, and, conse- 
 quently, all the angles bounded by OX and OQ are 
 included in the formula 
 
 ^mTT+TF-a (2) 
 
 where m is zero or any integer positive or negative. 
 
 The formulae (1) and (2) are together equivalent to the 
 sentence ; " Take any multiple of tt and add or subtract 
 the angle a according as the multiple of tt is even or odd"; 
 or, in symbols, 7i7r+( — l)^a (3) 
 
 Cor. 1. — If a = sin"%, the theorem becomes 
 Sin ~ ^a = riTT + ( -- 1 )"sin - ^a. 
 
 Gov. 2. — If a = l, P and Q coincide with F, and we get 
 
 Sin"il = 27i7r+|. 
 If a= — 1, P and Q coincide with Y', and we get 
 
 Sin-i(-l) = 27i7r-^. 
 
 If a = 0, P and Q coincide with X and X\ and we get 
 Sin-i0 = 7i7r. 
 
 Cor. 3. — In like manner, if a be one of the angles which 
 have a given cosecant ct, we can prove, by taking OH 
 equal to I/a, that all the angles having the given cosecant 
 are included in the formula 
 
 H 
 
114 
 
 FUNDAMENTAL P/iOPE/tT/FS OF THE 
 
 103. Example.— Solve the equation 
 
 cosec ^ — 4 sill $=% 
 Multiply both sides of the equation by sin 6, 
 theu, 1-4 sin2^= 2 sin d, 
 
 4sm2^+2siii(9-l=0, "^l**^* 
 
 ^ .w ^,. 
 
 6^ 
 
 -2±v^+16 
 
 8 
 ±J5-1 
 
 104. To find an expi^ession for all angles which have 
 a given tangent 
 
 Let a be the given tangent. 
 
 Describe a circle with centre 0, and radius OX equal to 
 unity. Through X draw TT 
 at right angles to OX, and let 
 XTy the direction in which the 
 intersection of TT and a line 
 rotating about in the posi- 
 tive sense would move, be the 
 positive sense of TT. On XT 
 take a length XA equal in 
 magnitude and sense to the 
 given quantity a. Through A draw the diameter PP'. 
 
 Then, all the angles bounded by OX and OP, and all 
 the angles bounded by OX and 0P\ and no other angles, 
 have their tangent equal to a. 
 
 If a be one of the angles bounded by OX and OP, then, 
 since any other of the angles bounded by these lines 
 diflfers from a by a multiple of four right angles, we see 
 that all the angles bounded by OX and OP are included 
 
 in the formula 2m7r + a, (1) 
 
 where m is zero or any integer positive or negative, 
 
CIRCULAR FUNCTIONS. 115 
 
 Again, since the angles XOP, X'OP' are geometrically 
 equal and of the same sense, it follows that one of the 
 angles bounded by OX and OP' is ir + a, and, conse- 
 quently, all the angles bounded by OX and OP' are 
 included in the formula 
 
 2m7r + 7r + a, (2) 
 
 where m is zero or any integer positive or negative. 
 
 The formulse (1) and (2) are equivalent to the single 
 formula nir + a (3) 
 
 Gov. 1. — If a = tan-^a, the theorem becomes 
 Tan - hi = nir-\- tan " ^a. 
 
 Cor. 2. — If a =00, 
 
 IT 
 
 Ifa = 0, Tan-i0 = 7i7r. 
 
 Cor. 3. — In like manner, if a be one of the angles which 
 have a given cotangent ot, we can prove, by taking XA 
 equal to 1/a, that all the angles with the given cotangent 
 are included in the formula 
 
 nir + a. 
 
 105. Example 1. — Solve the equation 
 
 V3 tair(9+ 1 =(1 +v/3) tan ^. 
 We have v^3tan2(9- (1 +V3)tan ^+1=0, 
 
 (V3tan(9-lXtan(9-l)=0, 
 
 tan 6= — or 1, 
 \'3 
 
 0=9177 + '^ or mr + '^. 
 6 4 
 
 Example 2. — Write down the four smallest angles which satisfy 
 the equation 3 cot-^ -1=0, 
 
 Since cot^=±-)-, 
 
 V3 
 
 the four smallest values of 9 are ^, -"^ ^, - ^^. 
 
 3 3 3 3 
 
116 FUNDAMENTAL PROPERTIES OF THE 
 
 106. In translating formulae expressed in terms of the 
 direct circular functions into the notation of the inverse 
 functions, attention should be paid to the many-valued 
 nature of the inverse functions. 
 
 The following relations are always true : 
 
 sec~iic = cos"i-, cosec"^ir = sin-^-, cot"^ir = tan"^-, 
 
 X XX 
 
 cos"^a;+sin"'^a;= »-, sec"^i:c + cosec'^aj = — , 
 
 but , tan~^a;+cot"%=,. or — ^, 
 
 according as x is positive or negative. 
 
 Again, from the formula cos20 + sin'^O = l, we deduce, by 
 putting sinO = aj, and taking the positive value of v 1 -^^ 
 
 sin-ia; = cos"^>v/l— a;^ when x is positive, 
 and sin"^a;= —q,o^~'^s/\—x^, when x is negative. 
 Similarly, if cos Q = x, we have 
 
 cos"^a; = sin~^/v/l— ic^ when x is positive, 
 and cos"^ic = 7r — sin"i>v/l— a;^, when x is negative. 
 
 Vivi Voce ExAii4>LES. 
 State, in degrees, the value of : 
 
 1. cos-U. 8. tan-X-V3). 17 cofiC— ^- 
 
 2. sin-ij. 9. sec-ix. ' V ^3 
 
 3. tan-i(-l). 10. sin-(-l). ' ^3 ^^^.,1 
 
 ^ 1 11. cot-loo. ^2 
 
 • ''' JS- 12. cosec-(-l). 10, eos-<-4-\ 
 
 5. sec-i(-l). 13. sec-n. ' \ J'^ 
 
 1 4 14 tan-V-x). 20. sin" i(- J) 
 
 6. cosec-i /- T -,- . n oi ^ 1/ . 
 
 V^ — 1 lo. sin-U. 21. tan-X + 00 
 
 7. cos-Y-l). 16. cosec-V^- 22. cot-i( + 0> 
 
CIRCULAR FUNCTIONS. 117 
 
 23. sec-i2. 26. tan-^l. 29. cofX-O). 
 
 Zo. sin ^-2 ;. 28. cos-i(-4). 
 
 Give, in circular measure, the values of: 
 
 32. Sin-H. 38. Sin-i(-i). ^''^ ^"^ V JiJ- 
 
 33. Tan->(-l). 39. Cos-U. 43. Sin-'O. 
 
 34. Sec-H-1). 40. Cosec->^i^.**- C°t-'^- 
 
 35. Cosec-12. sj^-^io. Tan "1(^3 -2). 
 
 36. Cot-n. 4i_ Sec-'(-A). 
 
 § 5. Curves of the Circular Functions. 
 
 107. In the follov^ng figures, let OX and OY be 
 any two lines at right angles to one another. Let 
 a line of any length measured from along OX in 
 the positive sense be chosen to represent one radian ; 
 then, if be a positive number, an angle containing 
 radians will be represented by a line OM measured 
 from along OX in the positive sense and 6 units in 
 length; and an angle containing —Q radians will be 
 represented by a line of the same length measured from 
 along OX in the negative -sense. From the end M 
 of the line OM, draw MP at right angles to OX, to 
 represent, both in magnitude and sense, any circular 
 
118 
 
 FUNDAMENTAL PliOPEliTlES OF THE 
 
 function of Q. The line chosen to represent a circular 
 function whose value is unity may be of any length, but, 
 in the diagrams here given, it is equal to the length of 
 the line representing one radian. Then, as M passes 
 alono^ the line OX, from an infinite distance in the 
 
 Curves of the Cosine and Secant. 
 
 Cosine 
 
 Secant 
 
 negative sense, through 0, to an infinite distance in 
 
 the positive sense, the point P traces out a curve, which 
 
 is the curve of the particular circular function considered. 
 
 By aid of the results of arts. 86-88 and 93-97, and 
 
 the numerical values of the ratios of the angles 0, 
 
 12' 6' 
 
CIRCULAR FIWCTIONS. 
 
 119 
 
 T, Z, iT. T» given in arts. 15, 16, 18, 19 and 28 (and, 
 4 o 12 2 
 
 if greater accuracy is required, of other intermediate 
 angles whose ratios are given in Mathematical Tables), 
 we may determine any number of points on the curves 
 of the circular functions, by means of which points the 
 curve may be drawn. 
 
 Curves of the Sine and Cosecant. 
 
 Sine 
 
 Cosecant 
 
 In the first of the figures, the continuous line belongs 
 to the cosine and the dotted line to the secant; in the 
 second figure, to the sine and cosecant, respectively ; and, 
 in the third, to the tangent and cotangent, respectively. 
 In each case two complete periods are given. 
 
120 
 
 FUNDAMENTAL PliOrERTlES OF THE 
 
 It is convenient to define some part of each curve 
 representing every value of the function without re- 
 petition as the principal part of the curve. The part 
 from to TT will be regarded as the principal part for 
 the cosine and secant (or even functions) ; that from 
 
 — ^ to ^ as the principal part for the remaining (or odd) 
 
 functions. In the figures, the principal parts of the 
 curves are indicated by broader lines. 
 
 108. It will be seen, 
 by inspection of the 
 figures, that a straight 
 line drawn through a 
 given point in OX and 
 parallel to OF cuts each 
 of the curves in one, 
 and only one, point; and 
 this represents graphi- 
 cally the fact, assumed 
 in drawing the curves, 
 that, corresponding to 
 a given value of the 
 angle, each of the cir- 
 cular functions has one, 
 and only one, value, or, 
 as in art. 85, the circular 
 functions are one- valued 
 functions of the angle. 
 
 Again, if a straight 
 
 ■I 
 
 Y 
 
 \ / 
 
 \ / 
 
 J 
 
 \ J 
 / \ 
 
 s^ 
 
 
 .-.-.- -< 
 
 ' 
 
 \ 
 
 \ / 
 
 A 
 / \ 
 
 / \ 
 
 / \ 
 / \ 
 / » 
 / \ 
 
 1 
 1 
 1 
 
 j 
 
 1 
 
 X 
 
 Curves of the Tangent and Cotangent 
 
 Tangent 
 
 Cotangent 
 
 line be drawn through a given point on one of the curves 
 and parallel to OX, it will cut the curve in an infinite 
 number of points; and this represents graphically the 
 
CIRCULAR FUNCTIONS. 121 
 
 fact that, corresponding to a given value of the function, 
 there are an infinite number of values of the angle. 
 
 The curves also illustrate the following facts : (1) the 
 continuity of the cosine and sine for all values of the 
 angle, the discontinuity of the tangent and secant in the 
 
 immediate neighbourhood of the angles ^, -^-, etc., and 
 
 of the cotangent and cosecant in the immediate neigh- 
 bourhood of the angles 0, tt, 2x, etc., and the continuity 
 of the last four functions for all other angles (art. 89) : 
 (2) the fact that the cosine and sine of an angle lie 
 between +1 and —1, while the secant and cosecant 
 have all values except those between +1 and —1, and 
 the tangent and cotangent have any values whatever 
 (art. 21). 
 
 If the unit angle and the unit circular function be 
 represented on the same scale, the graphical representa- 
 tion of the facts that the limits, when is zero, of 
 sin QjQ and tan QjQ are both unity (art. 75), is that the 
 sine-curve and the tangent-curve cut OX at at an angle 
 Itt ; for, if P be a point on either of these curves and PM 
 be perpendicular to OX, these limits shew that the triangle 
 PMO is ultimately isosceles when P coincides with 0. 
 
 Again, we may employ the curves to discuss the 
 number of solutions of such an equation as cos = 0, or 
 tan Q = kO, where k is a constant. 
 
 Using the cosine-curve, take any line OM in OX, and 
 draw MP parallel to OF and equal to OM, then we infer, 
 from the figure, that the indefinite line OP cuts the 
 curve in one point only; thus there is one solution of 
 
 the equation cos = 0, and that between and J. 
 
122 FCiVDAMtWTAL PlwrEUTlES OF Till-: 
 
 If we consider the tangent-curve, and make MP = k.OM, 
 we may similarly infer that the equation tanO — kO has 
 an infinite number of roots. 
 
 Examples XL a. 
 
 1. cosl05' + sinl05° = cos45°. 
 
 2. cos 30° + cos 60' -f cos 210° + cos 270° = I 
 
 ^ T?- A ^.u 1 ^ sin 495° -f cos 390° 
 
 3. Find the value of ^a-o . — •— on/Vo- 
 
 cos49o -f sin390 
 
 Solve the equations (4-8) : 
 
 4. 4cos20-|-2cos0=l. 
 
 5. sec30-2tan2O = 2. 
 
 6. sec2e(tan0-l) = 3tane-l. 
 
 7. cos20+2sin22e = l. 
 
 8. 7sin2(x + e) + 3sin2(^| + e) = 4. " 
 
 9. Find all the angles between and tt which satisfy 
 
 the equation 3 tan^O + 8 cos^O = 7. 
 
 10. Find all the angles between and tt which satisfy 
 
 the equation 3(tan220+cot220) = lO. 
 
 11. Find the general value of which satisfies simul- 
 
 taneously the equations 
 
 sin0= —"^ and cos0 = J. 
 
 12. Find the general value of which satisfies simul- 
 
 taneously the equations 
 
 tan 0= —I and cos = — ^. 
 
 13. Shew that the same series of angles are given by the 
 
 formula ('>^-i)7^-h(-l)4 and (2?i -f i)7r ± |^. 
 
 14. Shew that the same series of angles are given by the 
 
 formulc^ (27i-l)|-f(-l)"|and 2ti7r±|. 
 
^B CIlWULAli FUNCTIONS. ]23 
 
 ^H Examples XL b. 
 
 ^L cos 165° + sin J (35° = cos 135°. 
 
 2. Find the value of sin 30° + sin 60' + sin 210° + sin 300°. 
 
 o -r.- J ,1 1 p cos 435° + sin 315° 
 
 3. Find the value of -^-tw^-, i^^v-o. 
 
 sm435 +cos31d 
 
 Solve the equations (4-8) : 
 
 4. sec2O + 3cosec20 = 8. 
 
 5. V*^(cosec2^-2) + 2cot0 = O. 
 
 6. cot^0(cosec — I) = 1 + cosec Q. 
 
 7. 3(l+cos30) = 2sin23a 
 
 8. 6cos2(7r + 0) + 5sin(j + 0) + l = O. 
 
 9. Find all the angles between 0° and 500° which satisfy 
 
 the equation sin20 = j. 
 10.- Find all the angles between and tt which satisfy the 
 equation sec*0 - 6 sec^O + 8 = 0. 
 
 11. Find the general value of which satisfies simul- 
 
 taneously the equations 
 
 tan Q = ^3 and sec = — 2. 
 
 12. Find the general value of 6 which satisfies simul- 
 
 taneously the equations 
 
 cos 0= ——7^ and cotO= —1. 
 
 13. Shew that the same series of angles are given by the 
 
 TT T , TT 
 
 formulfe (27i±l;;7- and '}27r 
 
 7". 
 
 4 '-'-"-4 
 
 14. Shew that the same series of angles are given by the 
 
 formulae (2n + D7r± a and Oi-i)x+(-l)"(|^- a). 
 
CHAPTER VIII. 
 
 CIRCULAR FUNCTIONS OF TWO OR MORE VARIABLE 
 ANGLES. 
 
 109. To prove that cos(a + jS) = cos a cos ^8 — sin a sin ^. 
 Let OF be a line making an angle ^ in the positive 
 sense with OX. Let a line rotate about from the 
 
 position OXy through an angle a, to the position OP; let 
 it further rotate from the position OP, through an angle 
 13, to the position OQ; then, the angle XOQ = a + ^. 
 
 124 
 
TWO OR MORE VARIABLE ANGLES. 
 
 125 
 
 Let Op be a line making an angle -^ in the positive 
 
 sense with OP ; let if, N be the projections of Q on OP, 
 Op respectively, and H, K, L those of Q, M, N on OX. 
 
 ^ P 
 
 Since MQ is equal to, parallel to, and of the same sense 
 as, ON, 
 
 KH= OL in magnitude and sense. 
 We have, in all cases, 
 
 OH = OK+KH - '*" - -'^■ 
 = OK+OL 
 
 OK .,,. , OL ^,. 
 = OM-^^-^ON-^^ 
 
 = cosa. Oif+cosU + IJ . ON. 
 
 But, cosU + Z)= — sin a, 
 
 OH = cos a . OM-sin a . ON, 
 OH OM . ON 
 
 cos(a -h ^) = cos a cos ^ — sin a sin /5. 
 
126 
 
 CmCULA R FUXCTIONS OF 
 
 Cor.— Changing /3 into — /3, we get 
 
 cos(a — /3) = cos a cos( — y8) — sin a sin ( — /8) ; 
 but cos( — ^) = coS|8 and sin( — /3)= --sin/3, 
 
 cos(a — /3) = cos a cos j8 + si n a sin /3. 
 
 110. To prove that sin(a + /3) = sin a cos /8 + cos a sin 6. 
 Let OF be a line making an angle ^ in the positive 
 sense with OX. Let a line rotate about from the 
 
 position OX, through an angle a, to the position OP ; let 
 it further rotate from the position OP, through an angle 
 ^, to the position OQ; then, the angle XOQ = a + ^. 
 
 Let Op be a line making an angle ^ in the positive 
 
 sense with 0P\ let if, iV be the projections of Q on OP, 
 0^9 respectively, and H, K, L those of Q, M, iV on OY. 
 
 Since i/Q is equal to, parallel to, and of the same sense 
 as, ON, 
 
 KH=OL m magnitude and sense. 
 
TWO OR MORE VARIABLE A NG LBS. 
 
 We have, in all cases, 
 
 OH=OK+KH 
 = OK+OL 
 
 127 
 
 sina.0if4-sm(a + |).0i\^. 
 
 But, 
 
 sin(a+.^ ) = eosa, 
 
 0^=sin a . Oi/+eos a . ON, 
 
 OH . OM ^ ON 
 
 -^ = sma.-^.fcosa.^, 
 
 sin(a + ^) = sin a cos ^ + cos a sin ^. 
 Cor. — Changing /3 into — /3, we get 
 
 sin (a - /3) = sin a cos( -^) + cos a sin( - /5) 
 — sin a cos /5 — cos « sin /5. 
 
128 
 
 CIRCULAR FUNCTIONS OF 
 
 111. We may deduce either of the formulae for cos(a + /3) 
 and sin(a + /S) from the other. 
 
 Thus, in the formula for cos(a+/3), change a into 
 l'^ a, then 
 
 but 
 
 and 
 
 cos(^^ + a + /3J = cos(^^ + ajcos^-sin(^|' + ajsin^; 
 cos(| + a + /3) = - sin(a + ^), 
 cosg + a) 
 sing + a) 
 
 — sm a, 
 
 cos a ; 
 
 hence, we have 
 
 sin(a + /8) = sin a cos /3 + cos a sin ^. 
 The formulae for cos(a + )8), sin(a + |8), cos(a — ;8) and 
 sin(a — j8) are called the Addition Formulce for the 
 Circular Functions. 
 
 112. Since the Addition Formulce have been proved 
 for all angles, their consequences, as given in Chapter IIL, 
 are also, as before remarked (art. 24), universally true. 
 Thus we have, for all angles : 
 
 2 cos a cos ^ = cos(a — j8) + cos(a + ^y 
 
 2 sin a sin /3 = cos(a — ^5) — cos(a + P) 
 
 2 sin a cos ^ = sin(a + P) + sin(a — /5) 
 
 2 cos a sin |8 = sin(a + /3) — sin(a — 18) 
 , cos a + cos ^ = 2 cosh{a + /3)cos J(a — jSY 
 
 cos a — cos /3 = 2 sin J(a + /3)sin J(/3 — a) 
 
 sin a + sin/3 = 2sinJ(a + /S)cosJ(a — /3) 
 
 sin a — sin /3 = 2 cos J(a + /5)sin |(a — 13} ^ 
 
 J. / j./o\ tana±tan/3 
 tan(a±^)-__- -r— ^. 
 ^ '^ l + tanatan^ 
 
TWO OR MORE VARIABLE ANGLES. 129 
 
 fcos 2a = cos^a — sin^a = 2 cos^a --1 = 1 — 2 sin^a, 
 
 '2 cos^a = 1 + cos 2a, 2 sin^a = 1 — cos 2a, 
 
 sin 2a = 2 cos a sin a, 
 
 , ^^ n 2 tan a 
 tan2a = :j — t — r"' 
 1 — tan^a 
 
 T 
 
 i COS 3a = 4 cos^a — 3 COS a, 
 L sin 3a = 3 sin a — 4 sin^a, 
 3 tan a — tan^a 
 
 tan 3a 
 
 1 -Stanza 
 
 113. To find cos a and sin a in terms of cos 2a. 
 We have seen (art. 80) that 
 2 cos^a = 1 + cos 2a and 2 sin^a = 1 — cos 2a, 
 
 /H-cos2a -, . . /l 
 
 ;a/ — —— and sina=±A/ — 
 
 ^ . . - - _ _ , — cos2a 
 
 cos a ~'" 
 
 2 
 
 Hence, we have two values of cos a, and also two of 
 sin a, in terms of cos 2a; the two values of each pair 
 being equal in magnitude but of opposite sign. 
 
 That there should be two values of each may be shewn 
 as follows : — 
 
 (1) Algebraically. — If 2a be an angle which has a 
 given cosine, then all the angles which have this cosine 
 are included in the formula 2?i'7r±2a. 
 
 Hence, in finding cos a in terms of cos 2a, we are find- 
 ing the cosines of all angles included in the formula 
 J(2'7i7r±2a) or 7i7r±a. 
 
 Now, cos('}i7r±a) = cos a, if n be even, and —cos a, if 
 n be odd. 
 
 Hence, there are two values of cos a, and, similarly, 
 two of sin a, in terms of cos 2a, equal in magnitude and 
 of opposite sign. 
 
130 
 
 CIRCULAR FUNCTIONS OF 
 
 (2) Geometrically. — Take a circle of radius OX equal 
 to unity, and along OX make 
 ON equal to cos 2a, and, through 
 N, draw the chord PP' per- 
 pendicular to OX. 
 
 Then, by bisecting the group 
 of angles bounded by OX and 
 OP, we obtain two positions 
 of the second bounding line of 
 a, namely, OQ^ and OQ^ in the figure. 
 
 Also, by bisecting the group of angles bounded by OX 
 and OP', we obtain two more positions of the second 
 bounding line of a, namely, OQg ^^^ OQ^ in the figure. 
 
 It may be shewn that Q^Q^ and Q^Q^ are diameters. 
 Hence, we obtain two values of cos a, and two of sin a, 
 in terms of cos 2a, the values of each being equal in 
 magnitude and of opposite sign. 
 
 114. To find cos a and sin a in terms of sin 2a. 
 We know (arts. 10 and 31) that 
 cos2a + sin2a = l, 
 2 cos a sin a = sin 2a, 
 (cos a + sin a)^ = 1 + sin 2a, 
 (cos a — sin a)^ = 1 — sin 2a, 
 
 cosa + sina= ±>v/l+sin2a, (1) 
 
 ±>s/r-sin2a, ...(2) 
 
 and 
 
 and 
 
 and 
 
 cos a — sin a 
 2 cos a = ± V 1 + sin 2a ± /v/l — sin 2a, 
 and 2sina= ±x/l+sin 2a + x/l — sin 2a. 
 
 Hence, we have four values of cos a, and also four of 
 sin a, in terms of sin 2a ; the values occurring in pairs of 
 equal magnitude and of opposite sign. 
 
TWO OR MORE VARIABLE ANGLES. 
 
 131 
 
 That there should be four values of each, may be shewn 
 as follows : 
 
 (1) Algebraically. — If 2a be an angle which has a 
 given sine, then all the angles which have this sine are 
 included in the formula 7i7r+( — l)"2a. 
 
 Hence, in finding cos a in terms of sin 2a, we are finding 
 the cosines of all angles included in the formula 
 
 n-TT 
 
 J{7i7r + (-ir2a} or :^+(_l)-a. 
 
 Now. 
 
 eos|-y+i 
 
 ( — l)*^a[ = cosa, if '7i = 4m, 
 
 = sina, if 7i = 4m + l, 
 
 = —cos a, if ?i = 4m + 2, 
 
 = —sin a, if '7i = 4m + 3, 
 where m is zero or an integer. 
 
 Hence, there are four values of cos a in terms of sin 2a, 
 and similarly, four of sin a, the values in each case occur- 
 ring in pairs of equal magnitude and of opposite sign. 
 
 (2) Geometrically. — Take a circle of radius OX equal 
 to unity, and let OF be a radius 
 
 making an angle -x in the posi- 
 tive sense with OX. Along Y 
 make ON equal to sin 2a, and, 
 through Ny draw the chord PP' 
 parallel to OX. 
 
 Then, by bisecting the group 
 of angles bounded by OX and 
 OP, we obtain two positions of 
 the second bounding line of a, namely, OQ^ and OQ^ in 
 the figure. 
 
 Also, by bisecting the group of angles bounded by OX 
 
132 CIRCULAR FUNCTIONS OF 
 
 and OP', we obtain two more positions of the second 
 bounding line of a, namely, OQ^ and OQ^ in the figure. 
 
 It may be shewn that Q^Q^ and Q2O4 ^^^ diameters. 
 Hence, we obtain four values of cos a, and four of sin a, in 
 terms of sin 2a, the values in each case occurring in pairs 
 of equal magnitude and of opposite sign. 
 
 115. The proper signs to be taken before the radicals 
 in equations (1) and (2) of the preceding article may be 
 determined as follows : 
 
 If a lie between 2?i7r — t and 2n7r-{-T, cos a is always 
 
 positive and of greater magnitude than sin a ; 
 .'. the + sign must be taken in both equations. 
 
 If a lie between 2w7r+T and 2n'7r-{--T^, sin a is alwa,ys 
 
 positive and of greater magnitude than cos a ; 
 
 .*. the + sign must be taken in equation (1), and the 
 — sign in equation (2). 
 
 If a lie between 2n7r+-T- and 2ti7rH--j-, cos a is always 
 
 negative and of greater magnitude than sin a ; 
 .*. the — sign must be taken in both equations. 
 
 If a lie between 2'7i7r+-T- and 27i7r+ x, ^^^ « ^^ always 
 negative and of greater magnitude than cos a ; 
 
 /. the — sign must be taken in equation (1), and the 
 + sign in equation (2). 
 
 Example. — Given sin 210°= -^, find cos 105° and sin 105°. 
 
 The sine of 105° is positive and of greater magnitude than cos 105°, 
 
 cos 105° + sin 105°= +-^, 
 "^^ cos 105° - sin 105° = - ^f , 
 
 cos 105°=!^^ and sin 105° = 1±^. 
 
 2>J2 2 v/2 
 
TWO OR MORE VARIABLE ANGLES. 133 
 
 YiVA Voce Examples. 
 State the signs of cos a + sin a and cos a — sin a when a is : 
 
 1. 98°. 5. 235°. 27r ., Utt 
 
 2. 174°. 6. 300°. 3* ^^- 6 * 
 
 3. 87°. 7. 14°. 37r 12. 4x. 
 
 4. -12°. 8. 325°. 2' 
 
 116. Example 1.— If J +^+C=7r, then 
 
 ABC 
 
 sin ^ + sin ^ + sin C= 4 cos — cos — cos — . 
 2 2 2 
 
 sin B + sin C= 2 sin ^±^ cos ^^=^ 2 cos | cos ^^H^, 
 • 2 2 2 2' 
 
 sin^ = 2cos-siu- = 2cos-cos^+^; 
 2 2 2 2 ' 
 
 .-. sin^ + sin^+sin(7=2cos^fcos:?±^+cos:^Il^') 
 2\ 2 2 / 
 
 . ^ 5 C 
 
 = 4 cos — cos - cos — 
 
 2 2 2 
 
 . '- -^ 
 
 Example 2.— If ^+J5 + (7=7r, then 
 
 cos^ J + cos^^ + cos^(7+ 2 cos A cos J5 cos C= 1. 
 cos25 + cos^C- 1 = cos^^ - sin^C 
 
 = cos(5+ C)cos(5- C)= -cos.4cos(^ - C\ 
 . •. cos^^ + cos^i? + cos^C - 1 = - cos ^ { cos(5 + C) + cos(5 - C)} 
 
 = — 2 cos A cos B cos (7, 
 .*. cos^^ +cos25+cos2C+2 cos A cos 5 cos (7=1. 
 
 Example 3. — Shew that 
 
 cos-i|| + 2tan-ii=sin-^f. 
 Let a=sin-^f and yS = tan-^i 
 
 Then tan 2/3 = ^'\ = _5_ 
 
 COS 2^ = if and sin 2^ = /^, 
 cos(a-2^)=|.l| + |.-3-%=|f. 
 
 sin-i I - 2 tan-i -J- = cos-i f f . 
 
 M^ 
 
 *t,j] 
 
134 CIRCULAR FUNCTIONS OF 
 
 Example 4.— Shew that 
 
 l-xy 
 
 where 9i=l, or - 1, according as tan-^a? + tan-^y>|^, lies between 
 
 -J and ^, or< -^, t.e. as iry > 1 (^ and y being positiveX 
 2 2 2 
 
 ^ < 1, or ^ > 1 (:r and y being negative). 
 
 We have tan(tan-^^ + tan" H/) = -^^ ; 
 
 hence, tau~^,a;+tan~^y is one of the group of angles given by 
 
 Tan-i^±^-, or mr+ta.n-'-p^ . 
 \-xy \-xy 
 
 Now, each of the angles tan~^^, tan~^v and tan~^ ^ -^ lies 
 
 l-xy 
 
 between — - and -; therefore, 
 2 2' ' 
 
 if tan~^;r+tan~^y > J, we must have w = l, 
 2 
 
 if tan~-'^+tan~^y < -^, we must have w= — 1, 
 
 but if - 1^ < tan-^^ + tan-^y < ^, then n=0. 
 
 The three cases may be more readily distinguished by consider- 
 ing the values of the product xy : 
 
 If CO and y be positive and xy>lf we have 
 
 tan~^^+tan-^y > tan-^-p+tan"^-, i.e. > % ; 
 
 X 2 
 
 and, therefore, tan"^^ + tan-^y = tt + tan"^:^ — ^. 
 
 l-xy 
 
 Similarly, if x and y be negative and xy>l, we have 
 
 tan-^;r + tan" V < tan-^^ + tan"^ , i.e. <~ ; 
 
 X 2 
 
 and, therefore, tan-^a;+tan-V= -Tr + tau"^,^ — ^. 
 
 l-xy 
 
 In all other cases, 
 
 tan-^a: + tan- V = tan-^^:t2^. 
 
 1-Xlf 
 
TWO OR MORE VARIABLE ANGLES. 135 
 
 It follows that for all positive values of x and y, 
 tan- ^1' - tan- V = taD- \'^~^ . 
 
 Example 5. — To prove that 
 
 (1) tan-i^ + tan-i^=J (Euler's formula). 
 
 (2) 4 tan-^i + tan-^ g J-g = | (Machin's formula). 
 
 (3) tan-^YT9 = tan- VV - tan"^ ^V (Rutherford's formula). 
 (1) Let a=tan-H, ^ = tan-i^, 
 
 ^, . , , o\ tan a + tan/? \-\r\ f . 
 
 then tan(a + jS) = - — r -^ = —-y^ = -l = i- 
 
 ^ '^ 1 - tan a tan ^ 1 - i • f f 
 
 tan-i^ + tan-i^=|'. 
 
 (2) Let a=tan-^i, /?=tan-Vi-9: 
 
 , , X o 2 tan a 
 
 then tan 2a = . — - — 2 
 
 1-tan^a 1-oV 
 
 o 5 
 fan Ar,— ^-12 —120 
 
 tan 4a-- ^^ -TX95 
 
 tanaa-m- TT? " ¥¥9- _ 2 8 6 8 0-1 1 9 _ 2 8 5 6 1 
 tan(^4a P;-^ ^-^^-^ ,— 2844 1 + 1 2028561 
 
 4 tan-^l^ - tan 
 (3) Let a=tan-V^, /? = tan-V9J 
 
 2T9^ 
 
 fTiPn foT^/^^_/?N— 70 99 _ 99-7 _ 29 _ 1 
 
 tnen tan(^a /ij - — — - ggg^q.^ - btft - TT^-S"- 
 
 ^ + T (T • ■9 "9 
 
 tan-^a'W = tan-^y\^ - tan" V¥' 
 Hence, |= 4 tan-^^ - tan-^^ + tan" V9 • 
 
 Example 6. — Solve the equation 
 
 sin2^ + sin22 9 + sin23 9 + sin24 (9=2. 
 Multiplying both sides of the equation by 2, it becomes 
 - cos 2^ + 1 - cos 4^+ 1 - cos 6^ + 1 - cos 8(9=4, 
 cos 2^+ cos 4^ + cos 6^ + cos 8^=0. 
 
136 
 
 CIRCULAR FUNCTIONS OF 
 
 Now, 
 and 
 
 cos 2^ + COS 8^ = 2 cos bd cos 3^, 
 cos 4^+ cos 6^ = 2 cos bd cos ^, 
 2 cos 5^cos 3^+ cos ^)=0, 
 4 cos bS cos 26 cos 9=0. 
 
 66: 
 
 4-1 or 
 
 '=W7r + ^ or d=mr+ 
 
 2' 
 
 (9=^^!■+• 
 
 f^or.|-|.|or..+|. 
 
 It is obvious that all the angles given by the formula 
 
 -+| 
 
 are included in the formula 
 
 5 10 
 This is also shewn by the accom- 
 panying figure, in which XOX' and 
 F^OPg are lines at right angles ; 
 Pj, Pgj Aj ••• Ao ^^® *^® points in 
 which the second bounding lines of 
 the angles given by the formula 
 
 5 10 
 meet the circumference of a circle with as centre. Qi, Q2, $3, Qi 
 ■ are the corresponding points for the angles given by the formula 
 
 ""2^4' 
 and P3, Pg tliose for the angles given by the third formula 
 
 nir + l. 
 
 Thus, 6'=7i| + ^'^ or 
 given equation. 
 
 i- + - is the complete solution of the 
 
 Example 7. — Solve the equation 
 
 v/3cos^+sin ^=1. 
 Dividing both sides by 2, the equation becomes 
 
 ^cos(9+^sin(9=i 
 
 .(1) 
 
 4b&;% 
 
TWO OR MORE VARIABLE ANGLES. 137 
 
 cos J COS ^ + sin J sin 6 = cos J, 
 6 6 3 
 
 cosf ^-^J = cos 
 
 e = 2n7r + '^ or 2?i7r-| (2) 
 
 Equation (1) might, however, have been written 
 
 sin ^ cos $ + cos ^ sin ^ = sin ~, 
 3 3 6 
 
 e+|=«x+(-l)».|, 
 
 0=«x + (-l)''.|-|. (3) 
 
 It may easily be shewn that the formulae (2) and (3) give the 
 same series of angles. 
 
 For, if n be even and equal to 2m, the formula (3) becomes 
 
 2m7r+|-|or2m7r-|, 
 
 the second of the formulae (2). 
 
 If n be odd and equal to 2m + 1, the formula (3) becomes 
 
 (2m+l)7r-|-|or2m^ + |, 
 
 the first of the formulae (3). 
 
 The equation here solved is a particular case of the equation 
 whose method of solution is given in the next example. 
 
 Example 8.— Solve the equation a cos ^+6 sin 0=c. 
 Suppose c to be positive. Dividing both sides by sJd^+W, the 
 equation becomes 
 
 « cos ^ + —1= sin ^=. "" 
 
 Let 
 
 a 
 
 and sin a = , then tan a = - 
 
 b 
 
138 
 
 CIRCULAR FUNCTIONS OF 
 
 cos a COS ^ + sin a sin 6^= 
 
 cos(^-a) 
 
 Voi^+P 
 
 ^-a = 2n7r±cos~^- 
 0=2w7r±cos-^- 
 
 + a. 
 
 Va2 4-62 
 
 To determine a, we first obtain from the tables the angle a! whose 
 tangent is equal to the numerical value of hja. "We then have 
 a=a', if a is positive and h positive ; 
 
 a= — a', if a is positive and h negative ; 
 a=7r-a', if a is negative and h positive ; 
 a = — TT 4- a', if a is negative and h negative. 
 The following geometrical construction illustrates the solution of 
 this equation : — 
 
 From any line OX cut oflf a part OA equal to a in magnitude and 
 
 sense ; from A draw AB dit right 
 angles to OA and equal to h in 
 magnitude and sense ; so that 
 the angle XOB=a. With as 
 centre and OB as radius, describe 
 a circle. From OB cut off ON 
 equal to c ; and through N draw 
 (if possible) the chord PP' per- 
 pendicular to OB. 
 Then, 
 
 cos(^-a). 
 
 Hence, all the angles bounded by OX and OP, and all the angles 
 bounded by OX and OP, sa tisfy th e given equation. 
 
 If ON<OB, i.e. if c<\fa^+^, there are two series of values 
 of 6, real and different ; if c=^/a^-\-b^, there are two series of 
 values, real and equal ; and if c> JcfiTW\ the line PP does not 
 cut the circle in real points, and therefore there are two series of 
 values, but imaginary and different. 
 
 This is also evident from the solution obtained above, since the 
 cosine of an angle is never greater than unity. 
 
TWO OR MORE VARIABLE ANGLES. 139 
 
 Example 9. — Trace the changes in sign and magnitude of 
 cos 0+sin 0, as 6 increases from to 27r. 
 
 cos ^ + sin 0=j2(Ji—cos ^+-i. sin d 
 
 = V2[cos ^ cos ^+ sin J sin e\ 
 =v^2cos(^-|). 
 
 As increases from to — , cos^+sin^ is positive and in- 
 creases from 1 to ^^2 ; 
 
 As 6 increases from j to -— , cos^+sin^ is positive and de- 
 creases from ;^2 to ; 
 
 As 6 increases from -- to -—, cos^ + sin^ is negative and 
 decreases from to —^2; 
 
 As increases from -- to ■—, cos^ + sin^ is negative and 
 increases from - ^2 to ; 
 
 As 6 increases from - to 27r, cos^ + sin^ is positive and 
 
 increases from to 1. 
 
 The changes in the value of cos ^+sin ^ may be iUustrated by 
 a curve as in the case of the cosine and other trigonometrical 
 ratios. 
 
 Examples XII. a. 
 
 1. cosa + cosfa+~j + cosfa + -^j = 0. 
 
 2. cosec a + cosecf a + — ] + cosecf a + ^^ j = 3 cosec 3a. 
 
 ty ,a, ,a + 7r, .q + Stt „ , 
 
 3. cot ^ + cot — H — h cot — ^ — = 3 cot a. 
 
COS 
 COS 
 
 140 CIRCULAR FUNCTIONS OF 
 
 4. cos2a = 2sinfa + T)''^i"fa+^j. 
 i3a = 22sinfa+^Jsinfa+^jsinfa+^j. 
 
 5. 4 sin a sin ^ sin y = sin (^ + y — a) + sin(y + a — /8) 
 
 + sin(a + )8-y)-sin(a + ^+y). 
 
 6. Express 4(cos a cos ^ cos y cos ^ + sin a sin /3 sin y sin ^) 
 
 as the sum of four cosines. 
 
 7. sin(/34-y)+sin(y+a)+sin(a+/3) 
 
 . . a . B . y a + B-\-y 
 = 4 sm ^ sm ^ sm ^ cos ^ — '- 
 
 , A « /5 y • a + i8+y 
 + 4 cos ^ cos ^ cos ^ sm ^^ — ^. 
 
 ^ sin(0 — |8)sin(<^ — y) — sin(0 -- p)sin{0 — y) 
 sin(^-y) 
 _ sin(^ — y)sin(0 — a) — sin(0 — y) sin(0 — a) 
 ~" sin(y — a) 
 
 _ sin(0 — a)sin(0 — ^) — sin(0 — a)sin(^ — ^) 
 ~ sin(a — )8) 
 
 g sin(^-y) ■ 8in(y-a) ^ sin(a-/3) ^Q^ 
 cos |8 COS y cos y cos a cos a cos /3 
 
 10. cos2(/3-y) + cos2(y-a)+cos2(a -/3) 
 
 = 1+2 cos(/3 — y)cos(y — a)cos(a — P). 
 
 11. cos /3 cos y sin(/3 - y) + cos y cos a sin(y — a) 
 
 + cosacos/3sin(a-/3) + sin(^-y)sin(y-a)sin(a-/5) = 0. 
 
 12. sm -y- + sm — — sm y = 4 sm =: sm =- sm -=-. 
 
 If ^+5+a=7r, prove that (13-21): 
 
 13. sin 2J. + sin 25+ sin 2(7= 4 sin ^ sin 5 sin (7. 
 
TWO OR MORE VARIABLE ANGLES. 141 
 
 ABC 
 
 14. cos ^ + cos 5 + cos (7= 1 + 4 sin — sin ^ sin ^. 
 
 15. cos2^ + cos^^ + cos^^ = 2( 1+sin ^- sin ^ sin — ). 
 
 16. tan A + tan 5+ tan 0= tan A tan 5 tan G. 
 
 17. cot5cot(7+cot(7cot J.+cot J.cot5 = l. 
 
 18. sin 6J.+sin65 + sin 6(7=4 sin SJ: sin 85 sin 3(7. 
 
 19. (sin ^ +sin 5 + sin (7)(sin 5+sin G— sin J.) 
 
 X (sin (7+ sin ^ — sin 5) (sin A-i-sinB- sin C) 
 
 = 4sin2^sin2J5sin2a 
 sin 2^ , sin 25 . sin 2(7 
 
 20 
 
 l+COS^^ ■ l + C0S2i^ ■ l-\-G0S2U 
 
 sin 9.4 -Lain 9 7? -Lain OH 
 
 ' l+cos2^ + cos2jB+cos2a' 
 21. The three expressions sinM + cos A sin B sin (7, 
 sin^^ + cos 5 sin G sin J. and sin2(7+ cos G sin ^ sin B 
 are equal. 
 09 T£ >i_L R4_r' —'^ sin J. + cos 5 — sin G _ 1 + tan|5 
 ~2' sin^ + cos(7-sin5~l + tani(7' 
 
 23. If the sum of four angles be ir, the sum of the pro- 
 
 ducts of their sines taken two and two together is 
 equal to the sum of the products of their cosines 
 taken two and two together. 
 Prove geometrically the formulse (24-26) : 
 
 OA c^ l — tan^a 
 
 24. cos2a = .i-— — 2 . 
 
 l+tan^a 
 
 25. tana+tan^ = ii5Mj^. 
 
 cos a cos p 
 
 26. tan^ = g^"°+"'"^. 
 
 2 cos a + cos p 
 
 27. Prove the formula for cos(a + ^), when 
 
 a>Jand a + /3<7r. 
 
142 CIRCULAR FUNCTIONS OF 
 
 28. Prove the formula for sin(a+/8), when 
 
 a > TT and < — , and a + i8 > -^ and < 27r. 
 
 29. cos-iT^+sin-iy«^ = cos-iAV 
 
 30. cot-i2+cosec-VlO = ^- 
 
 31. 8in-ii+sin-i3^+sin-i-l^ = |. 
 
 32. 2tan-H + tan-i|=|'. 
 
 33. tan-iJ + tan-4+tan-H+tan-i^S=j. 
 
 nA -m- 1 j.1. J. i. i?i. 1 ajcosa , ,«;— sina 
 
 34. Find the tangent of tan-\- —--. tan"^ . 
 
 ^ 1 — aj sin a cos a 
 
 Solve the equations (35-46) : 
 
 35. cos50+cos30+cos0 = O. 
 30. 4 sine sin 30 = 1. 
 
 37. sin2/O+sin2s0 = cos(r~s)a 
 
 38. sin3a-sine = 0. 
 '39. 4sine = sec2a 
 
 40. tan + tan 20 = tan 3a 
 
 41. sin50=16sin5a 
 
 42. sin2r0-sin(r-l)0 = sin2a 
 
 43. tan(^ + 0) = 3tan(|-0). 
 
 44. cos — sin = — ^. 
 
 45. cos(a + 0) = sin(a + 0) + V2cos/3. 
 
 46. 3cos0-|-sin0 = 2. 
 
 Trace the changes in sign and magnitude, as increases 
 from to 27r, of (47-49) : 
 
 47. ^3 cos + sin a 
 
TWO OR MORE VARIABLE ANGLES. 143 
 
 sin 6 — ^3 cos 
 
 ^Ssir 
 
 cos 20 
 
 4g . 
 
 ;^3 sin 6 + cos 6 
 
 49. .- 
 cos 6 
 
 50. Find the values of cos 9°, sin52J°, sin 97J°, and 
 
 cos 195°. 
 
 51. Find the limits between which 2a must lie, when 
 
 2sina= — x/l + sin 2a + s/l — sin 2a. 
 
 52. Prove that tan a, when expressed in terms of tan 2a, 
 
 has two values. 
 
 53. Prove that cos a, when expressed in terms of cos 3a, 
 
 has three values ; and that sin a, when expressed 
 in terms of cos 3a, has six values. 
 
 54. Prove that tan a, when expressed in terms of sin 4a, 
 
 has four values. 
 
 55. Eliminate between 
 
 x = 2a sin sin 20 — a cos 0, 
 2/ = 26 sin cos 20 + 6 sinO. 
 
 56. If, in a triangle ABC, cos J. = sin 5 sin C, then the 
 
 triangle is right-angled. 
 
 57. sm\0-{-a)+sm\0 + l3)-2cos(a-l3)sm{0 + a)sm{0 + /3) 
 
 is independent of 0. 
 
 58. If (l + cosO)(l4-cos0) = sin0sin 0, then or or 
 
 + ^ = (2^ + l)7r. 
 
 59. Find sec(a + /3) in terms of sec a and sec/3, and prove 
 
 that sec 105° = - ^2(1 + V^)- 
 
 60. cos 12° + cos 60° + cos 84° = cos 24° + cos 48°. 
 
 61. tan 70° = tan 20° + 2 tan 40° + 4 tan 10°. 
 
 62. sin2l0° + cos220°-sinl0°cos20° 
 
 = sinnO° + cos240° + sin2l0°cos 40° = f . 
 
 4||^fc. 
 
144 CIRCULAR FUNCTIONS OF 
 
 Examples XII. b. 
 
 1. sma + sin(a + ^3^) + sin(a + ^) = 0. 
 
 2. COs2a + cos2(a + ^) + Co.s2(a + y) = f. 
 
 3. sm2a = 2sinasiD(a + ^), 
 
 sin3a = 22sin a sin(a + |)sin(a + ^\ 
 
 sin4a = 23sin a siD(a + ^)sin(a + |)sin(a + ^). 
 
 4. cos W — cos 2a = 2(cos Q — cos a)(cos — cos o + tt), 
 cos 30— cos 3a 
 
 = 22(cos 6 - cos a)rcos - cos a + -^) 
 
 X fcosO— cosa4- >r), 
 cos 40 — cos 4a 
 
 = 23(cos 6 - cos a)f cos - cos a + 5)(cos - cosa + 7r) 
 X fcosO — cosa + -^j. 
 
 5. tan(3O-gtan(o+-^^) = tan(0+^)tan(e-^). 
 
 6. 4 cos(/5 + y - a)cos(y + a - /5)cos(a + /3 - y) 
 
 = cos(a + /5 + y) + cos(^ + y _ 3a) + cos(y + a - 3^) 
 + cos(a + ^-3y). 
 
 7. sin /3 sin y sin(^ — y) + sin y sin a sin(y — a) 
 
 + sinasin/3sin(a-/3) + sin(/3-y)sin(y-a)sin(a-^)=0. 
 
 8. sin(^ + 2y) + sin(y+2a)+sin(a + 2/5) 
 
 + sin(2/3 + y) + sin(2y + a) + siD(2a + /S) 
 = 2sin(a + i8+y){4cosi(/3-y)cosi(y-a)cosKa-/3)-l}. 
 
TWO OR MORE VARIABLE ANGLES. 145 
 
 9. sin2(/5-y) + sm2(y-a) + sin2(a-/3) 
 
 + 4 sin(/3 — y)sin(y — a)sin(a — /3) = 0. 
 
 10. COS -y- + COS -=r + COS -=- + 4 COS -j^ COS -1^ COS -^- + 1=0. 
 
 If ^+5 + (7=7r, prove that (11-20) : 
 
 11. COS 2J. +COS 25-f-cos 2(7+4 cos A cos 5 cos C+ 1 =0. 
 
 12. sin-^ + sin -^ + sin ^ = 1 + 4 cos — 7 — cos — y- cos — j—. 
 
 13. sm2^- + sm^ + sin^-^ + 2 sm ^ sin -^ sin 9 = 1. 
 
 1 4. cot ^ + cot ^ 4- cot ^ = cot ^ cot -^ cot j^. 
 
 15. sin 4J. + sin 4j5 + sin 4(7+4 sin 2^ sin 25 sin 2(7= 0. 
 
 16. sin22^ + sin225+sin22a= 2(1 - cos 2A cos 25 cos 2(7). 
 
 17. sin*^+sin45+sin^(7 
 
 = f + 2 cos ^ cos 5 cos (7+ Jcos 2 J. cos 25 cos 2(7. 
 
 -„ . 2 _1— cosu4+cos5+cos(7 
 
 (7 1 — cos (7+ cos J. + cos 5 
 tan 2. 
 
 19. cos ^+ cos 5+ cos (7 
 
 . A B-C , . B 0-A , . A-B 
 = sin ^ cos — ^ ("Si'^ o" cos — ^ hsin -^ cos — ^ — . 
 
 20. cos 2^ (cot B - cot (7) + cos 25(cot (7- cot .4) 
 
 + cos 2(7(cot ^ - cot B) = 0. 
 
 21. If^+5+(7+i) = 27r, 
 
 . , J, , ^ , ;n . ^+C C+A A+B 
 cos^ + cos5 + cos G'+ cosZ) = 4cos— ^— cos— ^^— cos— ;i— . 
 
 ^ Zi Z 
 
 22. If ^+5+ (7= J, then 
 
 tan 5 tan (7+ tan (7 tan A + tan j! tan 5 = 1. 
 
146 CIRCULAR FUNCTIONS OF 
 
 Hence, shew that 
 2(tan2^ + tan25 + tan^O) - 2 
 
 = (tan5-tanC)2 + (tan(7-tanJ.)2 + (tan^-tan5)2, 
 and that the expression tanM + tan^JJ+tan^O is 
 never less than unity. 
 
 23. If the sum of four angles be two right angles, the 
 
 sum of their tangents is equal to the sum of the 
 products of the tangents taken three and three. 
 
 Prove geometrically the formulae (24-26) : 
 
 24. tan^° l^f^ii'. 
 
 2 1 + cos a 
 
 g„ tana + tan/3 _ sin(a-|-i3) 
 tan a — tan |8 ~ sin(a — ^)' 
 
 26. cot^±^ = "'°°-"'°^. 
 
 2 cos p — cos a 
 
 27. Prove the formula for cos(a-f/3), when a<~, and 
 
 a + /3>7r and<^. 
 
 28. Prove the formula for sin(a-f-/3), when a>'7r, and 
 
 29. sin-i| + sin-W = sin-i|-^, 
 
 30. 4(cot-i3 + cosec-V5) = 7r. 
 
 31. tan-4 + cot-4 + sin-i'^^ = 7r. 
 
 32. tan-iT2__|_2tan-i| = tan-i|. 
 
 38. tan-4 + tan-if + tan-H + tan-4 = |^. 
 
 34. Tan-i 7^^^"" +Tan-^^HL^ = r^7r + a, n being 
 l+w-cosa m + cosa 
 
 any integer. 
 
TWO OR MORE VARIABLE ANGLES. 147 
 
 Solve the equations (35-46) : 
 
 35. sin 0+sm3O+sino0 + sin 70 = 0. 
 
 36. cos cos 30 = cos 20 cos 6a 
 
 37. cosr0+cos(r-2)0 = cosa 
 
 38. sin 40 -sin = 0. 
 
 39. sin = cos 40. 
 
 40. tan0 + tan30 = 2tan20. 
 
 41. 27sin80 = cos80-4cos40 + 3. 
 
 42. sin 30-2 sin30 = f. 
 
 ^^' ^^^(252'^^^) = 'K272''^'4 
 
 44. cos + sin = c. 
 
 45. 1 + cos 20 -sin 20 = 0. 
 
 46. 2sin0-cos0 = J. 
 
 Trace the changes in sign and magnitude, as increases 
 from to 27r, of (47-49) : 
 
 47. cos — sin 0. 
 
 48. tan0-2cosec2a 
 V3-|-tan0 
 
 ^' V3-tan0* 
 
 50. Find the values of sin7J°, cos22J°, cos 127 J°, and 
 
 sin 1874°. 
 
 51. Find the limits between which 2a must lie when 
 
 2 cos a = — v/l + sin 2a — aJi — sin 2a. 
 
 52. Prove that sin 2a, when expressed in terms of sin a, 
 
 has two values of equal magnitude and opposite 
 sign ; and that cos 2a, when expressed in terms of 
 cos a, has only one value. 
 58. Prove that sin a, when expressed in terms of sin 3a 
 has three values ; and that cos a, when expressed 
 in terms of sin 3 a, has six values. 
 
148 CIRCULAR FUNCTIONS OF 
 
 54. Prove that sin a, when expressed in terms of tan 2a, 
 
 has four values. 
 
 55. Eliminate between 
 
 X = 2a sin W cos — a sin 20, 
 y = 2h cos SO cos — 6 cos 20. 
 
 56. Evaluate tanf|^+ a jtanf—+ a V 
 
 57. The tangents of two of the angles of a triangle are 2 
 
 and 3, find the third angle. 
 
 58. Prove that 
 cos(|8+y-a)+cos(y+a-/3)+cos(a+^-y)-4cosacos^COSy 
 
 vanishes when a + fi + y is an odd multiple of a 
 right angle. 
 
 59. cos20° + cosl00' + cosl40° = 0. 
 
 60. cos 12° 4- cos 108° + cos 1 32° = 0, 
 
 cos 108°cos 132°+cos 132°cos 12° + cos 12°cos 108°=-|, 
 
 cos 12°cos 108°cos 132° = 1±^. 
 
 lb 
 
 61. (2cos^+10cos|^y+(4sin^)'=7]. 
 
 62. If sin 3a = 71 sin a be true for any values of a besides 
 
 or a multiple of ^, then n must be less than 3 
 and not less than - 1. 
 
 117. Example 1. — If cos2a+cos^/3+cos^y + 2cosacos/?cos'y = l 
 find the relations which must exist between a, /? and y. 
 cos'^a+cos^jS + cos^y + 2 cos a cos /5 cos y - 1 
 
 = (cos a + cos /? cos y)'^ - cos^/? cos^y + cos^^ + cos^y - 1 
 
 =(cos a+cos /? cos y)^ - (1 - coa^/3){l - cos^y) 
 
 = (cos a + cos )8 cos y )2 - sin^/? sin^y 
 
 = (cos a + co=' /? cos y - sin /? sin y)(cos a + cos ft cos y + sin ^ sin y) 
 
TWO OR MORE VARIABLE ANGLES. 149 
 
 = {cos a + cos(/^4-y)}{cos a + cos(^ — y)} 
 
 a-\-B + y B + y-a y + a-^ a + /5-y ^ 
 2 A 2 Z 
 
 .'. either a + /3 + y, /3 + y-a, y + a- /3, or a + f3-y must be an odd 
 multiple of tt. 
 
 Example 2.— If v = taii-i^^^i±^i:i + tan-^-^,, 
 
 express x in. terms of y in its simplest form. 
 
 Suppose X numerically less than unity; let ^ = tan~^^, then 
 ^=tan 0. 
 
 vT+^-l__sec ^-l„l-cos6^^^^^ $ 
 
 X 
 
 tan e sin Q 2' 
 
 •1 'ix 2 tan Q x. r.n 
 and = - — — -— = tan 2^ ; 
 
 6'=tan-\r=^. 
 5 
 
 If X be positive and > 1, then it may be shewn that 
 
 .r = tan-|(y + 7r); 
 
 if X be negative and numerically > 1, that 
 
 ^=tan|(2/-7r). 
 
 If X have any other value, then 
 
 ^=tan-^. 
 5 
 
 Example 3. — Eliminate Q between the equations 
 
 (a + 6)(^ +^) = cos ^( 1 + 2 sin26'), 
 
 (a -h){x-y)= sin ^(1+2 cos26'). 
 
 We have (a + 6) (.^ + ?/) = cos ^ + sin Q sin 2 ^, 
 
 {a-})){x-y)—%va. ^ + cos ^sin 2^ ; 
 
 2(a^ + 6y)= (cos ^ + sin ^)(1 + sin 2^) 
 
 = (cos^ + sin6')3, 
 
 and %ciy + 6ji;) = (cos Q - sin ^)3 ; 
 
 (a^ + 6?/)^ + (a^/ + 6^)3 = 2-i-2^ = 2* 
 
150 CIRCULAR FUNCTIONS OF 
 
 Examples XIII. 
 1. Prove that 
 tan a -f tan 2a + tan 3a + tan a tan 3a tan 4a = 
 
 cos 2a cos 4a' 
 and verify this formula when a = t ^^^ when a = ^. 
 
 2. sin(^ + y — a)sin(^ — y)cos(^ — y) 
 
 + sin(y 4- a — /8)sin(y — a)cos(y — a) 
 + sin(a + /3 — y)sin(a — /3)cos(a — /3) = 0. 
 
 3. sin K/S - y)sin f (/? + y) + sin Ky - a)sin f (y + a) 
 
 + sinJ(a-/3)sinf(a+)8) 
 = 4 sin J(/3 - y)sin J(y - a)sin J(a - /3)sin(a + /3 + y). 
 
 4. COs(^ + y)cos(y+a)cos(a + i5) 
 
 = cos a cos /3 cos y cos(a + jS + y) 
 
 + sinasin^sinysin(a + /3+y). 
 
 5. cos2acos2(^ + y) + C0S 2/5cos2(y + «) + cos 2ycos2(a+y8) 
 
 = cos2acos2^cos2y + 2cos(^+y)cos(y+a)cos(a+^). 
 
 6. cos(a + /3)cos(a — |8)cos(y + ^)cos(y — ^) 
 
 — sin(a + /3)sin(a — /3)sin(y + (5)sin(y — (5) 
 = l-Jsin2(^+y)-Jsin208-y)-Jsin2(a + (5)-Jsin2(a-^). 
 
 7. cos(a + 18 + y)cos(^ + y — a)cos(y + a — /8)cos(a + )5 — y) 
 + sin(a + /3 + y)sin(/3 + y - a)sin(y + a - /3)sin(a + /3 - y) 
 
 = cos 2 a cos 2/3 cos 2y. 
 
 9. {sec a + cosec a(l + sec a)}(l — tan2Ja)(l— tan^Ja) 
 = (sec |a + cosec Ja)sec2Ja. 
 
 10. If^+5+C^=7r, then 
 
 sinM sin 2^ + sin^^ sin 2^+sin2(7sin 2G 
 = 2 sin J. sin B sin (7+ sin 2^1 sin 2B sin 2(7. 
 
 11. If ^+5+ (7= TT, then ^ 5 (7 
 
 tan^+tan^+tan^ = 4. ^^^^^,:,^B+.inG 
 
TWO OR MORE VARIABLE ANGLES. 151 
 
 12. Prove, geometrically, that 
 
 sin(a + fi)sin{a — /3) = sin^a — sin^/^. 
 
 13. If the sum of the sines of three angles is equal to the 
 
 sine of their sum, the sum of two of the angles 
 must be a multiple of four right angles. 
 Solve the equations (14-21), 6, and x being the 
 unknown quantities : 
 
 14. cosec 4a — cosec 40 = cot 4a — cot 40. 
 
 15. + 9^ = 240° and vers = 4 vers ^. 
 
 16. tan20 = 8cos20-cota 
 
 17. sin0-cos0-4sin0cos20 = O. 
 
 18. tan0 + tan(^^ + 0) = 2. 
 
 - Q sin a cos(/3 + 0) _ tan ^ 
 sin/3 cos(a-|-0) tana' 
 
 20. sin-i:r^, + tan-i ^^ 
 
 TT 
 
 1+x^ ' 1-x^ 2 
 
 21. tsin-'^x + t3i.n-\l-x) = 2t&n-'^Ajx-xi 
 
 22. Find all the values of u and 0, for which 
 
 _ tan 
 '^~l-ta.n20 
 changes sign, as 6 passes from a small negative 
 quantity, through zero, to four right angles. 
 
 23. Trace the changes in sign and magnitude of 
 
 sin + cos 
 sin — cos 
 as changes from to 27r. 
 Eliminate between the equations (24-26) : 
 
 24. cos20-hsin20 = cos0 + sin0 = (X. 
 
 OK • _ ^^^ " _ 1 
 
 '^^' '''' "" - 73^inY0 ~ 2-hV3cos20' 
 
 26. asin 04-6 cos = c and acosec0H-6sec0 = d 
 
152 CIRCULAR FUNCTIONS OF 
 
 27. Express as a single term 
 
 1 I - 1 . 
 
 ^2 cot |a — cosec Ja ^2 cot \a + cosec J a* 
 
 28. If sm(a + )8)cosy = sin(a + y)cos/3, then, either ^ — y 
 
 is a multiple of tt, or a is an odd multiple of ^. 
 
 29. Prove that 16 sin50 = sin 5^-5 sin 30 + 10 sin 0, and 
 
 deduce the value of 32 sin^0 in terms of cosines of 
 multiples of 0. 
 
 30. If atana + 6tani8 = (a + 6)tan^^^, then ?=^^. 
 
 '^ ^ ^ 2 b cos/3 
 
 31. If !!£i4^=!i5g±^, then either a and ^. or 6 
 
 sm(a + 0) sm(/3 + </)) '^' 
 
 and (p, differ by a multiple of tt. 
 
 32. If sec(0 — a), sec 0, and sec(0 + a) be in arithmetical 
 
 progression, then cos (p = ^1 . cos J a. 
 
 oo Tr- • . cos^O— sin0 J ^ sin^^ — COS0 ., 
 
 33. If sm = r jr-. — 7^ and cos <h = .; ^ . ^ , then 
 
 ^ 1— cos^sin^ ^ 1— cosOsm^ 
 
 . /^ cos^(^ — sin , ^ s\v?(h — cos 
 
 sin Q = - — -^ r-^- and cos 6 = z: ^ , . ^ . 
 
 1 — cos (p sin 1 — cos <p sm 
 
 34. If cos a = f and /3 — a = -r, find tan^ and tan(a + /3). 
 
 2 2 2 
 
 35. If -„ cos = ^ cos 0+^2 cos a, 
 
 a^ a^ 0^ ^ 
 
 and ^ - y - ^ 
 
 sin(0+0i) sin(0-0i) sin 20' 
 
 , , sin W 
 
 then ^_^ = 
 
 sin 0^ a^ 
 
 36. If tan a; = cos a tan y, 
 
 tan^" sin 2^/ 
 • then tan(2/ — a;)=■ 
 
 l+tan2^cos22/ 
 
TWO on MORE VARIABLE ANGLES. 153 
 
 cos U—e 
 
 87. If cosF= 
 
 1 — ecos V 
 
 U 
 
 then tan2 = yj^tan-2- 
 
 88. If cot = 71 cot(a-O), 
 
 39. If sin a and sin /3 be two values of sin satisfying 
 the equation a cos 20+ 6 sin 20 = c, 
 
 then cos^a — sin^-S = -^^-^ „. 
 
 40. cos8(a) + a)sin(/3 — y) + cos8(ir + ^)sin(y — a) 
 
 + cos Z{x-\- y)sin (a — /3) 
 = 4 cos(3cc + a + )8 + y)sin(/3 — y)sin(y — a)siu (a — /3). 
 
 41. (cos a + cos |Q + cos y) 
 x{cos2a+cos2/3-fcos2y-cos(|8+y)-cos(y+a)-cos(a+/3)} 
 — (sin a + sin /5 + sin y) 
 
 X {sin2a+sin2|8+sin2y-sin(/3+y)-sin(y+a)-sin(a+j8)} 
 = cos 8a + cos 8y8 + cos 8y — 3 cos(a + /3 + y). 
 
 42. cos A cos B cos G cos D + sin A sin 5 sin C sin i) 
 
 = cos a cos /3 cos y cos (5+ sin a sin /5 sin y sin ^, 
 where 2a=5 +(7 +i)-^, 
 
 2p = G^D+A-B, 
 2y = D+A+B-G, 
 ^6=A-\-B+G -D. 
 48. If a, i8, y be all unequal, and if no two of them 
 differ by a multiple of tt, 
 
 {tan(/3 - y) + tan(y — a) + tan(a — /3)} 
 
 X {cot(^-y) + cot(y-a) + COt(a-/5)} 
 = 1 — sec(|8 — y)sec(y — a)sec(a — P). 
 
154 CIRCULAR FUNCTIONS OF 
 
 ,. (sec g sec /3 + tan « tan /3)^ — (tan g sec /3 + sec a tan /3)^ 
 • 2(1 + tan^g tan2/3) - sec^g sec^^ 
 
 _ sec 2g sec 2/3 
 ~ sec^gsec^jS * 
 .p, 3sin3g 1 
 
 sing 'cos3g — cos3^ 
 
 = \ + 1 + \ 
 
 cosg-cos/3^ , /tt ^\ /27r ^V 
 
 '^ cosg + cos( K- — p) cosg — cos! -^ — /5) 
 
 46. (a sin ^ + 6 cos 0)(a sin 1/^ + 6 cos i/r)sin(0 — yp) 
 
 + (a sin \/r + 6 cos V^)(a sin + 6 cos 0)sin(i/r — 0) 
 4- (a sin + 6 cos 0){a sin ^ + 6 cos 0)sin(0 — 0) 
 + 4(a2 + 62)sin(9!> - V^)sin(^ - 0)sin(0 - 0) = 0. 
 
 47. 2 (cos y8 cosy -cos g)(cosy COS g-cos/3)(cos g cos^-cosy) 
 + sin^g sin^/? sin^y — sin2g(cos ^ cos y — cos g)^ 
 
 — sin2/3(cos y cos g — cos /3)2 + sin2y(cos g cos /5 — cos y)^ 
 = (1 — COS^g — COS^^ — COS^y + 2 COS g COS /3 COS yf. 
 
 48. If^+5 + C' = 7r, then 
 
 sin(^-a) mi\{G-A) ^m{A-B) 
 sin -4 sin B sin (7 
 
 4sin(^-(7)sin((7-^)sin(^~^) ^ 
 ■*■ sin 2^ + sin 2j5+sin 2(7 
 
 49. If ^+5 + (7= X, then 
 
 (2/4-2; COS ^ )(0 4-a5 cos 5)(fl?+2/ cos (7) 
 
 + (2/ cos A-\-z){z cos -B + a;)(a; cos (74- 2/) 
 vanishes, if x sin J. 4- ?/ sin 5 4-0 sin (7= 0. 
 
 50. If g, ^, y, ^ be the angles of a quadrilateral, then 
 
 tan a tan ;8 tan y tan ^^tan « + tan ^+tan y+tan 6 
 cot g4-cot p4-cot y4-cot S 
 
 51. If ^4-54- (7= TT, then 
 
 sin A cos(J. -5)cos(^ - (7)4- sin B cos(B - C)cos{B - A) 
 H-sin (7cos((7-^)cos((7-5) 
 = 3 sin J. sin B sin (7+ sin 2 J. sin 2B sin 2(7. 
 
TWO OR MORE VARIABLE ANGLES. 155 
 
 52. If J. +^+(7= TT, then 
 
 sin^^sin^a+ sin^asinM + sin^^ sin^^ - sinM sin^^sip^g 
 cos^ J. cos^^ cos^C 
 = (tan B tan (7+ tan G tan ^ + tan A tan -B)^. 
 
 53. If ^+5+(7=7r, and n be any integer, then 
 
 tan nA +tan 7i5 + tan nG= tan -Ji^ . tan nB . tan ^iC. 
 
 54. If J.+5+a=x, then 
 
 cos 2^ (tan 5 — tan G) + cos 25(tan G— tan ^) 
 + cos 2(7(tan J. — tan B) 
 2 sin(^- (7)sin(a-^)sin(^ -^) 
 ~ cos A cos J5 cos (7 
 
 •cos a; 
 cos a;* 
 
 trtr To ^ l/x «x ^\n COSa + ( 
 
 55. COS 2 tan-^tan rrtan ^ =t-; 
 
 L \ 2 2/ J 1 + cosa 
 
 56. tanr2 tan-^tan % tan(j-f )|1 =_ii5^£2i^. 
 
 L I 2 \4 2/JJ sm/3 + cosa 
 
 57. If 2^/ = aj + sin ~ \a sin ic), then 
 
 tan(a;-2/)=Y^'tan2/. 
 
 58. If It = cot ~ ^x/cos a — tan ~ ^Vcos a, then 
 
 sin u = tan^-. 
 
 59. If 2/ = tan"^ . , find the value of a; in 
 
 vl + aJ +v 1— a? 
 
 terms of 2/. 
 
 60. If sin(0+</)-\/r) = sin(a + /3), cos(V^ + ^-9!>)=cos(y + a), 
 
 and tan(0 + i/r — 0) = tan(y8 + y), find the general 
 values of 6, 0, yjr in terms of a, /5, y. 
 
 61. Find the general value of 20 from the equation 
 
 tan0+tan(j + 0) = 2. 
 
156 CIRCULAR FUNCTIONS OF 
 
 Solve the following equations (62-75), 0, cp and x being 
 the unknown quantities : 
 6 2. cos^^ 4- cos = 1 = sin^O + sin (p. 
 
 63. p sin^0 — q sin*0 =p, 
 p cos^O — q cos*0 = q. 
 
 64. ^3 cos 20 -^2 cos a = ^2 sine -sin 2a 
 ^„ ^ 2cos(0 + a)cos(O — a) 
 
 •cos(0 + a) + cos(0 — a) 
 
 66. cos(20 4- 3a)cos(2e - 3a) - 2 cos a cos 3a cos 20 + cos^a = 0. 
 
 67. 1 — cos 20 = 2(cos acos 0— cos 2a). 
 
 68. (1 + sin 0)(1 - 2 sin Of = (1 - cos a)(l + 2 cos of. 
 
 69. sec 40 -sec 20 = 2. 
 
 70. a tan 0+6 cot = c, by the aid of trigonometrical 
 
 tables. 
 
 K^ mtan(a--0) _J cos \^ 
 n tan ~ \cos(a — 0)/ 
 
 72. COS0 + COS ^ + cos a = sin + sin + sina, + = 2a. 
 
 73. cos30 - cos sin - sin30 = I. 
 
 74. cos(20 + 0) = sin(0-20), 
 cos(0 + 20) = sin(20-0). 
 
 75. sm ^ — hsin ^ — = ^. 
 
 76. If sin? = =^^ — ^, trace the changes in as in- 
 
 2 1 + COS0 ° ^ 
 
 creases from to ^- ; and find cos 0, sin and 
 
 tan <p in their simplest forms as functions of 0. 
 Trace the changes in sign and magnitude, as increases 
 from to 27r, of (77-82) : 
 ^y sin 30 
 '' cos 20* 
 78. sec 0-3 + 2 cos 0. 
 
TWO OR MORE VARIABLE ANGLES. 157 
 
 79. cos(7r sin 0). 
 
 80. sm(7r cos 20). 
 
 81. cos(7r sin 26). 
 
 82. cos(j7r cos 6) — sin( Jtt cos 6). 
 
 Eliminate 0, or 6 and (^, from the equations (83-88) : 
 
 83. tanO + tan^ = a, cot0 + cot0 = 6, — (p = c. 
 
 84. sin0 + sin^ = a, cos + cos = 6, (1 — c^)tan(O+0) = 2c. 
 
 85. c = acos + 5 sin = a cos{0 — a) + h sin{0 — a). 
 
 86. - = cos0 + cos20, | = sin0 + sin2a 
 
 87. a sec(0 -a) = h sec(0 - jS), a'sec(0 - a) = h'sec{e - /^O- 
 
 88. tan + tan ^ = a, tan 6 tan ^(cosec 20 + cosec 2^) = 6, 
 
 cos(0 + 0) = c cos(0 - 0) ; 
 
 and, if CL = —m and 6 = 2(^^2 — 1), find the least 
 positive values of 6 and 0. 
 
 89. If !ir^=!;z&=c, 
 
 cos t7 sm 
 
 and =——^^=:c, 
 
 cos sin 6 
 
 where c and c' are both positive or both negative 
 
 quantities, then (a — ay + {b — h')^ cannot be greater 
 
 than (c + c')2. 
 
 90. If a and /5 be two different values of satisfying the 
 
 , . cos . sin 1 
 
 equation — = — = -, 
 
 a c 
 
 then a cos — ~- = h sm '^ = c cos — q^- 
 
 A /!i A 
 
 91. If a and ^8 be two different values of satisfying the 
 
 ,• cos . sin 1 
 equation -+ 
 
 then ^-M^- 
 
158 CIRCULAR FUNCTIONS OF 
 
 92. Prove that 
 
 2 cos(/3 - y)co8(0 + ^)cos(0 + y) 
 
 + 2 cos(y — a)cos(0 + y)cos(0 + a) 
 
 + 2 cos(a - /3)cos(0 + a)cos(0 + /3) 
 
 -cos2(e+a)-cos2(O+)5)-co.s2(0 + y)-l 
 
 is independent of 0, and exhibit its value as the 
 
 product of cosines. 
 
 gg j£. sin(0 + g) _ 1 + e cos(0 + ffl 
 
 sin a lH-ecos/3 ' 
 
 /I 
 find tan - in terms ot a, /8, and e. 
 
 94. If a; = acos0 — rcosf^+^j, 
 
 2/ = 6 sin + r sin(^| + Y 
 
 ^ , /cos Q , sin 0\ 
 2cos<^ = r(— +-^-) 
 
 ^ . ^ fcoaO sin6\ 
 
 (a + h^Sf-^ib + ajSy-^- 
 
 95. If tan0tan0 = J(|^), 
 
 then (a - 6 cos 20) (a - 6 cos 20) = a^ - h^ 
 
 96. If ^3-tan0 = -l--tan0, 
 
 x/3 
 
 then tan(0-|) = 3tan(0-|). 
 
 97. If tan ^0 = tan^ J0, and tan ^ = 2 tan a, 
 
 then 0+0 = 2'M7r + 2a. 
 
 98. Extract the square root of 
 
 (2 + 2 sin + cos 0)2+ (2 + sin + 2 cos 0)2. 
 
TWO OR MORE VARIABLE ANGLES. 159 
 
 99. If Q and be acute angles, such that 
 3sin20 + 2sm20 = l, 
 Ssin 20-2 sin 20 = 0, 
 prove that + 20 = J; 
 
 and find sin Q and sin 0. 
 
 100. If ^2cos0 = cos0 + cos30, 
 
 and ^/2 sin = sin — sin^0, 
 
 then , ±sin(0 — O) = cos20 = J. 
 
 101. If^cos0 + 5sin0-(7vanishfora = a,0 = ^,0 = a+/3, 
 
 respectively, then A — G. 
 
 102. If cos0 + cos0 = a and sin0+sin0 = 6, 
 
 find the value of cos(0 — 0) and cos(O + 0) in terras 
 of a and h. 
 
 103. If 0^ aDd ^2 be two roots of the equation 
 
 Aco^e+B^me+G = Q, 
 such that the cosines of 0^ and Og ^^^ ^^^ equal, 
 then cos(0, + 0,) = ^!=J'. 
 
 104. 16 cos t;— cos -zr^ cos -=^ cos -=^ = 1. 
 
 15 15 15 15 
 
 105. If ; S = l, prove that both the numerator 
 
 sec a + sec /5 ^ 
 
 and denominator of this fraction must vanish un- 
 
 less 
 
 a = 27i7r + |, 
 
 or . 
 
 ^ = 2^7r + |, 
 
 or 
 
 a + ^ = (2n + l)7r. 
 
160 CIRCULAR FUNCTIONS OF 
 
 106. If a, P, y, S be all different, then 
 cot(a-/3)cot(a-y)cot(a-6) + cot(/3-y)cot(/3-^)cot(^-a) 
 
 + cot (y-^)cot(y-a)cot(y-^) + Cot(^-a)cot(^-^)cot(^-y) = 0. 
 
 107. If a, P, y, 6 be all different, then 
 Cot(a-y)cot(a-^) + cot(a-^)cot(a-/3) + cot(a-j8)cot(a--y) 
 
 + cot(^-^)cot(^-a) + cot (^-a)cot(/3-y) + cot(;8-y)cot(/3-^) 
 + COt(y-a)cot(y-^) + cot(y-^)cot(y-^) +COt(y-(5)cot(y-a) 
 + cot(^-^)cot((5-y)+cot(^-y)cot((5-a)+cot((5-a)cot(5-^)=-4. 
 
 108. If a, A y, S be all different, then 
 
 cos 2a , cos 2^ 
 
 . a-B . a-y . a-^ T~^-y . /3-5 . S-a 
 sin — ^ '^ sin —^ sm — ^ sm s^ sm ^— «— sm -'— ^ — 
 
 cos 2y , cos 2^ 
 
 . y — ^ . y — a . y — & - S—a . S — B . ^ — y 
 sin ^^-^ sm ^^-s— sm ^ '^ sm —^ sin — o^ sin — ^ 
 
 109. If (a+6)tan(e-0) = (a-6)tan(e+0), 
 
 and a cos 20 + 6 cos 2^ = c, 
 
 then fe2 _ ^2 _ 2c<x cos 2^ + a\ 
 
 110. If ajcos20+2/sin20 = a(cos0X, 
 
 and ccsin 20 — 2/cos2O = a(cos0)*'-isin0, 
 
 l-C+S)' 
 
 111. Eliminate from the equations 
 
 , , sin a cos — sin i8 sin 
 tan0 = ^ . ^ , 
 
 cos a cos t^ — cos p sm t7 
 
 , , sin a sin — sin i8 cos 
 
 tanY^ = r— 5 % -. 
 
 cos a sm t7 — cos p cos t7 
 
TWO OR MORE VARIABLE ANGLES. 161 
 
 112. Eliminate from the equations 
 
 cos^e 3sin2 e sin^e_ 4 
 a^ ac c^ ac 
 
 cos6 .8m6_ 2 
 c a J'oi'c 
 
 113. If cosOcos0 = sin(a — /3)sin(a + j8), 
 
 and sin(0 — 0)sin(04-0) = 4cosacos/3, 
 find cos and cos cj). 
 
 114. If a; + 2/+2; = 2sin0, a;2+2/H2;2 = cos20, 
 
 2(a;3 + 2/^+;s^) = 3 sin 30 and 2iC2/^ = 3 sin 0, 
 
 then f) = TiTT or TiTT ± J. 
 
 o 
 
 115. Find x from the equation 
 
 3 tan-i(aj + l) = 2 tan-X!:c-l)+tan-i 
 
 2-a5 
 
 116. Trace the chancres in siffn of — j ^ - . ' J. , as 
 
 ^ ^ cos(7r sm 0) 
 
 increases from to tt. 
 
 117. If cos i/r = cos 0(tan a sin ^ — cos 0) 
 
 = sin 0(sin —cot a cos 0), 
 
 prove that either yp- is an odd multiple of -^^ or else 
 
 i/r=(2'?i+l)7r±(0+^), where -n is an integer. 
 
 118. Prove that the expression 
 
 a cos^O + 26 cos sin + c sin^O 
 may be written in the form 
 a + c 
 
 where m=— ^ , 7i = o, tan0 = — . 
 2 ^ n 
 
 Hence, prove that the greatest and least values 
 
 of the given expression are the roots of the equation 
 
 {x — a)(x — c) = h^. 
 
162 CIRCULAR FUNCTIONS OF 
 
 119. If a' = acos20+2^cos^sia04-?>sm2a, 
 
 W ={h — a) cos sin ^ + A, cos 20, 
 })=a sin^^ — 2/i cos sin 6+h coa^O, 
 
 then 
 
 a'cos\<p -6) + 2h'cos{<p - 0)sin(9!) - 0) + 6'sin2(0 - 6) 
 = (X cos^^ + 2h cos sin ^ + 6 sin^^. 
 
 120. If y^-{-2yz cos e + z^ = p\ 
 
 z^ + 2zxcos<l) + x^ = q\ 
 x^+2xy cos \jr + y^ = r\ 
 and + ^ + V^ = 7r, 
 
 then 4(2/0 sin O + zxsiuip + xysm \]rf 
 
 = 2 (^V + ry +^2^2) _p4 _ ^4 _ ^_ 
 
 121. If a sin a = 6 sin ft a sin |8 = 6 sin 0, a — /3 = — (l), 
 
 and a, 6 be not numericall}^ equal, then a = /5 + '>?x, 
 = (p+n'7r, where ti is an integer. 
 
 122. If x = y cos Z+z cos Y, 
 
 2/ = cos X+ a? cos Z, 
 and if jr+ Y+Z be an odd multiple of tt, prove 
 that = oj cos F+ 2/ cos X. 
 
 Hence, prove that 
 
 C0SZ = ^^^— ^TT . 
 
 2yz 
 ^go Tf sin ra _ sin (9* + 1 )a _ sin(r + 2)a 
 
 I ~ 7n ~ n ' 
 , cos ?-a _ cos(r + l)a _ cos(r + 2)a 
 
 ^., 2m2 — ^(Z+'yi)~ 7n{n — l) ~n{l+n) — 2m^' 
 
 l24i. If tan (cot 0) = cot (tan 6), shew that the real values 
 1 of are given b}^ 
 
 sin 20 = 7^5 TT^' 
 
 72, being any integer, positive or negative, except 
 
TWO OR MORE VARIABLE ANGLES. 163 
 
 1 2o. cos*^ + cos^-^- + cos*-^ + cos^^ = jg. 
 
 126. sec*^+sec4^ +sec*-|" + sec'i^ = 1120. 
 
 10H. TT 27r 3x 47r Stt Btt Ttt /IV 
 
 127. cos ZTF cos vv cos 7^ cos :rr cos TV cos ^r^ cos T^ = 7: ) ' 
 
 15 lo 15 15 15 15 lo \2/ 
 
 1 28. Solve the equations 
 
 cos (0-\-a) = sin sin ^8, 
 cos(^ + /3) = sin sin a, 
 and shew that, if 0^, ^g ^c two values of not 
 differing by a multiple of tt, 
 
 , /^ , J \ sin 2^ 
 
 tan (0, + 02) =sIn2-^_cos2^s^^- 
 
 129. If -^ — h 7- 7) have its least positive value, prove 
 
 that 6 is greater than ^3 — 1. 
 
 130. If A, B, G be the angles of a triangle, and x, y, z 
 
 any real quantities satisfying the equation 
 
 2/ sin (7—0 sin J5 _ sin J. — ic sin C 
 aj— 2/ cos 0—0 cos ^ 2/~^cos^— ajcos C 
 
 then 
 
 y 
 
 sin A sin B sin C 
 
 131. If (sin^a — sin2^)(sin2a — sin^y) = sin2^sin2y cos*a, 
 
 then tan^a = tan^/? + tan^y. 
 
 132. If ^+5+0=7r, 
 
 and sin^ft) = sin (J. — w) sin {B — o)) sin ((7— o)), 
 
 then cot ft) = cot^+cot.B+cot 0, 
 
 and cosec^a) = cosecM+cosec^-B + cosec^C. 
 
 133. If C0s(^-y) + C0s(y-a) + C0s(a-/5)=-f, 
 
 then cos(a + 0) + cos(/3 + 0)+cos(y + 0) 
 and sin (a + 0) + sin (^8 + 0) + sin (y + 0) 
 vanish whatever be the value of 0. 
 
164 CIRCULAR FUNCTIONS, ETC. 
 
 134. If cos(/3-y) + cos(y--a) + cos(a-)8) = -f, 
 
 prove that cos na + cos ^1/3 + cos ny is equal to zero, 
 unless 71 is a multiple of 8, and that, if ti be a 
 multiple of 3, it is equal to 3cos j7i(a+/3-f y). 
 
 135. If cos2a(2/^cos2y8 + 2;2cos2y — aj^cos^a) 
 
 = cos2|8(2;2cos2y + aj^cos^a — y^cos^fi) 
 = cos2y(a32cos2a + y^co^^^ — z^co^^y), 
 and if cos^a + cos^/S + cos'^y = 1 , 
 
 then ±^ — = ± . o = ^— — • 
 
 sm a sin p sm y 
 
 136. If coso+cos/8+cosy+cosacos/3cosy = 0, 
 
 then 
 cosec^a + cosec^/3 + cosec^y ± 2 cosec a cosec /3 cosec y = 1 . 
 
 137. If ^+5+a=27r, 
 
 and if 
 2/^+2;2_22/2;cos-4=2;^+a;2 — 22^fl3cos5=a;2+2/^- 2£C2/cos C, 
 prove that each of these quantities is equal to 
 
 2 . 
 
 '^z sin J. +0£c sin B+xy sin 0). 
 
 V3' 
 
CHAPTER IX. 
 
 RELATIONS BETWEEN THE ELEMENTS OF A 
 
 TRIANGLE, SOLUTION OF TRIANGLES 
 
 AND PRACTICAL APPLICATIONS. 
 
 § 1. Relations between the Elements of a Triangle, 
 
 118. If ABC be any triangle, we denote, as in Chapter 
 v., the lengths of the sides opposite the angles A, B, and 
 G by a, h, and c respectively. 
 
 119. In any triangle, a = b cos (7+c cos B, etc. 
 
 Let ABC be the triangle, the angle C being either 
 
 C D B C 
 
 acute, obtuse or right. Draw AT) perpendicular to BG, 
 
 produced if necessary. 
 
 Then 5(7=^5 cos 5+^0 cos (7, if G be acute or right, 
 
 or AB cos B — AG cos(7r — G), if G be obtuse or right, 
 
 BG = AB GOB B + AG cos G, in every case, 
 
 i.e. a = 6 cos (7+ c cos B. 
 
 165 
 
166 RELATIONS BETWEEN THE 
 
 Similarly, 
 
 6 = c cos ^ + a cos (7, and c = a cos 5+ 6 cos A. 
 
 Cor. — If cZ, e, /denote the lengths of the altitudes AD^ 
 BE, OF respectively, it may be shewn that 
 
 2cZ = 6sin(7+csin5, 2e = c sin J. + a sin (7, 
 and 2/= a sin 5 4- 6 sin J.. 
 
 120. The sides of a triangle are proportional to the 
 sines of the opposite angles. 
 
 Using the figure and construction of the last article, we 
 have AD = ABsmB. 
 
 Also AD = AG sin G, if G be acute or right, 
 or AGsm(7r — G), if G be obtuse or right, 
 
 AD = AG ain G, in every case. 
 ABBmB = AGsmG, 
 6 : c = sin5:sinC. 
 Similarly, a:b = 8mA : sin B, 
 
 a _ b _ c 
 sin A sin B sin G 
 
 It will be seen, in the next chapter, that each of these fractions is 
 equal to the diameter of the circumcircle of the given triangle. 
 
 121. To find the cosines of the angles of a triangle in 
 terms of the sides. 
 
 (See figure and construction of art. 119.) 
 By Euclid II. 13 and 12, we have 
 A^ = AG^+BG^-2BG.GD, if G be acute or right, 
 or AG^+BG^+2BG. GD, if G be obtuse or right. 
 
 Now, GD = AG cos G, if G be acute or right, 
 or AG cos(Tr — G), if G be obtuse or right, 
 
 A^ = AG-^+ BG^ - 2BG .AG cos G, in every case, 
 i.e. c^ = a^ + h^- 2ab cos G.. 
 
ELEMENTS OF A TRIANGLE. 167 
 
 Similarly, 
 
 a2 = 62_j_c2-26ccos^ and h'^ = c^+a^-2caco^B, 
 
 and cos C= 
 
 2ab 
 
 It will be noticed that each of these expressions is a proper 
 fraction, if (taking the first) b^+c^-a'^<2bc, i.e. if })^ + G^-2hc<a\ 
 i.e. if (6-c)2<a2 or b-c<a or b<c+a, which is the case (Eucl. 
 I. 20). 
 
 Also, that cosil is positive or negative, according as a^< or 
 > b^+c^, i.e. as A is acute or obtuse. 
 
 122. To find the cosines of the semi-angles of a 
 triangle in terms of the sides. 
 
 2 cos^-^ = 1 + cos A 
 
 _ 62 + 02-02 
 
 ~ ^ "^ 26c 
 _2bc+^+f-a^ 
 ~ 26c 
 
 (6 + c)2-a2 
 
 26c 
 _(6 + c+a)(6 + c-a) 
 ~~ 26c 
 
 Let a + 64-c, or the perimeter of the triangle, be de- 
 noted by 2s. Then, 6 + c — a = 2s - 2^^, 
 ^A 2s.2(s-a) 
 
 C0S^-^= -~r -, 
 
 2 46c 
 
 A ls(s - a 
 
 o- -1 1 ^ Ms -6) , G ks-c) 
 
 Similarly, cos ^ = ^^ ^^ and cos^ = ^^ ^^ > 
 
168 RELATIONS BETWEEN THE 
 
 Each of the expressions under the radical sign is positive, 
 for «-a, 8-h and s-c are all positive (Eucl. I. 20). Again, each 
 is a proper fraction, for (taking the first), s(5-a)<6c, if 
 (6+c)2-a2<46c, i.e. if h^-'ibc+c'^Ka?^ i.e. \ih-c<a or 6<c+a, 
 which is the case. 
 
 Hence, these expressions give possible values for the required 
 cosines. 
 
 123. To find the sines of the semi-angles of a triangle 
 in teiins of the sides. 
 
 2 sin^-^ = 1 — cos ^ 
 
 26c 
 
 _ 26c-b^-cHa^ 
 26c 
 
 ^a^ -{h-cf 
 
 '2bc 
 _ (a—b+c)(a+h — c) 
 26c 
 ^^_ 2(8-6).2(8-c) 
 ^'"^ 2" 463 ' 
 
 =v 
 
 ^^^A=J(s-b)is-c) 
 
 2 \ 6c 
 
 cj- .1 1 • ^" l{s-c)is-a) , . G ks-a){s-h) 
 Similarly, sm ^ = ^ ^^ and sm ^ = y ^ -^^ — -'. 
 
 Also, as in the last article, it may be shewn that these expres- 
 sions give possible values for the required sines. 
 
 124. To find the tangents of the semi-angles of a 
 triangle in terms of the sides. 
 
 ^ A . A A 
 
 tan ^ = sin -^ h-cos -^ 
 
ELEMENTS OF A TRIANGLE. 169 
 
 'i 
 
 _ l(s -b)(s-c) 
 
 s(s — a) 
 
 Similarly, tan|=/-l^^|^ 
 
 and u.l=J^r^B. 
 
 2 \ s{s — c) 
 
 Each of the fractions under the radical sign is positive, for every 
 factor is positive ; and, therefore, these expressions give possible 
 values for the required tangents, 
 
 125. To find the sines of the angles of a triangle in 
 terms of the sides. 
 
 Sin A = 2 cos y sin -^ 
 
 _^ls{s-a) Us-hXs-c) 
 
 he V he 
 
 o 
 
 = 5^Vs(s-a)(s-6)(s-c). 
 
 The expression s/s{s — a){s — h){s — e) is usually denoted 
 hy& 
 
 Hence, sin A = 2Slbc, and similarly, sin B = 28/ ca and 
 
 sin(7=2>Sf/a6. 
 
 As before, the expression under the radical sign is positive. 
 
 Also, 2Slbc is a proper fraction, 
 if As{s-a){s-bXs-c)<bh\ 
 if {a-\-b + c){b + c-a){c + a-bXa + b-c)<Abh\ 
 if 26V + 2c2a2 + 2a^b^ - a^ - 6* - c* < 4b^c\ . 
 if aH6Hc4 + 26V-2c2a2-2a262^0, 
 if (a2 - 62 - c2)2 > 0, which is the case. 
 
 Hence the expressions obtained above give possible values for 
 the required sines. 
 
170 
 
 RELATIONS BETWEEN THE 
 
 126. To 'prove that, in any triangle, 
 
 B-G h-c ^A , 
 taii^- = j^^cot2,etc. 
 
 (1.) Algebraical proof . 
 
 6-c^sin^-sin G^ 2 cos i{B+G) sin J(^-C) 
 h+c sin^-hsin (7~ 2 sin 1{B+G) cos \{B-G) 
 
 ,B-\-C^ B-G 
 
 = cot — ^ — tan — ■= — . 
 
 B-G h-c, B+G 
 tan , =r-— tan ^ 
 2 6+c 2 
 
 6-0 ,^ 
 6+c 2 
 
 Similarly, tan 
 
 G-A c-a ,B 
 -2- = c-+^^^^2- 
 
 ind 
 
 tau — 7i — = — r-r cot 7j- 
 
 2 a + 6 2 
 
 (2.) Geometrical proof. — Let ABG be the triangle. 
 
 ^ From ^(7 and GA pro- 
 duced, cut off J.Z) and 
 AE equal to ^j5. Join 
 BD and 5^, and draw 
 D-Fperpendicular to BD, 
 meeting BG in F. Then, 
 BBE is a right angle, 
 and, consequently, BF 
 
 C F 
 
 and BE are parallel. 
 
 Now, angle ^5i) = angle^D5=p^£^: 
 
 B+G 
 
 and angle DBG= angle ADB — angle A GB 
 
 B+G ^_5-C 
 —2 ^-"2- 
 
Again, 
 
 ELEMENTS OF A TRIANGLE. 171 
 
 h-c_CD_DF_DF BE 
 h+c~GE~BE~DB'DB 
 
 ^ B-G , B+G 
 = tan — ^ — Htan — ^— • 
 
 B-G h-c ,A 
 tan — rr— = 7—— cot -^. 
 
 2 6+c 2 
 
 127. Example 1. — From the formulae a = bcoaC+ccosB, etc., 
 deduce the expressions for the cosine of an angle of a triangle in 
 terms of its sides. 
 
 We have a = 6 cos C+ c cos B, 
 
 b = c cos A-\-a cos C, 
 c=acosB+bcoaA. 
 Multiply both sides of these equations by «, 6, c respectively, and 
 subtract the first from the sum of the second and third : then, 
 b^-\-c^-a^ = be cos A+ab cos C + ca cos B+ be cos A-ab cos C-ca cos B 
 = 2bc COS A, 
 
 cos -4 = —} . 
 
 26c 
 
 Example 2. — Having given that the sides of a triangle are pro- 
 portional to the sines of the opposite angles, deduce the expression 
 for the cosine of an angle in terms of its sides. 
 We have a/sin A = 6/sin B = c/sin C. 
 
 Let each of these fractions equal d, so that 
 
 a=d8in A, b=^dsmB 3ind c=d sin C. 
 Then, 62+c2 - 26c cos A=d\smW + 8Ui^C- 2 sin B sin Ccos ^) 
 = d\sm B(sin 5 — sin C cos A)-\- sin (7(sin C— sin 5 cos A )] 
 = c?2[sin B{sm( 6^ + ^ ) - sin Ccos ^ } + sin r{sin( J + ^) - sin 5 cos ^ }] 
 =c?2[sin B cos Csin A +sin (7 sin A cos B] 
 = dhin A sin{B+ C) = dhin^A = a^. 
 Example 3. — If ^^' be a median of the triangle ABC, then 
 
 2cotJ^'^ = cotC-cot^. 
 
 
 Let 6 denote the angle AA'B; 
 
 then, 
 
 by art. 120, 
 
 
 ,«.T '(■>*<-] ■■ 
 
 wf:M 
 
 BA' 
 
 sin A'AB 
 
 _sin{d+B) 
 
 
 AA' 
 
 sin A' B A 
 
 sin B ' 
 
 
 and ^^'-. 
 
 _sinA'AG_ 
 
 _sin{e-C) 
 sin C 
 
 
 A A' 
 
 sin A'GA 
 
 
 A' AC -+C .: fi! i^t ^ ^ ' (L 
 
172 RELATIONS BETWEEN THE 
 
 But BA' = CA\ 
 
 s,\nCs,m{d-\-B)=BmB«m{d-C), 
 
 sin (7(8in 9co^B-\- cos 6 sin B) = sin 5(sin OcoaC- cos ^ sin C), 
 
 cos ^(sin 5 sin C+sin B sin C)=sin ^{sin 5 cos C- cos ^ sin C), 
 
 o«.^^■/^ sin jB cos C- cos 5 sin (7 
 
 2cot6/= : — - . -, , 
 
 sm BamC 
 
 2cotAA'B=cotC-cotB. 
 
 Example 4. — To find the relation between the lengths of the six 
 lines joining any four points in a plane. 
 
 Outline of proof : Let ABO be a triangle, a, b, c the lengths of 
 its sides, and let P be any point, either within or without the tri- 
 angle : join PA, PB, PC and let a, y8, y be the lengths of these 
 lines. Let the angles between each pair of the last three lines be 
 denoted by 6, <f>, yjr ; one of these angles being greater than two 
 right angles when P is without the triangle ABC. Then, 
 
 e + cfi + ylr = 27r, 
 and, consequently, 
 
 cos'-^ + cos^^ + cos^yf/ - 2 cos $ cos <^ cos i/r = 1 . 
 Express each of these cosines in terms of the given lines by art. 
 121 ; multiply both sides of the resulting equation by 4a^ /3^y^ ; and, 
 after reducing, the required relation will be obtained, namely, 
 
 a2a2(a2 + ^2 _ J2 _ ^2 _ ^2 _ ^2) + 52^(2,2 4. ^2 _ ^2 _ ^2 _ ^2 _ ^^2) 
 + C2y 2(c2 + y 2 _ ^2 _ 52 _ ^2 _ ^) + «2^2y2 + ^2 2^2 + c2a2/?2 + a262c.2 = Q. 
 
 Examples XIV. a. 
 
 1. Find the cosines of the angles of a triangle whose 
 
 sides are 10, 13 and 15. 
 
 2. The sides of a triangle are x, y and /^(x^ + oi^y + y^), 
 
 find the greatest angle. 
 
 3. If a = ^5, 6 = 2, c = JS, then 8 cos ^ cos 0= 3 cos B. 
 
 4. If ^=45° and 5=60°, then 2c = a(l + V'^). 
 
 5. If ^ = 30°, 5 = 45" and a = 8j2, find h and c. 
 
 6. Find the cosines of the semi-angles of a triangle 
 
 whose sides are 1, 4 and 4. 
 
ELEMENTS OF A TRIANGLE. 173 
 
 7. Find the sines of the semi-angles of a triangle whose 
 
 sides are 35, 15 and 34. 
 
 8. Find the tangents of the semi-angles of a triangle 
 
 whose sides are 25, 52 and 63. 
 
 9. Find the sines of the angles of a triangle whose sides 
 
 are 193, 194 and 195. 
 
 10. Find the tangents of the angles of a triangle whose 
 
 sides are 10, 35 and 39. 
 
 11. If 6 = 5, c = 3 and J. = 120°, find tan J(5-(7). 
 
 12. If a = 2, 6 = ^3 and C=30°, find A, B and c, 
 
 13. (6 + c)cos J.-l-(c + a)cos5+(a + fe)cosC=aH-6-fc. 
 
 14. asinJ[-f-6sinJ5+csin C=2(cZcosJ.+ecosJ54-/cosC). 
 
 15. sin {A-B)'. sin 0= a^-h^: c\ 
 
 16. cosi(^-5):cosi(^-f5) = a+6:c. 
 
 17. s^ = >S^ cot -^ cot ^ cot ^. 
 
 18. If D be any angle, 
 
 a sin (D - 5) -I- 6 sin (D + ^) = c sin Z>. 
 
 19. c = 6cos^±^(a2-62sinM). 
 90 cos 2 J. c os2^ _l 1 
 
 21. a(sin2|-fsin2g + 6(sin2^+sin2^) 
 
 + c(sin2| -l-sin^l) ^ J(a-f 6-Fc). 
 
 Examples XIV. b. 
 
 1. The sides of a triangle are 2, ^2 and ^^3 — 1 ; find 
 
 its angles. 
 
 2. The sides of a triangle are x^-\'Xy-^y^, 2xy + y^ and 
 
 x^ — y'^] find the greatest angle. 
 
174 RELATIONS BETWEEN THE 
 
 3. The sides of a triangle are 3, 4 and -s/38; shew, 
 
 without using tables, that the largest angle is 
 greater than 120°. 
 
 4. If 5 = 15° and (7=30°, shew that a^cj% 
 
 5. If^ = 15°, J5=105°and c = ^6, find a and 6. 
 
 6. Find the cosines of the semi-angles of a triangle 
 
 whose sides are 125, 154 and 169. 
 
 7. Find the sines of the semi-angles of a triangle whose 
 
 sides are 11, 25 and 30. 
 
 8. Find the tangents of the semi-angles of a triangle 
 
 whose sides are 25, 51 and 52. 
 
 9. Find the sines of the angles of a triangle whose sides 
 
 are 125, 123 and 62. 
 10. Find the tangents of the angles of a triangle whose 
 
 sides are 21, 89 and 100. 
 n. If a = 15, 6 = 8 and 0=90°, find tanJ(^-5). 
 
 12. If a=90° and a : 6 = ^3 + 1 : V3-l,find^,5, and c. 
 
 13. (6-l-c) sin ^ 4- (c-t-a) sin B-\-{a-\-h) sin 0= 2(c?+6+/). 
 
 14. a sin {B- G) + b sin {C-A) + c sin {A -B) = 0. 
 
 15. smi(A-B):smi{A + B) = a-b:c. 
 
 16. tan(4+5) = ^tan4. 
 
 \z / c — o 2 
 
 17. 8S= abc cos -^ cos -^ cos » . 
 
 Zi z z 
 
 18. bccosA-\-cacoaB+ahcosG=i(a^+¥+c^). 
 
 19. c2 = (a-6)2cos2^+(a+6)2sin2?. 
 
 20. a^cos 2B -f b^cos 2A = a^+b^- 4<ab sin A sin B. 
 
 ^ /sin A +sin 5-|-sinCY_a cos J. + 6 cos 5H-c cos (7 
 ' \ a + b+c / ~ 2abc * 
 
I 
 
 ELEMENTS OF A TRIANGLE. 175 
 
 Examples XV. 
 
 1. I{ ABGD be a quadrilateral, then 
 
 ABgosA-BGgos(A-\-B) + CDcosD = AD, 
 eindABsmA-BGsm(A + B)-CDsmD = i). 
 
 2. ^2 : fe2 : c2 : : cot 5 + cot : cot 0+ cot ^ : cot 4 + cot 5. 
 
 3. he sinM + ca sin^^ + ah sm^G 
 
 = {a+h + c){a cos B cos G+b cos GcoaA+c cos J. cos B). 
 
 4. a2+62+c2<2(6c + ca + a6). 
 
 5. The sides of a triangle are a, h, c, and the angle G is 
 
 120°. Shew that, if we form the triangles whose 
 sides are a, a + h, c and a + b, h, c, the angle 
 opposite c will in each case be 60°. 
 G, If a, h, c be in arithmetical progression, then 
 
 Stan ^tan^ = l. 
 
 7. If the sides of a triangle be in arithmetical progres- 
 
 sion, and if a be the least side, then 
 , 4C-86 
 
 8. If G=2B, then c^ = h(a + h). 
 
 9. If ^ = SB, then sin B = |a/(^^)- 
 
 10. If the sides of a triangle be in arithmetical progres- 
 
 sion, the tangents of its semi-angles are in har- 
 monical progression. 
 
 11. Shew, trigonometrically, that, if any angle of a tri- 
 
 angle be bisected, the segments of the base formed 
 by the bisecting line, are in the ratio of the sides 
 of the triangle. 
 
 1 2. ahc{l — 2 cos J. cos B cos G) 
 
 = a^cos B cos G-\- 6^cos G cos A + c^cos A cos B. 
 
176 RE LA TIONS BET WEEN THE 
 
 13. a^sin {B-C)^-hhm {C-A)^chm {A -B) = 0. 
 
 14. -^-T-cos^+ 1 , -cosjgH ri-cosC'=a* + 6* + c3. 
 
 6*c* c^a^ a^b^ 
 
 15. If a = (c-a)(a-6), /3 = (a-6)(6-c), y = (6-c)(c-a), 
 
 then 
 aV + 62/32+cV-22)C/8ycos^ - 2cayacos5-2a6a^cosO 
 = {aa + h^ + cyf. 
 
 acos^^ + ^cos^-x + c COSM- 
 IC. Express ^V-^ -77 r - "tt- i^ terms of the sides. 
 
 ^ cos^ + cos j5+cosC7 
 
 17. In a triangle ^5(7, ^D is drawn to meet BG, or BC 
 
 produced, in D, so that AD is equal to J-O; if the 
 sum of AB and AG is n times 5C, then their 
 
 difference is -th of BD. 
 n 
 
 18. If a cos J. = 6 cos jB, the triangle is either isosceles or 
 
 right-angled. 
 
 19. In any quadrilateral figure, the square on one side is 
 
 less than the sum of the squares on the other sides 
 by twice the sum of the products of these sides 
 taken two and two together and multiplied by 
 the cosine of the angle included between them. 
 
 20. If (a^ + b^) cos 2 A = b^-a\ the triangle is either right- 
 
 angled or two of its angles differ by a right angle. 
 
 21. P is any point in the base AB of a triangle ABG, 
 
 such that AP : PB = m : n, and the angle GPB is 
 6 \ shew that 
 
 (m+7i) cot = 71 cot A—m cot B. 
 
 22. On the base BG of a triangle ABG, two points, Q, R, 
 
 are taken, so that BQ = QR = RG; then 
 sin BAR . sin C^Q= 4 sin BAQ . sin GAR. 
 
ELEMENTS OF A TRIANGLE. 177 
 
 23. If a = 26, and A — SB, find the angles of the triangle 
 
 and the ratios otAB to BC and AG. 
 
 24. a^cos 2{B-C) = h'^cos 2B+c^cos 2C+2hccos{B-G). 
 
 25. Express the bisector of an angle of a triangle in 
 
 terms of the sides, and shew that the greatest 
 bisector is that which bisects the smallest angle ; 
 also, if a, /3, y denote the lengths of the bisectors, 
 
 that -cos-r- + ^cos-^+-cos^=-+r + -. 
 a zpzy z a c 
 
 26. If the squares of the sides of a triangle be in arith- 
 
 metical progression, the tangents of the angles are 
 in harmonical progression. 
 
 27. The measures of the lengths of sides of a triangle 
 
 are three consecutive integers, and the largest 
 angle is double of the least ; find the sides. 
 
 28. If a triangle be such that it is possible to draw a 
 
 straight line AD meeting BG in D, so that the 
 angle BAD is one-third of the angle BAG, and 
 also BD is one-third of BG, then 
 a262 = (62_c2)(62 + 8c2). 
 
 29. A triangle is turned round each of its three sides 
 
 successively till the vertex comes into the same 
 plane; if the vertices be joined forming a new 
 triangle whose sides are a', h\ d , then 
 
 ^=l + 8cos^sin5sina 
 
 A . ,B . .G 
 cot 
 
 on 
 
 • cot^-fcotj5-hcot(7 a^-{-¥-\-c^' 
 
 31. a^cos^-|-6^cos5-f c^cos G=abc(l + 4!CoaAcoaBcosG). 
 
 32. If the sides of a triangle be in geometrical progression, 
 
 and if the altitudes be taken as the sides of a new 
 
 COt^+COt-^+COt^ („ + j^.,)2 
 
178 RELATIONS BETWEEN THE 
 
 triangle, then the angles of this triangle will be 
 equal to those of the original triangle. 
 
 33. If the sides BC, GA, AB be divided in G, H, K, so 
 
 tha,tBG:GG=GH:HA=AK:KB; then 
 
 AG^+BR' + GK' 
 will be least when G, H, K are the mid-points of 
 the sides. 
 
 34. If p, q be the perpendiculars from Ay B on any arbitrary 
 
 line drawn through (7, then 
 
 a?-'p^ + ll^q^ — 2abpq cos G = a^bhin^G. 
 
 35. A triangle ABG, whose angles are given, is to be 
 
 drawn with its angular points on three given 
 parallel straight lines ; if the middle line, passing 
 through G, be at distances a, /3 from the other lines 
 passing through A, B, then the sides of the triangle 
 are determined by the equations 
 
 a _ h _ c 
 sin^~sini^ sinO 
 
 s/a^am^A + 2a^ sin A sin B cos G+ /Shin^B 
 ~ sin A sin B sin G 
 
 36. If A and B be the greatest and least angles of a 
 
 triangle, the sides of which are in arithmetical 
 progression, then 
 
 4(1 — cos ^)(1 — cos B) = cos ^ + cos B. 
 
 37. If the squares of the sides of a triangle be in arith- 
 
 metical progression, then 
 
 sin3^ ^/ a^-cV 
 sin jB ~ \ ac J ' 
 
 38. The altitudes of an acute-angled triangle meet in 0, 
 
 and OA, OB, OG are taken for the sides of a new 
 triangle. Find the condition that this should be 
 
ELEMENTS OF A TRIANGLE. 179 
 
 possible ; and, if it be, and the angles of the new 
 
 triangle be a, /?, y, then 
 
 _ , cosa , cos/3 , cosy . . „ n 
 
 l-\ -r-\ ^H ^=*secj4sec5sec(7. 
 
 cos ^ cos B cosG ^ 
 
 39. If P be any point in an equilateral triangle ABC, 
 
 PB^4-PG^ — PA^ 
 then co^iBPC- 60°) = 2PB PG 
 
 40. The sides of a given triangle are a, h, c, and the 
 
 angles A, B, G; if a point be taken within an 
 equilateral triangle, so that its distances from the 
 angles of the triangle are proportional to a, h, c; 
 then the angles between these distances will be 
 
 l+J, 1+5 and 1+0. 
 
 41. If X, y, z be the distances of any point in the plane 
 
 of an equilateral triangle, whose side is a, from 
 the angular points, then 
 
 2/%2 + 02^2 + rj^^yl _ ^4 _ ^4 _ ^.4 _^ ^2(^2 J^ylj^ ^2) _ ^4 ^ Q. 
 
 § 2. Solution of Triangles. 
 
 128. In every triangle there are six elements, namely, 
 the three sides and the three angles ; any three of which 
 may be independent of one another, except the three 
 angles. The relations that have been proved, in the 
 preceding section, between the sides and angles of a 
 triangle enable us, as a rule, to determine the remaining 
 elements, if three independent elements be given. 
 
180 SOLUTION OF TRIANGLES. 
 
 There are four cases to be considered, those in which 
 we are given : (1) one side and any two angles, (2) two 
 sides and the included angle, (3) two sides and the angle 
 opposite one of them, and (4) the three sides. 
 
 In the examples at the end of the last section, some 
 examples of the solution of triangles have already been 
 given, not however involving the use of logarithmic 
 tables. In the articles and examples which follow, we 
 shall consider more completely the method of solving 
 triangles with the aid of such tables. 
 
 129. Case I. — One side and any two angles being 
 given, to solve the triangle. 
 
 Let the side a and the angles B and G be given ; to 
 find 6, c and A. 
 
 By Eucl. T. 32, A^im^-B-C. 
 
 Also, &/a = sin5/sin^, (art. 120) 
 
 log 6 = log a + log sin J5 — log sin ^ . 
 Similarly, log c = log a + log sin (7 — log sin ^ . 
 
 130. Example 1.— Given a = 357, Br=AT 19' and C=58° 23'; to 
 find 6, c and A. 
 
 J = 180° -47° 19' -58° 23' = 74° 18'. 
 log 6 = log a + log sin B - log sin A 
 
 = log 357 + log sin 47° 19' -log sin 74° 18' 
 = 2-5526682 
 
 + 1-8663534 - 1-9834872 
 = 2-4190216 
 -i -9834872 
 = * 2-4355344. 
 
 2-4355418 = log 272-61, 
 2-4355344 = log(272-60 + S), 
 2-4355259 = log 272-60, 
 159 : 85 :: -01 : 8. 
 
SOLUTION OF TRIANGLES. 181 
 
 159) -85 (-005 . 
 795 
 
 55 
 
 6 = 272-605. 
 Similarly, log c = 2-4994036, 
 
 c = 315-794. 
 
 A=7i° 18' 
 
 b: 
 
 c- 
 
 = 74° 18' ^ 
 = 272-6051. 
 = 31 5-794 j 
 
 131. Case II. — Two sides and the included angle being 
 given, to solve the triangle. 
 
 Let the sides 6, c and the angle A be given; to find 
 a, B and G. ' ' . ^ 
 
 B—Gh — c,A , , ■,or>\" 
 
 tan — ^ — = J—- cot ^, (art. 126) 
 
 B—G A 
 
 log tan -y- = log(6 - c) + log cot ^ - log(6 + c). 
 
 From this equation, we can find ^(B—G), and we know 
 that 1(5+0) = 90°- J^. Hence, B and G can be found. 
 Again, as in Case I., 
 
 log a = log h + log sin ^ — log sin B. 
 
 132. We may, if we please, determine the side a, with 
 the aid of a subsidiary angle, from the formula 
 
 a^ = h^ + c^-2hc cos A, 
 without first finding the angle B or G. Thus 
 
 a^ = h^-2bc + c^-\-2bc(l-cosA) ■ 
 
 =:{b^c)^ + 4>bcsm^~ 
 
 =(^-Ki+A^^^4} 
 
 If, now, we put tsin^O = -pi r^sin^— , 
 
 ^ (6 '- c)2 2 
 
 the last equation becomes a = (6 -- c)sec 9. 
 
182 SOLUTION OF TRIANGLES. 
 
 133. Example 2.— Given 6 = 541, c = 2b% ^=48° 26'; to find 
 B, C and a. 
 
 B — C A 
 
 log tan = log(6 -c)~ log(6 + c) + log cot — 
 
 2 Z 
 
 = log 282 - log 800 + log cot 24° 1 3' 
 
 = 2-4502491 
 
 + 0-3470119 - 2-9030900 
 = 2-7972610 
 
 -2-9030900 
 = T-8941710. 
 
 1-8943715 = log tan 38° 6', 
 1-8941710= log tan 38° 5' S", 
 1-8941114= log tan 38° 5', 
 2601 : 596:: 60: S. 
 
 
 596 
 
 
 
 
 60 
 
 
 
 
 2601)35760(13-7 
 
 
 
 
 2601 
 
 
 
 
 9750 
 
 
 
 
 7803 
 
 
 
 
 19470 
 
 
 
 
 18207 
 
 
 
 
 |(5-C) = 38°5'14". 
 
 
 Now, 
 
 Ki?+O=65°47'0", 
 i=103°52' 14", 
 ^=27° 41' 46". 
 
 
 
 Again, as in art. 129, 
 
 
 
 
 log a = log b + log sin A — log sin B 
 
 
 
 
 = log 541 + log sin 48° 26' - log sin 
 
 103° 
 
 52' 14' 
 
 
 = log 541 + log sin 48° 26' - log sin 
 
 76°: 
 
 r46" 
 
 
 = 2-6200582, 
 
 
 
 
 a=416-925. 
 
 
 
 
 ^ = 103° 52' 14"" 
 
 
 
 
 C=27°41'46" ■• 
 
 
 
 
 a =416-925 
 
 
 
I 
 
 SOLUTION OF TRIANGLES. 183 
 
 134. Case III. — Two sides and the angle opposite one 
 of them being given, to solve the triangle. 
 
 Let the sides a, h and the angle A be given ; to find 
 c, B and C. 
 
 sin j5/sin A = hja, 
 
 sin jB = 6sin^/a; 
 log sin 5 = log 6 + log sin J. — log a. 
 This equation does not, however, always lead to a 
 definite result. For, since the angle is to be determined 
 from its sine, it is possible that there may in certain 
 cases be two angles satisfying the equation, both less 
 than two right angles and one the supplement of the 
 other. The following cases may occur : — 
 
 (1) a>h. Then A is greater than B, and, therefore, 
 the less only of the two values obtained for B is 
 admissible, for there cannot be two obtuse angles in a 
 triangle. 
 
 (2) a = 6. Then A is equal to B, and, again, only the 
 less of the two values obtained for B is admissible. 
 
 (3) a<h. There are here three sub-cases, according as 
 a is greater than, equal to, or less than 6 sin ^. 
 
 If a>6sin^, the expression for sin 5 is a proper 
 fraction, and both values obtained for B are admissible, 
 since B is greater than A. Thus, there are two triangles 
 which satisfy the given conditions, the value of B in one 
 being the supplement of that in the other. This is 
 known as the ambiguous case in the Solution of Triangles. 
 
 If a = b sin A, then sin 5 = 1, and B = 90°. Since the 
 supplement of 90° is also 90°, there is only one triangle, 
 or, rather, there are two coincident triangles, satisfying 
 the given conditions, the angle B in each case being a 
 right angle. There is thus no ambiguity in the solution. 
 
184 
 
 SOLUTION OF TRIANGLES. 
 
 If a < 6 sin A, the expression for sin 5 is greater than 
 unity, and there is no real value of B. Thus, there is no 
 triangle satisfying the given conditions. 
 
 We may sum up these three sub-cases (when a<h) 
 as follows: There are in each sub-case two triangles, 
 and they are real and different, real and coincident, or 
 imaginary and different, according as a is greater than, 
 equal to, or less than, 6 sin J.. 
 
 Having found one or both values of B, the correspond- 
 ing value or values of G may be obtained ; and, lastly, 
 the value or values of c, as before, from the equation 
 log c = log a -f log sin G— log sin A . 
 
 135. The third case in the Solution of Triangles may 
 be illustrated as follows : 
 
 Let AG=b', and angle GAB = A. With centre C and 
 radius equal to a, describe a circle cutting BA, produced, 
 if necessary, in B and B\ 
 
SOLUTION OF TRIANGLES. 185 
 
 (1) If a > 6, then B and B' are on opposite sides of A, 
 and there is only one triangle having the given angle A, 
 as well as the given sides a and 6, namely, the triangle 
 GAB. 
 
 (2) If a = h, then B' coincides with A, and there is 
 only one triangle. 
 
 (3) If a<h, then the circle will cut AB in two real 
 and different points, B and B\ on the same side of A, if 
 CB be greater than the perpendicular distance of G from 
 AB, i.e. if a is greater than 6 sin J. ; and there will then 
 be two triangles, GAB and GAB\ the angles GBA and 
 GB'A being supplementary. If a is equal to 6sin^, 
 the circle will cut AB in two real and coincident points, 
 i.e. it will touch it at 5 ; and the two triangles will be 
 coincident and right-angled at B. Lastly, if a is less 
 than &sin^, the circle will cut AB in two imaginary 
 points, and there will be no real triangle having the 
 given elements. 
 
 136. We may, again, investigate the third case in the 
 Solution of Triangles by considering the values of the 
 side c, instead of the angle B, in terms of the given 
 elements a, h and A. 
 
 For a- = ¥ + 6^ — 2hcQ0^A, 
 
 and, from this equation, which is a quadratic in c, two 
 values of c can be found, namely, 
 
 6 cos ^ ± VC^'cosM - 62 + a2), 
 or hQO^A±J{ci}-h^^m''A). 
 
 We have, as before, the following cases : 
 
 {I) a>h. Then, ^(a^-^^inM) is greater than 
 ;^(5'^ — 6%inM), or 6 cos J.. Hence, there is only one 
 positive value of c, and therefore only one triangle. 
 
186 SOLUTION OB' TRIANGLES. 
 
 (2) a = h. Then, ^(a^ — h^An^A) is equal to 6cos^. 
 Hence, the values of c are 26 cos A and : i.e. there is 
 only one triangle. 
 
 (3) a<h. Then, J{a?-h'^sm^A) is less than 6cos^. 
 Hence, there are two values of c, which are real and 
 different, real and equal, or imaginary and different, 
 according as a is greater than, equal to, or less than, 
 h sin A ; and, consequently, two triangles which are real 
 and different, real and equal, or imaginary and different, 
 according as a is greater than, equal to, or less than, 
 h sin A. 
 
 137. Example 3.— Given a = 273, 6 = 392 and ^ = 37n4'; to find 
 ByC^ and c. 
 
 Here, a being less than 6, it is possible that there may be two 
 values of B. 
 
 log sin 5= log sin ^ + log 6 - log a 
 
 =log sin 37°14' + log 392 - log 273 
 = T' 7818002 
 
 4-2-5932861 -2-4361626 
 = 2-3750863 
 - 2-4361626 
 = 1-9389237. 
 1-9389796 = log sin 60° 20', 
 1-9389237 = log sin 60° 19' 6", 
 T-9389076 = logsin60°19'. 
 720 : 161 : : 60 : S, 
 8=161x^^=13", 
 ^=60° 19' 13" or 119° 40' 47", 
 and C=82° 26' 47" or 23° 5' 13". 
 
 Corresponding to these pairs of values, we find, as in Case I., 
 
 c= 447-278 or 176-925. 
 Hence, the required elements are 
 
 ^=60° 19' 13"] i?=119° 40' 47" 
 
 C=82° 26' 47" V or C=23° 5' 13" 
 0=447-278 I c= 176-925 
 
SOLUTION OF TRIANGLES. 187 
 
 138. Case IV. — The three sides being given, to solve the 
 triangle. 
 
 tan4 = V(«-^>^«-«> - 
 
 9 
 
 s(s — a) 
 
 :. logtan2- = Hlog(s-??) + log(s-c)-logs-log(s-a)}. 
 Similarly, 
 
 75 
 
 logtan2=Hlog(s-c) + log(s-a)-logs-log(8-b)}. 
 
 From these equations A and B can be found, and C 
 from the equation 
 
 139. In this case, if more than one angle is to be found, 
 the formulse for the tangents of \A and \B are preferred 
 to those for the cosines or sines, as only four logarithms 
 have to be found, namely, those for s, s — a, s — h and 
 s — c. If the formulae for the cosines are used, six logar- 
 ithms must be found, those for s, s — a, s — 6, a, 6 and c ; 
 and six also if the formulae for the sines are used, namely, 
 those for s — a, s — 6, s — c, a, b and c. 
 
 140. Example 4— Given «=349, 6 = 521 and c=539 ; to find A, 
 B and C. 
 
 Here, s= 704-5. 
 
 log tan ^^ =\{\og{s - 6)+log(s - c) - log s - log(s - a)} 
 
 = ^(log 183-5 + log 165-5 -log 704-5 -log 355-5) 
 _ / 2-2636361 -2-8478810^ 
 ~*1 + 2-2187980 - 2-5508396/ 
 ^i/ 4-4824341 \ 
 ^V -5-3987206/ 
 = |(T-0837135) 
 =1-5418567. 
 
188 SOLUTION OF TRIANGLES. 
 
 T'5418747 = logtanl9° 12', 
 T-541 8567 = log tan 19" 1 1' 8", 
 l-5414678 = logtanl9° 11'. 
 4069:3889=60:8. 
 3889 
 
 60 
 
 4069 ) 233340 ( 57 3 
 20345 
 29890 
 28483 
 14070 
 12207 
 
 ^4 = 19° 11' 57-3", 
 ^=38° 23' 55". 
 Similarly, using the logarithms found above, we find 
 log tan f =1-8290603, 
 
 ^5=34° 0' 16-1", 
 t A ^=68° 0' 32", 
 ^]^'^ C=73° 35' 33". 
 
 -y' 
 
 \J\P ^1 = 38° 23' 55"] 
 
 ^ i • V^ ^=68° 0'32"l- 
 
 h i ^ f) 0=73° 35' 33" j 
 
 
 v*^ 
 
 Examples XVI. a. 
 
 Solve the triangles of which the following elements are 
 given : 
 
 1. a = 2992-95, 5 = 127° 54' 30", 0=33'^ 9' 10". 
 
 2. a = 5043-04, B = 84° 56' 14", (7= 58° 45' 33". 
 
 3. 6 = 7282-61, 0=28° 49' 5", ^ = 102° 40' 15". 
 
 4. 6 = 3572, c = 9147, ^ = 42° 15' 38"- 
 
 5. c = 3000, a = 1406, 5 = 120° 15' 40". 
 
 6. a = 304-532, 6 = 526-109, (7=78° 18' 44". 
 
 7. a = 5371-24, 6 = 2743-65, ^ = 49° 14' 30". 
 
SOLUTION OF TRIANGLES. 189 
 
 8. a = 4857, 
 
 6 = 6104, 
 
 ^ = 20° 19' 10". 
 
 9. a = 586, 
 
 6 = 987, 
 
 ^ = 60° 25' 25". 
 
 10. 6 = 807-8, 
 
 c = 1162-4, 
 
 5 = 41° 88'. 
 
 11. 6 = 7412-5, 
 
 = 9182-1, 
 
 0=64° 12' 20". 
 
 12. a = 54, 
 
 6 = 48, 
 
 = 86. 
 
 18. a = 329-4, 
 
 6 = 451-7, 
 
 = 154-2. 
 
 14. a = 620- 124, 
 
 6 = 711-005, 
 Examples XVI.b. 
 
 = 932-147. 
 
 Solve the triangles of which the following elements are 
 
 given : 
 
 
 
 1. a = 4686-50, 
 
 5= 122° 37' 45", 
 
 0=28° 37' 50". 
 
 2. 6 = 4670-13, 
 
 (7= 151° 52' 85'', 
 
 ^ = 18° 25' 15". 
 
 3. c = 6508-75, 
 H. 6 = 501-2, 
 
 ^ = 66° 39' 55", 
 
 J5=25°32'15". 
 
 = 398-5, 
 
 ^ = 68° 48'. 
 
 5. = 50-38, 
 
 a = 68-4, 
 
 j5 = 94°17'. 
 
 6. a = 891-204, 
 
 .6 = 172-537, 
 
 0=104° 14'. 
 
 7. a = 548-28, 
 
 6 = 1051-87, 
 
 ^ = 27° 12' 10". 
 
 8. a = 9621, 
 
 6 = 6758, 
 
 ^ = 59° 40' 40". 
 
 9. a = 742, 
 
 6 = 824, 
 
 ^ = 75° 10' 55". 
 
 10. 6 = 714-3, 
 
 = 958-2, . 
 
 0=87° 0' 4". 
 
 11. a = 2143, 
 
 = 4172, 
 
 ^ = 25°1'14". 
 
 12. a = 200-4, 
 
 6=295-8, 
 
 = 811-1. 
 
 ^18. a = 5102, 
 
 6 = 8074, 
 
 = 2314. 
 
 14. a = 5817-24, 
 
 6 = 345107, 
 
 = 2001-15. 
 
 Examples XVII. 
 
 1. In the ambiguous case, if a, 6, and A be given, and if 
 
 Oj, Cg be the values of the third side, then c^c^ = h'^-a^. 
 
 2. If one of the angles at the base of a triangle be 36°, 
 
 the opposite side 4, and the altitude ^o — 1, solve 
 the triangle. 
 
190 SOLUTION OF TRIANGLES. 
 
 3. The angles of a triangle are 35°, 65° and 80°, and the 
 
 difference between the longer sides is 1000 ; find 
 the sides. 
 
 4. If^ = 6r, j5 = 37° and a-6 = 372, find a and 6. 
 
 5. Given two sides and the included angle, find the 
 
 length of the altitude on the third side. 
 
 If this altitude be greater than the third side, 
 then sin 0+2 cos C> 2, where G is the included 
 angle. 
 
 6. If 6, c, and A be given, shew that a may be deter- 
 
 mined, without finding B and C, from the formula 
 
 A 
 a — (b + c)sin ~ sec <p, 
 
 where tan <h = y^- cot ^ • 
 
 b + c 2 
 
 B—G d) A 
 
 7. Prove that tan — ^ — = tan^^ cot -^, where cos (f) = clb) 
 
 and apply this formular to find B and G, where 
 ^ = 35°25'and6:c = ll:10. 
 
 8. If Q be an angle determined from the equation 
 
 cos = {a — b)/c, prove that 
 
 A-B (a + b)sme , A-^B csinO 
 
 cos — ;r — = ^^ — —4= — and cos — -i — = — j=' 
 
 2 2s/ab 2 2s/ab 
 
 9. In a right-angled triangle, the medians drawn from 
 
 the acute angles are a and /3 ; shew how to find 
 the angles. 
 
 Find the angles when a = 10, ^=14. 
 
 10. If c6s(5-0) = 31/32, and 6 = 50, c = 40, find a. 
 
 11. The tangents of the angles of a triangle are in har- 
 
 monical progression; given one of the sides and 
 the diflference of the first and third angles, shew 
 how to solve the triangle. 
 
SOLUTION OF TRIANGLES. 191 
 
 12. Given the mean side of a triangle, whose sides are in 
 
 arithmetical progression, and the angle opposite to 
 it, investigate formulae for solving the triangle, 
 and find the greatest possible value of the given 
 angle. 
 
 Solve the triangle when the mean side is 542, 
 and the opposite angle 59° 59' 59". 
 
 13. Given the area, the perimeter, and an angle of a 
 
 triangle, shew how to solve the triangle. 
 
 14. Given the lengths of the three medians, shew how to 
 
 solve the triangle. 
 
 15. Given the lengths of the three altitudes, shew how to 
 
 solve the triangle. 
 
 § 3. Practical Applications. 
 
 141. The application of the methods just explained 
 to the problems of Surveying forms the subject of this 
 section. 
 
 We have seen that, if three elements of a triangle, 
 including at least one of the sides, be known, we can 
 determine the magnitudes of the other three elements, 
 except in one particular case, when the solution is 
 ambiguous. 
 
 For the purposes of Surveying, therefore, we require 
 instruments : (1) for measuring the length of a line, called 
 a base-line or base, on the ground, and (2) for measuring 
 angles either in a horizontal or vertical plane. 
 
192 PRACTICAL APPLICATIONS. 
 
 142. Measurement of Base-line. — If the district to be 
 surveyed be not extensive, and if no very great accuracy 
 be required, the base-line may be measured with the 
 ordinary surveyors chain, known as Gunter's chain. 
 This chain is 22 yards long, and consists of 100 links 
 joined together by rings, having at every tenth link a 
 piece of brass attached to assist in counting the number 
 of links. 
 
 In more elaborate surveys extending over whole 
 countries, the base-line may be five miles or more in 
 length, and the greatest attainable accuracy is required. 
 Steel chains may be used, or rods of metal, glass, well- 
 seasoned wood, etc. ; the temperature must be observed 
 throughout, and allowance made for the expansion or 
 contraction of the rods, etc., resulting from changes of 
 temperature. This correction may, however, be avoided 
 by the use of compound rods, made of two bars having 
 different coefficients of expansion and connected at their 
 ends by cross-pieces, on which marks are made that 
 remain at the same distance apart for all ordinary 
 temperatures. 
 
 . 143. The Theodolite. — The accompanying, figure re- 
 presents a simple form of theodolite. It consists of an 
 achromatic telescope mounted so that it can turn round 
 two axes, one vertical and the other horizontal, and pro- 
 vided with graduated arcs by means of which angles in 
 horizontal and vertical planes may be measured. 
 
 A, B, C and D are four plates, the two upper ones, 
 G and D, being in contact with one another. At the 
 bottom of the lowest plate A is a screw by which the 
 theodolite is fixed to the head of a tripod stand. From 
 
PRACTICAL APPLICATIONS. 
 
 193 
 
 the plate G, a hollow axis projects downwards at right 
 angles to it, passes through the plate B, and ends in a 
 ball which works in a socket in the plate A. Inside 
 this axis, and co-axial with it, is another axis attached to 
 
 the uppermost plate D, called the vernier-plate, and 
 passing through the plate G. The theodolite can be 
 turned round as a whole about the outer axis, and can 
 be clamped in any position by the clamping-screw E, 
 which tightens a collar that surrounds the axis above the 
 plate B. When so clamped, a slow motion may be given 
 
194 PRACTICAL APPLICATIONS. 
 
 to the instrument by the tangent-screw F. The inner 
 axis allows the vernier-plate D, and the telescope, etc., 
 attached to it, to be rotated, while the rest of the 
 instrument is clamped by the screw E. By means of 
 the screw G, the vernier-plate can be clamped to the 
 plate below, and a slow motion can be given to the 
 former by the tangent-screw H. The four screws K 
 (of which only two are shewn in the figure) enable the 
 plate B and those above it to be set accurately level and, 
 consequently, one axis of the theodolite vertical. The 
 horizontality of the vernier-plate D is indicated by two 
 levels L fixed to it at right angles to one another. In 
 the centre of this plate is also fixed a compass M. The 
 edge of the plate G is graduated the whole way round to 
 every half degree, but angles can be read to one minute 
 by two verniers N on opposite sides of the vernier-plate, 
 exactly 180° apart. 
 
 The telescope T is carried by two F-supports 8, S, 
 attached to the ends of the bar R, which rotates about an 
 axis turning on the pillars, P, P. The axis is parallel 
 to the vernier-plate, and therefore horizontal when the 
 instrument is adjusted correctly. The bar R being at 
 right angles to this axis, the telescope turns in a vertical 
 plane w^hen the rest of the instrument is fixed by the 
 clamping-screws, E, G. A graduated semicircle Q is 
 attached to the bar P, and a slow motion can be given 
 to it by the screw Tf. The semicircle is graduated to 
 every half-degree, from 0° to 90° in either direction from 
 the middle point, and angles can be read to one minute 
 by the vernier X attached to the compass-bore M. A 
 level V is fixed to the telescope, and indicates when the 
 axis of the telescope is horizontal. 
 
PRACTICAL APPLICATIONS. 195 
 
 In the common focus of the object-glass and eye-glass 
 of the telescope are placed two spider-threads, called 
 cross- wires, at right angles to one another, and adjusted 
 so that, when the intersection of the cross- wires appears 
 to coincide with the centre of an object, the direction of 
 the axis of the telescope coincides with the direction of 
 the object as seen from the observer. 
 
 The verniers N and X are short fixed graduated arcs, 
 constructed so that 30 divisions of the verniers are equal 
 to 29 divisions of the circles; the difference in length 
 between a division of either is therefore equal to -sVth of 
 a division of the circle. The first or zero division of the 
 vernier, marked with an arrow-head, indicates the num- 
 ber of degrees and half-degrees on the graduated circle. 
 If the arrow-head coincides exactly with a division on 
 the circle, say 47J°, then this is the required reading. 
 But i£ it be between 47J° and 48°, it is then noted which 
 divisions of the two arcs coincide. If the division of the 
 vernier be the 14th from the arrow-head, then the arrow- 
 head is \^ of 80' beyond 47 J°, and the correct reading is 
 47i°-f-14>r47°44^ 
 
 In the above account the principal parts only of a 
 simple theodolite are described : no reference is made to the 
 details which are required for securing the correct adjust- 
 ments of the instrument, for it may be assumed that as 
 far as possible these adjustments are already made by 
 the maker. 
 
 144. In setting up the theodolite for use, the tripod is 
 firmly planted on the ground, its position being indicated 
 by a mark made below a plummet suspended from the 
 head of the stand exactly beneath the centre of the instru- 
 
196 PRACTICAL APPLICATIONS. 
 
 ment ; and the legs of the tripod are arranged so that the 
 vernier-phite is roughly horizontal. The instrument is 
 then turned so that the two levels L are parallel to the 
 line joining two of the levelling-screws K ', and, by 
 means of these screws, the vernier-plate is set exactly 
 horizontal, the bubbles of the two levels being then in the 
 centres of their ranges. If the vertical axes have been 
 accurately adjusted, the bubbles of the levels will keep 
 their positions while the instrument is turned completely 
 round. 
 
 The object-glass and eye-piece of the telescope are then 
 adjusted for the distance of the object to be observed and 
 for the eye-sight of the observer ; both the object and the 
 cross-wires must at the same time appear well-defined. 
 
 145. To measure the horizontal angle between any two 
 objects, i.e. the angle between the projections, on the 
 horizontal plane through the observer, of the straight 
 lines joining the observer to the objects, we proceed as 
 follows : 
 
 Turn the vernier-plate round until the zero-line of the 
 vernier (indicated by the arrow-head) coincides nearly 
 with the zero-line (marked 360°) of the horizontal circle. 
 Clamp the vernier-plate by the screw G, and make the 
 two lines coincide exactly by the tangent-screw H. 
 When this is the case, the zero-line of the second vernier 
 should coincide with the line marked 180° of the hori- 
 zontal circle. Loosen the screw E, and turn the instru- 
 ment round until the intersection of the cross-wires 
 coincides nearly with the centre of one of the objects; 
 then clamp the instrument by the screw E, and make the 
 intersection of the cross- wires coincide exactly with the 
 
PRACTICAL APPLICATIONS. 197 
 
 centre of the object by the tangent-screw F, Now, loosen 
 the clamp 0, and the rest of the instrument being still 
 fixed by the clamp E, turn the vernier-plate round until 
 the intersection of the cross-wires coincides nearly with 
 the centre of the second object ; clamp the vernier-plate 
 by the screw (r, and make the intersection of the cross- 
 wires coincide exactly with the centre of the object by the 
 tangent-screw H\ having previously, however, adjusted 
 the telescope, if the difierence of the distances of the two 
 objects is great compared with that of either. Read the 
 angles indicated by both verniers, and the mean of the 
 two readings will give the required angle. 
 
 If the instrument has been correctly adjusted, the zero- 
 lines of the vertical circle and the vernier X will coincide 
 when the vernier-plate has been set horizontal and the 
 axis of the telescope is also horizontal. Thus, the angle 
 of elevation of any object can be obtained by means of 
 the vernier X. 
 
 146. Trigonometrical Survey. — Let the length of a 
 base-line AB be measured on level ground, the ends A 
 and 5 being marked by flagstaffs or other prominent 
 and well-defined objects. c 
 
 Let (7 represent a similar 
 object on the same hori- 
 zontal plane with A and 
 By and let the angles 
 BAG, ABC be measured 
 with a theodolite. We D^ 
 
 have thus sufiicient data for calculating the angle AGB 
 and the lengths of the lines AC, BC (art. 129) ; though, in 
 practice, the angle AGB would be measured from the 
 
198 PRACTICAL APPLICATIONS. 
 
 station G to test the accuracy of the other measurements. 
 In a similar manner, if D represent another object in the 
 same plane, we can determine the lengths of the lines 
 AD, BD. Then, in the triangle AGD, knowing the 
 angle GAD and the sides AG, AD, we can find the length 
 of the line GD (art. 131). Now, proceeding to the points 
 represented by G and D, and selecting other suitable 
 objects E, F in the same plane as before, and measuring 
 the angles DGE, GDE, DGF and GDF, we can calculate 
 the lengths of the lines GE, DE, GF and DF ; and, from 
 these data, again, the length of EF can be found. 
 
 Proceeding in this manner, we may imagine a network 
 of triangles to be formed, increasing in size and spreading 
 over the whole surface of a country, the magnitudes of 
 the sides and angles determining the distances and bear- 
 ings of a series of conspicuous objects from one another. 
 
 This, briefly, is an outline of the manner in which a 
 trigonometrical survey of a country is carried out. 
 
 147. We have supposed, for simplicity, that the base 
 line AB is horizontal. This, however, is not necessary, 
 and, in practice, the base-line may be inclined at a small 
 angle to the horizon, but the requisite correction is easily 
 applied if the slope of the ground be known (art. 149). 
 
 Again, it will rarely, if ever, be the case that the 
 selected objects, G, D, E, etc., are in the same horizontal 
 plane with A and B. The angle GAB will not then be 
 the angle subtended at A by the line BG, but the angle 
 between the projections of the lines AG, AB on the 
 horizontal plane through A ; and the line AG represents, 
 not the actual distance between A and G, but the hori- 
 zontal distance between them, i.e. the projection of the 
 
PRACTICAL APPLICATIONS. 199 
 
 line AG on the horizontal plane through A. Thus, the 
 figure of the preceding article represents, in this case, the 
 relative positions of the points A, B, (7, etc., as they 
 would be indicated on a map of the country. 
 
 148. Now, it is obvious that, if the lengths of the 
 sides of the triangle be appreciable fractions of the 
 radius of the earth, the triangles will no longer be plane, 
 but spherical, or rather spheroidal, triangles. In all the 
 great trigonometrical surveys that have been carried out 
 
 ■ this is the case, the sides being sometimes as much as 50, 
 or even 100, miles in length. A detailed description of 
 such surveys must therefore lie beyond the scope of a 
 work on Plane Trigonometry. 
 
 If, however, the sides be short, say, not more than a 
 few miles in length, then, for all practical purposes, the 
 triangles may be treated as if they were plane triangles. 
 This is the case in the survey of the Mer de Glace and 
 tributary glaciers, executed in 1842 by Prof J. D. Forbes. 
 An account of this survey, though it was carried out 
 on a small scale and without many of the refinements 
 necessary in extensive operations, will illustrate some 
 of the more important features of a trigonometrical 
 survey. 
 
 149. Forbes' Survey of the Mer de Glace.— TAe Base- 
 Line. — The site chosen for the base-line was a road in 
 the valley of Chamouni, joining the villages of Les Praz 
 and Les Tines, and passing a short distance from the 
 foot of the glacier. At the time the survey was made, 
 this road was formed of dry, well-compacted gravel, and 
 its surface was apparently level, though in reality sloping 
 
200 PRACTICAL APPLICATIONS. 
 
 upwards towards Les Tines at an average angle of 44'. 
 Opposite the foot of the glacier, the road for a distance 
 of 1000 yards is absolutely straight, and along this 
 portion of it the base-line NO (see map) was measured. 
 
 The two ends of the line were marked by nails driven 
 into the to(>8 of long pins of hard wood fixed in the 
 ground, and at each end is an object visible from at least 
 one other station used in the survey. The station N is 
 exactly at the eastern end of the beam which forms the 
 south side of a wooden bridge near Les Praz ; and, close 
 to the station 0, there is a solitary tree with its lower 
 branches lopped off.- 
 
 The length of the base-line was measured with a ten- 
 metre chain and a steel tape divided into English feet 
 and inches. It was found to be 91 chains and a fraction, 
 the fraction being approximately two-fifths of a chain, 
 but determined more accurately by the steel tape to be 
 26 ft 2'50 ins. Thus, the length of the base-line was 
 
 91 chains -h 26 feet +2*50 inches. 
 
 The chain, being compared with the steel tape, was found 
 to be 32 ft. 10*675 ins. long, giving for the length of the 
 base-line 
 
 2992-95 English feet, 
 
 a result shewn, by re-measurement of part of the base, to 
 be probably correct to within about an inch, or about 
 1/36000 of the length of the base. 
 
 The road, however, being inclined at an average angle 
 of 44' to the horizon, this length should be multiplied by 
 the cosine of 44', or 099991 81 ; but as this would result 
 in shortening it only by about 1/12000, the correction was 
 not applied. 
 
PRACTICAL APPLICATIONS. 201 
 
 150. ThA Triangtdation, — The form of the glacier, 
 and the series of triangles by which it was deter- 
 mined, arc shewn in the accompanying map, which 
 is reduced from that of Professor Forbes. The sta- 
 tions, forming the angular points of the triangles, 
 and lettered /, L, F, G, 11, B, £, were chosen on the 
 rocks bounding the glacier and at some height above 
 it, HO that from each station two or more of the others 
 wam visible. 
 
 In many ways the survey was carried out under serious 
 disadvantages. " The walls of the glacier are excessively 
 rugged, often maccessible. The stations are difficult to 
 choose so as to be visible from one another, owing to the 
 intricate windings of the ice-stream and the enormous 
 height of the rocks. The fundamental triangulation 
 ijjust be carried up a valley, whose extremities, independ- 
 ent of mountains, differ in level by 4400 feet." The 
 triangles on these accounts ^re badly shaped and few in 
 number. In the triangle FOB, for example, two of the 
 angles are very small, and it is obvious that a very small 
 error made in the measurement of either of them would 
 '^'ive rise to disproportionately large errors in the lengths 
 of the calculated sides. A " well-conditioned " triangle 
 should be as nearly equilateral as possible, and in none 
 of the triangles employed in the survey are the sides 
 even approximately equal in length. With one excep- 
 tion, however, the three angles of every triangle are 
 irioasured, and in one only of the six other triangles does 
 the sura of the angles differ by more than a minute from 
 two right angles. 
 
 The third station /, forming with N and the first 
 triangle, is a rock above the Chapeau, distinctly visible 
 
202 PRACTICAL APPLICATIONS. 
 
 from both ends of the base-line. The observed angles of 
 this triangle were 
 
 NIO= 18° 55' 50'' 
 
 /OiV= 127° 54' 0" 
 
 ONI^ 33° 8' 40" 
 179° 58' 30" 
 The sum of the angles of this triangle being 1' 30" less 
 than 180°, each angle is increased by one-third of this 
 amount; and a similar correction is also made in the 
 other triangles. Since iVO = 2992-95 feet, angle ION 
 = 127° 54' 30" and angle Oi\^/ = 33° 9' 10", we have, 
 therefore, 
 
 i\r/= 7275-78 feet, 
 
 70 = 504304 „ 
 The fourth station L is a projecting mass of rock on 
 the ridge extending above the Montanvert towards the 
 Aiguille des Charmoz. From this point the stations / 
 and could be seen ; the west end N of the base-line 
 being, however, invisible. The observed angles were 
 OLI= 3G°18'20" 
 LIO= 84° 56' 21" 
 XQ/= 58° 45' 40" 
 180° 0'2r 
 giving (since 70 = 5 043 04 feet), 
 
 70 = 8484-49 feet, 
 77=7282-61 „ 
 The fifth station F is on the promontory of Les Echelets, 
 about 150 feet above the glacier, and from it the four 
 stations 7, 7, G and B were visible. The angles of the 
 triangle 777" are 
 
PRACTICAL APPLICATIONS. 203 
 
 LFI= 48° 30' 15" 
 ILF=10r W 10" 
 FIL= 28° 48' 50" 
 179° 59' 15" 
 giving IF= 9485-56 feet, 
 
 Zi^= 4686-50 „ 
 The sixth station G is marked by a pyramid of stones 
 on a large rock on the ridge of Tr^laporte ; it is about 
 300 feet above the glacier, and commands a view of the 
 three stations L, F, and B. Only two of the angles of 
 triangle LFG were measured, namely, 
 Zi^(^ = 122°37'45", 
 FLG= 28° 37' 50", 
 giving 6^i^= 4670-18 feet, 
 
 G^Z = 8208-28 „ 
 The seventh station 5 is a pyramid of stones built on 
 the promontory of the Tacul, and is seen from five of the 
 other stations. The angles of the triangle FGB are 
 GBF= 11° 42' 0" 
 FGB=15r52'25" 
 GFB= 16° 25' 5" 
 179° 59' 30" 
 giving i?'5 = 108531 feet, 
 
 OjB = 6508-75 „ 
 The eighth station H is at the foot of the Couvercle, 
 and is opposite station B. The angles of the triangle 
 GBH are 
 
 GBH= 66° 40' 10" 
 
 BHG= 87° 48' 5" 
 
 HGB= 25° 32' 30" 
 
 180° 0' 45" 
 
204 PRACTICAL APPLICATIONS. 
 
 giving GF= 5980-79 feet. 
 
 £ir= 280801 „ 
 
 The ninth, and last, station ^ is on a rocky promontory 
 high up the Glacier de L^chaud. The angles of the 
 triangle BHE are 
 
 BEH=- 21° 24' r 
 HBE= 67° 50' 40" 
 EHB= 90° 45' 40" 
 
 180° 0'2r 
 giving ^^=7127-97 feet, 
 
 J5i7= 7695-66 „ 
 
 151. The principal triangulation being completed, the 
 sides of these triangles were then used as base-lines for 
 determining the positions of several subordinate points, 
 and the outline of the glacier was drawn with the aid of 
 compass and tape measurements from known points and 
 lines. 
 
 152. Measurement of Heights. — In Chapter V., two 
 simple cases of the calculation of the height of a tower or 
 other object above a horizontal plane were explained : the 
 base-line in the first being measured on the plane from 
 the foot of the tower, etc. ; and, in the second, in the same 
 straight line with the tower, the foot of the tower being 
 supposed inaccessible. We shall now explain the method 
 by which the height of a mountain or any other object 
 may be determined, the base-line being measured in any 
 direction and not necessarily on level ground. 
 
PRACTICAL APPLICATIONS. 205 
 
 153. ^To find the height of a mountain. 
 
 (1) Let the length and bearing of the base-line AB be 
 measured on level ground, and 
 let G represent the summit of the 
 mountain. 
 
 From the two ends of the 
 base-line, the bearings of G are 
 measured. Let the directions of 
 these bearings be represented by 
 AD, BD ; these two lines meet 
 in a point D, vertically below (7, 
 and in the same horizontal plane with. AB (Eucl. XL 19). 
 
 At one end of the base-line, say A, measure the angle 
 of elevation GAD of the summit G. 
 
 Now, in the triangle ABD, the base AB and the angles 
 BAD, ABD are known ; and from these the side AD is 
 calculated. Also GD = ADi^^GAD. 
 
 This equation gives the height of the summit G above 
 the base-line AB, and, consequently, above the level of 
 the sea, if the height of the base-line above the same 
 level be known. 
 
 (2) Let the base-line be inclined to the horizon at a 
 known angle. 
 
 Let B and D now be the projections, on a horizontal 
 plane through one end A of the base-line, of the other 
 end of the base and of the mountain summit (7; so that 
 A, B and D represent the positions of the ends of the 
 base, and of the summit, as they would be depicted on 
 a map. 
 
 The same measurements are made as before, namely, 
 the length and bearing of the base, the bearings of the 
 summit from the two ends of the base, and the elevation 
 
206 PRACTICAL APPLICATIONS. 
 
 of the summit from one end A ; and, in addition, the in- 
 clination to the horizon of the ground on which the base 
 is measured. 
 
 Thus, AB is now equal to the length of the base-line 
 multiplied by the cosine of its inclination to the horizon ; 
 and, precisely as in the first case, we find the height of 
 the mountain G above the horizontal plane through A. 
 
 154. Exam/pie. — In addition to making the survey of 
 the Mer de Glace described above, Prof Forbes deter- 
 mined the height of every station and of many of the 
 neighbouring summits of the Mont Blanc range by the 
 method explained in the preceding article. In this part 
 of his work it was necessary to take into account the 
 effects of atmospheric refraction and the curvature of the 
 earth ; but, in most xBases, both effects were eliminated by 
 observing, not only the elevation of one station above 
 another, but also the depression of the first station below 
 the second, and taking the arithmetic mean of the two 
 angles. We give one example, the height of the station G. 
 
 The line LF, determined from the triangle ILF to be 
 4686-50 feet long, is here the projection of the base-line 
 on the horizontal plane through F. The bearings of G 
 from the ends of the base-line are given by the angles 
 LFG and FLG, which are 122° 87' 45" and 28° 37' 50" 
 respectively, giving 4670-13 feet for the length of the 
 line GF. 
 
 The arithmetic mean of the elevation of G above F 
 and of the depression of F below G was found to be 
 4° 48' 15''. 
 
 The height Qi) of G above F in feet is therefore 
 4670-13 X tan 4° 48' 15'; 
 
 I 
 
PRACTICAL APPLICATIONS. 207 
 
 log h = log 4670-13 + log tan 4^ 48' 15" 
 = 3-6693290 
 + 2-9245144 
 = 2-5938434 = log 892-51. 
 Hence, the height of the station G above the station F 
 is 392-5 feet. 
 
 In a similar manner, it was found that the height of F 
 above the Montanvert is 523-5 feet. The height of the 
 Montanvert above the observatory of Geneva was deter- 
 mined by barometric measurement to be 4960 feet, and 
 the height of the Observatory above the level of the sea 
 was known to be 1343 feet. Hence, the height of the 
 station G above the sea-level 
 
 = 392-5 + 523-5 -1-4960-1- 1343 feet 
 = 7219 feet. 
 
 155. Example. — The apparent dips of a stratum in directions 
 inclined to one another at an angle $ are a and f3, respectively ; 
 to obtain equations for determining the amount and direction of 
 the true dip. 
 
 Let OA and OB be the directions in which the apparent dips, 
 a and (3, are measured, A and B being Cj^ 
 points in the same horizontal plane 
 with 0, and a point in the plane 
 of the stratum vertically above 0. 
 Then, angle OAC = a and angle 
 0BC = f3. 
 
 Draw OB perpendicular to AB. 
 Then OD represents the direction of 
 the true dip, since AB is a. horizontal 
 line in the plane. Let 8 be the 
 amount of the true dip, and </» the angle made by its direction 
 with OA. 
 
 Now, 0D = OA cos (f) — OC cot a cos <^ / 
 
 = OB cos(^ -4>) = OC cot ^ cos((9 - </>), 
 
PRACTICAL APPLICATIONS. 
 
 tan /? cos </) = tan a(cos ^ cos ^ + sin ^ sin ^), 
 cos ^(tan /? - tan a cos 6) = sin ^ tan a sin ^ ; 
 
 tan<^=^^"^~^^"^^^"^=(tan/?cota-co3^)cosec^. ...(1) 
 tan a sm ^ 
 
 Also, tang=^= ,^'^^ ^=tanasec</> (2) 
 
 OD OC cot acoa<j> 
 
 Examples XVIII. 
 
 [Note. — Unless specially mentioned, the height of the 
 observer's eye above the ground is nob to be taken into 
 account in the following examples. See also the note at 
 the head of Examples VII. A.] 
 
 1. AB is a line 1000 yards long ; B is due north of A. 
 
 At B, a distant point P bears N. 70° E. ; at ^, it 
 bears N. 41° 22' E. ; find the distance from A to P. 
 
 2. J. ^ is a base-line 200 feet long, measured close to 
 
 one bank of a river and parallel to it ; is a mark 
 on the opposite bank; the angles BAG and ABC 
 are found to be 59° 15' and 47° 12', respectively; 
 find the breadth of the river. 
 
 3. The sides of a valley are two parallel hills, each of 
 
 which slopes upwards at an angle of 30°. A man 
 walks 200 yards directly up one of the hills from 
 the valley, and then observes that the angle of 
 elevation of the other hill above the horizon is 15°. 
 Shew that the height of the observed hill is 273 2 
 yards nearly. 
 
 4. A man, standing at the water's edge, finds the angle 
 
 of elevation of the top of a cliff to be a. He then 
 walks a feet directly up a beach, which slopes 
 upwards at an angle y, and, again measuring the 
 
PRACTICAL APPLICATIONS. 209 
 
 angle of elevation of the top of the cliff, finds it to 
 be /3. Find the height of the cliff above the sea- 
 level. 
 
 Find the height when .a = 60 feet, a = 42°, 
 /5 = 65°, y = 8°. 
 Two distant spires, P and Q, are seen from the two 
 ends of a base-line AB, 1200 yards long, measured 
 on a straight and level road, the spires being on 
 the same side of the road. The angles FAB, 
 PBA, QAB and QBA are found to be 108° 14', 
 89° 5\ 27° 30' and 114° 20', respectively. Find the 
 distance between the two spires. 
 
 6. AB is a straight and level road, 1500 yards long ; C 
 
 and D are church spires on either side of the road. 
 The angles BAG, ABC, BAD, ABD are measured 
 and found to be 43°, 57°, 29° and 37°, respectively. 
 Find the distance between the spires. 
 
 7. From one end of a horizontal straight road, running 
 
 N.W. and S.E., and 2000 yards long, the top of a 
 mountain is observed at an altitude of 23° in a 
 direction E. 4° N., and, from the other end of the 
 road, it bears N. 37° E. Find the height of the 
 mountain above the level of the road, and its 
 horizontal distance from the road. 
 
 8. In order to determine the height of a mountain, 
 
 a north-and-south base-line, 1000 yards long, is 
 measured ; from one end of the base-line, the 
 summit bears E. 10° N., and is at an altitude of 
 13° 14'; from the other end, it bears E.46°30'N. 
 Find the height of the mountain. 
 
 9. To determine the height of a steeple, a base-line, 150 
 
 feet long, is measured on a road running due east 
 
210 PRACTICAL APPLICATIONS. 
 
 and inclining upwards at an angle of 5°, the lower 
 end of the base-line being on the same level as the 
 foot of the tower. From this end of the base-line, 
 the top of the steeple bears due north, and its angle 
 of elevation is 1G° ; from the other end of the base- 
 line, the top of the steeple bears N. 28" W. Find 
 the height of the top of the steeple above the 
 ground to the nearest inch. 
 
 10. A ship, sailing uniformly and directl}'- towards a port 
 
 P, sights another sailing uniformly and directly 
 towards a port Q, and observes that the line join- 
 ing the two makes an angle a with its own 
 direction of motion. After a hours, the line joining 
 the two vessels points directly to P, and, after h 
 hours more, it points directly to Q ; and, at the 
 latter time, the distance PQ subtends an angle ^ 
 at the first ship. Compare the rates of sailing of 
 the two ships. 
 
 11. On the side of a hill inclined to the south at an angle 
 
 of 20°, a road is made which slopes upwards in 
 the direction E. 15° N. ; if the road be a mile 
 long, find the difference in height between its two 
 ends, 
 
 12. The shadow of a cloud at noon is cast on a spot 
 
 1600 feet due west of an observer. At the same 
 instant, he finds that the cloud is at an altitude of 
 23° in a direction W. 14° S. : find the height of the 
 cloud and the altitude of the sun. 
 
 13. The elevation of a steeple standing on a horizontal 
 
 plane is observed, and at a station a feet nearer it 
 its elevation is found to be the complement of the 
 former. On advancing in the same direction h feet 
 
PRACTICAL APPLICATIONS. 211 
 
 nearer still, the elevation is found to be double 
 the first ; shew that the height of the steeple is 
 
 {(«+^)-?}- 
 
 14. A person, walking along a straight road, observes the 
 
 greatest elevation of a tower to be a. From 
 another straight road, he observes the greatest 
 elevation of the tower to be /5. The distances of 
 the points of observation from the intersection of 
 the two roads are a, h, respectively; prove that 
 the height of the tower is 
 
 / (^2_52 Y 
 
 \cot2^-cotV' 
 
 15. A person walks from one end J. of a wall a certain 
 
 distance a towards the west, and observes that 
 the other end B then bears E.S.E. He afterwards 
 walks from the end B a distance a(^2 + 1) towards 
 the south, and finds that the end A bears N.W. 
 Shew that the wall makes an angle cot "^2 with 
 the east. 
 
 16. A man standing on an elevation can just see over the 
 
 surface of a calm sea the top of a mountain, the 
 height of which he knows to be 1650 feet above 
 the sea-level, and the summit of which is at a 
 distance of 70 miles from his own position. What 
 is the height of his eye above the sea-level, assum- 
 ing the radius of the earth to be 4000 miles ? 
 
 17. An observer, whose eye is h feet above the surface of 
 
 a lake, determines the angle of elevation of a point 
 on a cloud to be /3, and the angle of depression of 
 the image of the same point to be a ; find the 
 
212 PRACTICAL APPLICATIONS. 
 
 height of the cloud above the surface of the 
 lake. 
 
 Find the height of the cloud when A = 240 feet, 
 a = 30°, ^ = 17°. 
 
 18. Ay B, and C^are three points in a straight line. AB, 
 
 BC, and CA are 1000 feet, 2000 feet and 3000 
 feet respectively ; P is a point such that each of 
 the angles APB, BPG is 35°. Find the distance 
 AP. 
 
 19. A hill consists of two inclined planes sloping in 
 
 opposite directions at angles a and /3 to the 
 horizon. A cloud is driven with uniform velocity 
 and always at the same height in a line towards 
 the sun and at right angles to the axis of the hill. 
 The shadow of the cloud passes up one slope in t 
 seconds and down the other in f seconds. Find 
 the altitude of the sun. 
 
 20. At distances of 100 feet and 40 feet, measured in a 
 
 horizontal plane in the same straight line from 
 the foot of a tower, a flagstaff standing on the top 
 of the tower subtends an angle of 8° ; find the 
 length of the flagstaff. 
 
 21. A flagstaff stands on the top of a tower built on 
 
 a horizontal plane. A person observes the angles 
 subtended, at a point in the horizontal plane, by 
 the tower and the flagstaff; he then walks a known 
 distance towards the tower, and finds that the 
 flagstaff subtends the same angle as before. Find 
 the height of the tower and the length of the 
 flagstaff. 
 
 22. A person, walking along a straight road, observes the 
 
 greatest angle (a) subtended by two objects in the 
 
PRACTICAL APPLICATIONS. 213 
 
 same plane with the road. He then walks a 
 distance a along the road, and the objects appear 
 in the same direction making an angle ^ with the 
 road. Shew- that the distance between the objects 
 
 2 a sin a sin B • 
 
 IS 7-. 
 
 cos a + cosp 
 
 23. A, B, and C are three consecutive milestones on a 
 
 straight road, from each of which a distant spire is 
 visible. The spire is observed to bear N.E. at A, 
 E. at B, and E. 80' S. at C. Find the distance of 
 the spire from^, and the shortest distance of the 
 spire from the road. 
 
 24. At each end of a horizontal base of length 2a it is 
 
 found that the angular altitude of a certain peak is 
 a, and at the middle of the base it is /3. Prove 
 that the height of the peak above the plane of the 
 base is a Ana An fi 
 
 V sin(/3 + a)sin(y8 — a) 
 
 25. The angles of elevation of a balloon are observed from 
 
 two stations a mile apart, and from a point half- 
 way between them, to be 60°, 30°, and 45° respec- 
 tively. Prove that the height of the balloon is 
 440^6 yards. 
 
 26. Two stars, A and B, are so situated that, when A 
 
 is due south and at an altitude of a degrees, B is 
 setting at a point /3 degrees W. of S. ; find the 
 angle subtended by the two stars. 
 
 27. A straight flagstaff, leaning due east, is found to 
 
 subtend an angle a at a point in the plain upon 
 which it stands a yards west of the base. At a 
 point h yards east of the base, it subtends an 
 angle /3. Find at what angle the flagstafi' leans. 
 
214 PRACTICAL APPLICATIONS. 
 
 28. Two lines of straight railway, ABC, DEG, meet at 
 
 C, telegraph-posts being situated at A, B, D, E\ 
 the angles DAE, DBE are each equal to a; and 
 the angles EAB, EBG are /8 and y respectively ; 
 shew th^t 
 
 BC=AB . ;'°^f°S°l^ix V 
 sin(y - ^)sin(a + /3+y) 
 
 29. Two vertical faces of rock, at right angles to each 
 
 other, exhibit sections of a stratum ; the dips of 
 the sections so formed are found to be a and /3 
 respectively ; if ^ be the true dip of the stratum, 
 and its direction (measured from the vertical 
 plane corresponding to dip a), then 
 
 tan2^ = tan2a + tan2/3, 
 and tan 6 = tan /3 cot a. 
 
 30. A person, wishing to determine the dip of a stratum, 
 
 bores vertical holes at three of the angular points 
 of a horizontal square ; the depths of the stratum 
 at these points being a, h, c, and h the side of the 
 square, the dip of the stratum is 
 
 31. Find the amount and direction of the true dip of a 
 
 stratum, the apparent dips being 41° and 18° in 
 the directions N. 31° E. and E. 12° S. respectively. 
 
 32. Two lines traced on an inclined plane include an 
 
 angle a, and their inclinations to the horizon are 
 P and y. Shew that the tangents of their in- 
 clinations to a horizontal line traced on the plane 
 are 
 
 sin a sin /3 , sin a sin y 
 
 sin y - sin /3 cos a sin y8 — sin y cos a 
 
PRACTICAL APPLICATIONS. 215 
 
 S3. Two lines inclined at an angle y are drawn on an 
 inclined plane, and their inclinations to the horizon 
 are a and /3, respectively. Shew that the sine of 
 the inclination of the plane to the horizon is 
 cosec y(sin2a + sin^/S — 2 sin a sin /3 cos y)'^. 
 
 34. Two planes are inclined to the horizon at angles a 
 
 and /3 in directions which make an angle 6 with 
 one another : find the direction and inclination 
 to the horizon of their common section. 
 
 35. A tight rope connects the tops of two vertical poles, 
 
 a and h feet high, respectively, placed c feet apart 
 in an east and west line. If the sun be due south 
 and at an altitude of a degrees, find the direction 
 of the shadow of the rope. 
 
 Find the direction if a = 35 feet, 6 = 24 feet, 
 c = 30feet, a=35". 
 
 36. Given the angular distances (0, a) of the sun from the 
 
 planes of the meridian and horizon, find the breadth 
 of the shadow cast by an east and west wall of 
 height n on the horizontal plane through its base. 
 
 37. A wall, 20 feet high, bears 59° o' E. of S. ; find the 
 
 breadth of its shadow on a horizontal plane through 
 its base at the instant when the sun is due south 
 at an altitude of 30°. 
 
 38. A man, walking along a straight road which runs in 
 
 a direction 30° E. of N., notes when he is due 
 south of a certain house. When he has walked 
 a mile further, he observes that the house lies 
 due west, and that a windmill on the opposite 
 side of the road is N.E. of him. Three miles 
 farther on he finds that he is due north of the 
 windmill. Find the distance between the house 
 
216 PMACTICAL APPLICATIONS. 
 
 and the windmill, and shew that the line joining 
 them makes with the road an angle 
 
 39. A cloud, just grazing the top of a mountain a feet 
 
 high, is seen at an altitude a by a man at the 
 sea-level due south of the mountain. It is driven • 
 by the wind at the same height and with uniform 
 velocity, and, t seconds later, is seen by him at an 
 elevation /5 in a direction east of north. Find 
 the direction of the wind and its velocity in feet 
 per second. 
 
 40. On a plane inclined to the horizon at an angle of 30°, 
 
 a circle of radius 10 feet is described, and a post 
 is placed at the highest point of the circle per- 
 pendicular to the plane. At one end of the 
 horizontal diameter of the circle the angular 
 elevation of the top of the post is 45°. Find 
 the length of the post. 
 
 41. A and B are two places 12 miles apart, A being due 
 
 north of B. An observer at A determines the 
 altitudes of the two ends of the visible path of 
 a shooting star to be 82° and 70°, and the azimuths 
 of the same points to be N. 23° E. and N. 59° E. 
 An observer at B measures the azimuths of the 
 same points to be N. 12° E. and N. 30° E. Find 
 the length of the visible path of the shooting star, 
 and the heights of its two extremities. 
 
 42. A beacon is due west of a lighthouse and three miles 
 
 distant from it. The channel of a river is given 
 by the condition that a vessel shall enter due 
 south of the lighthouse, at such a point that the 
 
PRACTICAL APPLICATIONS. 217 
 
 lighthouse and beacon shall subtend an angle of 
 60° at the vessel, and shall continue to do so 
 until the beacon is north-west, when the channel 
 remains straight in the last direction in which the 
 vessel was sailing, until it is due south of the 
 beacon. Prove that the straight part of the 
 channel is ^3 + 1 miles long. 
 
 43. A curve on a railway, whose form is a circular quad- 
 
 rant, has telegraph posts at its extremities and at 
 equal distances along the arc, the number of posts 
 being n. A person in one of the extreme radii pro- 
 duced sees the^th and gth posts from the extremity 
 nearest him (from which his distance is a) in a 
 straight line. Find the radius of the curve. 
 
 44. A man, standing on a plain, observes a row of equal 
 
 and equidistant pillars, the tenth and seventeenth 
 of which subtend the same angles as they would 
 if they stood in the position of the first and were 
 respectively one-half and one-third of the height ; 
 shew that the line of pillars is inclined to the line 
 drawn to the first at an angle whose cosine is 
 nearly y%. 
 
 45. A man on a hill observes that three towers on a 
 
 horizontal plane subtend equal angles at his eye 
 and that the angles of depression of their bases 
 are a, a, a \ prove that, c, c\ and d' being the 
 heights of the towers, 
 
 sin(a — d ') sin(a'' — a) , sin(a — aQ _ q 
 c sin a d sin d c" sin d 
 
 46. A vertical pole of height a stands on a plane inclined 
 
 to the south at an angle S to the horizon ; if the 
 angular distances of the sun from the planes of the 
 
218 PRACTICAL APPLICATIONS, 
 
 meridian and horizon be and a, find the length 
 of the shadow of the pole on the inclined plane. 
 
 47. Tvyo ships are sailing uniformly in parallel directions, 
 
 and a person in one of them observes the bearing 
 of the other to be a degrees E. of N. ; p hours 
 afterwards its bearing is ^ degrees E. of N. ; and 
 q hours afterwards it is y degrees E. of N. Prove 
 that the course of the vessel is Q degrees E. of N., 
 where 
 
 tan ^— y^^QQsin(/3 — y) — gsinysin(a — ^) 
 p cos a sin(|8 — y) — 5 cos y sin(a — ^)' 
 
 48. A person in a balloon, which is travelling uniformly 
 
 eastward, and also rising uniformly, observes a 
 train travelling southwards. When it is seen in 
 the N.E., N., and N.W. its angular depressions are 
 a, )8, y, respectively ; shew that 
 
 tan a + tan y = ^^ 2 tan /3. 
 
 49. A person walking along a straight road observes that 
 
 the maximum angles of elevation of two hills on 
 the same side of the road are a, /3, and the distance 
 between the points of observation is a. Along a 
 road passing between them, inclined at an angle y 
 to the former, the maximum angles of elevation 
 are again a, /3, and are observed at points distant 
 h from one another. Find the heights of the 
 hills. 
 
CHAPTER X. 
 
 APPLICATIONS TO THE GEOMETEY OF TEIANGLES, 
 POLYGONS, AND CIKCLES. 
 
 156. A knowledge of the following geometrical the- 
 orems, in addition to those given in Euclid, will be 
 assumed in this chapter. 
 
 The straight lines which bisect the sides of a triangle 
 ABC at right angles pass through the circumcentre (8) of 
 the triangle. The straight lines which bisect the angles 
 of the triangle, both internally and externally, pass 
 through the incentre (/) and the three excentres (/j, ig* ^s)- 
 From, the latter theorem, it follows that any angular 
 point, the incentre, and the excentre opposite that angle 
 lie on a straight line ; that the same angular point and 
 the other two excentres lie on another straight line ; and 
 that these two lines are at right angles. The altitudes of 
 a triangle, i.e. the perpendiculars from the angular points 
 on the opposite sides, pass through a point (0), which is 
 sometimes called the centre of perpendiculars, but gene- 
 rally the orthocentre. The medians, i.e. the lines joining 
 the angular points to the mid-points of the opposite sides 
 pass through the centroid of the triangle. 
 
 If the perimeter of the triangle be denoted by 2s, the 
 
 lengths of the tangents from the angular points to the 
 
 219 
 
220 GEOMETRY OF TRIANGLES, 
 
 incircle are s — a, s — h, and s — c, respective!}'' ; and the 
 lengths of the tangents from each angular point to the 
 opposite excircle is s, the semi-perimeter of the triangle. 
 
 The lines joining the feet (D, E, F) of the altitudes 
 form the pedal triangle of the given triangle. Each 
 angle of the pedal triangle is bisected by the correspond- 
 ing altitude, and is equal to the supplement of twice the 
 opposite angle of the given triangle. Thus, the ortho- 
 centre of a triangle is the incentre of the pedal triangle. 
 Also, any triangle is the pedal triangle of that formed by 
 joining its excentres. 
 
 If the altitudes of a triangle be produced, the parts 
 intercepted between the orthocentre and the circumcircle 
 are bisected by the sides to which they are perpendicu- 
 lars. The parts of the altitudes intercepted between the 
 orthocentre and the angular points are double of the 
 perpendiculars from the circumcentre on the opposite 
 
 The nine-points-circle of a triangle passes through the 
 mid-points of the sides, the feet of the altitudes, and the 
 mid-points of the parts of the altitudes intercepted be- 
 tween the orthocentre and the angular points. Its centre 
 is the mid-point of the line joining the orthocentre and 
 the circumcentre, and its radius is half that of the circum- 
 circle of the given triangle. 
 
 157. To find the area of a triangle. 
 Let ABC be the triangle. Draw the altitude AD. 
 (1) In terms of two sides and the included angle. 
 Area of triangle ABG= \AD . EG 
 
 = i AB sin B. BO, 
 if B be acute or right. 
 
POLYGONS, AND CIRCLES. 
 
 221 
 
 or lAB sin(7r — 5) . BC, if B be obtuse or right, 
 
 = |oa sin B, in every case. 
 Similarljr, 
 
 area = J6c sin A = \ah sin G. 
 (2) In terms of the sides. 
 Area of triangle ABG 
 
 = lea sin B 
 
 B . B 
 
 = eacos.^sm^. 
 
 y ca ^ ca 
 
 = /s/s{s — a){s — h){s — c) = S. 
 
 158. To find the area of any quadrilateral in terms 
 of its sides and the sum of two opposite angles. 
 
 Let a, h, c, d denote the 
 lengths of the sides AB, BC, 
 CD, DA, respectively, of the 
 quadrilateral ABGD ; s the 
 semi-perimeter; 2a) the sum 
 of the angles A and G; and 
 Q the area of the quadrilateral. 
 Let A=()0 + a, G = a) — a. 
 
 Then Q = AABD+ABGD 
 
 = Jac? sin(ft) + a) + J?>c sin(ft) — a), 
 .-. 2Q = (ad-{-bc)cos a sin o) — (6c — ad)8m a cos w (1) 
 
 Again, by art. 121, 
 
 BD^ = a^ ^-d^- 2ad cos(ft, + a) = h^ + c^- 26c cos(a) - a), 
 (6c — ad) cos a cos w + (ad + 6c) sin a sin o) 
 
 = i(62 + c2-a2-c^2) ^2) 
 
222 
 
 GEOMETRY OF TRIANGLES, 
 
 Eliminating a from equations (1) and (2) by squaring 
 and adding, we get 
 
 {ad 4- 6c)2sin2ft, + {he - adf co?i'^w = 4Q2 + j(62 + c^^a^ - cV^)\ 
 :. 4>Q'- = aW+h^c^-2ahcd{2 cos^a)-l)-i{b^+c^-a^-d% 
 .-. 16Q2 = 4(acZ+6c)2-(62+c2-a2-cZ2)2-l6a6c(Zcos2ft, 
 = (2acZ + 26c + 62 + c2 - a2 - c?2) 
 
 x(2acZ+26c-6'-c'+a'+(Z')-16a6ccZcos'a) 
 = (6+c+a-(^)(6 + c-(x+c^) 
 
 x{a+d+b — c){a+d — h+c) — 16ahcdcos^(jt). 
 Q=^{{s — a){8 — 6)(s — c){8 —d) — abed cos^o} . 
 
 Cor. 1. — If the quadrilateral be cyclic, ft) = 90° ; and the 
 expression for the area becomes 
 
 \/{8 — a){s — b){s — c){s — d). 
 
 Cor. 2. — Hence, if the sides of a quadrilateral be given 
 in length, its area is greatest when it is cyclic. 
 
 159. To find the area of a regular polygon inscribed 
 in a given circle. 
 
 Let ABC... be a regular poly- 
 gon of n sides inscribed in the 
 given circle, of which is the 
 centre and r the radius. 
 
 Join OA, OB. Then angle 
 A0B = 2irln. 
 
 Area of triangle AOB 
 = lOA.OB^inAOB 
 
 = Jr^sin 
 
 area of polygon ABC. . . = ^^nr'^sin. 
 
 n 
 
 27r 
 n' 
 
POLYGONS, AND CIRCLES. 
 
 223 
 
 160. To find the area of a regular polygon described 
 about a given circle. 
 
 Let ABC... be a regular poly- 
 gon of n sides described about 
 the given circle, of which is 
 the centre and r the radius. 
 
 Let D be the point of contact 
 of the side AB. Join OA, OB, 
 OD. Then angle AOD = ir/n. 
 Area of triangle AOB^\OD.AB 
 
 = ^r . 2r tan 
 
 n 
 
 .'. area of polj^gon ABC. . . = nrHsna 
 
 n 
 
 161. To find the radius of the circumcircle of a 
 triangle. 
 
 Let R be the radius of the circumcircle of the triangle 
 ABC. Draw the diameter BA\ and join AV. 
 
 or 
 
 (1) In terms of a side and the opposite angle. 
 Since BCA' is a right angle, 
 BC=BA'BmBAV 
 
 = BA' sin A, if A be acute or right, 
 BA' sm{7r — A), if A be obtuse or right. 
 
 m: 
 
224 
 
 GEOMETRY OF TRIANGLES, 
 
 a — ^R^m A, in every case. 
 7?_ a h c 
 
 (art. 120) 
 
 2sm^ 2sin5 2sma 
 (2) In terms of the sides. 
 
 Byart. 125, sin^=2>Sf/6c, 
 
 R=:ahcl4<8. 
 
 162. To find the radius of the incircle of a triangle. 
 Let / be the centre, and r the radius, of the incircle 
 of the triangle ABC. Draw 
 IG, IH, IK perpendicular to 
 the sides, 
 
 (1) In terms of the sides, 
 ^ABC 
 
 = ABia-\-ACIA + AAIB 
 .-. S = \BG.IG+IGA.1H 
 -VlAB.IK 
 But IG = IH==IK=r, 
 
 S=r.i(a + b + c) = rs. 
 r = S/s. 
 (2) In terms of the angles and the radius of the 
 circumcircle. 
 
 BC = BG + GC = IG cot IBG + IG cot ICG, 
 
 ' B G 
 
 cos 2 cos-^ 
 
 sin- sm^^ 
 
 5.0, G . B 
 
 cosg sin^ + cos^-sm-g 
 
 ' . B . G 
 
 sin 2 sin- 
 
POLYGONS, AND CIRCLES. 
 
 225 
 
 _ sin 1(^+0) ' 
 
 . B . C 
 sm ^ . sin 7^ 
 
 a sm - sin ^ 2E sm A sm -^ sin ;^ 
 
 cos -^ COS g- 
 
 r = 4it sm ^ sm ^ sm ^. 
 
 163. To ^')i<i the radii of the excircles of a ti'iangle. 
 Let I^ be the centre, and r^ the radius of the excircle 
 opposite A ; rg and r^ the radii of those opposite B and (7. 
 
 (1) In terms of the sides. 
 
 AABC=AGI^A + AAI^B-ABT^G 
 
 S=iCA . I^H^-hiAB.I^K^-iBG. I^G^. 
 But JjG, = /i^i = I^K^ = r^, 
 .-. ^ = r^ J(6 + c — a) = ri(s — a), 
 .'. r^ = S/{s — a). 
 
 Similarly, r2 = >S/(s — 6), and 
 r^ = SI(s-c). 
 
 (2) In terms of the angles 
 and the radius of the circum- 
 circle. 
 
 BG=BG^+G^G 
 = GJ^cotG^BT^+G^I^coi Gfil^, 
 
 a = r^( tan^ + tan- j = ' 
 
 ^mlJB+G) 
 
 ' B G ' 
 cos— cos^ 
 
 B G ^^ . . B G 
 
 a cos 77 cos ^ 2R sm A cos ^ cos ^ 
 '2 2 2 2 
 
 cos 
 
 cos 
 
226 
 
 GEOMETRY OF TRIANGLES, 
 
 r^ = 4ic sin ^ cos ^ cos ^. 
 Similarly, 
 Vc, = 4it cos » sm - cos -^ and i\ = 4i2 cos ^ cos -^ sm ly 
 
 164. jTo yiTicZ, -i^i any triangle, the distances between 
 (1) the circumcentre and the orthocentre, (2) the cir- 
 cumcentre and the incentre and excentres, and (3) the 
 orthocentre and the incentive and excentres. 
 
 Let S be the circum- 
 centre, the ortho- 
 centre, and I the in- 
 centre, of the triangle 
 ABC. 
 
 (1) Since SA = R, 
 OA = 2R cos A, and 
 angle SAO = (90'' -B) 
 -(90° - C)-= C - B, 
 C therefore 
 SO^ = R\l -f- 4 cosM - 4 cos ^ cos(a- B)] 
 = R%1 - 8 cos J. cos B cos 6^). 
 
 A . B C 
 
 (2) Since SA =R, IA= r cosec — = 4i2 sin -^ sin ^, 
 
 and angle SAI=-^-{9Q'' -C) = \{G^B). 
 
 ..2^.;.2^ 
 
 J5 . G 
 
 :. SP = R'^-\QR^^\n^^m^^-2RAR^\n^f^m~co^\{G-B) 
 
 =:R^ — 2RAR sin -^ sin ;,^f cos — ^ — 
 = i22_2/?.4iisin|sin^cos^'^-, 
 
 Pt'^« 
 
 - 2 sin I sin ^), 
 
POLYGONS, AND CIRCLES. 227 
 
 = R^ — 2RAR sin -^ sin ^ sin -^, 
 
 = R^-2Rr. 
 Again, if J^ be the excentre opposite A, 
 
 I^A = r^cosec ^ = 4it cos ^ cos -^. 
 .-. 81^' = R^-2RAR cos | cos ^(cos ^^ - 2 cos | cos ^), 
 = R^ + 2RAR sin ^ cos | cos |' 
 
 = i^2_{_2i^7V 
 
 (3) Since /J. = 7^ cosec ^ = 4i^ sin -^ sin ^, 
 
 0^ = 2jRcos^, 
 
 and angle /^O = ^-(90°-(7) = KC'-^)- 
 .-. OP 
 
 AT>2r 2A . A ' 2^ • 2<^ A A ' ^ ' ^ C-^l 
 
 = 4it^ cos^^ + 4sin^-;TSin^j^ — 4 cos A sm ^ sin ^ cos — ^^ I 
 
 A r,o\~ o A , A • oB • oC A A ' B . G £ 
 
 = 4it^ cosM + 4 sin^- sin^^ — 4 cos -4 sin ^ si n -^ cos -^ cos ^ 
 
 — 4 cos A sin^ ^ sin^-^ I 
 
 = 4i?"^ cosM +4 sin^^ sin^^ — 4 cos A sin ^cos ^sin-cos-^ 
 
 — 4f 1 — 2 sin^-^ j sin^- sin^-^ I 
 
 = 4!RMcos^A — cos A sin B sin C+ 8 sin^— sin^^ sin^-^ j 
 = 2r2 _ 4i^2cog ^ cos ^ g^g (^ 
 
 Next, since A I^^r^cosec ~ = 4fR cos ^ cos — , we have, 
 similarly, 01^^ = 2r^^ — 4i?"^cos A cos 5 cos G. 
 
228 GEOMETRY OF TRIANGLES, 
 
 165. The nine-points circle of a triangle tovxihea the 
 incircle and excircles (Feuerhach's Theorem). 
 
 Let be the orthocentre of the triangle ABC, S 
 the circumcentre, F the nine-points-centre, / the incentre, 
 and /j the excentre opposite A. 
 
 We have to shew that IV^^R-r and l^V^^R + r^. 
 Since V is the mid-point of SO, 
 2IV^ = SP+IO^-2SV^ 
 
 = E^- 2Rr + 2r2 - 4i22cos A cos B cos G 
 
 - JE2 + 4722COS A cos B cos (7 
 = i(R-2rf. 
 IV=lR^r. 
 Again, from the triangle I^SO, 
 2/^ F2 = 81^^ + 0/^2 - 2>Sf F2 
 
 = i^ + 2Rr^ -f 2ri2 - 4i22cos A cos 5 cos C 
 - Ji22 + 4i22cos il cos B cos a 
 = J(i2+2r,)2, 
 I^V=\R+r^. 
 Similarly, /g F= JiJ+rg and /g F= Ji? -}- rg. 
 Hence, the nine- points-circle touches the incircle and 
 three excircles. 
 
 166. To find the area of a circle. 
 
 Let regular polygons of n sides be described in and 
 about the given circle of radius r. 
 
 The area of the former polygon is ^nt^s^n. — , and of 
 
 the latter 'Mr2tan -. The area of the circle lies between 
 n 
 
 these values, whatever be the value of n, and therefore 
 lies between them when n is infinitely great. 
 
 ^1' 
 
POLYGONS, AND CIRCLES. 
 
 229 
 
 Now, when n is infinity, 
 
 limit of Inrhm— = limit of A'nr^— ( sin — -4- 
 
 = 7rr2 (art. 75), 
 
 and limit of 'n.r^tan - = limit of nr^-i tan --r- - 
 n n\ n n 
 
 n 
 
 the area of the circle = irr^. 
 
 167. To find the area of a sector of a circle. 
 Let be the number of radians in the angle AOB 
 of the sector AOB, r the radius 
 of the circle. Then 
 area of sector A OB : area of circle 
 = angle A OB : 4 right angles 
 = e:27r, 
 
 area of sector = ■x-ttt'^ 
 
 168. Example 1. — If / be the incentre of the triangle ABC, and 
 /i'j, /i'2, ^3 the radii of the circumcircles of the triangles BIC, CIA, 
 AIB, then li^ Li^R^ = 272V. . 
 
 (See figure of art. 162.) 
 
 By art. 1 61, BC= 2 /?isin BIG 
 
 B C 
 
 = 2^i8in(7r - 
 
 -2/2iCOS^- 
 
 2/2sin^ = 2/?iCOs|:, 
 
 i2i = 2/2sin4, 
 
 RiR2R3=SB^sm^ sin^sin^ 
 2 2^ 
 = 2i2V. 
 
230 GEOMETRY OF TRIANGLES, 
 
 Example 2. — To fiud the area of a quadrilateral in terms of any 
 three sides and the two included angles. 
 Let AB=a, BC=b, CD = c, angle A BC = 6 &nd angle BCD=cf>. 
 
 E^ D 
 
 Draw BE parallel and equal to CD, and CF parallel and equal 
 to BA; and join AE, DF, BE and AF. Then ABCF, DCBE, 
 AEDF are parallelograms. Also, the triangles ABE, FCD are 
 equal, and A D bisects the parallelogram AEDF. Again, the angles 
 EBC, BCD are together equal to two right angles ; therefore, the 
 angle ABE ia the supplement oi 6 + 4>. 
 
 Now, twice the area of the quadrilateral A BCD 
 
 =par. ABCF+AADF-ADCF+p&r. DCBE- /\ABE-AADE 
 
 = par. ABCF-\-Y)a.r. DCBE-2AABE 
 
 = a6 sin ^ + he sin <f>-ac sin( 0-\-(f)). 
 .'. area of quadrilateral ABCD 
 
 = \ah sin ^ + ^6c sin </> - \ac sin( 6 + </>). 
 
 Example 3.— If the incentre of a triangle be equidistant from 
 the circumcentre and the orthocentre, one angle of the triangle 
 is 60°. 
 
 By art. 164, 
 
 0/2 = 2r2 - 4 A! '-« cos A cos B cos C, 
 and SP=R^-2Rr, 
 
 OP - SP =2r^ -4R^ COS A cos B cos C-R^ + 2Rr, 
 hence, using the relations 
 
 4 R C 
 
 r=4/2sin ^ sin „ sin- = ^(cos ^4-cos5+cos C— 1) 
 
 2i 2i 2t 
 
 and co'^A + cos^^ + co^C-\- 2 cos A cos B cos C= 1 , . 
 
 we get OP-SP= R\\-2cos A){\ - 2 cos B){\ - 2 cos C). 
 
 If 01= SI, one of the factors 1-2 cos J, l-2cos^, l-2cosC' 
 must vanish, and therefore one of the angles of the triangle is 60°. 
 
POLYGONS, AND CIRCLES. 231 
 
 Example 4. — If the incircle of a triangle pass through the cir- 
 cumcentre and the orthocentre, the angles of the triangle are 
 
 |, | + cos-V2-i) and ^-co^-\J2-\). 
 
 Since the incircle passes through the circumcentre and the 
 orthocentre, 
 
 .*. the incentre is equidistant from these two points, 
 .•. one angle of the triangle must be -, by the previous example. 
 
 Let A =f . 
 3 
 
 Again, since the incircle passes through the circumcentre, 
 
 .'. the radius of the incircle is equal to the distance between the 
 
 incentre and circumcentre, 
 
 .'. 16/j:2sin2i sin2^ sm^-=R^ - SRhin ^ sin - sin -. 
 
 2 2 2 2 2 2 
 
 AD C 
 
 Now, 4 sin — sin — sin - = cos A + cos B + cos C - 1 , 
 
 ju 2i 2i 
 
 . (cos ^1+ COS 5+ COS C-l)- = l-2(cos J+cos5+cosC-l), 
 
 . (cos A + cos B + cos C)2 = 2, 
 
 . cos A + cos ^ + cos C — ^/2, 
 
 . cos5+cosC'=v/2-i. 
 
 . 2cos^±f:'cos^-=^'=cos:^-<'^=V2-i, 
 
 .-. ^^=cos-\V2-i)and^=|, 
 
 .-. A=1, 5=^ + cos-Xv^2-|) and (7= J-cos--\^/2-|). 
 3 3 3 
 
 Examples XIX. 
 
 1. Find the areas. of the triangles whose sides are : (1) 
 
 15, 86, and 39 feet; (2) 198, 194, and 195 feet. 
 
 2. The sides of a triangle are 242, 1212, and 1450 yards, 
 
 shew that its area is 6 acres. 
 
 3. In the ambiguous case in the solution of triangles, 
 
 find the sum of the areas of the two triangles. 
 
232 GEOMETRY OF TRIANGLES, 
 
 4. In the ambiguous case, find the difference between 
 
 the areas of the two triangles. 
 
 5. Find the area of a triangle in terms of two angles 
 
 and the adjacent side. If 5 = 45°, (7=60°, and 
 a = 2(^3 + 1) inches, shew that the area of the 
 triangle is 6 + 2^3 square inches. 
 
 6. Prove that the area of a triangle is 
 
 Ja2sin25 + J62sin2^. 
 
 7. Prove that the area of a triangle is 
 
 lid'coiA + 62cot B + c^cot G). 
 
 8. If a, h, c be the sides of a triangle, the triangle whose 
 
 sides are m(b+c), m{c+a), mia+h) will be equal 
 to it in area if 
 
 2m* = sin -^ sm - sin -^. 
 
 9. If the sides of a quadrilateral be 23, 29, 37, and 41 
 
 inches, respectively, the greatest area it can have 
 is 7 square feet. 
 
 10. Find the area of a cyclic quadrilateral in terms of its 
 
 sides, without deducing it from the general ex- 
 pression of art. 158. 
 
 11. li ABGD be a cyclic quadrilateral, 
 
 j^(ji^ (^^ + 6cg)(acZ + he) 
 ab+cd 
 
 12. If ABGD be a cyclic quadrilateral, 
 
 **"2 V(s-c)(s-d)- 
 
 13. The area of a quadrilateral in which a circle can be 
 
 inscribed is ijabcd . sin co, where 2co is the sum 
 of two opposite angles ; and, if a circle can be also 
 circumscribed about it, its area is Jahcd. 
 
POLYGONS, AND CIRCLES. 233 
 
 14. If ABCD he a quadrilateral which can be inscribed in 
 
 a circle and also circumscribed about a circle, 
 
 , ^A he -I , o-D ah 
 tan^ ^ = — 7 and tan''-^ = — ^. 
 
 15. If a be the side of a regular polygon of n sides, and 
 
 -B, r the radii of its circumscribed and inscribed 
 circles, 
 
 16. The area of a regular polygon inscribed in a circle is 
 
 one-fourth that of the regular polygon of the same 
 number of sides described about the circle ; find 
 the number of sides in the polygons. 
 
 17. The area of a regular polygon inscribed in a circle is 
 
 to that of the circumscribed regular polygon of the 
 same number of sides as 3 : 4 ; find the number of 
 sides. 
 
 18. Compare the areas of regular decagons inscribed in, 
 
 and described about, a given circle. 
 
 19. Find the area of a regular dodecagon, the length of 
 
 whose side is a. 
 
 20. In any triangle, 4i^2sin A sin B sin G= 2S. 
 
 21. In the ambiguous case, the circumcircles of the two 
 
 triangles are equal. 
 
 22. If be the orthocentre of the triangle ABC, the cir- 
 
 cumcircles of the triangles BOG, CO A, AOB are 
 equal to that of the given triangle. 
 
 23. In any triangle, cos(5 — C) + cos A = bc/2R^. 
 
 24. If S be the circumcentre of the triangle ABC, shew 
 
 from the areas of the triangles ABC, BSC, CSA, 
 ASB that 
 
 sin 2J. + sin 25-j-sin2C=4 sin J sin jB sin C. 
 
234 GEOMETRY OF TIU ANGLES, 
 
 25. If S be the circumcentre of the triangle ABC, the 
 
 diameter of the circumcircle of the triangle BSC 
 cannot be less than the radius of the circunocircle 
 of the given triangle. 
 
 26. If the incircle of a triangle touch the sides in A\ 
 
 B\ G\ the square of the area of the triangle ABC 
 is equal to the product of BA\ CB\ AC multi- 
 plied by their sum. 
 
 27. Prove (1) algebraically, (2) geometrically, that 
 
 r = (s — a)tan — = (s — 6)tan — = (s — c)tan ^, 
 
 , A '^ ^ B , G 
 
 rj = stan^, 7'2 = stanj^, ^3 = 5 tan-. 
 
 2i Ji Z 
 
 28. Find the radii of the circumcircle, incircle and ex- 
 
 circles of the triangle whose sides are 25, 52 and 
 63 inches. 
 
 29. Find the radii of the circumcircle, incircle and ex- 
 
 circles of the triangle whose sides are 25, 101 and 
 114 inches. 
 
 30. A sphere whose radius is 2 inches rests on three 
 
 horizontal wires forming a plane triangle, whose 
 sides are 3, 4 and 5 inches ; find the height of the 
 top of the sphere above the plane of the wires. 
 
 31. r7vv'3 = ^2 
 
 32. r^rr,+r^r^ + i\r^ = s\ 
 
 33. 1+1+^=1. 
 
 n ^\ ^'3 ^' 
 
 34 ^^ = tan2^tan2ftan2^. 
 r^r^T^ 2 2 2 
 
 35. cot^^^^^i^. 
 
POLYGONS, AND CIRCLES. 235 
 
 36. sm2'^ = 
 
 37. a = {r,-\-r,)J'':^. . 
 
 \r2r3 
 
 38. (7'i - r)(r2 - r)(r3 - r) = 4r2i2. 
 
 (X 6 7'3 
 
 40. 7\ + r2 + r3 — r = 4i^. 
 
 41. The sum of the reciprocals of the altitudes of a 
 
 triangle is equal to the sum of the reciprocals of 
 the radii of the excircles. 
 
 42. The sum of the radii of the two excircles of a triangle 
 
 which touch the side a produced is equal to 
 
 a cot ^. 
 
 43. If the sides of a triangle be in arithmetical progres- 
 
 sion, then will the radii of the excircles be in 
 harmonical progression. 
 
 44. Each altitude of a triangle is a harmonic mean 
 
 between the radii of two excircles. 
 
 45. The distances of the excentres of a triangle from 
 
 the incentre are a sec ^, b sec ^, and c sec ^. 
 
 46. The distances between the excentres of a triangle 
 
 A , B , G 
 
 are a cosec -^, cosec ^, and c cosec ^. 
 
 47. If /g ^^d ^3 ^® ^^6 excentres opposite B and G, 
 
 48. The excentres of a triangle lie without the circum- 
 
 circle, and cannot be equidistant from it unless 
 the triangle be equilateral. 
 
GEOMETRY OF TRIANGLES, 
 
 49. The distances of the orthocentre of a triangle from 
 
 its angular points are 2i2cos^, 2i2cosJ5, and 
 2R cos G, or a cot A , h cot B, and c cot G. 
 
 50. Find the area of the segment of a circle of 14 inches 
 
 radius, the arc of the segment subtending an 
 angle of 30° at the centre of the circle (7r = 3|). 
 
 51. Each of two equal circles passes through the centre 
 
 of the other; shew that the area common to both 
 is ia2(47r-3V3), 
 
 where a is the radius of either circle. 
 
 52. If be the orthocentre of the triangle ABG, 
 
 OB . AB+OG . GA _ 0G . BG+OA . AB 
 BG ~ GA 
 
 OA.GA + OB.BG 
 AB 
 
 53. If /, /j, /g, Is be the incentre and excentres of a 
 
 triangle ABG, 
 
 lI,^ + IJ^ = II,^ + I,I,^=:IIi^-IJl 
 
 54. Also r3 . //, . 11^ . lis = I^' • I^^ ' I(^- 
 
 55. IfDEFhe the pedal triangle of a triangle, the lengths 
 
 of EF, FD and DE are i2sin2^, i^ sin 25, and 
 B sin 2C, or a cos A, 6 cos B, and c cos (7 ; and the 
 perimeter of the triangle DEF is 
 4i2 sin A sin B sin G. 
 
 56. The area of the pedal triangle of the triangle ABG is 
 
 2S cos A cos B cos G. 
 
 57. The radius of the incircle of the pedal triangle of the 
 
 triangle ABG is 2R cos A cos B cos G. 
 
 58. The area of the triangle formed by joining the 
 
 excentres of a triangle ABC is ahcj^r or sa/sin J.. 
 
POLYGONS, AND CIRCLES. 237 
 
 59. If be the orthocentre and S the circumceatre 
 
 of a triangle, 80^ = 9R^ -a'-h''- c\ 
 
 60. If be the orthocentre of a triangle ABC, and DEF 
 
 the pedal triangle, shew, from the areas of the 
 triangles ABC, AEF, BFD, CDF, and DEF, that 
 Gos^A + cos^^ + cos^O + 2 cos A cos B cos C = 1 . 
 
 61. If P be a point on the circumcircle of the quadri- 
 
 lateral ABCD, the products of the perpendiculars 
 on the following pairs of sides, BC, AD ; GA, BD ; 
 and AB, CD; are each equal to 
 PA.PB.PG.PD 
 
 62. If (J) be the angle between the diagonals of a quadri- 
 
 lateral ABCD, its area is 
 
 63. li ABCD be a quadrilateral such that the lines joining 
 
 the mid-points of opposite sides are equal, then 
 ac cos(A -\-D) = hd cos( J. -f B). 
 
 64. If a regular pentagon and a regular decagon have the 
 
 same perimeter, prove that their areas are as 2 : ^5. 
 
 65. If P and Q be the areas of two regular polygons 
 
 inscribed and circumscribed respectively to the 
 same circle, and if P' and Q be the areas of the 
 inscribed and circumscribed regular polygons, in 
 the same circle, with double the number of sides, 
 then 
 
 l = Vp-J and 1 = 1(1+;^). 
 
 Hence, find an expression for the area of the 
 
 octagon circumscribed to a circle whose radius is r. 
 
 QiQ. If the radius of the circumcircle of a triangle be equal 
 
 to the perpendicular drawn from one of the angles 
 
238 GEOMETRY OF TRIANGLES, 
 
 on the opposite side, the product of the sines of the 
 angles adjacent to that side is h 
 
 67. If be the orthocentre of the triangle ABC, and K 
 
 the incentre of the triangle BOG, the radius of the 
 
 A 
 circumcircle of the triangle BKG is ^R cos -^• 
 
 68. ABC is a triangle inscribed in a circle of radius R, 
 
 and G, H, K are the mid-points of the arcs 
 
 BC, GAy AB\ prove that the radius of the incircle 
 
 of the triangle GHK is 
 
 ,^ . B+G . G+A . A+B 
 4>M sin — 7 — sin — ^ — sin — -j — . 
 
 69. The internal bisectors of the angles of the triangle 
 
 ABG meet the circumcircle in G, H, K ; if S' be 
 the area of the triangle GHK, then 
 S'IS = Rl2r. 
 
 70. If p be the radius of the incircle of a triangle whose 
 
 sides are 6 + c, c+a, a +6, where a, h, c are the 
 sides of a given triangle, then 
 p^ = 2Rr. 
 
 71. / is the incentre of a triangle, and G, H, K its points 
 
 of contact with the sides. If p^, p.^, p^ be the radii 
 of the circumcircles of the triangles HIK, KIG, 
 GIH, respectively, then 
 
 72. /is the incentre of the triangle ABG ; R, R^, R^, R^ 
 
 are the radii of the circumcircles of the triangles 
 ABG, IBG, IGA, TAB, respectively; prove that 
 
 73. If a, /3, y be the altitudes of a triangle. 
 
POLYGONS, AND CIRCLES. 239 
 
 74 If I, m, n be the distances between the excentres of a 
 triangle, 
 
 Imn sin A sin 5 sin C= Sr^rg^'g. 
 
 75. If I^, ig, -^3 be the excentres of a triangle ABC, the 
 
 distances between those of the triangle /j/^a^s ^^® 
 
 8it cos — :r- , 8it cos — -. — and 8B cos —.--. 
 4 ' 4 4 
 
 76. In the ambiguous case in the solution of triangles, 
 
 the sum of the radii of the two incircles and of the 
 two excircles opposite the given angle, is equal to 
 twice the common altitude of the triangles. 
 
 77. In the ambiguous case, HA, a, h be given, and if 
 
 S, S' be the areas of the two triangles, the con- 
 tinued product of the radii of the incircles and of 
 the excircles opposite B, is equal to SS'. 
 
 78. The area of the triangle formed by joining the points 
 
 of contact of the incircle of a triangle with the 
 sides is 2rB'^lahc. 
 
 79. Find the radii of the circles which touch two sides of 
 
 a triangle and the incircle. 
 
 80. The radius of the incircle of a triangle can never be 
 
 greater than one half that of the circumcircle. 
 
 81. The rectangle under the segments of any chord of the 
 
 circumcircle drawn through the orthocentre is 
 greater than twice the rectangle under the seg- 
 ments made on a chord of the incircle drawn 
 through the same point by twice the square on. 
 the radius of the incircle. 
 
 82. Two circular sectors are of equal area, and the chords 
 
 of their arcs are equal ; their angles are as 2 : 1 ; 
 find the angles. 
 
240 GEOMETRY OF TRIANGLES, 
 
 83. The circumference of a semicircle is divided into two 
 arcs, such that the chord of one is double that of 
 the other. Shew that the sum of the areas of the 
 two segments cut off by these chords : area of the 
 semicircle = 27 : 55. (tt = 3|.) 
 
 84. The altitudes of a triangle intersect in the point ; if 
 
 Pv /°2' P3' P4» Pb> Pe ^® ^^^ radii of the circles taken 
 in order, inscribed in the six triangles of which 
 is the common vertex, then 
 
 PiP3P5 = PiPiPe- 
 
 85. AL, BM, ON, the medians of a triangle ABC, inter- 
 
 sect in G; if /Op pg- Ps' Pi> Pb> Pe ^® ^^® TS^d^x of the 
 circles inscribed in the triangles BGL, LGG, CGM, 
 MGA, AGN, NGB, then 
 
 Pi Pz P5 Pi Pi Pe 
 
 86. The points of contact of each of the four circles 
 
 touching the three sides of a triangle are joined ; 
 if the area of the triangle thus formed from the 
 incircle be subtracted from the sum of the areas of 
 those formed from the excircles, the remainder will 
 be double of the area of the original triangle. 
 
 87. If, in a triangle ABC, AG, BH, GK are cut off from 
 
 the sides AB, BG, GA, and respectively equal to 
 m . AB, m . BG, m . GA, and if R, R^, R^, R^, be the 
 radii of the circumcircles of the triangles ABG, 
 AKG, BGH, GHK, then 
 a?R^^ + y'R^ + cm^^ = (a2 + 6^ + c^)R%Sm^ Sm + l). 
 
 88. If 8 be the circumcentre, and / the incentre, of the 
 
 triangle ABG, then 
 
 a . A^^/= i2V(cos B ^ cos G). 
 
POLYGONS, AND CIRCLES. 241 
 
 89. A triangle is formed by joining the points at which 
 
 the lines bisecting the angles of a given triangle 
 meet the opposite sides. Shew that the area of 
 the new triangle is to that of the given triangle in 
 the ratio of 2abc to {h-\-c){c-\-a){a-{-h). 
 
 90. If / be the incentre of a triangle ABC, IG, IH, IK 
 
 perpendiculars on the sides, p^, p^, p^ the radii of 
 the circles inscribed in the quadrilaterals AHIK, 
 BKIG, GGIH, then 
 
 Pi _|_ P2 _^ Ps = f ^ 
 
 r-p^ r-p^ r-p^ r 
 
 91. F is the nine-points-centre of a triangle ABC, and 
 
 A\ B\ G' are the mid-points of the sides. If a, /3, 
 y be the angles subtended by VG, VA and VB, 
 respectively, at A\ B\ and G\ then 
 a cos a + 6 cos /3 -f- c cos y = 0. 
 
 92. If a\ h\ c' be the sides of the triangle formed by join- 
 
 ing the excentres of a given triangle ABG, then 
 g^ 6^ c^ ^ahc _ ^ 
 a^ h'^ c'^ (Jbh'd 
 
 93. If the incircle of the pedal triangle of a given triangle 
 
 touch the sides of the former in A\ B\ G, then 
 
 B G G^A' A'B' a A -n n 
 
 ^^77 = -7YT = -li^FT = " COS A COS B COS G. 
 
 BG GA AB 
 
 94. Through A, B and G are drawn straight lines A^B^, 
 
 B-fi^, G^A.^ perpendicular respectively to the sides 
 AB, BG, GA of the triangle ABG, forming the 
 triangle AJi^G^-, the triangle A^BJO^ is formed in 
 a similar way from the triangle A^B^G^; if AnBnGn 
 be the n^'^ triangle so formed, the radius of the 
 circumcircle of this trianofle is 
 
 \ 2 sin A sin BninG ) ' 
 
242 GEOMETRY OF TRIANGLES, 
 
 95. Two spherical surfaces, whose radii are p^, p^, cut at 
 
 an angle of 60°, and p is the radius of their circle 
 of intersection ; prove that 
 
 3^1 1 _ 1 
 
 V Pi^ P2 PlP'2 
 
 96. From a point at a distance c from the centre of a 
 
 circle whose radius is a, two tangents are drawn, 
 and a second circle is described touching the first 
 circle and the tangents, a third circle touching the 
 second and the same straight lines, and so on. 
 Shew that the sum of . the areas of all the 
 circles: the area of the first circle ={c — af-Aac. 
 
 97. On the sides BC, CA, AB of the triangle ABC are 
 
 described the three triangles A'BC, AB'C, ABC, 
 equal in every respect to the triangle ABC. Shew 
 that the sides of the triangle A'B'C are 
 aisjl + 8cos J. sin5sin G, h\/l + 8smA cos^sin C, 
 
 Cs/ 1 + 8 sin J. sin J5 cos G, 
 and that 
 area of A^'5'0' : area oiLABG= 3+8 cos^ cos^cosO : 1. 
 
 98. G is the centroid of a triangle ABG, and A', B\ G' the 
 
 mid-points of the sides ; if p^, p^, p^ be the radii of 
 the circumcircles of the triangles B'GG', CGA', 
 A'GB\ then 
 GA^j_pI^GBKpI^GG^ .^^1 a^+b^-\-c^ 
 a2 62 ^-2 3'a^^ c^^* 
 
 9' 9 ' 9 
 
 Pi P2 PZ 
 
 99. Also, 
 
 2 ' 9 ' 2 
 
 P\ Pi Pi 
 
 100. The equation giving the length x of the diagonal 
 
 AG f)f the quadrilateral ABGD, is 
 
POLYGONS, AND CIRCLES. 243 
 
 {x\ah + cd) - (ac + hd){ad + bc)Y 
 = 4 abed Gos^(o{{x^ -a?- b^)(x^ ~ c^- d'^) + 4a6ccZ sin-o)}, 
 2ft) being the sum of two opposite angles. 
 
 101. If chords of the circumcircle of a triangle be drawn 
 
 through the points in which the line joining the 
 centres of the circumcircle and incircle meets the 
 incircle, the product of the rectangles under their 
 segments is equal to 
 
 102. The radii of the excircles of a triangle are the roots 
 
 of the equation 
 
 x^ - x\4R + r) + xs^ - rs2 = 0. 
 
 103. If the sides of a triangle be roots of the equation 
 
 x^ — Ix^ + mx — n = 0, 
 the altitudes of the triangle are roots of the 
 equation 
 
 SR^x^ - 4^mR^x^ + 2lnRx - oi^ = 0. 
 
 104. If be the orthocentre of the triangle ABC, OA, 
 
 OB and OC are roots of the equation 
 
 x^ - 2(E+r)ic2+(r2 - 4>R^+s^)x - 2R{s^ - (r+2Rf] = 0. 
 
 105. A circle can be inscribed in a quadrilateral, three of 
 
 whose sides taken in order are 5, 4, 7; and the 
 quadrilateral itself is inscribed in a circle. Shew 
 that the sine of the angle between 'the diagonals is 
 Ss/70/67. 
 
 106. If 78 and 50 be the lengths of the diagonals of a 
 
 quadrilateral inscribed in a circle of radius 65 and 
 sin"^f be the angle between them, the sides of the 
 quadrilateral are llx/26, 5^26, 5^/26 and 19>v/26. 
 
 107. A circle touches two sides of a triangle and the 
 
 circumcircle, find its radius. 
 
244 GEOMETRY OF TRIANGLES, 
 
 108. If S be the area of a triangle ABC, and >Si' the area 
 
 of the triangle formed by joining the points in 
 which the bisectors of the angles of ABC meet the 
 opposite sides, prove that 
 S' _ 2 sin J. sin 5 sin 
 
 S~(sin5+sin0)(8in(7+sin^)(sin^+sin5)* 
 
 109. If the incentre and circumcentre of a triangle be at 
 
 equal distances from one side, the cosines of the 
 angles adjacent to that side will together be equal tol. 
 
 110. A hexagon, two of whose sides are of length a, two 
 
 of length 6, and two of length c, is inscribed in a 
 circle of diameter d ; prove that 
 
 and that the difference between the square of the 
 area of the hexagon and the square of the area of 
 a triangle whose sides are a^2, 6^2, c^J'!, is 
 ahcd+\d\ 
 
 111. If Xy y, z be the perpendiculars from the angular 
 
 points of a triangle on any straight line, then 
 a\x-y){x-z)^h\y-z){y-x)^-c\z-x){%-y)={'Lb.ABQ)\ 
 if the proper sign be given to the perpendiculars. 
 
 112. If "p, q, r be the lengths of the bisectors of the 
 
 angles of a triangle produced to meet the circum- 
 circle, and ii, v, lu the lengths of the altitudes of 
 the triangle produced to meet the same circle, then 
 p\v — w)-{- q\w — u) + r\u — t^) = 0. 
 
 113. If ^, q, r be the bisectors of the angles of a triangle, 
 
 and p', q', r these bisectors produced to meet the 
 circumcircle, then 
 
 cos \A , cos JJ5 , cos \C _ 1,1,1 
 p q r a h c 
 
 and ^'cos J^-f^'cos J5+r'cos \G=a-\-h + c. 
 
POLYGONS, AND CIRCLES. 245 
 
 114. On the sides of a scalene triangle ABC as bases 
 
 similar isosceles triangles are described, either all 
 externally or all internally, and their vertices are 
 joined so as to form a new triangle A'B'C'\ prove 
 that, if A'B'C be equilateral, the angles at the 
 bases of the isosceles triangles are each 30°; and 
 that, if it be similar to ABC, they are each 
 
 tan- ^^ 
 
 115. A straight line AB is divided at C into two parts of 
 
 length 2a and 26 respectively. On A C, CB, and 
 ^jB as diameters, semicircles are described so as to 
 be on the same side of AB. If be the centre of 
 the circle which touches each of the three semi- 
 circles, shew that the radius of the circle is 
 ab{a-\-h) 
 
 and that its diameter is equal to the altitude of 
 the triangle AOB. 
 
 116. If, in a triangle, the feet of the perpendiculars from 
 
 two angles on the opposite sides be equally distant 
 from the mid-points of those sides, the other angle 
 will be 60° or 120°, or else the triangle will be 
 isosceles. 
 
 117. Find the relation which exists between the angles 
 
 of a triangle whose orthocentre lies on the incircle. 
 
 118. A triangle is formed by joining the feet of the per- 
 
 pendiculars from any point P on the sides of a 
 triangle ABC', if 8 be the circumcentre of the 
 triangle ABC, and S the distance of P from S, 
 shew that twice the area of the new triangle is 
 (E2_(52^siQ^sin5sina 
 
246 GEOMETRY OF TRIANGLES, 
 
 Prove what it becomes when P is (1) at the 
 centre, (2) on the circumference of the circumcircle. 
 
 119. If twice the square on the diameter of the circum- 
 
 circle of a triangle is equal to the sum of the 
 squares on the sides, then the triangle is right- 
 angled. 
 
 120. The alternate angles of a regular pentagon are joined 
 
 by straight lines which form another pentagon ; 
 the alternate angles of this pentagon are joined, 
 and so on continually. Given a side of the first 
 pentagon, find the sum of the areas of all the 
 pentagons continued ad infinitum. 
 
 121. A circle of radius p touches externally three circles 
 
 which all touch each other externally, and whose 
 radii are p^, p^, p^ ; prove that 
 
 JP2±P3±P _{_ JPs±P2+_P + J pi + P2 + P = JP1+P 2 + P3 
 
 ^ Pi ^ P2 ^ Ps ^ P ' ' 
 
 If the first circle touch the other three and 
 include them all, find a similar relation between 
 the radii of the four circles. 
 
 If the three circles be each of radius a, the radii 
 
 of the other two circles are (2^3 ±3).,. 
 
 122. If /be the incentre, S the circuracentre, and the 
 
 orthocentre of the triangle ABC, the area of the 
 triangle ISO is 
 
 — 2R^sin — ^ sin — ^ — sin — ^ — . 
 
 123. The triangle D^^i^ circumscribes the excircles of the 
 
 triangle A BC, prove that 
 
 _EF ^ FD _ DE 
 a cos A h cos B c cos G 
 
POLYGONS, AxVB CIRCLES. 247 
 
 124. If c, c be the diagonals of a quadrilateral which is 
 
 incyclic and circumcyclic, D, d the diameters of 
 the circumcircle and incircle respectively, then 
 
 (P cc 
 
 125. Three circles, touching each other externally, are all- 
 
 touched by a fourth circle including them all. If 
 a, b, c be the radii of the three internal circles, and 
 a, |8, y the distances of their centres from that of 
 the external circle, respectively, prove that 
 
 \0G ca ao/ a^ b^ c^ 
 
 126. Two points A, B are taken within a circle of radius 
 
 p whose centre is G. Prove that the diameters of 
 the circles which can be drawn through A and B 
 to touch the given circle, are the roots of the 
 equation 
 x\p^c^ - amf^m^G) - 2xpc^(p^ - ah cos C) 
 + cHp^ - 2p^ah cos G+ a^¥) = 0, 
 where the symbols refer to the parts of the 
 triangle ABC. 
 
 Miscellaneous Examples. IT. 
 
 a. 
 
 If an arc of ten feet on a circle of eight feet diameter 
 subtend at the centre an angle of 143'' 14^22", find 
 the value of tt to four places of decimals. 
 
 If^+-B + 0=7r, then 
 
 sinM + sin25- sin2(7= 2 sin A sin B cos G. 
 
248 GEOMETRY OF TRIANGLES, 
 
 3. Solv3 the equations : (1) cos 30 = cos 0, 
 
 (2) sia4e + sin20 = cosa 
 
 4. Find the cosines of the least and greatest angles of a 
 
 triangle whose sides are 7, 14, 15 ; and apply the 
 formula a^ =, ^2 _^ ^2 _ 26c cos J. to prove that, if the 
 straight line which bisects the vertical angle of a 
 triangle also bisects the base, the triangle must be 
 isosceles. 
 
 6. It is observed that the altitude of the top of a mountain 
 at each of the three angular points J., 5, (7 of a 
 horizontal triangle is a ; shew that the height of 
 the mountain above the plane of the triangle is 
 Jatanacosec^. 
 
 6. Shew that four times the area of a triangle is 
 62sin2a+c2sin25; 
 and interpret the result geometrically. 
 
 1. 8(cos^a + sin^a) = 5 + 3cos4a. 
 
 2. cos-iff+2tan-H = sin-H. 
 
 3. Find the conditions under which it is possible that the 
 
 expressions sin(a + /3)cos y and sin(a + y)cos/3 may 
 be equal. 
 
 4. If the sines of the angles of a triangle are as 13 : 14 : 15, 
 
 then the cosines are as 39 : 33 : 25. 
 
 5. If the sides of a triangle be 4219, 5073, 3104, find the 
 
 greatest angle. 
 
 6. Shew that (1) r= ^^^'^^^ • 
 
 /2> 1+1+1= 1 
 ^^ hccaah Mr 
 
POLYGONS, AND CIRCLES. 249 
 
 y- 
 
 1. If, in a triangle, each of the angles J. and 5 is double 
 
 of the third angle G, then 
 
 A+B A + B-{-G A^-B+G 
 
 cos 7^ COS = COS* -. • 
 
 2 4 
 
 2. If^ + 5+(7=27r, then 
 
 sin 2J.+sin 2i?4-sin 2(7= — 4sin^ sin J5 sin G. 
 
 3. Solve the equations: (1) sin 80 = 2 sin 0, 
 
 (2) tanO-|-tan20 = tan3a 
 
 , T- . . 1 a^ + y^ l + cosM-5)cosO 
 
 4. In any triangle, ^ , o =i— ; t-j — t^t »• 
 
 •^ * a^ + c^ 1 + cos( J. — C/)cos 5 
 
 5. When the sun is 20° E. of S. and at an altitude of 25°, 
 
 the shadow of the top of a church spire falls at a 
 point A on the level ground on which it is built. 
 At a point B, 60 feet north of A, the bearing of 
 the top of the spire is 15° E. of S. Find the 
 height of the spire. 
 
 6. The sides of a triangle are 11, 90, and 97, find its area, 
 
 and the radii of its circumcircle, incircle, and 
 excircles. 
 
 S. 
 
 1. cos^a + cos^f J + a j + cos^f I" + a j + cos^f -^ + a J = 2. 
 
 2. tan-Xl + r.H^)-tan-i(l+r^^i.r) = tan-i ^^,^^J,^^ . 
 
 3. If ABG be a triangle^ and sin 2A, sin 25, sin 2(7 be in 
 
 arithmetical progression, then tan A tan (7=3. 
 
 4. If ABO be a triangle, find cos G, having given 
 
 sin^ m , tan J. p 
 
 -. — D = — and D= • 
 
 sm B n tan B q 
 
250 GEOMETRY OF TRIANGLES, 
 
 5. If ^=49^5' 30", 5 = 64** 15^20", and 6" = 5127, find C 
 
 and a. 
 
 6. lip, q, r be the lengths of the perpendiculars from the 
 
 circumcentre of a triangle to the sides, then 
 
 4^^ + ^ + ^^ = ^. 
 \p q rJ pqr 
 
 €. 
 
 1. Shew how to construct an angle when its sine is given, 
 
 and apply to the construction of an angle whose 
 sine is 3/(2 + ^5). 
 
 o 
 
 2. lfA+B+G=-^^, express cos 2A 4-cos 25+cos 2(7-1 
 
 as a single term. 
 8. Find the general values of the limits between which A 
 lies, when sinM is greater than cos^J.. 
 
 4. If a, /3, y be the lengths of three straight lines AB, BC, 
 
 CD, and they be so placed that A, B, C, D are on 
 a circle whose diameter is ^D ; then will the 
 length oi AD be the positive root of the equation 
 a;3 - a;(a2 + /32 + y ) - 2a;5y = 0. 
 
 5. A tower stands on a slope inclined at an angle a to 
 
 the horizon. At the foot of the slope, directly 
 beneath the tower, the angle of elevation of the 
 top of the tower is 2a, and a feet further up the 
 slope it is Za. Find the height of the tower, and 
 the distance between the base of the tower and 
 the toot of the slope. 
 G. Between two concentric circles lies a series of circles, 
 given in number, each of which touches the two 
 nearest of the series and also the two concentric 
 
POLYGONS, AND CIRCLES. 251 
 
 circles ; find the ratio of the areas of the con- 
 centric circles. If the number of the series of 
 circles be six, the area of the outer of the con- 
 centric circles is nine times that of the inner. 
 
 1. sin3a + sin3(12(r-ha) + sin3(240°H-a)= -f sinSa. 
 
 2. ABCDE is a regular pentagon, the middle point of 
 
 the arc AE of the circuracircle ; if a be the radius 
 of the circle, shew that 
 
 (1) OB-OA = a, 
 
 (2) OA.OB = a\ 
 
 3. If sin a — a cos a = a, find the value of tan a. 
 
 4 ABC is a triangle, and K is the middle point of AB ; 
 AD is drawn perpendicular to BG cutting GK in 
 L ; prove that AL is equal to 
 ah sin G 
 a~\-hcosG' 
 
 5. If ct = ^3-l, 5 = ^3 + 1, and J. = 15°, solve the tri- 
 
 angle. 
 
 6. In any triangle, R = ^ \\ ^^^/ ^ , ^ . ■. 
 
 Y]. 
 
 1. A ring, 10 inches in diameter, is suspended from a 
 
 point one foot above its centre by six equal strings 
 attached to its circumference at equal intervals. 
 Find the cosine of the angle between two con- 
 secutive strings, 
 
 2. If^-f5+C=7r, then 
 
 2cos(^ + 40)sinU + 20) + sin2(5~a) 
 = -4sin0sin(5-C')sin(5-2C). 
 
252 GEOMETRY OF TRIANGLES, 
 
 3. Prove that cos 9° = iVs + Jb + \Jb - ^5, 
 
 sin 9° = iV3 + V5 - J V5£75, 
 cos 27° = W 5"4-V^ + J V3^V^', 
 sin 27° = \Jh + V5 - i V3 - Jh. 
 
 4. Solve the equation 
 
 sina + sin(0-a) + sin(2e + a) = sin(^ + a)H-sin(20-a). 
 
 5. A church tower BCD with a spire above it, stands on 
 
 a horizontal plane, B being a point in its base and 
 G being 9 feet vertically above B. The height of 
 the tower is 289 feet and of the spire 35 feet; 
 from the extremity JL of a horizontal straight 
 line BA it is found that the angle subtended by 
 the spire is equal to the angle subtended by BG\ 
 prove that AB is 180 feet, nearlj^ 
 
 6. If a', h\ c' be the sides of the triangle formed by the 
 
 external bisectors of the angles of a triansie, then 
 
 "^ 6 "^ c " r 
 
 h'-^ c'^ abc 
 
 1. 2.an-i(J^tan|) = cos-<^ + ^^"^\ 
 
 \Va + 6 2/ \a + o cos x/ 
 
 2. ABODE is a regular pentagon and any point on the 
 
 arc AE of the circumcircle. Prove the formula 
 cosa + cos(72° + a)+cos(72°-a) = cos(36° + a) + cos(36°-a) 
 and apply it to shew that 
 
 OA + OG+OE=OB-{-OD. 
 
 3. Eliminate between x = acos(0 :\- a) and y = h cos(0 + /3). 
 
 4. 1( {a^ + h^)sm(A-B) = (a^-h^)sm{A+B), the triangle 
 
 is either isosceles or right-angled. 
 
 5. If a : 6 = 379 : 214 and C = 40" 24' find A and B. 
 
POLYGONS, AND CIRCLES. 253 
 
 6. AD, BE, OF, the altitudes of a triangle ABC, are 
 produced to meet the cireumcircle in d, e, /; 
 shew that 
 
 ^DEF : Adef : AABG= 2 : 8 : sec ^ sec 5 sec C. 
 
 1. Find the cosine, sine, and tangent of the angle between 
 
 two faces of a regular tetrahedron ; also of half the 
 angle between two adjacent faces of a regular 
 octahedron. 
 
 2. If ABC be a triangle, and if 1 — cos J., 1 — cos 5, 
 
 1— cosO be in H.P, then sin^, sinj5, sin (7 are 
 also in H.P. 
 
 3. The number of grades in an angle of a regular polygon 
 
 is to the number of degrees in an angle of another 
 as 5 : 3 ; find the number of sides in each, and shew 
 that there are only three solutions. 
 
 4. Solve the equation tan 80 = 5 tan 0. 
 
 5. Find the smaller value of c, having given ^ = 10°, 
 
 a = 2308-7, 6 = 7903-2. 
 
 6. In any triangle, dR^ is not less than a^ + h'^ + c^. 
 
 X. 
 
 1. Given (p) the sum of the three tangents, and (q) the 
 
 sum of the three cotangents, of the angles of a 
 triangle ; find an equation whose roots will be the 
 three tangents. 
 
 2. ABCDEFG is a regular heptagon, the middle point 
 
 of the arc AG of the cireumcircle; if a be the 
 radius of the circle, shew that : 
 
 (1) OC-OB+OA=a, 
 
 (2) OA .OB . OC=a^ 
 
234 GEOMETRY OF TUT ANGLES, 
 
 3. Find the relations which must exist between a, /5, y in 
 
 order that 
 
 (1) tan a + tan/8+tany = tanatan^tan y. 
 
 (2) tan /3 tan y 4- tan y tan a + tana tan/3=l. 
 
 4. ABC being a triangle, express 
 
 a . h c 
 
 cos A cos B cos C 
 in a form adapted to logarithmic computation. 
 Find the numerical value of the expression when 
 a = 1000, A = 35° 4', jB= 10° 30'. 
 
 5. All vertical sections of a hill from the base to the 
 
 summit are alike, and consist of two equal arcs of 
 equal circles, of which the lower has its convexity 
 downwards, and the upper has its convexity up- 
 wards, the highest and lowest tangents being 
 horizontal ; shew that a person who goes right 
 over the hill traverses a less distance than one 
 who goes half round it. 
 
 6. Lines drawn parallel to the sides of a triangle ABC 
 
 through the excentres form a triangle A'B'C. 
 Shew that the perimeter of the latter triangle is 
 
 ,^ .A B n 
 
 4iK cot cot -^ cot » . 
 
 fi. 
 
 1. Find two regular polygons such that the number of 
 
 their sides may be as 3 : 4, and the number of 
 degrees in an angle of the first to the number of 
 grades in an angle of the second as 4 : 5. 
 
 2. If X satisfy the two equations 
 
 x^+a^-2xacose = b^, x^+a^-2xacos(60° - e) = c\ 
 then 2^2^a2 + 62 + c2±4^3.>Sf, where >Sf denotes the 
 area of the triangle whose sides are a, h, c. 
 
POLYGONS, AND CIRCLES. 255 
 
 8. Solve the equations: (1) tan + sec 20 = 1. . 
 
 (2) tan - hix + J sec " '^hx = -y. 
 
 4. If P be any point within a triangle ABC, prove that 
 cotPi?(7.AP5(7+cotP(7^.APa^ + cotP^5.AP^j5 
 
 is independent of the position of P. 
 
 5, Prove that three times the area of a triangle is equal 
 
 to 
 
 where I, on, n are the lengths of the medians. 
 
 V. 
 
 1. The angles x and y vary subject 'to the relation 
 
 smx = ksiny, where Jc is a constant greater than 
 unity. Shew that, as x changes from zero to a 
 
 right ancjle, continually increases, and find 
 
 "= ° ' tan^/ ^ 
 
 the values of this expression, when x = 0, and 
 when x = ~. 
 
 „ T-,. , ^ sin a sin .t, - . a sin a sin 
 
 2. It tan (/) = ^ — , prove that tan = : ^^. 
 
 ^ cos t^ — cos a cos </) ± cos a 
 
 3. Solve the equations : 
 
 ,-. V sin I cos 1 
 
 cos — cos a sin a — sin sin(a — 0) 
 (2) lOcos0 = 2cosec0 + seca 
 
 4. Two lines of length f, q are drawn from the same 
 
 point, and are inclined to one another at an angle a. 
 From their extremities perpendiculars are drawn 
 to each of them : find the area of the parallelogram 
 so formed. 
 
 5. Each of three circles within the area of a triangle 
 
 touches the other two and also two sides of the 
 
256 
 
 triangle ; if a be the distance between the points 
 of contact on one of the sides, and 6, c be like 
 distances on the other two sides, prove that the 
 area of the triangle, of which the centres of the 
 circles are the angular points, is equal to 
 
 IT. 
 
 (Geometry of the Quadrilateral.) 
 
 1. In any quadrilateral 
 
 g^sinM - chm'^C hhin^ G- d% m^A 
 
 m\\A+B) "^ 8m\B+C) 
 
 2 sin AmuG , j j^ i dx 
 
 = -.— 7-i y^^-^—y-n — TvN (cd cos i) — a6 cos B). 
 
 8in(^+i^)sin(i^ + 6y ^ 
 
 2. If a, b, c be three sides of a quadrilateral, being the 
 
 angle between a, h, and (/> that between h, c ; also, 
 if ah cos 6 = bc cos (p = ac cos{0 + 0) ; shew that the 
 quadrilateral is inscribable in a circle of which the 
 fourth side is the diameter. 
 
 3. If, in a quadrilateral, a + b==c + d, the difference be- 
 
 tween the areas of the triangles ABC and CD A is 
 equal to 
 
 area of quadrilateral x -.—i: ~ r^l 
 sin h{B + I)) 
 
 4 If the sum of the opposite angles of a quadrilateral be 2ft), 
 
 and if the angle between the diagonals be a, then 
 
 . 2 _TiQ{(8 — a)(8 — b){s — c)(s — d) — abed cos^o)} 
 tan a (^2 _ 52 4. ^2 ^ ^2y2 • 
 
 5. The length of the line joining the points of intersec- 
 tion of pairs of opposite sides of a cyclic quadri- 
 lateral is 
 
 (ad+bc)^(ab + cd)^bd{c^ - a^f -h acQ)' - d''f]h 
 (b^'-d^Xc'-a^) 
 
CHAPTER XI. 
 
 HYPEEBOLIC FUNCTIONS. 
 
 169. Circular Functions in relation to the Sector of 
 a Circle. — Let a point move from X on the circumference 
 of a circle, whose centre is 0, to 
 the position P ; let ^ be the 
 number ot radians in the angle 
 XOP, A the area of the sector 
 XOP ; then, if a be the radius 
 of the circle, we have 
 
 2A 
 A = ^a^O, and therefore 6 = —^' 
 
 Hence, 
 
 cos 6 = cos 
 
 2A 
 
 2A 
 
 „, sinO = sin ^ 
 
 and so on for the other circular functions, i.e., we may 
 regard the circular functions as functions of a sector of a 
 circle, and the results obtained will be identical with 
 those for angles measured in radians, provided that the 
 unit of area, in terms of which the sector is measured, 
 is the square whose diagonal is the radius of the 
 circle. 
 
 The sense of the sector XOP is the same as that of the 
 angle XOP, and is denoted by the order of the letters. 
 
 R 257 
 
258 
 
 HYPERBOLIC FUNCTIONS, 
 
 Thus, for all positions of X, P, and Q on the circumference 
 we have 
 
 sector XOQ = sector ZOP+ sector POQ. 
 
 170. Definitions of the Hyperbolic Functions. — Let 
 a point move along the curve from the vertex A of one 
 
 bi-anch of a rectangular hyperbola, whose centre is and 
 semi-axis equal to a, to the position P ; let J. be the area 
 
 of the hyperbolic sector A OP, and let u= — ^ » ^-^ ^^^ ^'^ ^^ 
 the measure of the sector AOP, the unit of measurement 
 being the square whose diagonal is the semi-axis. 
 
 Take OF a line making an angle of 90** in the positive 
 sense with the transverse axis OAX, and let OM. ON be 
 
HYPERBOLIC FUNCTIONS. 259 
 
 the projections of OP on OX, OY respectively, then the 
 
 r«atio 
 
 OM : OA is called the liyperholic cosine of u, 
 
 ON : OA the hyperbolic sine of u, 
 
 ON : OM the hyperbolic tangent of u, 
 
 OA : 0-M the hyperbolic secant of u, 
 
 OJ. : ON the hyperbolic cosecant of tt, and 
 
 Oif : OiV the hyperbolic cotangent of u. 
 
 The abbreviations for these hyperbolic functions are 
 cosh ^6, sinh u, tanh Uy sech u, cosech u, coth u.. 
 
 Inverse Hyperbolic Functions.— If a; = coshu, then 
 we write inversely, as in the case of the circular functions, 
 u = cosh " '^x. Similarly, we denote the other inverse func- 
 tions by sinh-^a;, tanh-^aj, etc. The symbol 'cosh~^aj' 
 may be read: 'the sector whose hyperbolic cosine is x, 
 the unit in terms of which the sector is measured being 
 the square whose diagonal is the semi-axis. 
 
 If u be determined from the equation a; = cosh u, where 
 a; is a given number greater than unity, u is a two- valued 
 function of x, the values being equal in magnitude and 
 opposite in sense ; we define cosh~^a; as the positive value 
 oi 11. 
 
 Similarly, sech'^a; is defined as the positive value of n, 
 which satisfies the equation a; = sechu, x being a given 
 positive number not greater than unity. 
 
 The sign of each of the other inverse functions is the 
 same as the sign of x. 
 
 Hence, each of the quantities cosh~^a;, sinh~^a;, tanh~^a;, 
 sech~^a;, cosech -^aj, coth~^aj is a one- valued function 
 of X. 
 
260 
 
 HYPERBOLIC FUNCTIONS. 
 
 171. Elementary Relations between the Hyperbolic 
 Functions. — We have, by definition, 
 
 cosh it-= 
 
 and seen u = — ^, 
 
 Similarly, 
 and 
 
 OA 
 sech u = 
 
 cosech u = 
 
 coth u = 
 
 OM' 
 
 cosh It' 
 
 1 
 sinh u' 
 
 1 
 tanh u' 
 
 .(A) 
 
 Again, by definition. 
 
 tanh. = g5 
 
 So also. 
 
 coth 16 : 
 
 OA ' OA coshtfc" 
 coshu 
 
 (B) 
 
 sinh u 
 
 From the property of the rectangular hyperbola, we 
 have (Taylor's Elem. Geom. of Conies, art. 53) 
 
 OM^-PM^ = OA\ I 
 
 OM^-ON^=OA^ 
 fOMV_fONY_, 
 \0AJ \0AJ ^ ' 
 
 Similarly, 
 and 
 
 cosh^u — sinh% = 1. 
 1 — tanh^u = sechht, 
 coth^u — 1 = cosech%. 
 
 •(C) 
 
 172. To determine the value of any Hyperbolic Func- 
 tion in terms of any other. 
 
 The formulae of the last article furnish five inde- 
 pendent relations between the six hyperbolic functions, 
 from which, in the same manner as in the case of the 
 
HYPERBOLIC FUNCTIONS. 
 
 261 
 
 circular functions, we may deduce the value of any func- 
 tion in terms of any other. The hyperbolic cosine and 
 secant of any sector are, hj definition, positive ; the hyper- 
 bolic sine, cosecant, tangent and cotangent of a sector 
 are all positive or all negative. Hence, we must write 
 
 r--^o ^ 1 . Vl + cosech% 
 
 cosh u=-\- V smhm + 1 ; cosh u=± ■ — ;t- > 
 
 the upper or lower sign being taken according as cosech u 
 is positive or negative; and so on. The values and 
 proper signs are given in the following table : 
 
 cosht^ 
 = c. 
 
 sinhu 
 
 = 8. 
 
 tanh w 
 
 sechM 
 
 = 05. 
 
 cosech u 
 
 cothw 
 
 = 2. 
 
 c 
 
 1 
 
 s 
 
 t 
 
 ~ c 
 s 
 
 1 
 
 c 
 
 1 
 
 **- 
 
 1 
 s 
 
 t 
 
 + ^ 
 
 + ^ 
 
 s 
 
 1 
 t 
 
 -Jl-x' 
 
 t 
 
 ±sf\-x^ 
 
 + 1 
 
 X 
 
 z 
 
 1 
 X 
 
 ^sli-x" 
 
 X 
 
 1 
 y 
 
 -^l-x^ 
 
 y 
 
 ±s/z^-l 
 
 
 I 
 
 z 
 
 z 
 
 173. Even and Odd Functions.— From the definitions 
 and figure of art. 170, we have immediately 
 
 cosh( — u) = cosh u, 
 and sech( — u) = sech w ; 
 
 thus, the hyperbolic cosine and secant are even functions 
 of u. 
 
262 
 
 HYPERBOLIC FUNCTIONS. 
 
 Also, sinh( — u) = — sinh u, 
 
 tanh( — u)= — tanhif, 
 
 cosech( — }i)=— cosech u, 
 
 coth( — u) = — coth u, 
 
 i.e. the hyperbolic sine, tangent, cosecant, and cotangent 
 
 are odd functions of u. 
 
 174 If P, Q, R, S he points taken in order on a branch 
 of a rectangular hyperbola, and if QR and PS be 
 parallel, then will the sectors POQ, ROS be equal ; and, 
 conversely, if the sectors POQ, ROS be equal, then will 
 QR and PS be parallel. 
 
 Let QR be produced to meet the asymptotes in Q\ R', 
 and let PS be produced to meet them in P\ S'. 
 
 Since the intercepts on any chord between the curve 
 and its asymptotes are equal {E.G.G. art. 50), therefore 
 
HYPERBOLIC FUNCTIONS. 263 
 
 the curvilinear areas PP'QQ and BR'B'8 can be divided 
 into an infinite number of pairs of equal strips by draw- 
 ing chords parallel to PSj and therefore these areas are 
 equal. 
 
 We have also ^OQQ' = ^ORR\ (Eucl. I. 38) 
 
 and AOPP' = AOSS\ 
 
 ... AOQQ' + PP'Q'Q - AOPP' = AORR + PR'S' 8 - AO;Sf>S', 
 i.e. sector POQ = sector RO 8. 
 
 Hence, conversely, by a reductio ad ahsurdum, it 
 follows that, if the sector POQ = the sector R08, then 
 will QR and P8 be parallel. 
 
 175. If POQ, R08 he equal sectors of a rectangular 
 hyperbola, and if p, q, r, s he the projections of the points 
 P, Q, R, 8 on an asymptote, then will 
 
 0p:0q = 0r:0s. 
 Since the sectors POQ and R08 (see figure of art. 
 174) are equal, it follows that the chords QR and P8 
 are parallel (art. 174), 
 hence Op:Oq = PP':QQ' 
 
 = 88':RR 
 = 8s:Rr. 
 But, since R and 8 are on the hyperbola, we have 
 
 Or. Rr = Os . 8s, (E.G.G. art. 49) 
 
 and, therefore, 8s:Rr = Or : Os. 
 
 Hence, 0p:0q=0r:0s. 
 
 176. If RV he an ordinate to any diameter OQ of a 
 rectangular hyperhola, and if u= rTAT — > ^^^^ ^^^ 
 
 OA'^ 
 
 , OF , . , VR 
 
 cosh n= jy^ and smn u^-z^- 
 
264 
 
 HYPERBOLIC FUNCTIONS. 
 
 Let P be a point on the curve such that sector 
 2y .4 OP = sector QOR; let a, p, q, 
 T be the projections of A, P, Q, 
 R on the asymptote OT \ then 
 by art. 175, we have 
 
 Op'.Oa^^Or.Oq. 
 
 But, since OM and MP are 
 equally inclined to the asymp- 
 tote, 
 
 OM-^MP _Op 
 OA Oa 
 
 AM 
 
 Similarly, since V and VR are equally inclined to the 
 asymptote (E. G. C, art. 54), 
 
 OV+VR^Or 
 Oq 
 
 _OV±VR 
 OQ 
 
 Hence, 
 
 OQ 
 OM+MP 
 
 OA 
 
 In like manner, since Pp :Aa = Rr: Qq, 
 
 OV-VR 
 
 (art. 175) 
 
 we have 
 
 /. by addition, 
 
 and, by subtraction, 
 
 OM-^MP 
 OA 
 
 OM 
 OA' 
 
 MP 
 
 OA' 
 
 OM 
 
 OQ 
 
 qv 
 
 OQ' 
 
 VR 
 
 OQ' 
 
 MP 
 
 But, by definition, cosh u = -y-^, and sinh u = ^yj; 
 
 , 07 , . , VR 
 
 cosnu = -rj^, and smh u= jj^- 
 
HYPERBOLIC FUNCTIONS. 
 
 265 
 
 177. To "prove that 
 
 cosh (if ■\-v) = cosh u cosh v + sinh u sinh v. 
 
 2A0P 
 
 Let P. Q be points on the curve such that u= ^.^ 
 
 ±POQ 
 
 Draw QV, the 
 
 ordinate of Q to the dia- 
 meter PV, and draw VL 
 perpendicular to OA and 
 VW to Qi\^. 
 
 Since lAOP = lVQW 
 {E. G. a, art. 54), the tri- 
 angles QFF and 0PM are 
 
 similar. 
 Now, 
 
 AM L N 
 ^, - , OiY OL + LN 
 cosh(u+^) = (^=-^X- 
 
 ^a¥ qL_ MP_ VW 
 
 OA' OM OA MP' 
 
 But, 
 
 and 
 
 OX OF , 
 W=OP = ^^^^^' 
 FIT FQ . , 
 
 (art. 176) 
 
 cosh (it + ?;) = cosh u cosh 'y-f- sinh u sinh ^...(D) 
 
 In like manner, 
 
 . ,, ^ ^ NQ MP LV.OM WQ 
 smh(^ + ^) = ^=^.^+^.^ 
 
 ^MP OV OM VQ 
 OA' OP OA'OP 
 = sinh u cosh v + cosh usinh v (D) 
 
HYPERBOLIC FUNCTIONS. 
 
 178. Hence we have 
 
 sinh(u + i;) 
 
 i&.n\\{u + v) = 
 
 cosh(^ + v) 
 
 sinh u cosh ^;+cosh u sinh v 
 cosh u cosh '?;+sinh u sinh v 
 
 , ,, n i. 1./ . \ tanh t/, + tanh -y 
 and th erefore tanh (u-\-v) = q— -7 — r-'-— — ^— . 
 
 1 + tanh u tanh v 
 
 Again, putting v = Uy we get 
 
 cosh2u = cosh2ifc4-sinh% ^ 
 
 = 2 cosh% — 1 
 
 = 2sinh2u+l. 
 
 sinh 2^ = 2 cosh u sinh u, 
 
 2 tanh u 
 
 ,(D) 
 
 tanh 2it 
 
 l + tanh^tt" 
 
 .(E) 
 
 179. Since the geometrical proof of art. 177 is inde- 
 pendent of the sense of u and v, we have, by changing v 
 into —V and attending to the results of art. 173, 
 cosh(u — v) = cosh n cosh v — sinh u sinh v, 
 sinh(u — v) = sinh u cosh -y — cosh u sinh -y, I _ m^ 
 
 . , , , tanh u — tanh V 
 
 tanh (16 — ?;) = :; — 7 — \ 7 — r— • 
 
 ^ 1 — tanh u tanh v 
 
 From the addition formulae 
 
 cosh(u + 'y) = cosh u cosh y+sinh u sinh v^ 
 cosh(u — v) = cosh 16 cosh v — sinh i6 sinh v I 
 8inh(i6 + i^) = sinh 16 cosh v + cosh u sinh v I 
 8inh(i6 — -y) = sinh i6 cosh v — cosh u sinh i;J 
 we obtain, as in the case of the circular functions, 
 2 cosh 16 cosh V = cosh(i6 + v) + cosh(i6 — vy 
 2 sinh 16 sinh v = cosh(i6 + -y) — cosh(i6 — v) 
 2 sinh u cosh i; = sinb(u + v) + sinh (u — v) 
 2 cosh 16 sinh v = sinh(K, -\-v) — sinh (u — v)> 
 whence, putting ;S for 16 + ^; and D for u — v, we have 
 
 ,(F) 
 
HYPERBOLIC FUNCTIONS. 
 
 267 
 
 cosh >Si + cosh Z) = 2 cosh — ^ cosh — ^ 
 cosh 8 — cosh D = 2 sinh — ^ — sinh — g— 
 sinh >Sf + sinh D — 2 sinh — ^ — cosh — ^ 
 sinh >Sf — sinh D = 2 cosh — ^^^ — sinh — ^ — 
 
 .(F) 
 
 180. Gudermannian Function. — Let P be a point 
 
 on a rectangular hyperbola, 
 
 FM the ordinate of P, MT 
 
 a tangent to the auxiliary 
 
 circle, P and T being on the 
 
 same side of the axis. 
 
 ^, .„ 2 sector J. OP 
 ihen, II u = 
 
 and 
 
 e= LAOT, 
 
 . OM OM 
 we have sec6> = jyn — q-j> 
 
 and .'. 
 
 Hence, 
 and 
 
 Also, since 
 
 . ^ I ta 1 x/Oi/2-0^2 ^p 
 
 tan = s/sec^O — 1= y^-, = tt-t- 
 
 OA OA 
 
 sec = cosh u^ 
 
 tan = sinh. uJ 
 
 tan 6 ^ , 1 u sinh u 
 
 tan - = IT- 7^, and tanh ^ = .r- ^ — . 
 
 2 l + sec0 2 1 + coshu 
 
 tan ^ = tanh -^ 
 
 •(G) 
 
 (G) 
 
 The angle 6 is called the gudermannian of u, and the 
 relation between and ^t is written 
 
 ^ = gdit, or u = gd-'^0. 
 
268 
 
 HYPERBOLIC FUNCTIONS. 
 
 181. Curves of the Hyperbolic Functions. — It will 
 be shewn at a later stage that 
 
 cosh'M, = l + n7 + 7T+ ...ad inf. 
 
 11 ^ 1L 
 
 and sinh u = u+.-jr-{-r-=+ ...ad inf. 
 
 [3 [5 -^ 
 
 From these series the values of cosh u and sinh u for given 
 values of u may be calculated. Then, by division, we 
 obtain the values of the remaining hyperbolic functions 
 of u. If 6 = gdu, we may employ the equation 
 sec = cosh u and a table of natural secants to find 
 when u is known. 
 
 Table of Approximate Values of Hyperbolic 
 Functions. 
 
 
 d 
 
 sec^ 
 
 cos^ 
 
 tan^ 
 
 COt0 
 
 8in^ 
 
 cosec 6 
 
 u. 
 
 gdu. 
 
 cosh u. 
 
 sechu. 
 
 sinh u. 
 •00 
 
 cosech u. 
 
 tanh u. 
 
 coth u. 
 
 0-0 
 
 0° 
 
 1-0!) 
 
 1-00 
 
 00 
 
 0-00 
 
 00 
 
 0-2 
 
 11° 
 
 ro2 
 
 •98 
 
 •20 
 
 4-97 
 
 -20 
 
 507 
 
 0-4 
 
 22" 
 
 108 
 
 •92 
 
 •41 
 
 2-43 
 
 •38 
 
 2-63 
 
 0-6 
 
 32' 
 
 119 
 
 •84 
 
 •64 
 
 157 
 
 •54 
 
 1-86 
 
 0-8 
 
 42° 
 
 1-34 
 
 •75 
 
 •89 
 
 1-13 
 
 •66 
 
 1-51 
 
 10 
 
 50" 
 
 1-54 
 
 •69 
 
 1^18 
 
 •85 
 
 •76 
 
 1-31 
 
 1-2 
 
 56° 
 
 1-81 
 
 •55 
 
 1-51 
 
 •66 
 
 •83 
 
 1-20 
 
 1-4 
 
 62° 
 
 2-15 
 
 •46 
 
 1-90 
 
 •53 
 
 •89 
 
 1-13 
 
 1-6 
 
 67° 
 
 2-58 
 
 •39 
 
 2-38 
 
 •42 
 
 •92 
 
 1-09 
 
 1-8 
 
 71° 
 
 3-11 
 
 •32 
 
 2-94 
 
 •34 
 
 •95 
 
 1-06 
 
 2-0 
 
 75° 
 
 3-76 
 
 •27 
 
 3-63 
 
 •28 
 
 •96 
 
 1-04 
 
 30 
 
 84° 
 
 10-07 
 
 •10 
 
 10-02 
 
 •10 
 
 •995 
 
 1-005 
 
 40 
 
 88° 
 
 27-29 
 
 •04 
 
 27-27 
 
 •04 
 
 •999 
 
 1-0007 
 
 50 
 
 89° 
 
 74-74 
 
 •01 
 
 7473 
 
 •01 
 
 •9999 
 
 1-0001 
 
 00 
 
 90° 
 
 CO 
 
 0-0 
 
 00 
 
 0^0 
 
 1^0 
 
 10 
 
 By aid of the above table we may readily draw the 
 curves representing the functions. 
 
HYPERBOLIC FUNCTIONS. 
 
 269 
 
 i:^^ 
 
 O Co 
 
270 
 
 HYPERBOLIC FUNCTIONS. 
 
 Curves of the Hyperbolic 
 
 Tangent and Cotangent 
 
 Tangent • 
 
 Cotangent 
 
 
 
 
 jxf 
 
 Q 
 
 / 
 
 ^ 
 
 
 
 I 
 
 
 
 
 / 
 
 ^ 
 
 
 u 
 
 -6 
 
 '» 
 
 •I 
 
 -pcf 
 
 O 
 
 e' 
 
 / 
 
 
 -3 
 
 Curve of the Guderniannian 
 
[ 
 
 HYPERBOLIC FUNCTIONS. 271 
 
 Examples XX. 
 
 L Shew that the area included between a branch of a 
 rectangular hyperbola and the asymptotes is 
 infinitely great. 
 
 2. 1- 
 
 sinh^a , „ /., tanh^a 
 
 sinh 2fl3 ^ , J sinh ^x , , 
 
 4-. — r^^ — r^ = tanh x, and — r-s r = coth x. 
 
 cosh2a; + l cosh2£C — 1 
 
 ^ . 1 "-, 2 tanh a; 
 5. sinh 2x = 
 
 6. cosh 2x = 
 
 1-tanhV 
 1 + tanh^o; 
 
 1 — tanhV 
 
 1 + tanh X 
 
 7. cosh2ir + sinh2cc-^ ^ , . 
 
 1 — tanhir 
 
 8. 2 cos^iT cosh-y + 2 sin^a; sinh^i/ = cos 2x + cosh 2y. 
 
 9. 2 coth 2cc - coth x = tanh a). 
 
 10. tanh «;+ tanh y= ^'f t^' + y) . 
 
 •^ cosh a? cosh 2/ 
 
 11. cosh(a;+2/)cosh(i:c — 2/) = cosh2cc + sinh22/ 
 
 = cosh^y + sinh^ic. 
 
 12. sinh(aj+ 2/)sinh(aj — y) = sinh^cc — sinh^^/ 
 
 = cosh^ix; — Gosh^y. 
 
 18. cosh{x+ y + z) 
 
 = cosh X cosh y cosh + S cosh a; sinh y sinh 0, 
 
 sinh(aj + 2/ + 2;) 
 
 = sinh X sinh 2/ sinh 2; + 2 sinh a? cosh 3/ cosh z. 
 
 , ^ , , , . , tanh £C tanh y tanh + S tanh x 
 14 tanh(a;+2/ + «) = i + vfanh ytanh^ ' 
 
272 HYPERBOLIC FUNCTIONS. 
 
 16. cosh 3tt = 4 cosh% — 3 cosh -M, 
 sinh 3t6 = 4 sinh^u + 3 sinh u. 
 
 tn J. 1 « tanh% + 3tanlia; 
 1 + 3 tanh^a; 
 
 17. If = gd'M<, then 
 
 cos = sech u, sin = tanh u, 
 cot = cosech u, cosec Q = coth -a. 
 
 18. cosh - la = sinh "Vo^^. 
 
 19. sinh-ia= ±cosh-i/\/a2+l, the upper or lower sign 
 
 being taken according as a is positive or negative. 
 
 20. tanh-ia+tanh-i6 = tanh-i:^^. 
 
 1 + ah 
 
 21. If PT, the tangent to a rectangular hyperbola at P, 
 
 meet the radius OQ in T, then 
 
 PT , , 2P0Q 
 ^ = tanh-^^. 
 
 22. If from i\^, the foot of the ordinate of a point Pon a 
 
 rectangular hyperbola, NQ be drawn to touch the 
 auxiliary circle at Q, then the bisector of the angle 
 AOQ will bisect the hyperbolic sector AOP. 
 
 23. If 
 
 24. If 
 
 -^2:- = 1, and y^ + -r^- = 1, 
 
 sin^aj cosmic ' cosh^y sinh^^/ 
 then a^ = sin^a; cosh^y and y8^ = cos^a; sinh^^/. 
 u -y 1 
 
 sin 2a; si nh 2y cos 2a; + cosh 2y 
 then u^+v^+2ucot2x = '[, 
 
 and u^ + 1;2 — 2^ coth 2y= —1. 
 
 25. (cosh u+sinh u)(cosh v+sinh v) = cosh(tt+i;)+sinh(u+v). 
 
 26. (cosh Uj+sinh u^Xcosh Ug+sinh u^). . .(cosh Un+sinh. Un) 
 
 = cosh(u^+U2+. . .+Un)-\-smh(u^+U2+. . .+Un). 
 
 27. If n be any positive integer, 
 
 (cosh u + sinh u)^ = cosh nu + sinh nv,. 
 
HYPERBOLIC FUNCTIONS. 273 
 
 28. If t'= 2 cosh u, Vn = 2 cosh nu, prove that 
 
 Apply this formula to shew that 
 v^ = v^ — 2v, 
 
 29. sinh - ^a - sioh - 1& = sinh " \as/b^ + l - bs/cf^). 
 
 •^0 ^^^-if ^^^^O-^^'^^^^^ \ . X i/tan0-tanh0 
 • ^^ Vtan20-tanh2^7'^^^'' VtanO + tanh0 
 = tan - i(cot coth 0). 
 31. Prove geometrically that 
 
 . , . , - tanhu+tanhv 
 
 tauh(u+iO=rT-i — tt'— I — T— • 
 1 + tanh'i^tanhi' 
 
CHAPTER XII. 
 
 INEQUALITIES AND LIMITS. 
 
 § 1. Inequalities. 
 
 182. In Chapter VI. (arts. 74, 76) the following in- 
 equalities have been proved, being the number of radians 
 
 in an acute angle : sm$<6< tan 0, 
 
 0-2 
 cos > 1 — -^ , 
 
 4 
 
 183. If 6 he the number of radians in an acute angle, 
 
 sme>e-^,co8e<l-^ + -^, and tsin6>e + ^- 
 
 (1) Let the arc AB subtend 
 an angle of 6 radians at 0, 
 the centre of a circle of any 
 radius (a) ; let G be the middle 
 point of the arc AB, D that of 
 the arc AG, E that of the arc 
 AD, and so on. Then, OG cuts 
 AB at right angles in N. 
 
 Now, the area of the segment 
 AGB — area of sector A OB - area of triangle A OB 
 
 = ^-(0-sin0). 
 
 274 
 
 J 
 
INEQUALITIES. 21 h 
 
 Also, since the area of a circle is the limit of the area 
 of a regular polygon inscribed in the circle, when the 
 number of its sides is infinitely great (art. 166), it follows 
 that the area of the segment ACB is 
 
 = ^AGB+''1^ADG+^''^AED+ adinf. 
 
 Now, AAa5 = J^5.(7i\^ = asin|.a(l-cos|) 
 
 = 2a%n-sin222<2a^-2*y ^'ir^ 
 
 .,., ...i,c<.5(|)-.,|,«* 
 
 ^^^AED < I. • ^', and so on. 
 2* lo 
 
 a20V. .1.1. \ . .a^ 4 
 
 16 '3' 
 
 -(0-sin0)<-^(^l+-2 + ^^+-->^-e-< 
 
 e-sin0<-^. 
 
 
 
 sin e> 0-77- 
 o 
 
 (2) cos0 = l-2sin2| 
 
 <^ JO e^v 
 
 
 V4 48^23047' 
 
 <l-^+^. 
 2^24 
 
 (3) tan0=^, 
 
 cosO 
 
 2''"24 
 
276 
 
 INEQUALITIES. 
 
 %,e. 
 
 2 "^24 
 
 , by division. 
 
 Now, since ^ ^ ^, 
 
 ^ 07 05 
 
 8" ~ 72' °^ 72^^ ~ ^^^' ^^ P^s^^^^® J 
 and l-|V~,or^V{(6-a7-12}, is positive; 
 
 tana>0+|-. 
 o 
 
 184. In the accompanying figure are drawn the curves 
 of sin 6, 0'-\6^ and — ^6^, between the values 6 = and 
 
 TT 
 
 = — ; and the figure thus represents graphically the 
 
 id 
 
 inequalities we have been considering. It also shews 
 how closely the expression Q — \Q^ may be regarded as an 
 
INEQUALITIES. 
 
 277 
 
 approximation to the value of sin 0, so long as Q is an 
 
 acute an^le. 
 
 The dotted line is the curve of — f 
 
 e^an 
 
 expression which may be shewn to be greater than sin 0, 
 while Q lies between and -^. 
 
 185. To shew that sinh x>x> tanh x, 
 
 and 
 
 cosh a? > 1 + ' 
 
 If A denote the area of the sector, we have 
 
 sinh x:x: tanh x 
 _PM. 2A .PM 
 OA'OA^'OM' 
 _PM. 2 A .AL 
 OA'OA^'OA' 
 = iPM.OA:A:iAL.OA, 
 = triangle J. OP : sector AOP 
 : triangle A OL. 
 
 sinh x>x> tanh x. 
 
 Also, 
 
 cosh a? = 1 + 2 sinh^-, 
 >l + 2 
 
 I.e. 
 
 >^+r 
 
 186. Example 1. 
 angle, 
 
 -If be the number of radians in an acute 
 
 sin d > 
 
 > ''' vers I 
 
 Let AOP be the given angle containing 6 radians, AOB a right 
 angle, APB an arc of a circle with centre 0. 
 
 Draw PM perpendicular to OA, and, with if as centre, and MP and 
 MA as radii, describe the quadrantal arcs PN and AL. 
 
278 
 
 INEQUALITIES. 
 
 Then (art. 68), arc PiV> arc ^P> arc AL. 
 
 Z-PM>BxcAP>l- AM, 
 
 sin6>>^>^versa 
 
 O AT 
 
 "When ^ = ^, these three quantities are ultimately equal, for the 
 arcs NP. AP and AL coincide in the limit with the arc AB. 
 
 Example 2. - 
 greater than 1. 
 
 If ABC be a triangle, 8 sin— sin— sin— is not 
 
 2 ^ ^ 
 
 (1) We have sin — sin— = \{ 
 
 B-C B+C' 
 
 cos — - — — COS — 
 
 = i(co.^-sin|). 
 
 and this is greatest, for a fixed value of A, when B=G. 
 Hence, if any two of the three angles A^ J?, C be unequal, we 
 
 ABC 
 can increase the product sin — sin — sin by making the two angles 
 
 2 2 2 
 equal without changing the third angle. 
 
 It follows that the product sin - sin sin— is greatest when the 
 
 triangle is equiangular, in which case it is equal to sin^ 30°, or \. 
 
 ABC 
 
 8 sin — sin - sin -^ ^ 1- 
 
 2 2 2 
 
 (2) Oeometrical proof, — If S be the circumcentre, and / the in- 
 centre, of the given triangle, 
 
 .S72 = /22_2i?r (art. 164). 
 2r:j>i2, 
 
INEQUALITIES. 279 
 
 Z A A 
 
 8sm:^smf siii^4-l. 
 Z " Z 
 
 Example 3. — If ABC be an acute-angled triangle, 
 
 sin ^ + sin ^4- sin (7>cos A +cos jB+cos C. 
 
 sin ^ + sin ^ -1- sin C— cos ^ — cos 5 - cos C 
 
 = sin^-sin(|-^^ + sin B - sin^|-Z?)+ sin C - siuf^ - C\ 
 
 = 2cos^|sin(^-j)+sin(^-|)+sin(C--)}. 
 
 Now, sin a + sin /? + sin y - sin(a + /5 + y ) 
 
 = 4sintosinr+^sin^, 
 z z z 
 
 whatever be the values of a, /?, y ; 
 .•.sin(^-j) + ri„(iJ-|) + sm((7-^) 
 
 .77,.- /5 + C Tr\ . IC+A jr\ . /A+B t\ 
 
 and each factor of the last expression is positive, since all the angles 
 A, B, C are acute. 
 
 sin J. + sin B + sin C - cos A — cos B — cosC is positive, 
 
 sin ^ + sin ^ + sin (7 > cos ^ + cos ^ + cos C. 
 
 Examples XXI. 
 
 1. If be the number of radians in an acute angle, 
 
 tan e> 30-2 sin a 
 
 2. Draw the curves of cos 0, l-JO^ and l-^O'^ + ^O^ 
 
 f between the values = and = -^- 
 
 3. If m and n be integers, 
 
 . TT 
 
 n sm — 
 n 
 
 7r> : — ^. 
 
 TT TT TT 
 
 cos ,,— cos -,— ... COS ^~— 
 
 271 4?i 2"*'^i 
 
280 INEQUALITIES. 
 
 4 If ABC be a triangle, 8 cos ^ cos jB cos O 1. 
 
 5. The sum of the cosines of the three angles of a 
 
 triangle cannot be less than 1 or greater than IJ. 
 
 6. If ABC be a triangle, 
 
 sin2^+sin25+sin2C 
 
 < 2 sin jS sin (7-f 2 sin C sin ^ + 2 sin A sin B. 
 
 A B 
 
 7. If ABC be a triangle, the least value of tan^-j^ + tan^^ 
 
 Q 
 
 + tan2- is unity. 
 
 8. The sum of the acute angles which satisfy the 
 
 equation cos2a + cos2y8 + cos2y = l is less than tt. 
 
 9. The sum of the positive angles which satisfy the 
 
 equation cos2a + cos2/3 + cos'^y = 2 is greater than ^. 
 10. If ABC be a triangle, 
 
 tan^^ tan2-4 tan^^tan''^— H-tan^— tan^^ 
 
 is always less than 1 ; and, if one angle approach 
 indefinitely near to two right angles, the least 
 value of the expression is J. 
 ] 1. The value of the expression 
 
 V(a'cos-^^ + hh\n^<f>) + ^(a^in^^ + ^^cos^) 
 is intermediate to a + 6 and ^(2a^ + 26^) ; also, the 
 value of the expression 
 
 11 4 
 
 is intermediate to — [-7- and 
 
 12. The geometric mean of the cosines of n acute angles is 
 never greater than the cosine of the arithmetic 
 mean of the angles. 
 
\ 
 
 INEQUALITIES. 281 
 
 13. If 6 be the number of radians in an angle less than 
 two right angles, 
 
 . 2 sm - 
 
 sm 6 2 
 
 < 
 
 2+COS0 „ , e 
 
 S + cos^ 
 
 Hence, prove that 
 
 3 sin e 
 
 > 
 
 2 + COS0 
 
 14. If cos{a + 0)-\-m cos = n, shew that n^ cannot be 
 
 greater than l + 2mcosa + m^. 
 
 15. If (J), cf), yfr, xj/ be all acute angles, and be greater 
 
 than i//-, and sin = /x sin 0', sin i/r = ^ sin yj/, where 
 /x > 1, then ^ — i/r > <^' — >//■'. 
 
 16. sin(cos Q) < cos(sin 0), for all values of 0. 
 
 17. If -A^C be a triangle, 
 
 . „B . „G , . Ju . 9^ , . ^A . J^ 
 sm'^g sin^— + sm^— sm^ „ + sin^- sm'^^ 
 
 Li A Z Z Li li 
 
 is not less than 
 
 iV (sinM + sin^i^ + sin-^O). 
 
 § 2. Limits. 
 
 187. We have already defined a limit, and illustrated 
 the definition by finding the limits of cos 0, sin 0, etc., for 
 
 the values = and 0= „ (arts. 18. 19). It has also been 
 
 shewn that, if Q be the measure of an angle in radians, 
 the limits, when Q is zero, of sin 0/0 and tan 0/0 are both 
 
282 LIMITS. 
 
 unity (art. 75). We express these results briefly by the 
 
 notation : 
 
 Xt(cos0) = l, Lt{^\ne) = l, Lt.(sme/e) = l,etc. 
 e=Q a_lL e=o 
 
 The following important algebraical propositions will 
 be frequently used : 
 
 and Lt(l-^Y = \ 
 
 where e is the sum of the infinite series 
 
 1+1+1+1+ 
 
 188. // A and B he functions of a quantity 0, luhose 
 limits for a given value of 6 are known, to find the limits, 
 for the same value of 0, of A±B, AxB, A^B and A^. 
 
 Let a and b be the limits of A and B for the given 
 value of Oy and, when 6 has any other value, let A=a + x, 
 and 5 = 6 + 2/, where x and y ultimately vanish as Q 
 attains the given value. 
 
 (1) Lt{A±B)^Lt{a^x±{h-{-y)) = a±h ' 
 
 = Lt.(A)±Lt.(B). 
 
 (2) Lt.(A X B) = Lt.{{a-\-x) x (6 + 2/)} 
 
 = Lt.{ah+hx-\-ay+xy) 
 = ab = Lt(A)xLt{B). 
 Hence, 
 
 Lt(A xBxCx...) = Lt.{A) x Zt.(B) x Lt{C) x . . . . 
 
 (3) ^=^-^' 
 
 LtiA)^Lt.{^xLt{B), by (2), 
 
LIMITS. 283 
 
 (4) Suppose a > 0, and therefore, for values near to a, 
 A>0; then A^ = (e^''s Ay = e^iog a ^ 
 
 Now Lt.(B log A) = Lt.B. Lt.Qog A), by (2) 
 
 = h log a, 
 since log ^ is a continuous function of A, 
 
 B log A = h log a-\-z, where z ultimately vanishes. 
 Hence J.^ = e&ioga+z^g&iogaxe^. 
 
 but, ultimately, e^ = e^ = 1, 
 
 189. To find the values of 
 
 ^sinhcc\ ,^^/tanhcc' 
 
 Lt\ ) and Lt\- 
 
 :0\ X / a;=0\ ^ 
 
 It has been shewn (see art. 185), that 
 sinh x'.x'. tanh x = triangle A OP : sector A OP : triangle A OL. 
 
 Now, triangle A OP : triangle AOL=OP:OL = OM : OA. 
 
 Let P move along the curve towards A. Then, ulti- 
 mately, M coincides with A and OM is equal to OA. 
 
 Hence, the triangles AOP, AOL, and the sector AOP 
 (which is intermediate to them in area) are ultimately 
 equal, 
 
 190. To find the values of 
 
284 
 
 LIMITS. 
 
 Now, Z«. {(l-sin2|)«-^T} = l, 
 
 by the theorem referred to in art. 187, and 
 
 , / sm - 
 
 71=00 ^^ ^^ n = oo 
 
 2n\ a 
 
 \ n 
 
 n 
 
 = 0, 
 
 therefore, by art. 188, 
 
 i,(eos°)"=(lY = l. 
 n=«,\ w ve/ 
 
 Again, since 
 
 sin-<- <tan-, 
 n 
 
 sm — 
 
 a 
 
 sm 
 
 71 n 
 a 
 
 —<1, and cos— < — 
 ^ a 
 
 a 
 
 71 
 
 or 
 
 sm 
 
 C0S°<_J'<1, 
 71 
 
 This is true for all values of n, 
 
 a\n 
 
 = 1. 
 
 Cor. — In the same way, it may be shewn that 
 
 / a\'^^ -°^ ( aS^ 
 
 • X^. I COS - ) —e 2 Xf. I cos - ) = 0, etc., 
 
 n=oo\ ^/ n=^\ "Tl/ 
 
 for Lt (""- sin2 '') = "', and Lt ("^ sin^-') = oo , etc. 
 
 n=oo\2 71/ 2 n:=xV2 7?/ 
 
 J 
 
1 ^2 
 
 !a\8inh2" 
 
 n 
 
 ) n 
 1/ 
 
 sinh2. 
 
 LIMITS, 285 
 
 191. To find the values of 
 
 I sinh -\ 
 Lt. ( cosh - ) and Lt. ' • 
 
 n=ooV W n=co\ a j 
 
 We have 
 
 n 
 
 XI (cosh-T = Z^. (l+smh2-)' 
 
 71 = 00 ^ ^Z 71 = 00 ^ '^/ 
 
 = Lt. ifl+sinh^^ 
 
 n = oo [\ 71 
 
 1 
 
 Now, XI (l+sinh2-)^^^^'" = 6 (art. 187), 
 
 71 = 00 ^ 71'' 
 
 and Xt(^sinh^«) = xJW!^ 
 
 therefore, by art. 3 88, 
 
 Lt fcosh-y = eO = i. 
 
 Affain, since sinh — > — > tanh -, 
 n n n 
 
 sinh — 
 
 cosh- > > 1, 
 
 n a 
 
 = 0. 
 
 This is true for all values of n, 
 
286 LIMITS. 
 
 sinh 
 
 Lt\ Vl 
 
 n=cp\ ct 
 
 n 
 Cor, — In the same way, it may be shewn that 
 
 Lt. ( cosh - ) = e 2 and Lt ( cosh — ) =00, etc., 
 
 for Lt (-^ sinh^- ) =^, and Lt ( — sinh^- ) = go , etc. 
 n=ooV2 nJ 2' n=«\2 nJ 
 
 Examples XXII; 
 
 1. Prove that the ultimate vahae of the ratio sin : 6, 
 
 when is zero, may be made equal to any quantity 
 whatever by adopting a suitable unit-angle. Find 
 the unit-angle in order that the ultimate value 
 of the ratio may be tt. 
 
 2. Find the value 0^ Lt(sinpO/ta.nqO). 
 
 6 = 
 
 3. If sin = 2 sin (0 — ^), find the limit of smO/sm<p, 
 
 when 6 and apj^roach to tt. 
 
 4. Find the value of Zl{(7r-20)/sin 20). 
 
 V=2 
 
 5. Find the value of X?^.|(tane-l)('l-tan|)|. 
 
 .. -D ,x. ^ T^ { ^ ^ ^^ sine 
 
 6. Prove that Lt (cos ^ cos -^^... cos -^j = —^. 
 
 / Q rfh'^ 
 
 7. Prove that Lt cos - + sin — =6^0. 
 
 -Find the value of X^.(cos 0)0"". 
 
 / 6=0 
 
LIMITS. 287 
 
 9. If u = (cossa^y^°«^°'-< 
 
 -ii 
 shew' that X^.(u) is either 1, e srs^ or 0, according 
 
 as k is less than, equal to, or greater than, 2. 
 
 10. What is the ultimate value of the logarithm of the 
 
 sine of an angle with the tangent of the angle as 
 base, when the angle is diminished indefinitely ? 
 
 11. Regular polygons, each of n sides, are inscribed in 
 
 and 'described about each of a series of equal 
 circles, the number of which is n times n. Prove 
 that the difference between the sum of the areas 
 of the inscribed and the sum of the areas of the 
 circumscribed polygons, when n becomes infinite, 
 is equal to ttM, where A is the area of one of the 
 circles. 
 
 12. Two circular arcs have a common chord and lie on the 
 
 same side of it ; prove that, if the radii approach 
 indefinitely near to equality, the quotient of the 
 area contained between the two arcs divided by 
 the difference of their lengths is ultimately equal 
 to the radius of either of them. 
 
 13. If a and h be positive quantities, and if a-^ = ^{a-{-h), 
 
 \ = {a-p)^, a^= lia^ + h^, h^=={aj)-^^, and so on; 
 
 prove that a^ = b^ = -^ ; 
 
 cos 
 
 ■ r 
 
 b 
 
 and shew that the value of tt may be calculated 
 by means of this theorem. 
 
CHAPTER XIII. 
 SERIES. 
 
 §1. The Addition Formulae Extended. 
 
 192. Notation. — The symbol l>GrSn-r is used as an 
 abbreviation for " the sum of all possible terms that can 
 be obtained from n angles, each term being the product 
 of the cosines of r angles and the sines of the remainin": 
 {n — r) angles." Thus, with two angles A^, A^, we have 
 (72 = cos A-^coa A^, 110^8^ = cos A^ sin ^2+ cos A^ sin A^, 
 /S'2 = sin J.isin^2 5 ^^^» ^^^^ three angles J.^, A^, A^, we 
 have 
 
 Cg = cos A^coa A2COS A^, 
 SCg/Sj = cos J.2C0S J-gsin J.^ + cos J.3COS ^.^sin A ^ 
 
 + COS J-^cos J-gsin A^, 
 1,0^82 = cos ^jsin A2sin ^3+ cos ^2sin A^^'m A^ 
 
 + cos ^gsin A-^sm A 2, 
 Sq = sin J-jSin J.2sin A^. 
 
 The number of terms in llCrSn-r is the number of 
 combinations of n things taken r at a time, or 
 
 — ^ , ,^ / — ^^ — -• This number is denoted by 
 
 1 .2.3. ...r -^ 
 
 the symbol (n)r. 
 
 288 
 
ADDITION FORMULAE EXTENDED 289 
 
 193. Formulae for the Cosine and Sine of the sum of 
 11 Angles. — To ijvove that 
 
 and 
 
 We know that Q,o^(A^ + A^=^C^ — S2y 
 
 ^m{A^^Al) = ^GX. 
 and that co&{A^-{- A^-\-A^ = G^ — Y.G^Sc^, 
 
 sin(A, + ^2 + ^) = 2aA->Sf3. 
 Assume that 
 
 COS(^l + ^2+...+^„_l) = a„_i-S(7«_36f2 + 2Cn_6/Sf4-..., 
 sin(^, + ^2+"- + ^n-l) = 2C„_2^1-SCn-4^3 
 
 + E(7n_6'Si6— .... 
 
 Then we have 
 
 C0S(^l + ^2+--+^n-l + ^n) 
 
 = C0S(^1 + J.2+ . . . +^7i-l)C0sJ.„ 
 
 -sin(^i + ^2+ ••• +^«-i)sin^n 
 = ((7„_i-SC,_3<Sf2+2a,_5^4-...)cos^„ 
 
 - (2C,_2^Sfi - Ea,_4*Sf3+ . . .)sin ^„ 
 
 = Gn-\CO^ An — (^Gn-AQ0^An-{-^Gn-2Si^\nA^ 
 
 + (2(7n-5>S^4COS J.„ + E(7«_4/Sf3Sin J-n) — . .. 
 = Gn — 2C^_ 2^02 + 26.^_4>S4 — ... 
 
 Also, ^m{A^+A^+...-{-An-i+An) 
 
 = ^m{A^+A^-\-...+An-i)(to^An 
 
 + cos(^i + J.2+...+J.„_i)sin^n 
 
 = (EC„_2^1-2C,_4^3 + 2C„_6^5-...)COS^„ 
 
 + {Gn-i-J.Gn-A+ ...)^inAn 
 = (2C„_2^iCOs ^n+ C„_isin ^„) 
 
 — (E(7«_4>Sf3Cos J.,i + 2(7„_3>Sf2sin J.„) 
 
 + (26\_6>Sf5Cos ^„ 4- 20„ _ s^Sf^sin J.„) — . . . 
 
 = 2C7,i_lOl — ZjGn _303 + 2Cyi_ 5^5 — 
 
 Thus, if the formulfe be true for {n — 1) angles, they are 
 
290 ADDITION FORMULAE EXTENDED. 
 
 also true for n angles. But we know that they are true 
 for two and three angles, hence they are true for any 
 number of angles. 
 
 Gov. 1, — Hence, it follows that 
 
 Dividing the numerator and denominator of this frac- 
 tion by C„, and writing ^T^ for "the sum of all possible 
 terms that can be obtained from n angles, each term being 
 the product of the tangents of r of the angles," we have 
 
 tanu,+^.+...+^„)= Yii:iysl!i:;: - 
 
 Got. 2. — In like manner, it may be shewn that 
 
 cosh (Uj -h 1^2 + • • • + '^«) = ^^^n + ^Ghn - 2^A'2 + 2C/l„ . Sh^^ + ... , 
 
 sinh(iti-|-U2+ . .. -f-tf'n) = I'Ghn-iShi + 'EChn-sShs 
 
 + 2GK-,8h+..., 
 
 tanhK+tt,+ ...+^.)= i + ^ThJ^Th,+ ... ' 
 
 where 'EGhrShn-r stands for " the sum of all possible 
 terms that can be obtained from n sectors, each term 
 being the product of the hyperbolic-cosines of r of the 
 sectors and the hyperbolic-sines of the remaining (n — r) 
 sectors " ; and XThr for " the sum of all possible terms 
 that can be obtained from n sectors, each term being the 
 product of the hyperbolic-tangents of r of the sectors." 
 
 194. To prove that 
 
 2"C0S JjCOS J.2 ... cos J.n = 2cOS(± J.j±^2- ... ±^n- 
 
 We know that 
 
 2 cos ^icos A^ = cos(i4i + J-g) +cos(^i — J.2), 
 and therefore that 
 
ADDITION FORMULAE EXTENDED. 291 
 
 22cOsJ.^COsJ.2 = COs(+^l + ^2) + ^<^s( + ^l~^^) 
 
 + C0S( — ^l + ^2) + C0s(-J.i — J.2), 
 
 or 22n2(cos^ ) = 222Cos( ± A^ ± ^2). 
 
 where 1X2(008 J.) is an abbreviation for "the product of 2 
 
 factors of the type cos J.," 
 
 and S22COs(± ^.^±^2) ^*^^ "the sum of 2^ terms, each 
 term being the cosine of one of the 2^ angles ( ±^^ ± A^!' 
 This formula may be shewn to be true for any number of 
 angles. 
 
 Assume that 
 
 2"-in,,_l(C0S^) = S2n-iC0S(±^l±^2±---±-^n-l). 
 
 Multiplying each side of the equation by 2 cos An, and 
 expressing each double product on the right side as the 
 sum of two cosines by aid of the formula 2 cos a cos /3 
 = cos(a + /3) + cos(a — /3), we get 
 
 2^nn(C0sJ.) = E2nC0s(±J.i±^2±---±^n-l±^«)- 
 
 Thus, if the formula be true for ('^ — 1) angles it is also 
 true for n angles. But we know that it is true for two 
 angles, hence it is true for any number of angles. 
 
 Got. — In like manner it may be shewn that 
 
 2*^nn(C0sh U) = S2nC0sh( ±Wj ± 1^2 ± . . . ±Un)- 
 
 195. Formulae for the Cosines and Sines of Multiple 
 Angles. — By art. 193, we have 
 
 cos(^i+^+...+^i)=c,,-sa^_2/Sf2+2C„-A-.... 
 
 Let each of the angles A^, A^, A^, ..., An be equal to 0, 
 then each term in 'ZCn-rSr becomes cos^"^0 sin*'^, and the 
 
 number of such terms is -^ \ <^ ^ — ^ or in)^, 
 
 \. L. 6 ...r ^ 
 
 2(7,,_ A-= Wrcos"-'*^ sin'-a 
 
 Hence, we get 
 
 cos 710 = cos^0 - ('njgcos^ - ^0 sin^O + ('Ji^cos^^ - ^Q sin^^ - ... (1 ) 
 
292 ADDITION FORMULAE EXTENDED. 
 
 So also, from the equation 
 
 we deduce the equation 
 
 sin7i0=('M)iCOS»*-iasin0-(7i)3COs«-30sin3e+ (2) 
 
 Cor. 1. — Dividing (2) by (1), we get 
 
 1 - {n\%d.n^e + {n)^t2in^e -...' 
 Cor. 2. — From art. 193, Cor. 2, it follows, in like man- 
 ner, that 
 
 cosh nu = cosh**i^ + (ti)2C0sh*^ - ^u sinh% 
 
 + ('?i)4Cosh" - *u sinh*tt + . . . , 
 sinh nu={n\cosh.'^-'^ii sinh u + (7i)3COsh**"^i6 sinh%+ ..., 
 ta.nhnu== Witanhu+W3tanh%+... 
 1 + ('^i)2tanh% + (rij^tanh^ + . . . 
 
 196. To prove that, when n is even, 
 2" - i cos'^e = cos nO + {n\QO^{n - 2)0 + (n)2Cos(n - 4)0 
 
 + ... + iWn. 
 
 2 
 
 and that, when n is odd, 
 
 2« - icos»0 = cos nO + (7i)iCos('M - 2)6 + ('m)2Cos('7i - 4)0 
 
 + ...+Wn-lCOS0. 
 
 2 
 
 By art. 194 we have 
 
 2"nn(cOS^) = S^^COS(±^l±^2±---±-4«)- 
 
 Let A^ = A2= ...=An = 6, then the left side becomes 
 
 2"cos"0. On the right side collect the 2" terms into 
 
 n 
 groups. Take as the (r + l)th group, where r not > ^, all 
 
 the terms in which the expression for the angle con- 
 tains r and only r plus signs, or r and only r minus signs. 
 Each term of this group will be equal to cos {n — 2r)0, and 
 the number of terms in the group will be 2{n)r, except in 
 
ADDITION FORMULAE EXTENDED. 293 
 
 the case for which n is even and r = ^, when the number of 
 terms will he {n)^ only, the half-group which contains - 
 
 2 ^ 
 
 and only ^ plus signs being also the half-group which con- 
 
 tains - and only ^ minus signs. 
 
 Thus,thefirstgroup = 2cos')i^,the second 2(t^)iC0s(7i — 2)0, 
 the third ^{n)^Q)^{n — '^)Q, and so on, the final group when 
 n is even being (7i)„, and when n is odd 2(?i)„_icos0. 
 
 Hence, dividing by 2, we have 
 2n-icos"0 = COS nO + (^)iCos(7i - 2)0 + ... + \{n)n. 
 
 2 
 
 or 2'^ - ^cos'^O = COS nQ + (71)^008(^1 - 2)0 + ... + (7i)h-iC0S 0, 
 
 2 
 according as n is even or odd. 
 
 Cor. — From art. 194, Cor., it follows, in like manner, 
 
 that 1^ - ^cosh"i6 = cosh nu -h {n)^Q^{n — 2)u -f- . . • , 
 
 the last term on the right side being 
 
 \{n)n or (7i),^^iC0sh u, 
 2 2 
 
 according as n is even or odd. 
 
 197. To express 2^"^sin*^0 as a series of cosines or sines 
 of multiples of 0. 
 By art. 196 we have 
 2/1 - icos^O = cos nO + (n\cos{n — 2)0 -f {n)2C0s(n — 4)0 -f . . . 
 
 Change into ., — 0, and reject from each angle the 
 
 multiple of a right angle. 
 
 There are four cases, according as n is of the form 4m, 
 
294 ADDITION FORMULAE EXTENDED. 
 
 when n is called an even even-number, since such num- 
 bers occupy the even places in the series 2, 4, 6, 8 ..., 
 of the form 4m + 1, an odd odd-number, 
 of the form 4m +2, an odd even-number, 
 or of the form 4m +3, an even odd-number. 
 
 Observing that cos( r'^-A ) is equal to cos A,ainA,-coaA, 
 
 or —sin A, according as r is even even, odd odd, odd even, 
 or even odd respectively, we have the following results — 
 
 n even even, 2^-^sin"0 
 
 = cos nO — (n\cofi(n — 2)0 + (n)2Cos(n — 4<)0 
 
 2 
 
 n odd odd, 2^-hm^O 
 
 = sin nO — {n)-^sm(n — 2)0 + (n)2sm{n — 4)0 
 -\-,..-\r{n)n-ismO, 
 
 2 
 
 n odd even, 2^-hm^O 
 
 = — cos nO + (n)jCos(n — 2)0 — {n)2C0s(n — 4)0 
 
 + ...-{-i{n)n, 
 
 2 
 
 n even odd, 2'^-hm^O 
 
 = — sin nO + (n)^sm(n — 2)0 — {n)2sin{n — 4)0 
 + . . . + ('^),^isin 0, 
 
 2 
 
 the last term being positive in every case. 
 
 The method of this article depends on the periodicity 
 of the circular functiono, and is not, at the present stage, 
 applicable to hyperbolic functions. 
 
 198. When n is not very large the series for 2""^cos*^0 
 and 2""^sin"0 may be conveniently obtained by aid of 
 
ADDITION FORMVLAE EXTENDED. 295 
 
 Pascal's Triangle, 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 2 
 
 1 
 
 
 
 
 
 
 8 
 
 3 
 
 1 
 
 
 
 
 
 4 
 
 6 
 
 4 
 
 1 
 
 
 
 
 5 
 
 10 
 
 10 
 
 5 
 
 1 
 
 
 
 6 
 
 15 
 
 20 
 
 15 
 
 6 
 
 1 
 
 
 7 
 
 21 
 
 35 
 
 35 
 
 21 
 
 7 1 
 
 
 8 
 
 28 
 
 oQ 
 
 70 
 
 56 
 
 28 8 
 
 etc. 
 Taking the coefficients from the Triangle, and observing 
 that a coefficient which occurs once only in a row is to 
 be halved, we get 
 cos — cos 0, 
 2cos2e = cos20+l, 
 4cos30 = cos30 + 3cos0, 
 8 cos*0 = cos 40 + 4 cos 20 + 3, 
 16 cos50 = cos 50 + 5 cos 30 + 10 cos 0, 
 32 cos60 = cos 60 + 6 cos 40 + 15 cos 20+10, 
 64 cos70 = cos 70 + 7 cos 50 + 21 cos 30 + 35 cos 0, 
 128 cos80 = cos 80 + 8 cos 60 + 28 cos 40 + 56 cos 20 + 35, 
 etc., 
 and 
 
 sin = sin 0, 
 2sin20=-cos20 + l, 
 ■ 4 sin30 == _ sin 30 + 3 sin 0, 
 
 8 sin40 = cos 40 - 4 cos 20 + 3, 
 16 sin50 = sin 50-5 sin 30+10 sin 0, 
 32sin60= _cos60 + 6cos40-15cos 20 + 10, 
 64 sin70 = -sin 70 + 7 sin 50 -21 sin 30 + 35 sin 0, 
 128 sin80 = cos 80- 8 cos 60 + 28 cos 40- 56 cos 20 + 35, 
 etc. 
 
296 SERIES OF PO WEliS 
 
 199. Example. — Express siii''^ cos^^ as a series of sines of multi- 
 ples of 6. 
 By art. 198, we have 
 
 16 sin5^ = sin 5^-5 sin 3^+10 sin 6. 
 Multiplying by 2 cos ^, we get 
 
 32 sin^^ cos $= 2 sin 5^ cos ^ - 5 . 2 sin 3^ cos 6^+ 10 . 2 sin 6 cos 6 
 =sin Q9 + sin Ad-b sin 4^-5 sin 2^-f 10 sin '2.9 
 =sin 6^-4 sin 4^+5 sin 19. 
 Again, multiplying by 2 cos 9., and proceeding as before, we get 
 64 sin^^ cos2^=sin 7^-3 sin 5^-fsin 3^ + 5 sin ^, 
 and, again, 
 
 128 sin6^cos3^=sin 8^-2 sin 6^-2 sin 4(9 + 6 sin 29. 
 
 § 2. Series of Powers of a Cosine or Sine. 
 
 200. From the series of art. 195, namely, 
 cos nO = cos^a - (7i)2Cos" - ^0 sin^O + (t^^cos" - *^sin*0- . . . , 
 and 
 
 we may, by substituting 1 — cos^O for sin^^, obtain series 
 for cos nO and sin nO/sin in powers of cos only. The 
 general form of the coefficients may be determined in 
 this manner, but the method adopted in arts. 201 and 202 
 is somewhat simpler, and depends only on the elementary 
 identities 
 
 sin('7i + 1)0 -\- sin(n — 1 )0 = 2 sin nO cos 6 
 and sin('M -\-l)0 — sm(n — 1)0 = 2 cos nO sin 6. 
 
 The series may be arranged either in descending or 
 ascending powers of cos 0, and we may also obtain similar 
 series in descending or ascending powers of sin 0. It will 
 
OF A COSINE OR SINE. 297 
 
 be seen hereafter that, except in the case of the series in 
 descending powers of cos 0, the form of the series will 
 vary according as n is even or odd. There will accord- 
 ingly hQ fourteen series of this type in all. In arts. 201 
 and 202, we investigate the two series in descending 
 powers of cos Q ; in arts. 203 and 204, we obtain, by 
 re-arrangement of the terms of the series in descending 
 powers of cos 6, four series in ascending powers of cos 0. 
 
 By changing into ^ — we may without difficulty 
 
 deduce four series in descending powers of sin 0, and 
 
 four in ascending powers of sin 0. 
 
 The group of series here considered is an important 
 
 and a natural one ; but the order in which the series are 
 
 taken, and the method of demonstration, are to a great 
 
 extent arbitrary. The series in descending powers of cos 
 
 are the simplest of the group, since their form is the 
 
 same for even as for odd values of n, and for this reason 
 
 they are here made the fundamental ones ; the expan- 
 
 sin nQ 
 sion for — ; — ^r- is taken before that for cos?i0 because the 
 sin t^ 
 
 coefficients of the terms of the equivalent series are simpler 
 
 for the former. 
 
 201. To prove that 
 
 ^^^ = (2cose)"-l-(7^-2X(2cos0)^-3 + O^-3)2(2cosef-5 
 
 -... + (-iy(n-r-l)r(2Gosey'-^'-'^+.... 
 Let Un = sin nO/s'm 0, v=2 cos 0. 
 
 From the identity 
 
 Bin{n + 1)0 + sin(7i — 1 )0 = 2 sin nO cos 0, 
 
 it follows that Un^l = UnV — Un-l (1) 
 
 and from the definition of ^(^ and U2 that 
 
298 
 
 SERIES OF POWERS 
 
 u. 
 
 1, 
 
 and u^ = v. 
 
 Hence, by the use of (1), we have in succession 
 
 etc. 
 
 The numerical va^ 
 following table : — 
 
 ue of the coefficients is shewn in the 
 
 Uc 
 
 u. 
 
 u. 
 
 Un 
 
 10 
 
 etc. 
 
 If these coefficients are read obliquely from left to right 
 downwards, we obtain the coefficients of a binomial series, 
 and this law of the coefficients is general, since the 
 process of formation of the successive numbers by aid 
 of (1) is the same as that employed in forming the 
 
OF A COSINE OR SINE. 299 
 
 binomial coefficients, the (r + l)th number in any oblique 
 line being the sum of the (r+l)th and rth numbers in 
 the preceding oblique line. 
 
 Now, the coefficient of the (r4-l)th term of Un is the 
 (> + l)th number of the {n — l)i\i row in the table, and 
 therefore the (r+l)th number of the {n — T — \)th. oblique 
 line, and consequently is equal to {n — r—\\. 
 
 Thus, we get 
 
 + (-iy('M~r-l),i;"-2^--i+..., 
 or 
 
 sm iiif) 
 
 -^—^ = (2cosa)" - ^-{n-2\(2coHey - ^ + (71-8)2(2 cos 0}" -'-,.. 
 sm (7 " 
 
 + (-l)^(7i-r- 1)^2 cos 0f-2'-i+... 
 Cor. — In like manner, it may be proved that 
 
 '^Sr^= (^ '"^^ u)-l-(.^~2X(2 cosh ^)"-3 
 + (^-8)2(2 cosh uy-^- ... 
 + (- 1/(72 -r- 1)^2 cosh i(,)"-2»-i+... 
 202. To prove that 
 
 2 cos nO = (2 cos Oy^ - ^^2 cos 0)" - 2 + -^^^^2 cos 0)" - ^- . . . 
 
 From the identity 
 
 sin(7i + 1)6- sin(7i - 1 )0 = 2 cos nO sin 6, 
 
 it follows that 
 
 ^ ^ sin (71 + 1)0 sin(7i — 1)0 
 
 2 cos Old = — \ ' ^ ^ — —^^, 
 
 sm 6 sm 
 
 and, by art. 201, we have, 
 
 !in|+l)^ = (2 cos Br-{n-lU2 cos 0)"- 
 + (n-2\(2 cos ey-*- ... + {-iy{n-r),{2 cos e)»-='-|- ■ ■■ 
 
300 SERIES OF POWERS 
 
 and 
 
 + (-l7(?i-r-lX_i(2cos0)"-2^+... 
 Now, (n — 1)1 + 1 = '^, 
 
 (7i-r-l)...(ii-2? - + l) 
 "^ 1.2.3...(r-l) 
 _ 7i(?i — r — l)(7i — 7' — 2) . . .(ti — 2r + 1) 
 ~ 1.2.3...r * 
 
 Hence, 
 
 2cosrie = (2cos0y^-7i(2cose)^-2+^^:^^?l^\2c^^ 
 
 J.2.3...r 
 Cor. — In like manner, from the identity 
 
 Sinn 26 smhu 
 
 and from art. 201, Cor., it follows that 
 
 2 cosh nu = {2 cosh u)" — ')i(2 cosh u)" - ^ 
 
 +!<!L^)(2coshu)-*-... 
 
 + (-l). "0^-^-lX^->-2)-(^-2>-+l) (2co3hu)»-^-+- 
 203. To prove that, when n is even, 
 
 = n cos ^—r^ — ^cos^^ H — ^^ r^ cos^0 — . . . 
 
 and that, when n is odd. 
 
OF A COSINE OR SINE. 301 
 
 ^-^^ -ST 
 
 By art. 201 we have 
 
 ^^ = (2cos^)^-i-(7i-2X(2cos0f-3. 
 
 that for 
 equal to 
 
 am (7 
 
 + (- lX(?i-r- 1)^2 COS 0)"-2»--H 
 When n is even, the last term of this series is 
 
 which 91 — 2r — 1 = 1, orr=^ — 1, and is therefore 
 1.2.3...g-l) 
 
 — -1 
 = ( — 1)2 71 cos 0; 
 
 the last term but one 
 
 , ,t- (^')i(i--)- 
 
 ,4 
 
 — — (2cos0)^ 
 
 1.2.3...g-2) 
 
 = -(_l)l-!^gzf)eos3e, 
 the last but two 
 
 = (-!> r (2cos0)= 
 
 1.2. 3. ..(1-3) 
 
 and so on. 
 
302 SERIES OF POWERS 
 
 Hence, when n is even, 
 sin nO 
 sin0 
 
 = (-1)2 j ncosO — ^ — ^-cos^6+— j^ -^cos^^-. . . k 
 
 and therefore 
 
 . .^.1+1 sin tie 
 ^ ^^ sin 
 
 ^ n(n^-2^) 3^ , 71(7^.2-22X712 -42) ,. 
 [3 [5 
 
 Next, let n be odd, then the last term of the series for 
 
 — ; — ^ in descending powers of 2 cos 6 is that for which 
 
 n — 1 
 71 — 2r— 1 = 0, or ?'= , and is therefore equal to 
 
 n—1 71—3 
 1.2.3...^ 
 
 (-1J 
 
 the last term but one 
 n + l 71-1 
 
 n-3 2 2 ■' «-l772__-|2 
 
 = (-1)"^-^ ^ ^(2cos0)2= -(-1)— !L-J:-cos2(9, 
 
 1.2.3...^ ^-^ 
 
 the last but two 
 
 71+3 71 + 1 „ 
 n-5 2 ' Q •••^ 
 
 = ( - 1)^ ^;r3T(2 cos e)* 
 
 1.2.3...^V^ 
 
 / ,,^-^H^'-82)(7l2-l2) 
 
 = (-1) ' 1.:^ 3.4 ^ Q"^> 
 hence, when 7i is odd. 
 
 and so on 
 
OF A COSINE OR SINE. 303 
 
 sin nO 
 sinO 
 
 = (_]) 2 |1_____COS20 + ^ j'^- - ^COS^0-...|, 
 
 and therefore 
 
 ^~^^ TirTe 
 
 = 1 r^-- cos^O + —ri — - cos*0 — . . . . 
 
 [Z [4 
 
 Cor. — In like manner it follows from art. 201, Cor., that 
 when n is even, 
 / |/|+i sinhjTiu 
 ^ sinhi6 
 
 , '71(7^2-22) , - , 7^(n2-22)(^2_42) 
 
 = n cosh It ^ — ^coshm -\ — ^ rj~ ^coshm - . . . 
 
 [3 [5 
 
 and that, when n is odd, 
 
 , _ .VlzI sinh nu 
 (_1) 2 . 
 
 smh u 
 
 ^l-^^C0sh2.+ (-^-^y-^^ 0sh%-....- 
 
 |2 [4 
 
 204. To 'prove that, when n is even, 
 
 ( - 1)2 cos -TlO = 1 - ,— COS20 H ^ , ^ COS^O - . . . , 
 
 and that, when n is odd, 
 
 n-l 
 ( — 1) 2 COS 110 
 
 . n{n^-l^) „. , n(n^-V)(n^-S^) ,. 
 = ncosO ^^-^^r ^ cos^O + ^ -^ '^cos^O - . . .. 
 
 [S_ |5 
 
 By art. 202, we have 
 2cos7i0 = (2cos0)^-~(2cos0)"-2 + ^^^^^:::^^ 
 
304 SERIES OF PO WERS 
 
 If n be even, the last term of this series is that for 
 which r = ^, and is therefore equal to 
 
 ^e-i)e-2)... 
 
 (-1)' ;; (2cos0)'>=(-i)2.2, 
 
 1.2.8...| 
 the last term but one 
 
 « ,™-2-l2-V-^ « n^ 
 
 1.2.3...g-l) L2 
 
 the last but two 
 
 , ..■,. '(i^')i(;-o-' ,. ,, 
 
 = ( - 1)22 . — ^-^ -^ COS^0, 
 
 and so on ; hence, when n is even, 
 
 2cos7i0 = (-l)^2.|l-^cos2e+^?^^^^^cos*0-...|, 
 
 and therefore 
 
 ( - l)^cos ne=l-^ cos^e + ^'^'^'^~ ^'^ cos^^ - . . . . 
 
 Next, let n be odd, then the last term of the series for 
 2 cos nO in descending powers of 2 cos is that for which 
 
 -, and is therefore equal to 
 
 2 
 
 n — 1 ti — 3 
 n. 
 
 9 
 
 (-l)V ? ^ (2cos0) = (-l)''2 .7i.2cose, 
 
 1.2.3...^- 
 
OF A COSINE OR SINE. 
 
 the last term but one 
 
 71 + 1 n — 1 , 
 
 1.2.3...^ 
 
 = -(-1)2 -A_^ ^.2 008^0, 
 
 the last but two 
 
 71 + 3 n + 1 
 
 = (-1)'^ t-T-(2cose)^ 
 
 1.2.3... -~- 
 
 , ^ «-Zi 7l(ri2 -12) (^2 _ 32) 
 
 = ( - 1) 2 — ^^ ^ i 2 cos^O, 
 
 and so on ; hence, when n is odd, 
 
 2 cos 710 
 
 = (-1) 2 2J71COS0 — ^-T^ — ^cos^O+-^ r^^ ^cos^0-... K 
 
 and therefore 
 
 n-l 
 ( — 1) 2 cos 710 
 
 . 7l(n2-12) ^^2_12)(^2_32) 
 = -^1008 ^— rs — -^008^0+-^^ -iP ^cos^O— ... 
 
 Cor. — In like manner from art. 202, Cor., it follow^s that 
 when 71 is even, 
 
 ( — 1)2 cosh Tiu = 1 — — cosh^it H ^- -^cosh% — . . . , 
 
 \2 [4 
 
 and that, when oi is odd, 
 
 M-l 
 
 ( — 1) 2 cosh Tilt, 
 
 ;= 71 cosh u ^— , rr ^cosh^^t + -^ r-^^ ^cosh ^u-,., 
 
 |3 \o 
 
 u 
 
306 SUMMATION OF SERIES. 
 
 By changing into f,— ^ we may derive from the 
 
 series of arts. 201-204 four series in descending powers 
 of 2 sin 0, and four series in ascending powers of sin 0, 
 but we cannot, at the present stage, apply this method to 
 the series of powers of the hyperbolic cosines. 
 
 § 3. Summation of Series. 
 
 205. To find ike sum of the cosines of a series of angles 
 in arithmetical 'progression. 
 
 Let a be the first angle of the series, /3 the common 
 difference, then the (r + l)th term of the series of cosines 
 is cos(a + r^). We have 
 
 2cos(a + r/3)sin| = sin{a + (r+i)/3}-sin{a + (r-i)/8}. 
 
 . Putting r = 0, 1, 2, . . . {n — 1) we get 
 
 2 cos a sin 'I = sinf a + ^ ) — sinf a — ^ j, 
 
 2 cos(a + /3)sin ^ = sinf a + -~ j — sinf a + ^\ 
 
 2cos(a + ^i^.)5)sin| = sin{a + (n-J)^}-sin{a + 0i-f)/3}. 
 
 Hence, by addition, 
 2sin^{cosa + cos(a + ^)+cos(a + 2^) + . . . +cos(a -\-n-l.^)} 
 
 = sin{a + (-^ — J)/5} - sinf a - ^ ) 
 
 = 2cos{a + (7i-l)f}sin^^, 
 
SUMMATION OF SERIES. 307 
 
 and therefore 
 
 cosa + cos(a + /3) + cos(a + 2/3) + ...+cos(a + 7i-l./3) 
 
 sin I 
 
 This result gives the following Rule : — 
 
 '' To find the sura of the cosines of a series of n angles 
 in an arithmetical progression whose common difference 
 is p, multiply the cosine of the average value of the 
 
 angles by the ratio sin -^ /sin ^." 
 
 The average value of the angle is readilj^ obtained by 
 taking half the sum of the first and last angles. 
 The result may be written in the form 
 
 S cos(a + 7^/3) = cos 1 a-\-{n — iy^\ .sin -7j-/sin^. 
 
 Cor. — In like manner it may be shewn that 
 
 2 cosh(a + 7'^) = cosh j a-{-{n — l)'^\ . sinh -^/sinb ^. 
 
 206. To find the sum of the sines of a series of angles 
 in arithnetical 'progression. 
 
 Let a be the first angle of the series, /3 the common 
 difference, then the (r + l)th term of the series of sines is 
 sin(a + rP). 
 
 We have 
 
 2sin(a + r/3)sin| = cos{a + (r-J)/3}-cos{a + (r + J)/3}. 
 Putting ?' = 0, 1, 2 ...(71 — 1), we get 
 
 2 sin a sin ^ = cosf a ~ 9 ) ~ cosf a + ^), 
 
308 SUMMATION OF SERIES. 
 
 2sio(a+/3)sin| = cos(a + f)-cos(a + ^), 
 
 2sin{a + (7i-l)^}sin| = cos{a+0i-f)/3}-cos{a+(7i-i)/3}. 
 Hence, by addition, 
 
 2sin|{sina+sin(a+/3)+sin(a+2;8)+...+sin(a+7i-l/3)} 
 
 = cos(^a - ^ j - cos {a + ('^?' - J)/?} 
 
 = 2 sin j a + {n — 1)^ [sin -^, 
 and therefore 
 
 sina + sin(a + ;8) + sin(a + 2^) + ... + sin{a + ('>r-l)^} 
 
 3in j a + (71 - Vf^ Uin -2^ 
 
 sm 
 
 This result gives the following Rule : — 
 
 " To find the sum of the sines of a series of n angles in 
 arithmetical progression whose common difference is /3, 
 multiply the sine of the average value of the angles by 
 
 the ratio sin -|^ /sin ^." 
 
 The result may be written in the form 
 
 2 sin(a + r/5) = sm j a + ('^ — l)^- f • sm -^/ sm ^. 
 Cor. — In like manner it may be shewn that 
 '"lr'sinh(a + r/3) = sinh|a4-('^-l)5| • sinh ^^-/sinh |. 
 
SUMMATION OF SERIES. 309 
 
 207. Many series may be summed by aid of the method 
 or results of the two preceding articles. 
 
 The following list of difference-forms will be found 
 useful : — 
 
 cot r^ — cot X = cosec x. 
 
 tan X — tan ^ = tan= sec x. 
 tan 36 - tan 9 = 2 sin sec Sa 
 
 tan-i(l + r.?Tl)-tan-Xl+^^i.r) = tan-L-^j^^2- 
 
 cot 33 — 2 cot 1x — tan x. 
 
 tan 2cc — 2 tan x — tan^cc tan 1x. 
 
 2 coth 2x — coth x = tanh x. 
 
 sin a 
 
 cotra — cot(7^+l)ct 
 
 sin7'asin(r+l)a* 
 
 sin^ 
 
 cos(a + r — l/3)cos(a + ^'/3) 
 
 tan(a + t/3) - tan(a + r - 1 /5) = 
 
 2sin| 
 cosec 2x — l cosec x = —. — ^i — 
 
 cosec X — cosec Sx = 2 cos 2x cosec Sx. 
 cosec^a:; — cosec^^a^ = 8 cos 2x cos^o) cosec^3fl3. 
 
 sin^O — 2 sin^j^ = 2 cos sin^^- 
 
 cof^2aj — I cot^ic = I tan'^a; — h. 
 
 2n 
 tSiii~hi(n-\-l) — tan"^(?i — lW = tan"^ , . , ^ — tt~^* 
 
 cosec^o; — 2 cosec-2ic = 2 cos 2x cosec22a;. 
 tan 8^ — 3 tan x = S sin^x sec Saj. 
 sec rO sec(?' + 1 )0 — sec(7' — 1 ) sec rO 
 
 = 2 sin sec(r- 1)0 tan rO sec(r ^- 1)0. 
 
310 SUMMATION OF SERIES. 
 
 The list may easily be extended by the observation or 
 invention of the reader. The discovery of the difference 
 form, by means of which a given series may be summed, 
 furnishes indirectly a valuable exercise in the manage- 
 ment of trigonometrical formulae. 
 
 208. Example 1. — Sum the series 
 
 tan x+i tan l+^L^tan 1 + ... +-L, tan g^. 
 
 "We have tan x=cotx-'2, cot 2;r, 
 
 1 , X \ ,x , 
 - tan - = - cot - - cot .j;, 
 2 2 2 2 
 
 lx.37l,.rl,.r 
 22^^^22=P'"'p-2'"*2' 
 
 1 -^ ^ 1 cot-A,, 
 
 271-2' 
 
 gSTitan 2;^^ = ^^^cot-^j-^^2^ 
 
 Sn=-^ cot -^j - 2 cot 2x. 
 
 Example 2. — Sum the series 
 
 tan^a tan 2a + i tan^Sa tan 4a + ... + J^, tan22"-^a tan 2"a. 
 •6 2**-! 
 
 We have tan^a tan 2a = tan 2a - 2 tan a, 
 
 \ tan^2a tan 4a = ^ tan 4a - tan 2a, 
 
 — tan24a tan 8a = — tan 8a - ^ tan 4a. 
 
 -1-. tau^2'»-^a tan 2"a = -^ , tan 2"a - JL tan 2"-^a. 
 2n-i 2"-^ 2"~2 
 
 Sn = o^zY tan 2"a - 2 tan a. 
 
 Example 3. — Find the sum of n terms of the series 
 tan-.2+tan-.^^ + tan-'^g + tan-'j^^ + ... 
 
SUMMATION OF SERIES. 
 
 311 
 
 We have tan~^- 
 
 2r 
 
 tan 
 
 .^ r(r+l) -(/'-!> 
 
 = tan- V(r + 1 ) - tan-\r - l)r. 
 Putting r = l, 2, 3 ... w we get 
 
 tan-^2 = tan-U.2-0, 
 
 tan-i ^— =tan-i2 . 3 - tan-U . 2, 
 
 1+3.4 
 
 tan" 
 
 27i 
 
 :tan-^7i(7i + 1) - tan-^^t - l)n, 
 
 >S'„=tan-^?i(?i + l). 
 Example 4. — Sum to n terras the series 
 
 cos*^ + cos*2(9+ 008*36'+ etc. 
 We have 8 cos*.r = cos 4.^; + 4 cos 2^ + 3, 
 
 . • . s'^s' cosV^ = Ycos 4r (9 + ^1. cos 2r (9 + 3?^ 
 
 r=l r=\ r=l 
 
 = cos2(?i+l)^sin2?i^/sin2^ + 4cos(?i+l)^sin7i^/sin^+3n, 
 2 cosV^= 1 cos 2(w + l)^sin 2n(9/sin 2(9 
 
 + ^ cos(?i + 1 ) ^ sin ?i^/sin B + ^n. 
 
 Examples. — Let Aq, A^, A2...An-\ be n points symmetrically 
 ranged on the circumference of a 
 circle whose centre is ; then shall 
 the sum of the projections of OA^^ 
 OAi...OAn-i on any line OX be 
 equal to zero. 
 
 Let LXOAQ=a, 
 
 lAqOAi = lA^OA^ = etc. = /?, 
 then 7i/3 = 277. 
 
 Also, if r be the radius of the circle, 
 the sum of the projections of OA^, 
 OA^...OAn-i on OX 
 
 = r{cos a + cos(a + 13) + cos((x + 2/3) + 
 
 = rcos-(a + (w-l)^}sin -^/sin^ 
 ^ 2 -* 2 / 12 
 
 = 0, since sin — - = sin tt = 0. 
 
 ^^«.z 
 
 + cos(a + 7i-l . /S)} 
 
312 
 
 GONVEROENCY AND 
 
 Example 6. — If a regular polygon of n sides be iuacribed in 
 a circle, and if I be the length of the chord joining any fixed point 
 on the circle to one of the angular 
 points of the polygon, then 
 
 where a is the radius of the circle, 
 and m any positive integer less than n. 
 
 Let be the fixed point, J ,. a vertex 
 of the poly on, LOCAQ=a. 
 
 Then 
 27r> 
 
 0Ar=2a Bin Ua+r.^\ 
 
 Now 
 
 ( - 1)'"(2 sin </)f~ = 2 cos 2m</) - (2m)i2cos(2?n -2)cf>+ ...+(- l)"'{2m),„ 
 
 r=o *- \ n J 
 
 -(2m)i2cos(m-l)(a + r?^)+... + (-l)'«(2mV}. 
 But iim<7i the sums of the cosines vanish by art. 205, 
 
 = na — -^' 
 Qrny 
 
 § 4. Convergency and Continuity of Series. 
 
 209. In dealing with an infinite series of terms we 
 inquire, in the first instance, whether the series is con- 
 vergent, i.e., whether, however great the number of terras 
 
CONTINUITY OF SERIES. 313 
 
 may be, their sum is a finite quantity tending to some 
 fixed limit; and, in the second, whether the series is 
 continuous, i.e., whether an indefinitely small change in 
 the variable involved in the series produces an in- 
 definitely small change in the limit to which the sum 
 approaches as the number of terms is continually in- 
 creased. 
 
 The following classification and terminology will be 
 adopted. 
 
 Infinite series are either convergent, oscillating, or 
 divergent. 
 
 A convergent series is either absolutely convergent, or 
 semi-convergent. 
 
 A convergent series has already been defined as one 
 the sum of whose terras is a finite quantity tending to 
 some fixed limit, however great the number of terms may 
 be. Tf the terms of such a series are all positive, or if 
 the series remain convergent when the terms are made 
 positive without change in their numerical value, the 
 series is said to be absolutely convergent. 
 
 Thus ^ + iT + ^ + To + • • • ^or all values of x, 
 Li Lt l£ 
 and 1+iccos O + o^^cos 2O + i^^cos30+... when x<\ 
 are absolutely convergent series. 
 
 If a series is convergent, but does not remain conver- 
 gent when all its terms are made positive, it is said to be 
 semi-convergent. 
 
 Thus, 1-I + 5-1 + - 
 
 and cos-^ + ^cos -^-hgCos -g--F... 
 
 are semi-convergent series. 
 
314 CONVERGENCY AND 
 
 If the sum of a series never exceeds a certain finite 
 quantity, however great the number of terms may be, but 
 at the same time the sum does not tend to a fixed limit, 
 the series is said to oscillate. 
 
 Thus, 1-1 + 1-1 + 1-..., 
 
 and cos -^j + cos ., + cos -^ + . . . 
 
 O tJ o 
 
 are oscillating series. 
 
 If the sum of a series increases without limit with the 
 number of terms, the series is said to be divergent. 
 
 Thus, 1 + 1+I+1 + ... 
 
 is a divergent series. 
 
 For the fundamental Theorems on Convergency and 
 Divergency the reader is referred to Todhunter's Algehxi, 
 chap. XL., or C. Smith's Treatise on Algebra, chaps. 
 XXI., XXV. 
 
 With respect to continuity, it should be observed that 
 the continuity of an infinite series is not a necessary con- 
 sequence of the continuity of its several terms, for the 
 sum of an infinite number of indefinitely small changes in 
 the terms may be a finite quantity. 
 
 210. If c&o, ttj, a^...he a series of constantly decreasing 
 'positive quantities, and if Lt an = 0, and /3 be not equal 
 
 to zero or any multiple of ^ir, then will the series 
 
 a^cos a + aiCos(a + /8) + a^C0B{a + 2^) + . . , 
 be convergent. 
 
 Let 8n = a^cos a + a^QO^{a + /5) + . . . + anCOs(a + n^), 
 
CONTINUITY OF SERIES. 315 
 
 then 2^m^Sn = aA sin fa + ^j-sin («-§)[ 
 -\-aA sin (^a + ^j-sin (a + |)| 
 
 + (Xn{sin(a + 7i + i . p)-^m{a-\-n-\ . /3)}. 
 Therefore 
 
 2 sin ^AS^ji+ctoSin (a - ? ) — ajisin(a + '?v+i • jS) 
 
 ^-{an-i-an)^m{a-n-l.^) (1) 
 
 Now ((Xq - a^ + («! - ag) + . . . + (a„_ i - an) ="%- o^n, and, 
 by hypothesis, Lt an = 0, 
 
 n=oo 
 
 .-. the series (a^ — a^) + (ttj - Wg) + • • • is convergent. 
 
 Also, since ao>ai>a2..., its terms are all positive. 
 Hence, the series 
 
 {aQ-aj)sm\^a + ^J + {a^-a2)8m{a + -^j + ... 
 
 whose terms are numerically less than the corresponding 
 
 terms of (a^ — a^) + ((Xj - cig) + . . . is also convergent. 
 
 Hence, observing that the limit of ansin{a + n + ^. ^) 
 
 is zero, and that sin ^ is not equal to zero, we conclude 
 
 from (1) that Sn tends to a finite limit as n is indefinitely 
 increased, and therefore the series 
 
 a^cos a + aiCos(a + P) + a.2Cos{a + 2^) + . . . ad inf. 
 is convergent. 
 
 Co7\ — Similarly, it may be shewn that if a^, a^, ofo . . . be a 
 series of constantly decreasing positive quantities, and if 
 
316 CONVERGENCY AND 
 
 Lt. a„ = 0, and /3 be not equal to zero or any multiple of 
 
 'Z-TT, then the series 
 
 a^siu a + aisin(a + ^) + a2^in{a + 2^) + . . . ad inf. 
 is convergent. 
 
 211. // the series aQ+a^x + a^^+ ...ad inf. he abso- 
 lutely convergent for all values of x not greater than 
 some fixed quantity r, then for all values of x less than r 
 the series will he a continuous function of x. 
 
 First, suppose x and each of the coefficients a^, a^, 
 a^... positive. 
 
 Let ccj, x^^ be two adjacent values of x less than r, of 
 which a?! is the greater, 
 let ^j = a^ + a^x^ -f a^fc^ + ...ad inf. 
 
 and ^2 = %+ a-^x^ + a^x^ + ...ad inf., 
 
 then we have to prove that, when x^ — X2_ is indefinitely 
 small, so alscJ is 8^ — 82- 
 
 By subtraction, we get 
 8^-S2 = a^{x^-X2)-\-a2(x^^-X2^') + ...+an(x^''-X2'')+... 
 
 Now -1^^2"^^«-l_,_^n-2^^^n-3^^2_^_^ _|.^^«-l^ 
 
 1 2 
 
 and therefore, since a?i>a?2> we have 
 
 ■a?. 
 
 <vx^^~'^ and >iia;2*^"^ 
 
 Hence, S^-'82< (x^ — ccg) (a^ + 2c/2a5i + SagCt^^ + . . . 
 
 and > (iCj — a?2)(^i + 2a2a'2 + ^a^x^^ + . . . 
 
 Now, the ratio of the (n + iy^ to the n^^' term of the 
 series a^ + 2a^ + Sa^x^ + . . . + nanX^ ~ ^ + 
 _ n-{-l an+iX 
 ~ n an 
 
 r l\an+ix^ 
 \ n) an 
 
CONTINUITY OF SERIES. 317 
 
 and, by sufficiently increasing n, this ratio can be made 
 as nearly equal to ""'"^ as we please, and therefore, if x 
 
 be less than r, this ratio can be made less than _^!±2_ ^ g^ 
 
 an 
 
 less than the test ratio of a convergent series ; therefore 
 
 the series a-^-{-2a^x-\-^a^x^-\- ... is convergent when x<t. 
 
 Hen ce, % + 2a^x^ + ^a^x-f + ...ad inf. 
 
 and a^ + ^a^x^ + ?ta^x.^ -\- ...ad inf. 
 
 are finite quantities. 
 
 Therefore, when x^ — Xc^ is indefinitely small, 8-^ — Sc^ lies 
 between limits which are also indefinitely small, and, 
 consequently, 8-^ — 82 is indefinitely small, hence the 
 series a^ + a^cc + ctgCt;^ + . . . ad inf. is a continuous function 
 of X for all values of x less than 7\ 
 
 Secondly, if some of the terms of the series be positive 
 and some negative, we can arrange all the terms of the 
 same sign into a group, and apply the theorem just 
 proved to each of the groups separately. Then, since the 
 change in each group is indefinitely small when that in x 
 is indefinitely small, the algebraical sum of the two 
 changes will aiso be an indefinite!}'" small quantity. 
 Hence, the theorem holds for negative as well as for 
 positive values of %, ^g' ^3 • • • ^^^ ^• 
 
 212. The argument of art. 211 is not valid when x^^Vy 
 for the series aj + Sctgr+Sagr^H-... may be divergent. 
 For example, let r=l, and let the series «! + ag + %+••• 
 
 be J2 + 22 + 32+--- > 
 
 then a^ + 2a2r + Sa^v^ + . . . becomes t + :^ + « + • • •. 
 
318 CONVEROENCY AND 
 
 and this is divergent. Thus, the reasoning of art. 211 
 does not sliew that tlie limit of the series 
 
 when X increases up to unity, is the series 
 
 The theorem of the following article may be applied in 
 such limiting cases. 
 
 213. If the series a^-^- a^ + a^-\- . . . ad inf. he convergent, 
 and if x be less than unity, then the limit of 
 
 aQ + ajX + dgic^ + . . . ad inf. 
 as X approaches the value unity will be equal to 
 aQ-{-a^-\-a.2+...-ad inf. 
 Let s = aQ+a^+a2+... ad inf., 
 
 S = aQ+a-^x + ^2^^ + • • • ^^ '^V-» 
 and let x = l—h, where h is a. small positive quantity, 
 then we have to shew that 
 
 Lt.{s^S) = 0. 
 
 h=0 
 
 Let 8 = ao+^i+... + an+?'n, 
 
 S=aQ-\-a^x+...+anX''-\-Rn, 
 then s — S 
 
 = a^(l-x) + a^{l-x^) + ...-han{l-x'') + rn-Rn 
 = (\-x:){a^+a2{l+x)+aQ{l-hx+x'^)+.. +an{l-\-x-\-...+x'^-'^)] 
 
 +r„-R„ 
 = h{a^-\-a2+a^+...+a„ 
 +x(a^ + a^-\-...+a„) 
 
 h(u^ + u^ + u^-{-...-\-u„) + r„-B„, 
 
CONTINUITY OF SERIES. 319 
 
 where Ur = x^' ''^{ar + ar+i + . . . + a„), 
 
 hence, if U be the average value of the quantities 
 
 we obtain the result 
 
 s — S=hn U+ Tn — Rn- 
 Now, in consequence of the convergency of the series 
 
 C6o + ai + «2+--- 
 
 and of the limitation of the value of x to numbers :t- 1, 
 it follows that each of the quantities denoted by Ur is 
 hnite (or indefinitely small) however great n may be ; 
 and therefore that U the average value is finite (or in- 
 definitely small) ; also as n increases indefinitely, Vn and 
 Rn diminish without limit. 
 
 Let h = ~j, then s — >Sf = — [-Tn — Rn- 
 
 Now let h diminish without limit, and n increase without 
 limit, then Lt{s-8) = 0. 
 
 h=0 
 
 Cor. — If the limiting value of x for which the series is 
 convergent be R, where R is any fixed number, the limit 
 o^ ctQ + a^x + a^pc^ -}-... as x increases up to R will be 
 aQ+a^R + a^R^ +.... 
 
 For if we put bn for cinR"', and p for x/R, we may write 
 the series in the forms 
 
 and 60+61 + fe2+---, 
 
 and apply the theorem of the present article. 
 
 214. The argument of the preceding article depends 
 on the conditions that a^ + ag+.-. + ttn is finite for every 
 value of n, and that Vn and R^ vanish for any indefinitely 
 
320 CONVERGENCY AND CONTINUITY, ETC. 
 
 great value oi n. In Example 1, all these conditions are 
 satisfied ; in Example 2, some of them only. 
 
 Example 1. — When x increases up to the value 1, the linjit of 
 
 the series -^ - „ + a" ~ • • • ^^ infinitum = 1-- + -^ — ... ad inf. 
 2 3 2 3 
 
 Here ai + a2 + ... + a„ is less than 1 and greater than \ whatever 
 
 n may be ; also Tn is numerically < -, and therefore a foi'tiori 
 
 n-{- 1 
 
 Rn is numerically < ; hence, the conditions of art. 213 are 
 
 satisfied, and the proposition is true. 
 
 Example 2. — Consider the series 
 
 \-x-\-x^-x^+... ad inf. (1) 
 
 in relation to the series 
 
 1-1 + 1-1 + .. . adinf (2) 
 
 The series (1) is convergent for any value of x, if x<\, and the 
 limit to which it converges is - — — ; hence, as x approaches the 
 
 value unity, the series (1) approaches the value ^. 
 
 The series (2) oscillates, its sum being or 1 according as the 
 number of terms is even or odd. 
 
 The equation s- S=hn U-\- ?•„ - Rn 
 
 still holds; a-^-^-a^-V ...■\-an is always finite, and therefore C^ is 
 
 finite ; hence, if h = -^., the term hn U becomes indefinitely 
 
 small, when h becomes indefinitely small. But r„ does not vanish, 
 
 nor does Rn vanish, for, with the assigned relation h= -^, the 
 
 {n + iy-^ term of (1) is still finite, since 
 
 Thus, we are unable to infer that 
 
 Lt.(s-S)=0. 
 
 h=0 
 
INFINITE SERIES FOR THE COSINES, ETC. 321 
 
 5. Infinite Series for the Cosines and Sines of x in 
 ascending powers of x. 
 
 215. To prove that 
 
 X ■, ^ . / ^x^^' 
 
 2r 
 
 cos« = l-|+^-... + (-l)'j2^+(-l)'-+'iJ, 
 
 where 0<R< 
 
 2r+2 
 
 If n be any even integer we have, by art. 1 95, 
 cos nO = cos«e - --^^^^cos^- 20 sin^O 
 
 + ... + (_ l)'^^|^^^|^^cos--0 sin-e 
 
 + ••• + 1 ^^i.2...(m-l)m^ '^' 
 the number of terms in the series being o + 1- 
 Hence, cos n0 = cos«0{l _-«-l^-^)(ta^y 
 
 , , . - ,, 'yi0('^0 - 0). . . (ti0 - 2^^^! . 0) ^ tan V>- 
 + ...+(-1) l.2.3...2r \ e J 
 
 + ... to f^ + lj terms h. 
 
 In this equation, let nO remain constant and equal to 
 X, and let n be indefinitely increased, and consequently 
 indefinitely diminished ; then, since 
 
 Lt. cos«0=il(cos-Y = l, (art. 190), 
 
 we have, by art. 188, 
 
322 INFINITE SERIES FOR THE 
 
 ,,/' a;(aj-e)/tan0\2 
 
 -h...i-^ i) i.2.3...2r \ e ) 
 
 + ... to f^ + lj terms K 
 
 Let kr denote the absolute value of the ratio of the 
 (7* + l)th to the rth term of this series, 
 
 then k r; (^'-^^--^ ■ e){x-iT-l . eV tan fly 
 then k,-Lt (2r-l)2r V d /* ' 
 
 Hence, for &i\ finite values of r, kr= „ ._i\9 » *id, 
 for values of r such that - is finite, ^r<7o tto"' until 
 
 at the end of the series, when ''"=-^. 
 
 7 _ 20.6 /tan0\ 
 
 n ""^^^'"'^^(2r-l)2r' 
 
 ~2' 
 
 2 2a;2 
 
 {n-'l)n\ 6 J {n-l)n^ 
 It follows that the terms of the series for cos x increase 
 in absolute value with r so long as {2r — l)2T<x^, and 
 that, from and after the greatest term, each term is less 
 than the preceding term, and that the terms ultimately 
 vanish. Also, the terms alternate in sign. Hence, we 
 have for all finite values of r such that (2r—l)2r>x\ 
 
 cos.=i-|+|-...+(-i)^+(-ir'ij, 
 
 where R = a. series of diminishing terms, alternately 
 positive and negative, and therefore R is numerically less 
 than the first of these terras, viz.. 
 
 x(x-e)...(x-2r+l. 0) /tan e\2^+2 
 1.2.3...(2r+2) \ J ' 
 
COSINES AND SINES. 323 
 
 and therefore, since 
 
 we have, a fortiori, R < r-r -. 
 
 ' -' |2rH-2 
 
 Cor. — In like manner, it may be shewn from the for- 
 mula of art. 195, Cor. 2, that 
 
 x^ x^ x^^ 
 
 where i2 = a series of diminishing positive terms, of which 
 
 the first is less than j^ — —^ and the ratio of each to the 
 
 |2r + 2 
 
 and.-. . -^< [-2^q:2 /r'"(2r + l)(2r+2y' 
 
 ^.2r+2 
 
 or R< 
 
 |2r{(2r+l)(2r+2)-a;2}- 
 216. To prove that 
 
 /y.2r+l 
 
 where 0<R< 
 
 [2r-hl 
 If n be any even integer, we have, by art. 195, 
 
 sin 710 = 71 cos"-iO sin 0- ^^^7i^^o^^^ cos^"^^ sin30+ .., 
 
 +(_ i)r-i<^zi%:%r:!r+?)cos'.-^+i0 sin2-'e+ , 
 
 1. , A , o . . . i^r -^ i^ 
 
 . , ^.Vl-1 n(n — l)...S .2 /> • „ i/^ 
 
 +<- ^>^ i.2:..(».-2)(^-ir ^ '"^ ^ ^' 
 
 the number of terms in the series being -^. 
 
324 INFINITE SERIES FOR THE 
 
 Hence, 
 
 -20)/tan0y 
 
 smn0 = cos-0|n0.-g 1,2.3 \-e-) +•• 
 
 ■*"^ ^ 1.2.3...(2r-l) V e / 
 
 + . . . to ^ terms K 
 
 Let 710 = Xy and let a; remain constant while n is 
 indefinitely increased, and therefore indefinitely di- 
 minished ; then, since 
 
 Lt cos'^0= Lt. (cos ^Y= 1 (art. 190), 
 
 we have, by art. 188, 
 
 r^ ( tane a;(a;-0)(£C-20)/tan0V , 
 
 sin a? 
 
 a;(a3-0)...(a;-2r-2.0 )/tan0Y^-i 
 "^^ ^ 1.2.3... (2r-l)" V / 
 
 + . . . to ^ terms !-. 
 
 Let kr denote the absolute value of the ratio of the 
 rth to the (r — l)th term of this series, then 
 
 , _ , (g; - 2r - 3 . 0)(a; - 2r ~ 2 . 0) / tan Q V 
 '^^-r. (2r-2)(2r-l) \ J' 
 
 and therefore, for all finite values of r, kr= 
 
 (2r-2)(2r-l) 
 
 and, for values of r such that - is finite, A;y<- 
 
 Sl 
 
 71 ' ^ (2r-2)(2r-l)' 
 
 when ^=9^, 
 
 , 20.20 / tan^ Y^ 3.2.a;^ 
 
 '^'~(ri-2)(n-l)\ ) {n-2){n-\)v? 
 
 7h 
 
 until at the end of the series when ^= «, 
 
COSINES AND SINES. 325 
 
 It follows that the terms of the series for sin x increase 
 in absolute value with r so long as (2r— 2)(2r— l)<a;^ 
 and that, from and after the greatest term, each term is 
 less than the preceding term, and that the terms ulti- 
 mately vanish. Also, the terms alternate in sign. 
 Hence, we have for all finite values of r such that 
 (2r-2)(2r-l)>i:c2 
 
 sin^ = a)-|+... + (-iri^^+(-iyE, 
 
 where jR = a series of diminishing terms, alternately posi- 
 tive and negative, and therefore R is numerically less 
 than the first of these terms, 
 
 and therefore, since Lt ( -^— ) =1, 
 
 we have, a fortiori, R < .^ -. • 
 
 Cor. — In like manner, it may be shewn from the 
 formula of art. 195, Cor. 2, that 
 
 0? /p2r-l 
 
 sinha; = a; + ,-^+... + ,- =-|-jK 
 
 (3 \iT-\ 
 
 where R < 
 
 L2r-l(2r.2r+l-a;2) 
 
 /y%Z /yi4 /yiD 
 
 217. The series l±^+^±^+... 
 x^ . x^ , x^ 
 
 and ^±_+_±_+... 
 
 are absolutely convergent for all values of x, and there- 
 fore, by art. 211, they are continuous functions of x for 
 all values of x. 
 
326 INFINITE SERIES FOR THE 
 
 It will be observed that the series for cos x and cosh x 
 contain even powers only of x, a result in agreement with 
 the known theorem that cos a; and cosh a; are even 
 functions of x. Similarly we observe that the series for 
 sin X and sinh x contain odd powers only of x. 
 
 The series for cos x and sin x are equal to these func- 
 tions for all values of x, and therefore the series must be 
 periodic, i.e., they must converge to the same limit when 
 X has values differing by a multiple of 27r, a result that 
 may be verified by actual computation. 
 
 Thus, to take a simple case, let x have the values 
 
 ^ and -y^, then we must have 
 
 1 
 
 (fo)7(i+(fo)yii--'^-/- 
 
 ^-r^^k^-mk-.'^^i^i 
 
 MO 
 
 Working to two places of decimals we get 
 first series =l-05 + -00- ... =-95, 
 
 and second series 
 = 1 - 21-76 + 78-93 - 114-52 + 89-01 
 
 -43-05 + 14-19-3-39 + -62--09 + -01--00 + 
 = 183-76 -182-81 =95. 
 
 218. To prove that cosh a; = J (e^^ + e " *) 
 and that sinh x = ^{^—e-^). 
 
 By arts. 215, 216, we have 
 
 x^ aj* 
 cosh a; = 1 +777 + r-r + . . . , 
 
 [2 [4 . 
 
 and sinh a; = a;+,— + .—+..., 
 
 [3 [5 
 
COSINES JiND SINES. 327 
 
 cosh oj + sinh a; = 1 +a;+— + j-^ + . . . 
 
 and coshic — sinncc = l— cc+r^ — ro + --- 
 
 Hence, cosha; = J(e*+e-«), 
 
 and sinhaj = |(e^— e-^^). 
 
 It is to be observed that in the exponential values of 
 cosh a? and sinho; the single arithmetical value of e* is 
 always to be taken ; thus, if cc = J, then e* = + s/^i or if 
 
 a;=-i then e^ = ^^^ 
 
 219. To 'prove that, if Q — gd u, then will 
 
 0' 
 
 . = logtang+|) 
 
 By art. 21 8, e^ = cosh u + sinh u. 
 
 But cosh u = sec 6, and sinh u = tan 6 ; 
 
 ,, /I , X /I 1 + sinO 
 
 e« = secO + tan = —^ — ^r— 
 cos 6 
 
 = . = *^^4+2> 
 cos « — sin ^ 
 
 u = logtan(|+|). 
 
 220. The hyperbolic cosine and sine might have 
 been defined by the equations cosh aj = J(e*+e"*) and 
 sinh a; = J(e* — e~^) or the equivalent series. 
 
 From these definitions we at once obtain the results 
 cosh!53 + sinh£c = e* and cosher — sinh cc = e~*, 
 hence, by multiplication, we have cosh^aj — sinh2aj=]. 
 
828 INFINITE SERIES FOR THE 
 
 Again, cosh(a;+2/) = K«''"^^+6"''"^) = JC^'"- e^+e"*-^'^) 
 
 = cosh X cosh y + sinh x sinh y. 
 
 Similarly, the other formulae relating to the hyperbolic 
 functions may be established. Thus, we obtain a purely 
 algebraical treatment of the hyperbolic functions. 
 
 It will further appear in Part III. that the circular func- 
 tions might be defined and their properties investigated 
 in a similar manner, without any reference to geometry. 
 
 221. By aid of the exponential values of coshoj and 
 sinh X, series involving these functions may frequently be 
 reduced to known algebraical forms. 
 
 Example 1. — To prove that 
 
 cosh a: + cosh(:r +y) + cosh(a;+ 2y) + . . . to w terms 
 
 =co8h(^+^y ) sinh ^/sinh |. (See art. 205.) 
 
 We have cosh(.2: + r^/) = |(e*+'-^ + e"*"*^), 
 
 's" cosh(a; + n^)=^*^ 2~V+'^ + ^ '^S'e-*-'^. 
 
 r=0 r=0 r=0 
 
 Hence, hy the formula for the sum of n terms of a geometrical 
 progression, 
 
 = coshf ;r + ^^^^— y jsinh ^ /sinh-|. 
 
COSINES AND SINES. 329 
 
 Example 2. — Find the sum of the series 
 
 ^inh u + n sinh '■lu -h ~ ^ sinh 3w + . . . to (?i + 1 ) terms. 
 1 . ^ 
 
 Let AS'=the sura of the series, 
 then 2AS'=e"+?ie2« + ^^^^^)e3M4., , to {n+l) terms 
 
 _|e-«+ne-2«+'^^^"-^V3«+... to (w + 1) termsj, 
 
 and therefore, by the Binomial Theorem, 
 
 2>S = e"( 1+ e")" - e-"(l + e-"r 
 
 = 2sinh^|+lV. (2cosh|V. 
 aS'= 2"cosh"| sinh^l + 1 V 
 
 Examples XXIII. 
 
 1. Find the sum of n terms of the series 
 
 cos + cos 20 + cos 30+ .... 
 
 2. Sum the series 
 
 siu a + sin 3a + sin 5a + . . . to n terms. 
 
 3. Prove that 
 
 TT , 37r , Stt , Ttt , Qtt 1 
 cos Y^ + cos ^ + cos jY + cos Yi + cos YY = 2' 
 
 and that 
 
 27r , 47r , Gtt , Stt , lOx 1 
 
 cos — + C0S jy + cosyy + cosyy + cos-y^= -^• 
 
 4. Sum the series 
 
 008(71— 1)0 + cos('M — 2)0+ cos('M — 3)0+... to 271 terms. 
 
 5. Sum the series 
 
 cos a — cos(a — /3) + cos (a — 2/3) — ...+( — l)"'Cos(a — n^). 
 
330 SERIES. 
 
 6. Sum the series 
 
 sin Q cos 30 + sin 20 cos 60 + sin 40 cos 1 20 + . . . to n terms. 
 
 7. Find the sum of the series 
 
 cosec0+cosec20+cosec40+... to n terms. 
 
 8. Sum the series 
 
 tan^ + 2tan2J[ + 22tan2M + ... + 2'^-itan2«-i^. 
 
 9. Find the sum of the series 
 
 tan0sec20+tan20sec220+... + tan2«-i0sec2«0. 
 
 10. Shew that 
 
 3«sin |, - sin = 4|sin3|+ 3 sin3|+ . . . + 3*^- isin^l^j-. 
 
 11. Sum the series 
 
 8in20+sin2204-sin230+...+ to n terms. 
 
 12. Shew that 
 
 cos*g + cos*-g- + cos*-g + cos*-g- = 2- 
 
 13. Expand cos^0sin20 in a series of cosines of multiples 
 
 of0. 
 
 14. Prove that when m is an odd integer 
 
 r . m2-l . „ , (m2-l)(m2-9) . . "I 
 8minx=m\ sma;- ^ sm^fl;4- . 9. i a, k sm^a;-... . 
 
 15. Shew that when 71 is a positive integer 
 
 ^ — 172"*^''^+ 1.2.3.4 *^"^"--:^^^- 
 
 16. Find all the vahies of determined by the equation 
 
 sin + sin 30+sin 50+... +sin(27?,-l)0 
 = cos + cos 30 + cos 50 + ... + cos(2'?i - 1)0. 
 
 17. Sum the series 
 
 sin0 , 2 sin 20 , 22sin 2^0 , 
 
 + ^ S7^ — T + ^ S7?i — T + ... to 71; terms. 
 
 2cos0~l 2cos20-l ' 2cos 2^0-1 
 
SERIES. 881 
 
 18. Shew that 
 
 cosO — cosf + 3-) + cosf 0+-^j — cosf^+n?) 
 
 +cos(e+^)-cos(0+|^) + cos(e+^) = O. 
 
 Interpret the equation geometrically. (See Ex. 39.) 
 
 19. .AO is a diameter of a circle of radius unity, AP^ any 
 
 arc of the circle. If arc J.P„ = ti.arc AP^, and 
 chords OPi, OP«_i, OPn, OPn+i are drawn, shew 
 that the formula 
 
 2cos(7i+l)0 = 2cos7i0.2cos0-2cos(n-l)0 
 may be written in the form 
 
 OPn+l=OPn.OP,-OPn-l. 
 
 Prove the formula geometrically by aid of Eucl. 
 
 VI. D. 
 
 20. A series of points are distributed symmetrically round 
 
 the circumference of a circle. Shew that the sum 
 of the squares of their distances from a point on 
 the circumference is twice that from the centre. 
 
 21. Find the sum of the series 
 
 cos h COS f- COS [-••• + cos^ : 
 
 n n n n 
 
 22. If ^ = -^, shew that 
 
 lo 
 
 cos0 + cos3^ + cos50+... + cos 110 = J. 
 
 23. Find the sum of the series 
 
 cos a + cos 3a + cos 5a + ... + cos(27i — l)a. 
 
 24. If 71 be a positive integer, and sinJ0=^— , shew that 
 
 cosJ0 + cos|0 + cos|0+... + cos — ZL^Q = n^m7i6. 
 
332 SERIES. 
 
 25. Find the sum of 
 
 sin 2a + sin 5a + sin 8a + ... to 7i terms. 
 
 26. Find the sum of 
 
 cos2a + cos^2a + cos23a+... to n terms. 
 
 27. Sum to n terms 
 
 cos3a4-cos3(a + ^) + cos3(a + 2/3)+.... 
 
 28. Sum to n terms 
 
 sin^a + sin2(a + ^) + sin2(a + 2^8) + . . . , 
 and hence find 12+22+32+.. .+^2. 
 
 29. If n be an even integer, prove that 
 sinti^ 
 
 COS0 
 
 = (2sme)»-i-^(2sine)''-H^^^^|^— \2sin0)''-5-... 
 
 30. If n be an even integer, prove that 
 
 n 
 (—1)2 2 cos TlO 
 
 = (2 sin QY- ^(2 sin 0)^-2 + ^-^\2 sin 0)^-*- ... 
 
 31. If ti be an odd integer, prove that 
 
 ^ = 1 j-5— sin20 + ^ f} -^sm*0— .... 
 
 008 [2 [4 
 
 32. Prove that 
 
 , X ,, X X , , X X , 7'/.. 
 
 tan ^ sec a; + tan ^^sec ^ + tan ^sec ^-\-...aa inj. = tan x. 
 
 33. Sum the series 
 
 1, a.lj a,li. a. 
 
 to 71 terms, and also to infinity. 
 
SERIES. 333 
 
 84. Find the sum of 
 1 
 
 cosacos(a + )8) cos(a + )8)cos(a + 2^) 
 
 +- 
 
 cos(a + '3^ — 1 . /8)cos(a + n/3) 
 35. Sum the series 
 
 ^ 1 2.1 ,, , 2.2 ,, , 2.3 
 
 144.12.^2^ 2H22 + 2^ 3*4-32 + 2 
 
 2 4 
 +^^^" V + 4H2 "*"-- ^ '^ *®^°^^' 
 
 36. Sum the series 
 
 ^sec2|+-2sec2|+^sec2|+... to n terms. 
 
 37. If n be even, find the sum of the products of the 
 
 sines of pairs of angles equidistant from the 
 beginning and end of the series 
 
 a, (a + ^), (a + 2/3) ... (a + ^i^=a/3). 
 
 38. From a point within a regular polygon perpendiculars 
 
 are drawn to all the sides : find the sum of the 
 squares on these perpendiculars. 
 
 39. If A^,. A^, A^... A^n+i be the angular points of a 
 
 regular polygon inscribed in a circle, and any 
 point on the circumference between A^ and A^n+i) 
 prove that the sum of the lengths of OA^, OJ.3, 
 J. 5, ... OAin+i will be equal to the sum of OA^, 
 
 0^4, 0^...0^2n. 
 
 40. If P1P2' ^2^z> • • • PnPi he equal arcs round the cir- 
 
 cumference of a circle, and if P^M^^, P^^i • • • Pn^n 
 be drawn perpendicular to any diameter ABy 
 shew that the arithmetical mean of the rectangles 
 AM^.BM^, AM^.BM^,...AMn.BMn is half the 
 square on the radius of the circle. 
 
334. SERIES. 
 
 41. If 71 be an odd integer, prove that 
 
 ^ ^ COS0 ^ 1 ^ ^ 
 
 + <^-y^-^> (28me)n-»— ■ 
 
 +(-i /^-^-if-;-2)-(^-2^) (2sine)n-v-i+.... 
 
 42. If n be an odd integer, prove that 
 (-l)V2sin7i0 
 
 == (2 sin er- 1:(2 sin 0'*-2+^?:^^i:i?l(2 sin 0)^-*- 
 
 +(-i)- "<-"7y;3(;";^-+^> (2siner-^+.... 
 
 43. If 71 be an even integer, prove that 
 
 COS0 1 |3^ |5 
 
 44. If 71 be an even integer, prove that 
 
 cos nQ = l — Tg-sin^^ H — ^ - — ^sm^0 
 
 If Iz 
 
 7l2(7l2-.22)(7l2-42) . „. , 
 
 45. Prove that 2 2 sin(pa + g^) 
 
 p=l g=l 
 
 _ sin|masin|?i^sin|{(m + l)a + ('y^+l)i3} 
 ~ sin la sin J/3 
 
 46. Sum the series 
 
 cos 20 cosec 30+ cos 60 cosec 3^0+ cos 180 cosec 3^0+ . .. 
 to n terms. 
 
 47. Sum the series 
 
 2 cos sin2-+ 2^003 ^ sin2-2+ 2^cos^sin2^^3+ ... to 7i terms. 
 
SERIES. 336 
 
 48. Sum the series 
 
 cos W cosec220 + 2 cos 2^0 cosec2220 + . . . 
 + 2«-icos2'^0cosec22»^a 
 
 49. Find the sum, to n terms, of 
 
 sinO . sin 20 . sin 30 
 
 + 
 
 cos + cos 120 ' cos 20 + cos 220 ' cos 30 + cos 320 
 
 50. Sum the series 
 
 1 1 
 
 sin^a; sec 3a; + ^ sin^3fl3 sec Z^x + -^ sin332ic sec 3^ic + . . . 
 
 to n terms. 
 
 51. Sum the series 
 
 tan 0sec 20 + sec tan 20sec30+sec 20tan 30sec 40+ ... 
 to n terms. 
 
 52. A regular polygon of n sides is inscribed in a circle, 
 
 and from any point on the circumference chords 
 are drawn to the angular points ; if these chords 
 be denoted by c^, ^2' • • • ^» (beginning with the 
 chord drawn to the nearest angular point, and 
 taking the rest in order), prove that the quantity 
 
 ^1^2 ' ^2^3 + • • • + ^n - l^n ^rfil 
 
 is independent of the position of the point from 
 which the chords are drawn. 
 
 53. From a point a straight line OA is drawn, making 
 
 an angle «(<— xt) with a fixed straight line AB^ 
 
 and n other straight lines OA-^, OA^, ... OA^ are 
 drawn to it making the angles A OA^^, AfiA^, ... all 
 equal and each equal to a ; if B^, B^... Bnhe the 
 radii of the circles circumscribing the triangles 
 OAA^y OA^A^ ..., find the value of 
 i^i + jRg + -'^s + • • • + ^- 
 
SERIES. 
 
 54. Shew that 
 
 and give the coefficient of 0^". 
 
 55. From the equation 
 
 2 sin ra — sin ^(n + l)a sin J^ia/sin Ja 
 
 r=l 
 
 deduce the sum of the first n natural numbers, and 
 also the sum of their cubes. 
 
 56. Sum to n terms the series 
 
 JL . ^ J. . Z . o 
 
 57. Find, by aid of the exponential value of the hyper- 
 
 bolic sine the sum of the series 
 sinh u + sinh(u + v) + sinh(u + 2v) + ... to n terms. 
 
 58. Sum the series 
 
 (n-l )cos e+{n- 2)cos 26 + (n- 3)cos 3^ 
 + . . . + 2 cos(n -2)0+ cos(7i - 1)0. 
 
 59. Sum the series 
 
 (I **" i) +(i**° p) +-+(^ta°|i)'' 
 
 and shew that the sum to an infinite number of 
 terms is 
 
 (tana)2 a'^S 
 
 60. The sum of cosecaj + cosec 2ic + cosec 2^33 + . . . + cosec 2" " ^a; 
 
 is zero i^ ^ = o^i— f> w, and n being integers. 
 
 61. Find the sum, to n terms, of 
 
 sm^O cosec 2^0 + 2 sm^20 cosec 2^0 
 
 + 22sin222^ cosec 2^0+.. .. 
 
 62. Sum the series 
 
 sin sec S^ + sin 3^ sec m+. . . + sin 3«-i0 sec 3*^^ 
 
SERIES. 337 
 
 63. Find the sum, to n terms, of the series 
 
 COS d Q0&\ cosec"-^ + cos W cos^— cosec^-^ 
 
 + COS 3^0 cos^' gj cosec^-^ + - 
 
 64. Find the sum of all the values of cos(j0a + g'|8), where 
 
 f and q may have any positive integral values 
 between and ?i — 1. 
 
 65. Shew that 
 
 sin{a- 271^} +sin{a- 2(71- l)/3}+sin{a- 2(71-2)^}+... 
 
 + sin{a + 27i/5}=^^sin(27?, + l)^. 
 ^ sin/5 ' 
 
 66. Prove that 
 
 vi=M n=N p~P 
 
 S 2 E ... cos(ma+7i/5+_29y4-...) 
 
 m-O 71=0 p=0 
 
 = cosi(ifa+i\r/3 + Py+...)sinK^+l)asinJ(iV'+l)/3 
 X sin J (P + 1 )y . . . X cosec Ja cosecJ^S cosec Jy . . . . 
 
 fiX g ~ * 
 
 67. If (p{x) denote ^ _^ , shew that 
 
 6 ~)~ 6 
 
 9n+l -I 
 
 9!>(a;) + 2(p(2x) + 220(22^^) + . . . + 2"0(2«aj) : ^ -^ 
 
 9^(2-+ia)) 0(^)- 
 
 68. Prove that the sum of the series 
 
 - log tan 20 + "2 log tan 2^6-{-...to n terms 
 = log(2sin20)-l^log(2sin 2^+i0). 
 
 Zi 
 
 69. Find the sum of 
 
 cos -^+ cos —^ — I- cos -^ + ... to 71 terms ; 
 
 and apply the result to prove that, if one angle of 
 a triangle be n times another angle, the side 
 opposite the former angle is less than n times the 
 
338 SERIES. 
 
 side opposite the latter angle, where n is any 
 integer. 
 70. Straight lines whose lengths are successively pro- 
 portional to the numbers 1, 2, 3, ...,n form a 
 rectilineal figure whose exterior angles are each 
 
 equal to — . If a polygon be formed by joining 
 
 the extremities of the first and last lines, its area is 
 
 — ~ ^ cot — h-r^ cot - cosec^-. 
 
 24 . n 16 n n 
 
CHAPTER XIV. 
 FACTOES. 
 
 § 1. Fundamental Theorem on Trigonometrical 
 Factors. 
 
 222. The object of this chapter is to shew that the 
 resolution of trigonometrical expressions into factors can, 
 in a great number of cases, be made to depend imme- 
 diately on a single fundamental theorem. 
 
 The fundamental theorem may be enunciated as 
 follows : — 
 
 Ifvn denote any one of the functions 2 cos nx, 2 cosh nx, 
 or x^^—, and if Un denote the function 2 cos na, then 
 
 X 
 Vn — Un 
 
 = {Vi-2cosa}|vi-2cos(« + ^)||vi-2cos(a + -;^)|... 
 to n factors. 
 
 The theorem may be written in the form 
 
 r=n-l C / 27r\l 
 
 ^^71 — ^11= n i'^i — 2cosf a + r . — jk 
 
 where 11 \v. — 2 cosi a-\-r . — )}■ is an abbreviation for 
 r=o I ^ \ n/j 
 
 339 
 
340 FUNDAMENTAL THEOREM ON 
 
 " the product of the n factors obtained by assigning to r 
 the values 0, 1, 2, ... (n— 1) in the expression 
 
 Vj — 2cosfa + r . — j. 
 
 It will appear in the following articles that the proof 
 of the fundamental factor theorem depends only on the 
 elementary facts that the functions 2 cos tix, 2 cosh nx and 
 
 a;"+ — severally obey the law 
 
 X 
 
 '^in^n ^^ '^m+n "i I'm - nj 
 
 and that 2 cos na obeys the same law and is also a periodic 
 function of nay of period 27r, so that for cos na we may 
 substitute cos(7ia + 27r), cos{na + 47r). . . or cos('7ia + r . 27r), 
 where r is any integer. 
 
 223. If Vn and Un he two functions, each of which obeys 
 the law f(m)xf(n)=f(m-{'n)+f(m — n)for all integral 
 values ofm and n, and if Vq=Uq, then will Vi^--u^ be a 
 factor of Vn-Un. 
 
 We have VnV^ = Vn+i + Vn- 1, 
 
 Vn+l^VnV^-Vn-i. 
 So also, Un+1 = UnU^ — Un-i ', 
 
 .'.Vn^l-Un+i = {Vn-Un)v^-]rUn{v^-U^)-(Vn-i-Un-\)...{l) 
 
 Hence, if v^ — u^ is a factor of Vn — Un for any two con- 
 secutive values of n^ it is also a factor for the next higher 
 value of n. 
 But, from (1), 
 
 v^ — u^=^ {v^ — u^v^ + u^{v-^ — 'W'l) , since Vq — Uq = Q\ 
 v^ — u^ is a factor of -y^ — u^ and v^ — u^ 
 and therefore of v^ — u^, and therefore, by successive in- 
 ferences, of Vn — Un. 
 
TRIGONOMETRICAL FACTORS. 341 
 
 224. Since 
 
 2 cos mx . 2 cos nx = 2 cos(m + n)x + 2 cos('m. — n^x, 
 2 cosh mx . 2 cosh nx = 2 cosh(m + n)x + 2 cosh(m — ^1)33, 
 
 »d (a=».+l.)(cr» + l) = (^"'■^»+^J + (^'"-'•+^0. 
 and since, when yi = 0, each of the functions 2cos?ia;, 
 2 cosh nx, and rc^+ — is equal to 2, we see that 2 cos na?, 
 
 2cosh7icc, or x^ + — may be substituted for Vn or u,i in 
 
 the theorem of art. 223 ; thus, we infer that 
 
 cos a — cos /5 is a factor of cos na — cos n^, 
 cosh X — cos a of cosh, nx — cos na, 
 cosh cc — cosh y of cosh nx — cosh ti^/, 
 
 03 H 2 cos a of 03'^+— -—2 cos Tia, 
 
 X x^ 
 
 and so on. 
 
 225. // Vn denote any one of the functions 2 cos nx, 
 2 cosh nx, 0;"+ — , then will 
 
 X 
 
 r=n-l C / 27r\1 
 
 Vn — 2Q,o^na=^ 11 j-yi — 2 cosf a + r — j\. 
 
 By the theorem of art. 223, we know that v^ — 2 cos a is 
 a factor of -yri— 2 cos na. 
 
 Hence, if ?' be any integer, v^ — 2cosia + r'^] is a 
 
 \ n ' 
 
 factor of -Vyj— 2 cos (71a + r. 2ir), 
 
 i.e. of I'n — 2 cos na. 
 
 Assigning to r the succession of values 0, 1, 2,...{n — V) 
 we obtain n factors of 'y^ — 2 cos na, and these factors are, 
 
342 FUNDAMENTAL THEOREM ON 
 
 except for special values of a, all different, hence we may 
 write 
 
 Vn — 2cos7ia 
 
 = X{Vi — 2 cos a}\ v^ — 2 cosf aH — -j \.-' 
 
 x{..-2cos(«+(2^-)} 
 
 where X has to be determined. 
 
 From the formula Vn+i = VnV-^ — Vn-i, it follows, by- 
 repeated inferences, that Vn is an integral function of v^ 
 of the n\h. degree, and that the coefficient of v^ in the 
 value of Vn in terms of v^ is unity, hence we get X = 1, and 
 therefore 
 
 Vn — 2cos'^a 
 
 Jn;{.,-2cos(„+.^)}. 
 
 r=0 
 
 Cots. — cos nx — cos na 
 
 = 2^-i{cosa; — cosajMcosaj — cosfa + — H 
 
 X ] cos a; — cosf a + — j [ . . . to n factors 
 
 _2n-i n jcosa; — cosfa+r—jk 
 
 cosh nx — cos na 
 
 = 2"~^{cosh £C — cos a} j cosh x — cosf a H j f 
 
 X I cosh X — cosf a H — - ) [. . . to ?i factors 
 
 r=n-lf / 27rM 
 
 _2n-i jj -^coshaj — cos(a + r — ) h. 
 r=o I \ n/j 
 
TRIGONOMETRICAL FACTORS. 
 
 X'^-\ - — 2 008 710 
 
 X^ 
 
 = \x-\ 2 COS a M ^ H 2 cos ( a + — 
 
 y X ) y X \ n 
 
 x\x-\ 2cosf a + — j L.. ton factors 
 
 ='l[~'|aj+i-2cos(a + r— )|. 
 
 226. Demoivre's Property of the Circle. — If 
 
 Aq, A^, A^ .., An^x he n points ranged symmetrically on 
 the circumference of a circle whose centre is 0, P any 
 point in the plane of the circle, then shall 
 PA,\PA,\PAi..,PA:_, 
 
 = 0P2"- 20P« . O^o^^cos nPOA^^ 0A;'\ 
 Let 
 
 OP = X, OAq = a, /. POAq = a. 4 
 
 By art. 225, we have, writing 
 
 X p 
 
 -- for X, 
 a 
 
 '=^-Hx a . / 27r\\ 
 
 = n \'-\ 2cos( a+r — jK 
 
 ,.=0 W X \ n/j 
 
 multiplying by a'^x^ we get 
 
 a;2n - 2a;"a"cos na + a^" 
 
 ='^~n'|x2-2a;acos(a + r^)+a4 (1) 
 
 From the triangle POAr we have 
 
 PA,' = x^- 2xa cos(a + r^) + a\ 
 
 and therefore, from (1), 
 
 PAS.PA^.PAl.. P^„li = a;2«-2a;Wcos7ia + a2« 
 
344 • FUNDAMENTAL THEOREM ON 
 
 or 
 
 PA^ . PA^ . PAl . . PAl, = OP^^' - 20P^' . OA^cos n POA^ 
 ■hOA;\ 
 
 Core. — If the angle POAq = 0, i.e., if P lie on a radius 
 through one of the n points, then 
 
 PAo .PA^.PA^.., PAn-i = OP^ - OA^. 
 
 If the angle POAq = -, i.e., if P lie on the bisector of 
 
 the angle between radii through two consecutive points 
 of the system, then 
 
 PA(,.PA^.PA^...PAr,.i = OP''+OA^. 
 These results are known as Cotes s Properties of the 
 Circle. 
 
 227. The wide applicability of the fundamental factor 
 theorem arises from the consideration that by assigning 
 special values to x, or a, or both, and making elementary 
 transformations, we can deduce an indefinite number of 
 factorial forms. Thus, if factors of cos na are required, 
 we may write v,i = 2cos(?ia + 7r), then v^^Un becomes 
 — 4cos7ia, and we obtain a factorial form of cos-na; 
 or, if factors of sin-nO be required, we may put x = 0, 
 a = W, then Vn — Un becomes 2 — 2 cos -ti. 20 = 4 sin^nO, and 
 we obtain a factorial value of sin^nO. 
 
 In deriving particular cases in which the factors are 
 given from the general theorem, the necessary substitu- 
 tions and transformations may often be inferred from 
 a careful consideration of the number of factors, and of 
 the limits between which angles involved in the product 
 lie. A geometrical representation of such a series of 
 
 angles as that denoted by a + r — will be of service; 
 
TRIGONOMETRICAL FACTORS, 345 
 
 thus, if XOAq be the angle a, and if r range from to 
 
 27r . 
 (n — 1), a-\-r — will denote a series of angles obtained 
 
 by drawing radii to n points A^, A^,A^, ...,An-h succeed- 
 ing one another at equal distances on the whole circum- 
 ference of a circle ; while, if we have such a series of 
 
 angles as a, a H- — , a H , . . . , a + ^^ — , the representative 
 
 points Aq, A^, A^, ..., An_i are n points on a semi-circle. 
 228. Example 1. — If n be a positive integer, shew that 
 
 sinw(/) = 2«-^sin(/)sin('(^ + !^)sin('(^+^V..sin(<^+^!i::l^y 
 By the fundamental theorem, 
 
 cosn6-cos7i.2cl> = 2''-^ U ]cos ^-cos( 2</) + r— ) [. 
 Let ^=0, then 
 
 l-cos7i.2(/) = 2"-i'^'n |2sin2^<^ + r-U, 
 
 Extracting the square root we get 
 
 sin?i</) = 2'*-i'^~n |sin^(^ + r-U. 
 
 The positive sign is taken for all values of <^, since 
 
 if <n(f)<7r , sin ntf) and all the factors are positive, 
 
 if TT <ncf)< 27r, sin ncf) and the last factor only are negative, 
 
 if 2ir<n(f)< Sir, sin n(f> is positive, and the last two factors 
 
 only are negative^ ; and so on. 
 
 Example 2. — Prove that, when n is an odd integer, 
 :r" + l=(^+l)/'.v2_2^cos-+ yA^2_2^cos^^ + l 
 
 x(^2_2a;cos^i:i^7r + lY 
 
 "We have, by art. 225, 
 
 I r=n-l f I / 27r\"\ 
 
 ^" + -„-2cos?ia= n -^ A- + --2cos(a+r — ]}. 
 
346 FUNDAMENTAL THEOREM ON 
 
 Multiply by af*y and let na=7r, then 
 
 r—w— 1 f 9r4- 1 ^ 
 
 (^" + 1)2= n {a^-^XCOB^^-^TT+l}. 
 r=0 '^ n i 
 
 Since n is odd, there is a middle factor, viz., that for which r= t_. 
 This factor = a^'-^ - 2^ cos tt + 1 = (^ + 1 f, 
 
 and the factors equidistant from the middle factor are equal, 
 hence (^"+l)^ = (^-2xcos- + lWar^-2^cos— +1^... 
 
 
 Ix^ - ZX cos ^^TT + 1 ) (a?+ 1)2, 
 
 and therefore, since :*?'*+ 1 and :r + l have the same sign, we obtain 
 the result 
 
 ^"+l = (jp+l)f^2_2A^cos-+lVa;2-2^cos — + l)... 
 X Ix"^ - ^x cos — — TT + 1 y 
 Example 3. — Shew that 
 
 coalcos H^ ... cos(?2jdk=(::i)lzi 
 
 n n n 2""^ 
 
 By the fundamental factor theorem, we have 
 
 cosm^-coswa=2'"-^ 11 -jcos ^-cosf a + r--j j-. 
 
 Ijetm=2n, 6='^, a=0, then 
 
 »-=2n-l ^ A.,r"^ 
 
 cosw7r-l=22'»-i n -^-cos — y 
 
 =0 
 
 since the number of factors is even. 
 
 = 2-' n {cos^}. 
 
 Hence, <rl)lll=cos?r. cos?2:...cos(?^:il)f. 
 
 22n-l 
 
 =cos- . cos 
 
 n n 
 
 Example 4. — Qj^, Q^ . . are n points ranged symmetrically round 
 a circle of radius a and centre 0. P is a point such that OP=c, 
 
 FOQi = 0. If c = a cos - sech (j>, prove that 
 
 71 
 
TRIGONOMETRICAL FACTORS. 347 
 
 2aVsiii"-(cosh «</> + cos nO) 
 
 sin QiPQ^ . sin Q<,PQz ... sin Q^PQi 
 
 Draw QiNj^, $2^2 perpendicular to 
 OP, then 
 
 A^iP$2 = ^OQ,Q., + AOPQ^ - AOPQ2, 
 .'. PQ^.PQ^. Bin Q,PQ., 
 
 = a cos - . 2a sin - + ( a iV, - §2^2)^ 
 n n 
 
 = ac|2sin-cosh^ + sin^-sin(^+— H 
 
 = 2ac sin — -{ cosh </> - cosf ^ + - ) }■• 
 Similarly, 
 
 PQ2 . PQz sin $2^^3= 2ac sin --! cosh (/> - < 
 
 and so on. 
 
 Hence, {PQ^ .PQ^... PQnf sin Q^PQ^ . sin Q^PQ^ ... sin QnPQx 
 
 = 2"a'*c"sin"-'^ fl I cosh d) - cos f ^ + - + r-^ ) ) . 
 
 But we know that 
 
 {PQx .PQ.2... PQnf = a''' - 2c«a"cos %^ 4- (r% 
 
 and that 2'»-i''~ff '{cosh </> - cos^ (9 +^ +r^) | 
 
 = cosh w<^-cos(?i^ + 7r)=cosh 7i<f) + cos nO. 
 
 2a"c'*sin'*-(cosh 7i<f) + cos nO) 
 
 •t)} 
 
848 
 
 PRODUCTS. 
 
 o o •n-.^j,,^*^ *^ zi sin 71^ , sinh-MU 
 
 § 2. Products for cos nO, — r—^, cosh nu, 
 
 n sin 0' 
 
 71 sinh 16* 
 
 229. To prove that 
 
 cos7id = cos"0"ff'[l4— ^^^ 
 
 tan 
 
 27Z, 
 
 COS 710 -cos 710 = 2*^-1 n jeos^-cosr^+r-^U. 
 
 We have, by art. 225, 
 cos Tiff) — cos nO = 2^- 
 
 Let nd) = n0+'7r, and therefore (h = 0-\--, 
 
 ^ n 
 
 then - 2 cos 716 = 2^- ^'~li 'jcos^^ + '^) - ^^^(^ + ^— ) } 
 
 on i^'^-iV'/o • fn 2r+l \ . 2r-l 1 
 
 In this identity put = 0, 
 then -2 = 2«-i'^~n'|2si] 
 
 r=0 I 
 
 .'. by division, 
 
 2r+l . 2r 
 sm -7^ — X . sm 
 
 271 
 
 2n 
 
 M. 
 
 r=n-l 
 
 cos 710 = n 
 
 r=0 - 
 
 sin0 
 
 . 2r + l 
 
 tan — t:: TT 
 
 + COS0 
 
 271 
 
 or 
 
 COS710 = COS"0 IT 
 r=0 
 
 1 + 
 
 tan0 
 
 . 2r+l 
 tan^ — TT 
 
 271 
 
 It should be observed that, as r passes through the 
 series of values 0, 1, 2...(?i — 2), (ti— 1), the angle 
 
 -^ — TT passes through the series of values ^, ^, ~ ..., 
 2n ^ ^ 2n 2n' 2n ' 
 
 ( TT — 2^J, ( TT — ^ ), and that, the nearer r is to the middle 
 
PRODUCTS. 349 
 
 2'r+l 
 of its series of values, the less does the angle — 7^ — tt differ 
 
 from a right angle, and therefore the more nearly is the 
 
 factor 1 H — equal to unity ; hence, in estimating 
 
 tan — ^ — TT 
 
 2-71 
 
 the value of the product we must remember that the 
 factors at the beginning and end of the series have the 
 greatest weight. 
 
 The reader will find it useful to represent the factors 
 by a rank of equidistant ordinates, the corresponding 
 abscissae being the values 0, 1, 2...('7i — 1) of r. The 
 Corollaries of arts. 229-232 may be illustrated by folding 
 the rank of ordinates so as to bring the extremes together. 
 
 Cor. — Taking together factors equidistant from the 
 
 beginning and end of the product, and observing that, 
 
 7?/ — 1 
 when n is odd, the factor for which r= — — is equal 
 
 to unity, since — ^ — tt is then a right angle, we get 
 the results : 
 
 ft 
 
 n even, cos n6 = cos^O U J 1 — 
 
 .=0 I .....2r4-l 
 
 tan^- 
 
 2n 
 n odd, cos nd = cos'^0 U J 1 - 
 
 r=0 
 
 tan^^ TT 
 
 '2n 
 230. To prove that 
 
 sin 710 ^ i/=A"^fi , tan^ 
 
 I n 
 
850 PRODUCTS. 
 
 We have, by art. 225, 
 
 Let 710 = nQ + ^, and therefore = + g- ; 
 
 and let Tia = 7i0 — ^ , and therefore a = 0—^\ 
 then 
 
 -2 sin ne = 2"-' n"{cos(0+^) -eos(0+*^x)} 
 
 = 2»-/n;{2.n(.+3sin?^.}, 
 -2sin7i0 ^,, /=W-^f • (^ , ^7r\V=;W^ . 2r-l 1 
 
 SmO r=l I \ '^/J r=0 I 271 J 
 
 In this identity put = 0, then since Xt -^ — 7r = n, we 
 have -2'M = 2^-i H Uin — [ H ■^2sin^^x[; 
 
 r=l I '?«' J r=0 I 2W J 
 
 therefore, by division, 
 
 sin'7i0 _^=Jl-^r sinO 
 '7^sinO 
 
 I I hcos^l, 
 
 "^ I tan — I 
 
 I tan — I 
 
 or — ^-p^=cos^-iO n J1+- 
 
 -Tismt^ • ,.=1 i 
 
 ^- n 
 
 We observe, as in art. 229, that the factors near the 
 
 middle of the product approach to unity, and also that, 
 
 til 
 if n be even, the factor given by 'r = - is equal to unity. 
 
 A 
 
 Got. — Taking together factors equidistant from the be- 
 ginning and end of the product, we get the results : \ 
 
PRODUCTS. 351 
 
 -yz, even, — ^ — 7; = cos^"^t? II I 
 
 7ismt^ ,.=1 -\ x...9'^ 
 
 n-l 
 
 71 odd, — ^7, = cos^-iO n 1 — 
 
 ^ n 
 
 231. To prove that 
 We have, by art. 225, 
 
 r=n-lf / 27rM 
 
 cosh, nv - COS na = 2^-'^ H scosht' — cos(a + '?"- — • 
 
 r=0 I ^ '^/J 
 
 Let 71a = TT, and therefore a = -, 
 
 r=7i-l r 27'+ 1 1 
 
 then cosh 71V+ 1 = 2^-1 11 ■{ cosh v — cos -ttK 
 
 2 cosh2!|^=2^-i'^rf '|2smh2^ + 2sin2?^x|. 
 
 Let ^ = 11, then 
 
 r=n-ir 2r-4-l 1 
 
 cosh27iu = 22^-2 n jsinh%+sin2-^^7rK 
 
 In this identity put u = 0, 
 then 1 = 2^^-2 H W^^-^^^h 
 
 .-. by division, cosh^nu = II J I H ^ -, I 
 
 1 «'°^;r'^/ 
 
362 PRODUCTS. 
 
 And, by a simple transformation, 
 
 - , sinh% , „ f, , tanh^ti) 
 
 cosh^Tiu = cosh^^it n IH ft-^ I 
 
 Gov. — Taking together factors equidistant from the 
 beginning and end of the product, and observing that 
 cosh nu is positive, and that each of the factors is positive, 
 we get the results : — 
 
 n even, cosh nu = cosh**u II 1 1 + 
 
 r=0 
 
 {^ tanh^ 1 
 
 n odd, cosh oiu = cosh"i6 11 1 + 
 
 ^ ^ tanh% 
 
 232. To prove that 
 
 \nsmhuJ .=, I ^^^,r^^ 
 
 We have, by art. 225, 
 
 r=n-l f / ^_ 
 
 cosh Tiv — COS 71a = 2*^-1 11 jcoshi; — cosf a + r— 
 r=o I \ n . 
 
 Let a = 0, then 
 
 cosh'/i-y— 1 = 2""^ n i cosh i; — cos r — \, 
 
 2 sinh2'^'" = 2»^-i^~lf ' J2 sinh2^ + 2 sin2M. 
 
PRODUCTS. 
 
 353 
 
 Let Q = i^, then 
 
 smh2rm = 22«-2 n -^smh^i^ + sin^— [, 
 
 — ^-.j = 2^" - 2 n i sinh% + sm^ — \. 
 
 \ smh u / r=i L "^i J 
 
 In this identity put u = 0, then since Lt. . , — = 7i, 
 
 M=0 
 
 we get 7i2=22«-2 n isin2-^k 
 
 therefore, by division, ( — ^-, — ) = 11 J 1 H 1, 
 
 -^ Vti smh uJ r-\ 1 • 2'^7r I 
 
 ^ n ^ 
 
 and applying the transformation 
 
 ^ , sinh^t^ 1 9 /-, , tanh^wN 
 
 1+ -2/3 =cosh2i6( 14-— — 2^-) 
 
 sm^O \ tan^O / 
 
 to each of the factors, we have finally 
 
 Cor. — Taking together factors equidistant from the 
 beginning and end of the product, and observing that 
 sinh nu and sinh u have the same sign, and that all the 
 factors of the product are positive, we get the results : — 
 
 n even, 
 
 n odd, 
 
 n _ , 
 
 sinh nu , ^ , ~|, 
 
 — r-T— - = cosh"-% n 
 71 smh u r-i 
 
 1 + 
 
 tanh^u 
 
 tan^ — 
 tanh^u 
 
 smh nn . , tt 1 1 . 
 
 — r--r— - = cosh'^-iu n Jl + 
 
 71 smh u ^^1 \ ^^^2T2L 
 
 n 
 
354 INFINITE PRODUCTS. 
 
 § 3. Infinite Products for the Cosines and Sines of x- 
 233. To j)rove that 
 
 and to find limits between which Fr lies for a given 
 finite value of r. 
 
 By art. 229, Cor., we have, for any even value of n, 
 
 Let n6 = x, and let x remain constant while n is in- 
 definitely increased, and consequently 6 indefinitely 
 diminished. Then 
 
 Lt cos«0 = Lt (cos -)" = 1 ; (art. 190) 
 
 n=oo n=oo ^ n/ 
 
 and, so long as r is finite, 
 
 2r+l \2 
 tan2a 
 
 cos nO = cos' 
 
 n=co,__o^r-\- 1 
 
 -l(2r+l)xj ' 
 hence, cos», = (l-^Xl-3^,j ... (1-^^--^^^)F, 
 
 where /', = it/l ^^\(l- *^^'' ^ 
 
 to (rt— '^) factors. 
 
INFINITE PRODUCTS. 365 
 
 We shall now shew that 1> Fr> 1 — > tt— o for all 
 
 ^ (r-l)7r2 
 
 values of r such that (2r + 1)^^ > x. 
 
 A 
 
 Whatever finite value x may have, we may take r 
 
 «j. 2t4-1 
 
 such that (2r + 1)^ > x, and therefore —z^ — ir > 6. There- 
 
 foie, from and after this value of r, every factor of Fr 
 h positive and less than unity, and therefore jP,.< 1. 
 Again, 
 
 ^->1- --;2rTr + , ,2r+3 +-to(^-r)terms I 
 
 therefore, a fortiori, 
 ^ ^ , f tan^e . tan20 , . ("n \ , 1 
 
 , /27itan0Vr 1 1 J. fn \^ 1 
 
 ^'•^- > 1 - (-^^- j l(-2^Tl?+(2^^T3P^- *" W -^) *^^^^|- 
 
 tan — 
 Now, Lt.{ni2ijnQ) = Lt ,x = x', 
 
 n=oo n=o3 X 
 
 n 
 ^^ (2r+l)2 + (2r + 3)2+--- 
 
 ^(2^2+(2r+2)2+-" 
 
 4l(r-l)r^r(r+l)^'"'r 
 
 <4 
 
 *For if Oi, ttg, ... a^ be positive, and each < 1, then (l-ai)(l-a2) 
 >!-(«! + a2)> ^^^ therefore (1 -%)(! -a2)(l -a3)>(l -aj + ag)!! -aj) 
 > 1 - (tti + ttg + a^) ; and therefore, by successive inferences, 
 (l-ai)(l^a2) ...(l-a^)>l-(ai + a2+...+a^). 
 
356 INFINITE PRODUCTS. 
 
 UJ. 11 1_4. \ 
 
 <4\r-\ r"*"r r+1 J' 
 
 1 1 
 
 and therefore, < 
 
 Hence Fr>l- 
 
 4 r— 1 
 
 (r-l)7r2 
 
 Thus, we have 1 > i^^ > 1 - 
 
 (r-l)7r2 
 
 And since the latter limit can be made as nearly equal 
 to 1 as we please by sufl&ciently increasing r, we may 
 
 write cosa; = (^1 — ^j(l "svA^ ""svj - «^ ^V- 
 
 234. To prove that 
 sin X (^ x^ 
 
 =(.-5)(>-^)-('-i.)^. 
 
 X 
 
 and to find limits between which Fr lies for a given 
 
 finite value of r. 
 
 By art. 230, Cor., we have, for any even value of n, 
 
 =--1 
 sinnO . ,^ ^ [^ tan^^ ^ 
 
 nsinO 
 
 = cos"-i^ 11 1 
 
 -^ 1 tan^^ 
 I n . 
 
 -]■ 
 
 Let nO — x, and let x remain constant while n is 
 indefinitely increased, and consequently indefinitely 
 diminished. 
 
 . X 
 sm- 
 
 Then Lt (n sin 6) = Lt. x = x, 
 
 ■n=oo TO=oo ^ 
 
 n 
 and Lt. cos^ - ^^ = X^. ( cos - ) =1, 
 
 n=oo n = oo ^ "'' • 
 
 also, so long as r is finite, 
 
INFINITE PRODUCTS. 357 
 
 n \ nl \n J 
 
 Hence, ?i^^ = (l-5)(l-^J...(l-^>„ 
 / tan^e \ / tan^e \ 
 
 to f— — r-1) factors. 
 
 n=oo 
 
 iC" 
 
 We shall now shew that l>Fr>l ^ for all values 
 
 Tir- 
 
 of r such that (r + l)7r > x. 
 
 Whatever finite value x may have, we may take r such 
 
 that (r+l)7r>cc, and therefore —>Q. Therefore, 
 
 from and after this value of r, every factor of F^ is 
 positive and less than unity, and therefore Fr<l. 
 A • 1P^^ { tan^e , tan^^ , 
 
 n . n 
 
 to f ^ — r — Ijtermsl, 
 
 , ^. . „ , r tan^e , tan20 . 
 .-., <.fomori, ^,> 1-1^3^-^, + ^^^-^^+... 
 
 ^^ f:^ — r — ijtermsl, 
 _, /?itan0\2r 1,1, 
 
 to f^ — r— Ijtermsi. 
 
368 
 
 INFINITE PRODUCTS. 
 
 Now, £1.(71 
 
 n—oo 
 
 tan 
 ^tan0)= Lt £C = a3; 
 
 
 n 
 
 and 
 
 (r+l)^' (r+2)2+-- 
 
 
 ^r(r + l) ' (r+lXr+2) ' - 
 
 i.e. 
 
 ^ ^+1 • r + 1 r+2^*"' 
 
 and therefore, < -• 
 
 r 
 
 Hence 
 
 ^'^l-rl^- 
 
 Thus, we have 1 > i^^ > 1 — —^. 
 
 And since the latter limit can be made as nearly equal 
 to 1 as we please by sufficiently increasing r, we may 
 write 
 
 235. To 'prove that 
 
 and to find limits between which F^ lies for a given 
 finite value of r. 
 
 By art. 231, Cor., we have, for any even value of n, 
 
 . ''X^fi , tanh% 
 cosh Tiu = cosh" 16 ii j ^ -\ iyZTTT 
 
INFINITE PRODUCTS. 359 
 
 Let nu — Xf and let x remain constant while n is in- 
 definitely increased, and consequently u indefinitely 
 diminished. 
 
 Then Lt cosh'^i^ = Lt ( cosh - j = 1 ; (art. 191) 
 
 X 
 
 n. 
 and, so long as r is finite, 
 J. tanh^u 
 
 2r + l 
 _ i^tanh^YI 2n '^ | ^ u s^' ^ | 2x 
 
 -^ ^ ^Uan^^n2r+lJ 1(2.+ 1K. 
 
 2^ / \ 2?i 
 Hence, cosh x 
 
 where F, 
 
 ff- 
 
 TT^ JV'^S'VJ"'V'^{2r-l)V. 
 
 T^ , ^ , tanh% \ /^ , tanh% \ , fn 
 
 2n /\ 2n I factors. 
 
 We shall now shew that 1 < Fr<e^''-^^'^\ 
 Since every factor of F^ is positive and greater than 
 unity, we have at once F^>\. 
 
 Again, i^^ < e '^ ^n 2^ j 
 
 r_tauh2^ + tanh2u to(r^-r )termsl 
 
 .:, a fortiori, Fr<e^ '^'' / \ 2n / J^ 
 
 i.e. <^^ "" "^ l(2;-+l)2^^(2r+3)2 V2 ' / i 
 
 *For if ttj, a^ ... a,„ be positive, then l+ai<l+— V-— + -.af^ inf., 
 2.e. <e''i ; and therefore (l + ai)(l + a2)...(l+aw)<e«i+«2+--+«OT, 
 
360 INFINITE PRODUCTS. 
 
 tanh - 
 Now Lt. (n tanh u) = Lt .x = x; 
 
 n=oo n=ao X 
 
 n 
 
 and -^ — 7-TT9+7^ — riv:9+---< 
 
 (2r+l)2 ' (2r+3)2 ' •"^4 r-1 
 Hence Fr< eC'-DTS. 
 
 Thus, we have 1< i^^ < e(r-i)7r2. 
 
 And since the latter limit can be made as nearly equal 
 to 1 as we please by suflSciently increasing r, we may 
 write 
 
 cosh a; = (l + -^)(l + pfij^^ + 52^2) • • • «^ W- 
 
 Cor. — In like manner it may be proved that 
 sinhic A , ic^\A , x^ \ /-, . a;^ \_, 
 
 where l<Fr<e'''^'', 
 
 and therefore = n (l+-?-o). 
 
 236. Example 1. — Prove that 
 
 J=2 2 4 4 6 6 ^^ ^. . /pallia's Theorem). 
 2 13 3 5 5 7 
 
 We have «_l£^=fl -?Vi _ JlVi _ ^ \ .... 
 6 \ irVV 2VA 3V2r • 
 
 Let 6=^, and therefore - = -, 
 
 _ 1.3 3^ 5/7 
 2.2' 4.4' Q.Q""' 
 7r224466 ,.. 
 
 2133557 -^ 
 
INFINITE PROD UGTS. 361 
 
 EXAMPLT5 2. — Prove that the sum of the products of the recipro- 
 cals of the fourth powers of every pair of positive integers is 
 
 3847r8 
 
 Equating known values of — ^— , we have 
 u 
 
 Taking logarithms, we get 
 
 Expanding by the logarithmic series, and equating coefficients of 
 e^ and ^, we get 
 
 r^ I [5^2 ((3)2/ 
 
 and 2^=4^{-|+l_l_+_|_^^-__|^+^^^^^^ 
 
 Hence 22-\. A=(4y-4 = 4.^(r + -^"r— 1 
 
 =i^.l92, 
 ^|9 
 
 2I i = 3847r8 
 
 r4 s4 
 
 L519 
 
 Example 3. — Prove that 
 
 1 + 3+6 10 
 
 The n^^ term of the series =M^L±1) 
 
 (27i+l)4 
 
 64V"^ Vi 
 
 . 4.71^ + An _!/• 1 1 ^ 
 
 '8(271+ 1)^ 81(271 + 1)'^ (27r+Ty/" 
 
3G2 INFINITE PRODUCTS. 
 
 we we «>s^=(l-|?)(>-|?)(l-^^).... 
 and coa^=I-,^+,^-.... 
 
 Equating coefficients of ff^ and ^ in these expressions, we get 
 
 V^^2:' b'' 8' 
 
 and i . i + 1 .1 + 1.14-. =Z!.. 
 
 X2 g-^-T-p 5-^^32 -52 42j4' 
 
 hence, ^ +1 +1 + ... = (!LY_ gJEl^'^l. 
 
 1* 3* 54 V8/ 42|4 96 
 
 Therefore the given series = l{(^^ l,)-(^- 1,)} 
 
 64\ 12/ 
 Example 4. — To prove that 
 
 sin (9 see , . . 
 
 — TT— = cos - cos -, cos - . . . aa iwr. 
 ^ 2 4 8 -^ 
 
 We have sin (9 = 2 cos - sin 
 
 2 2 
 
 „ cos— i sin-; 
 2 22 22 
 
 ^ (9 (9.6' 
 2COS2,co823sm- 
 
 = 2«cos^cos2^,cos|3...cos|sin|, 
 '''" ,-S=^^^|-^2'-«2-3---2-- 
 
 Now, Lt. 2''sin -^ =Z^. S • ^= ^' 
 
 2" n-=co ^ 
 
 2" 
 
 and consequently 
 
 sin <9 ^ e 6 . . . 
 
 __=cos - cos^-jCOSga ... ad mf. 
 
FACTORS. 363 
 
 Examples XXIV. 
 
 1. Prove that, when n is an even integer, 
 x^-\-l = \x^ — '2.x COS —+\\\x^ — 2x cos f- 1 ) • • • 
 
 2. Prove that, when n is an even integer, 
 
 a;^_l = (a;2_l)(^ic2-2cccos — + lV£c2_2^cos — + lY.. 
 
 x(^2~2^cos^^^ + l) 
 8. Prove that, when n is an odd integer, 
 
 x'^ — l={x — V)\x^ — 2ic cos — + 1 jf a?^ — 2fl? cos — + 1 j . . . 
 
 4. Write down the factors oi x^-\-\,x^ — l, x^ + \,x^+l, 
 
 x' — 1, x^-\-\,x^^ — \; illustrating each case by the 
 division of the circumference of a circle. 
 
 5. Shew that, when n is even, 
 
 2 / 
 
 cos -Jia = n / 1 ^ , ^ 
 
 sm" „ 
 
 zn 
 
 n 
 r=^ - 1 
 
 and — -. — = cosa 11 /I- 
 
 sin^ — 
 
 n 
 
 6. Shew that, when n is odd, 
 
 n—5 
 
 ' 2 
 
 cos7ia = cosa 11 / I 
 
 '■=0 1 sin^'^. 
 2^1 
 
364 FACTORS. 
 
 and fiEL^^J'iTA-^. 
 
 7. Shew that 
 
 . X sinj + (-^i — 1)— [ = cos 4^ — cos Tif + 2 )• 
 
 8. Shew that, if n be odd, 
 
 (--l)~2"sin'n,0 = 2"-ism0sin(^+— )sin('0+— v.. 
 
 xsin(9!> + ^i:^.^). 
 
 9. If 71/3 = 27r, prove that 
 
 cos a cos(a + /3)cos(a + 2/3). . .cos(a + 7i — 1.8) 
 (-l)Y nir \ 
 
 10. Prove that 
 
 11. If w be odd, shew that 
 
 cos 719!) = ( - 1)^" 2»» - 1 cos (f) cos(d) + '^)cos((f> + — ) . . . 
 
 xcos( 0H tt)- 
 
 12. Shew that, when n is even, 
 
 2^ sm-sin — sm — ...sm^ ^ =1. 
 
 71 7i 71 71 
 
 13. Shew that, when n is odd, 
 
 / o^^ -^ Stt 57r (7i-2)7r 
 
 jjn = 2 2 cos -^ cos ^r— cos ;:i— . . . cos — • 
 
 ^ 2n 271 271 2n 
 
FACTORS. 365 
 
 l^. Find the value of 
 
 .7r.27r.87r . (n—V)7r 
 
 sm — sm — sm — ... sm^ 
 
 n n n n 
 
 where 71 is a positive integer. 
 
 15. Shew that, when n is even, 
 
 J^ . 7r . 3x . 5x . (n~l)7r ^ 
 
 16. If a = -r/ — r^iT> prove that 
 
 4(71 + 1) ^ 
 
 cosec a cosec 5a cosec 9a ... to 7^+1 factors = 2"^2. 
 
 T7 -P +1. . 1 1 3 5 7 9 11 , . . 
 
 17. Prove that ^ = ^.^.g.g.^.j^...ac^ tT./. 
 
 18. Sum the series 
 
 l + 32+52 + f2+--^^^V-> 
 
 and 1 + 34+54+^+... ad inf. 
 
 19. Shew that the sum of the squares of the reciprocals 
 
 of all positive integers is -^ . 
 
 20. Prove that ji-2+^+3ig+^ + ...=g. 
 
 „, T, ,, , 3 5.7 11.13 17.19 
 
 21. Prove that - = -g^. ^2^ .^g^ 
 
 22. Find the sum of the products two together of 
 
 1 1 i i 
 
 12' 32' 52' 72"" 
 
 23. Find the sum of the products three together of 
 
 I i i 1 
 
 12' 22' 32' 42 " " 
 
 24. Shew that p-^+^-^+...=|^- 
 
366 FACTORS. 
 
 25. Sum to infinity the series -^-\--ki — f""!!" + • • • • 
 ^^ „ 2 10 26 50 , . . 
 
 , 5 17 37 . . . 
 ^ = ^4' 16' 36 •••^^'^•^•' 
 then will 4a2-62=4. 
 
 27. If ^ = 2n^x > prove that 
 
 2"cos Ocos 20 cos 220 ... cos 2«-i0 = l. 
 
 28. Prove that 
 
 2^ V2 J{2+J2) V(2V(2+V2)) J(2+J{2+J(2^J2))) 
 TT 2 ' 2 * 2 ' 2 
 
 29. Find to n terms the sum of the series 
 
 log(l + cos 0) + \og(l + cos I) + log(l + cos I) + . . . . 
 
 30. Prove that 
 
 (2cos0-l)(2cos20-l)...(2cos2«-i0-l) = ^^^"^"^/. 
 ^ ^ 2 cos 0+1 
 
 31. Prove that 
 
 (l_tan^f)(l-ta.^|)(l-tanf,)...a^i./. = ii-,. 
 
 32. From the identity sin0 = 2sin-sin— ^— , deduce in 
 
 succession : 
 (i.) sin0 
 
 ^„ ■ . e . e + TT . + 27r . 0+(^-l)7r 
 
 = 2^-ism-sin sin ... sm — -^^ ^— 
 
 p p p p 
 
 where p=2^*; 
 
 (ii.) sin0 
 
 = 2^ " ^ sin ( sin^ sin^- ) ( sin^— — sin^- ) . . . ; 
 
 p\ p p/\ p p) 
 
FACTORS. 367 
 
 (iii.) sin 6 
 
 = e(l-5)(l-2&)(l-3&)-«'^^™/- 
 
 33. A^, A^y A^ ... A^n+i, are the vertices of a regular 
 
 polygon inscribed in a circle of radius a, OA^+i is 
 a diameter, prove that 
 
 OA^.OA^.OA^...OAn = a^. 
 
 34. A^,A<^,A^...A2n are the vertices of a regular polygon 
 
 inscribed in a circle of radius a, is the mid-point 
 of the arc A-^A^n, shew that 
 
 OA^.OA^.OA^...OAn = s/^ 
 
 a' 
 
 35. AB is a diameter of a circle, Qq any point on the 
 
 circumference ; if Q^, Q^, Q3 • • • Qn be the points of 
 bisection of the arcs AQq,AQj^,AQ2 ..., prove that 
 
 36. Shew that if A-^A^A^ ... A2n, B^BJB^... B2n be two 
 
 concentric and similarly situated regular polygons, 
 then 
 
 PA^.PA^...PA2n-l _ PB,.PB,...PB2n-l 
 PA,.PA,...PA2n ~PB^.PB,...PB2n ' 
 
 where P is anywhere on the concentric circle 
 whose radius is a mean proportional between the 
 radii of the circles circumscribing the polygons. 
 
 37. Prove that 
 
 sin 910 = 2" " ^ sin ^f cos ^ — cos — j f cos r/> — cos 
 
 X (cos ^ — cos^ ^- 1. 
 
 38. From the equation 
 
 sin?i0 = 2^^-i'"ff^|sin(0-|-^)|, 
 
 n 
 
368 FACTORS, 
 
 deduce, when n is even, 
 
 (-l)^sin'n,0 = 2»-i'r['Vcos(0+^-^)|, 
 
 and'n;{tan(04-3} = (-l)^ 
 
 39. Prove that 
 
 -(''+-l)=7-^(-+?)(-?J(-^S-- 
 
 40. Deduce the factorial expression for cos0 from that 
 
 for sin Q by aid of the formula cos = f> . ^ - 
 •^ 2sm6 
 
 41. Prove that 1 —cos a is a factor of 
 
 71 sin('?i + 2)a — (371 + 2)sin(7i + l)a + (371 + 4)sin na 
 
 — (71 + 2)sin(7i — l)a ; 
 and find the other factor. 
 
 42. If 71 be odd, shew that sin 7i0 + cos 7i0 is divisible 
 
 either by sin + cos or by sin 6 — cos 6. 
 
 43. For what integral values of n are sin nQ and cos nQ 
 
 divisible by cos ? 
 
 44. From the formula 
 
 sin(7i +1)0+ &in{n — 1)0 = 2 sin nO cos 0, 
 shew that sin7i0 is divisible by sinO for all 
 integral values of n. 
 
 45. Prove that if = -— and — = 7: , 
 
 271 <p TT 
 
 , 2sm^esm^2e...sm%n-l)0 _<l) 
 sin'^0 sin'-^^^ . . . sin2(?i — 1)0 ~ 6' 
 
 46. If 04-0 + i/r = 7r, shew that 
 
 ^^ ^2n+1^^2n+l.|-^ 2n-fn 
 I 271+1 / 
 
 n=or (27i+l)V2"*" (27i+l)V (2r^+l)Vr 
 
 71 = 00 r 
 
 2 (-1)' 
 
CHAPTER XV. 
 APPEOXIMATIONS. 
 
 § 1. Approximations and Errors. 
 
 237. In many theoretical calculations, it is not always 
 necessary that the result should be perfectly accurate, an 
 approximate value being sufficient for the purpose in 
 view. In all calculations based on measurements (which, 
 however carefully made, cannot be free from error), an 
 approximate result only can be obtained; and it is 
 useless giving to that result an appearance of greater 
 accuracy than the character of the measurements will 
 allow. 
 
 For example, if we wish to find the square of 30001 
 correctly to four places of decimals, the term ('0001)^ may 
 be omitted from the equation 
 
 (3-0001)2 = 9 + 2 X 3 X 0001 + (0001)2, 
 and we obtain 90006, a result which differs from the 
 true value only by -00000001. 
 
 Again, if x and y be quantities so small compared 
 with a that their products and second and higher powers 
 may be neglected, we have approximately : — 
 {a + x){a + 2/) = ^2 + a(a; + y), 
 (a + a^)-..(a + 2/)=l+^, 
 
 2 a 369 
 
870 APPROXIMATIONS AND ERRORS. 
 
 X _x 
 a±x~ a 
 
 V(a^±a,) = a±^, 
 
 cos(a ± ic) = cos a + a; sin a, 
 sin(a±a;) = sin a±x cos a ; 
 
 X in the last two cases being the measure of an angle 
 
 in radians. 
 
 Definition. — If, in measuring or calculating the magni- 
 tude of any quantity whose real magnitude is a, an 
 error x be made, the absolute error is x, and the relative 
 error is xja. 
 
 From the third of the above approximations, it follows 
 
 that the relative error may be taken as either - or —7—. 
 
 '^ a a±x 
 
 238. Example 1. — If A be the length of the chord of a circular 
 arc, B that of the chord of half the arc, the length of the arc is 
 approximately equal to {8B-A)ld. 
 
 Let 6 be the number of radians in the angle subtended by the 
 arc at the centre of the circle, a the radius of the circle : then 
 
 A = 2a sin _ = 2a( - - -—-\-— — - . 
 2 V2 48^3840 
 
 2 \2 48 ' 3840 
 and 5=2asinf=m-^ + ^^^-...), 
 
 Let \ and /n be determined by the equations 
 These give X= -1, ^=|, ^^,^X+^,I^= -^, 
 
 8B-A /, e^ _^ 
 
 = /l- 
 3 V 7680 
 
 where I is the length of the arc. 
 
I 
 
 APPROXIMATIONS AND ERRORS. 371 
 
 Hence, approximately, 
 
 ;=(8J5-J)/3. 
 
 This is known as Huyghens' approximation to the length of a 
 circular arc. (See also Example xxv. 23.) 
 
 The closeness of the approximation may be shewn by finding the 
 length of an arc which subtends an angle of 45° at the centre of a 
 circle whose radius is one mile. We have 
 
 ^ = 2 X 63360 X sin 22°30' = 2 x 63360 X 0-3826834 
 
 =48493-64 inches, 
 5=2 X 63360 X sin 11°15'=2 X 63360 x 0-1950903 
 = 24721-84 inches. 
 The values, using Huyghens' approximation, give for the circular arc 
 a length of 49760-36 inches. 
 
 The real length is 63360 x 3-14159265 x J or 49762-83 inches. 
 
 The absolute error in this case is therefore a little less than 2^ 
 inches, and the relative error about 1/20,000. 
 
 Example 2. — The height of a tower standing on level ground 
 is determined by measuring a base-line from the foot of the tower, 
 and the angle of elevation of the top of the tower from the other 
 end of the base-line. If a small error be made in observing this 
 angle, to find : (1) the absolute error in the calculated height of the 
 tower, and (2) the angle of elevation for which the relative error is 
 a minimum. 
 
 Let CB represent the tower, BA the base-line. Let A be the true 
 angle of elevation of the top of the 
 tower, and B radians (represented by 
 CAC) the error made in observing 
 this angle, the error being either in 
 excess (as in the figure), or in defect, 
 of the true angle. 
 
 Let a be the true height of the 
 tower, X the error in the calculated 
 height due to the angular error 6, c the length of the base-line. 
 
 Then, a=ct3iii A, a+x=cta.n{A +6), 
 
 "^ :tan(^-f^)-tan^= ^^" ^ 
 
 c cos(^ -f ^)cos A 
 
 = ^sec2u4, neglecting 6'^ and higher powers of 6, 
 ,r=^csecM. 
 
372 A PPROXIMA TIONS A ND ERRORS. 
 
 Agaiii) the relative error is xja^ which 
 
 ^ec&ed^A^ e ^ J'd 
 
 c tan A cos A sin A sin '2, A' 
 
 Hence, for a given error 6 made in observing the angle of eleva- 
 tion, the relative error in the calculated height of the tower is least 
 when sin 2 J is greatest, i.e. when the angle of elevation is 45°. 
 
 239. In any right- angled triangle the number of 
 degrees in the smallest angle divided by 172 is very 
 nearly equal to the smallest side divided by the sum of 
 the other side and twice the hypotenuse. (Ozanam's 
 Formula.) 
 
 In the right-angled triangle ABC, let G be the right 
 angle, and A the smallest angle ; let A be the number of 
 degrees, and a the number of radians in this angle, so 
 that a = 7rjl/180 = 3J./l72, approximately. 
 
 "N" ^ _ csin J. _ sing 
 
 6 + 2c~2c + ccos J.~2H-cosa 
 
 = ^, approximately, 
 
 ^ 2" 
 
 = l = lf2^ approximately 
 
 This proves Ozanam's formula, when A is not large. 
 Writing / for the fraction 
 
 ^(2 + cos^) 
 sin J. 
 we see then that, for small values of A, J does not differ 
 greatly from 172. In the following table, the value of J 
 is given to three places of decimals for every five degrees 
 from 0' to 45*' ;^ 
 
APPROXIMATIONS AND ERRORS. 373 
 
 A. 
 
 J. 
 
 A. 
 
 /. 
 
 0° 
 
 171-887. 
 
 25° 
 
 171-923 
 
 5° 
 
 171-887. 
 
 30° 
 
 171-962 
 
 10° 
 
 171-888. 
 
 35° 
 
 172-026 
 
 15° 
 
 171-892. 
 
 40° 
 
 172128 
 
 20° 
 
 171-.902. 
 
 45° 
 
 172-279 
 
 The degree of approximation may be shewn by solving the 
 triangle in which 6'= 90°, c=4156, a =2537. 
 We find 6 = 329r8, and, by Ozanam's formula, 
 
 - ^^ 2537x172 ^ 436364 ^3^0 3^. ^g. 
 
 3291-8 + 4156x2 11603-8 
 The correct value of A is 37° 37' 17", so that the absolute error in 
 this case is only 59". 
 
 240. Any three elements of a triangle (except the three 
 angles) being increased by given small amounts, to find 
 the consequent changes in the other three elements. 
 
 Let the increments of a, b, c, A, B, C be denoted by 
 ^y y, z, 0, 0, V^, respectively, 0, (p, and yfr being measured 
 in radians. 
 
 Then a sin B = b sin A, 
 
 and (a + x)sm(B + 0) = (6 + y)sm(A + 0). 
 
 Now, since all terms of the second and higher degrees 
 of X, y, z, 0, (p, \lr may be neglected in comparison with 
 the first, 
 
 (a 4- x)sm{B + 0) = (a + x){sm 5 + cos B) 
 
 = asin J5 + a;sin5 + 0acos5. 
 Hence, making use of the first of the above equations, 
 the second may be written 
 
 ic sin 5 + 0a cos B = y sin A-\-Qb cos A. 
 And, dividing the left side of this equation by a sin 5 
 
374 APPROXIMATIONS AND ERRORS. 
 
 and the right side by the equal expression 6 sin J., we get 
 
 --h0COt5 = ^ + 0COt2l, 
 
 a 
 
 or --dcot^=^-0cot5. 
 
 a b ^ 
 
 Similarly, from the equation a sin (7= c sin ^, we obtain 
 6 cot A = — yfrcotC. 
 
 |-dcot^=|-0cot^ = ^--V^cota (1) 
 
 Again, A+B+C^tt, 
 
 and A + e-\-B+(p + G-{-\[r = 7r, 
 
 e^<p+xlr = (2) 
 
 Hence, if any three of the six quantities x, y, z, 0, <p, 
 xfr (except 0, <p, yj/) be given, the other three can be 
 determined by means of the three equations (1) and (2). 
 
 There are obviously four cases to be considered, cor- 
 responding severally to the four cases in the solution of 
 triangles : — (1) given x, <p, \[r, to find y, z, 0; (2) given 
 y, z, 6, to find x, <f>,-^; (3) given x, y, 0, to find z, <l),\lr; 
 (4) given x, y, z, to find 6, 0, xfr. (See Examples xxv. 
 13-16.) 
 
 241. The last proposition may be proved geometrically 
 as follows : — 
 
 Let the triangle ABC be changed into the triangle 
 A'B'C by changes in any three elements (except the 
 three angles). 
 
 Draw BL parallel to B'A\ AN and BP perpendicular 
 to FA\ AQ parallel to GA\ and AM and QR perpen- 
 dicular to GA\ 
 
 Let LABL = <f), LACM=\lr, and BB' = x; then y = A'M, 
 a.nd z = A'N+PB'. 
 
APPROXIMATIONS AND ERRORS. 375 
 
 A 
 
 \ 
 
 C B B 
 
 Since ^ and x/^ are very small angles, we have AL — 
 C(f) and AM=h\lr', also, if sin(5 + 0) be multiplied by 
 any of the small quantities conside'red, we may replace 
 sin(-B+0) by sin J5, etc. 
 Now, y = MA' = AQ + RA' 
 
 ^ANco^Qc{A + e) + RQcoi{A+e) 
 
 = AL cosec J.+5Pcosec^+J.if cot J. 
 
 = c0 cosec A+x sin(5 + 0)cosec J. + 6\/r cot ^, 
 
 2/ sin J. = c^ + 6i/r cos ^ + aj sin 5 (1) 
 
 Again, z = A'N-^PE = A'Q + QN^PB' 
 = QRco^Qc{A + e)+ANcoi{A-\-6)+BFco^{B+cl>) 
 = 6i/r cosec -4 + C0 cot A +£c sin B cot J. +aj cos B, 
 z^inA — h-yp--\-C(l)Q,o^A-\-xs,m.{A-\-B) 
 
 = 6\/r + c0 cos J. + a? sin (7 (2) 
 
 Also, as before, + ^ + \/r = O, and, by means of this 
 equation, equations (1) and (2) may be transformed to 
 the equations (1). of the preceding article. 
 
376 APPROXIMATIONS AND ERRORS. 
 
 Examples XXV. 
 
 1. It -~vj = TTTTT. then = 4 24 , approximately. 
 
 2. If be nearly a right angle and n>l, then 
 
 (sm dr= — , ., , ; Tv-i — 7i, approximately. 
 
 3. If be less than a radian, then 
 
 ^_^/ 3-3cos0 \^ 
 
 ^-"^VS + cos^y * 
 very approximately, the approximate measure of 
 the error on the left-hand side being 07^^80 radians. 
 
 4. What value should be given to the constant m, in 
 
 order that the formula 
 
 /I tan^+msin^ 
 
 ^= — THi^: — 
 
 shall be the best approximation to the number of 
 radians in a small angle Q in terms of its sine and 
 tangent ? 
 
 5. If u — esinu = t;, where e is small, shew that, if 
 
 powers of e above the first may be neglected, we 
 
 have tan| = (l + e)tan| 
 
 6. If u = ^ 4- ^ sin n^ where e is small, then 
 
 u = + e sin + terms of the second order, 
 
 t6 = 0-|-esin0-|-^sin20H-terms of the third order, 
 
 u = + e sin + ^ sin 2^ + ^ (3 sin 30 — sin <^) 
 
 + terms of the fourth order. 
 
 7. Solve the triangle in which c = 5793, a = 1489, G= 90^ 
 
 by means of Ozanam's formula. 
 
APPROXIMATIONS AND ERRORS. Til 
 
 8. If, in a triangle ABC, ^ = 30°, c=l, and a = 250, find 
 
 approximately the number of minutes and seconds 
 in the other angles. 
 
 9. If, in a triangle ABC, c, A and B be given, A and B 
 
 being measured in radians and very small, find the 
 other sides, and shew that 
 
 a-\-h = c{l-\-lAB). 
 
 10. If, in a triangle ABC, a, h and C be given, when 
 
 C='7r — 0, being measured in radians and very 
 small, find c, A and B in terms of a, b and 6, 
 
 11. If a parallelogram, formed by four jointed rods, be 
 
 slightly deformed, find the change in its area. 
 
 12. In the ambiguous case in the solution of triangles, if 
 
 a, h and A be the given parts, and there be a small 
 
 error y in the value of b, prove that the error in 
 
 either of the corresponding values of c is 
 
 y(G cos A — b) 
 
 c — b cos A 
 
 13. If, in a triangle ABC, a, B and C be increased by the 
 
 small quantities x, <p and yfr, find the resulting 
 changes in b, c and A. 
 
 14. If b, G and A be increased by the small quantities 
 
 y, z and 0, find the resulting changes in a, B and G. 
 
 15. If a, b and A be increased by the small quantities x, 
 
 y and 0, find the resulting changes in c, B and G. 
 
 16. If a, b and c be increased by the small quantities x, 
 
 y and z, find the resulting changes in-^, B and G. 
 
 17. If a triangle be solved from the observed parts 
 
 (7=57°, a = ;^6, 6 = 2, shew that an error of 10" in 
 the value of G would cause an error of about 3"'66 
 in the calculated value of B. 
 
378 APPROXlMATIOyS AND ERRORS. 
 
 18. The sides of a triangle are observed to be a = 5, 6 = 4, 
 
 c = 6, but it is known that there is a small error in 
 the measurement of c ; examine which angle can 
 be determined with the greatest accuracy. 
 
 19. Three vertical posts are placed at intervals of one 
 
 mile along a straight canal, each rising to the same 
 height above the surface of the water. The visual 
 line joining the tops of the two extreme posts cuts 
 the middle post at a point 8 inches below the top. 
 Find, to the nearest mile, the radius of the earth. 
 
 20. On the top of a spire is an iron cross, the length of 
 
 whose arms is a, and whose plane lies east and 
 west. A person standing due north* observes that 
 the horizontal and vertical arms subtend small 
 angles of a and /3 radians respectively. Find the 
 height of the tower and its distance from the 
 observer. 
 
 21. Ay B, G are three given points in a straight line ; D is 
 
 another point whose distance from B is ascertained 
 by observing that the angles ADB, CDB are equal 
 and of an observed magnitude 6 ; prove that the 
 error in the calculated length of BB, consequent on 
 a small error S in the observed magnitude of Q is 
 _ 2ah{a+hfs.m 6 
 {a^-\-b^-2abcos2e)^' ' 
 approximately, where a and h are the distances 
 between A, B and B, G respectively. 
 
 22. The side c and the angles A and B, of a triangle ABC, 
 
 are measured, but, on measuring the angles, equal 
 small errors are made. If the resulting relative 
 errors of the sides a and b be equal, shew that the 
 triangle must be isosceles. 
 
APPROXIMATIONS AND ERRORS. 379 
 
 23. If A be the length of the chord of a circular arc, B 
 
 that of the chord of half the arc, and C that of the 
 chord of a quarter of the arc, the length of the arc 
 is approximately equal to 
 
 ^-40^ + 256(7 
 45 
 Find the absolute error in the length of an arc 
 determined by this formula, the circle being one 
 mile in radius, and the angle subtended by the arc 
 at the centre being 45°. 
 
 24. Shew that the area of a small segment of a circle is 
 
 very nearly f base X height, and that the error is 
 very nearly Q'^J^O of the area, where is the num- 
 ber of radians in the angle subtended by the arc at 
 the centre. 
 
 Calculate the numerical value of this fraction 
 when the angle contains 5°. 
 
 25. The sides of a triangle, a, h„ c, are increased by small 
 
 amounts x, y, z; shew that the radius of its cir- 
 cumcircle is increased by 
 
 ^cotAcotBcotG{xsecA-^y8ecB+z&ecC). 
 
 26. If, in a triangle, a, h and B be given, and if the true 
 
 value of the angle exceed the measured value B by 
 -a small angle of 6 radians, the cube and higher 
 powers of which may be neglected, shew that the 
 diameter of the circumcircle is 
 
 h cosec B[l - cot B + iO^cotW + cosec^B)]. 
 If the square of 6 may be neglected, how is the 
 third side affected by ? 
 
 f 
 
w 
 
 380 THEORY OF PROPORTIONAL PARTS. 
 
 § 2. Theory of Proportional Parts. 
 
 242. We shall now examine the principle stated in 
 Chapter IV. for finding the logarithm of a number, a 
 circular function of an angle, or its logarithm, when the 
 number or angle lies between two given in the tables. 
 
 The approximate nature of this principle, which is 
 known as the principle of proportional parts, will be 
 evident from the following geometrical illustration for 
 the case of the sine of an angle. 
 
 Let OL, OM and ON, measured along a line OX, be 
 ^ proportional to the angles 6, 
 e + S, and e + S\ S and 6' being 
 -, small. From L, M and N draw 
 LP, MQ and NR perpendicular 
 to OX, and proportional to the 
 sines of 6, Q+S and 6-\-S' ', so 
 lif N H that P, Q and R are three points 
 near to one another on the curve of the sine. Draw PK 
 perpendicular to RN, cutting QM in H. • 
 
 Now, according to the principle of proportional parts, 
 we must have 
 
 siD(0+^)-sin : sin^O +S')- sin 6 = 8 : S\ 
 i.e. QH:RK=PH:PK, 
 
 i.e., since the angles PHQ, PKR are equal, PQR must 
 be a straight line. 
 
 Hence, all that is implied in the principle in this case 
 is that if we take two points very near to one another 
 on the curve of the sine, the part of the curve between 
 them does not differ from a straight line. 
 
 We shall suppose the tables to give the logarithms to 
 
THEORY OF PROPORTIONAL PARTS. 381 
 
 the base 10 of all numbers from 1 to 100,000 correct to 
 seven places of decimals, and the circular functions and 
 their logarithms to the same number of places for every 
 minute from 0° to 90°. 
 
 243. Logarithms of Numbers. — Let n be any number 
 and S a number small compared with n. Then 
 
 \og{n + ^) - log ^ = log(l + -) = M log.(l + - 
 
 S 1 ^2 1 ^ 
 
 
 where fj. is the modulus of the common system of loga- 
 rithms. 
 
 Now, if S be so small compared with n that the squares 
 and higher powers of Sjn may be neglected, we have 
 
 \og{n -\-S) — log n = jjiSIn, approximately, 
 i.e., the change in the logarithm varies approximately as 
 the change in the number. 
 
 In the case in which the principle of proportional parts 
 is usually applied, we have n<t 10,000 and ^ = 1 ; also 
 since />t = -43429448, i.e. < |-, we have 
 
 ^ < 7 • Va9 ^-e- < -0000000025 ; 
 2n^ 4 10^ ' 
 
 hence, to at least 7 places of decimals, we have 
 
 \og{n-\-§)- \ogn •.\og{n-\-\) — \ogn = S'. 1, 
 
 S being <1. 
 
 To find the smallest number n whose logarithm can 
 
 be obtained correctly to 7 places of decimals by means 
 
 of the principle of proportional parts, S being 1, we have 
 
 ^' = •0000001, 
 and 91 = 1474, approximately. 
 
382 THEORY OF PROPORTIONAL PARTS. 
 
 Again, if the difference between the logarithms of two 
 consecutive numbers, each containing six digits, be 
 •0000100, then a difference of 0000001 between the 
 logarithms of two numbers lying between the former 
 pair will correspond to a difference of "01 between the 
 numbers ; i.e., given the logarithm, we can in such a 
 case find the corresponding number correctly to two 
 places of decimals. Now, w^hen yu. l/7i= 00001 we have 
 71 = 43429. Hence, if the logarithm be given, we can 
 find the number correctly to two places of decimals if 
 the number be less than 43429, and correctly to one 
 place of decimals if the number be greater than 43429. 
 
 244. Circular Functions. — We shall examine the prin- 
 ciple of proportional parts fully in the case of the sine, 
 and briefly in the cases of the other circular functions. 
 
 Sine. — We have 
 
 sin(0 + (5) — sin = cos ^ sin 5 — sin 6(\ — cos S) 
 
 = Scose-iS^sme-i6^cose+.... 
 
 The ratio of the third term of this series to the first 
 being — ^<5^ and therefore very small, the third and suc- 
 ceeding terms may be neglected. 
 
 The ratio of the second term to the first is — ^S tan 0. 
 
 (1.) If tan 6 be not great, i.e. if be not nearly 7r/2, this 
 ratio is very small, and therefore the second term may be 
 neglected in comparison with the first, and the above 
 equation becomes 
 
 sin {6-\-S)— sin 6 = Scos 0, approximately, 
 i.e. when the change in the angle is very small, the change 
 in the sine of the angle varies approximately as the 
 change in the angle. Hence 
 
THEORY OF PROPORTIONAL PARTS. 383 
 
 (2.) If, however, be very nearly 7r/2, the ratio 
 - \S tan may be finite. If this be the case, the second 
 term cannot be neglected in comparison with the first, 
 and the above equation becomes 
 
 sin(^ + ^) — sin = ^ cos — J^^sin 6, approximately ; 
 consequently, the change in the sine does not vary as the 
 change in the angle ; the change in the sine is said to be 
 irregular. 
 
 Again, the second term, though it cannot be neglected 
 in comparison with the first, is less than the first, for the 
 
 greatest value which J(5tan0can have is J^tanf^ — ^j 
 
 or J^/tan S^ and this is never greater than J (art. 74) ; 
 also, Sco^O is vei:}" small compared with 6, since 6 is 
 very nearly 7r/2 ; hence, sin(0+^) — sin is very small 
 compared with S, and the change in the sine is said to 
 be insensible. A small change in the sine will there- 
 fore correspond to a great change in the angle ; in other 
 words, several successive angles differing by 1" will have 
 the same tabular value of the sine, and, consequently, 
 an angle cannot be found exactly from its sine when it is 
 very nearly a right angle. 
 
 Eeferring to the figure of art. 89, it will be seen that, if POP' be 
 a small constant angle, the difference between P'M' and PM, and 
 therefore the difference between sin J. OP' and sin ^ OP, diminishes 
 as the angle AOP increases, and becomes infinitely small, i.e. insen- 
 sible, when AOP is nearly a right angle. 
 
 245. Cosine. — The cosine of an angle being the sine of 
 its complement, it follows that, when the change in the 
 angle is very small, the change in the cosine varies 
 approximately as the change in the angle, except when 
 the angle is very small; and then the change in the cosine 
 becomes irregular ; in this case it is also insensible. 
 
384 THEOR Y OF PROPORTION A L PA RTS, 
 
 These results may also be deduced from the equation 
 cos — cos(^ + ^) = ^ sin + \^co^ 0, approximately. 
 Tangent. — We have 
 tan(0-4-^)-tand 
 
 _ sin^ 
 
 ""cos(0-f^)cosO 
 
 = sec20 tan S(l- tan tan ^) - 1 
 
 = sec2a(^+~+...)(l + (5tan0+^tan2a+...) 
 
 = 8 sec2^ + ^Han 6 sec^O +S^{i+ t&n^e.jsec^O + . . . . 
 
 As before, the third and succeeding terms may be 
 neglected in comparison with the first. The ratio of the 
 second term to the first is ^tan^. Hence, when the 
 change in the angle is small, the change in the tangent 
 varies approximately as the change in the angle ; except 
 when the angle is nearly a right angle, and then the 
 change in the tangent is irregular. It is never insen- 
 sible, for S sec^O is always > S. 
 
 Cotangent. — The cotangent of an angle being the tan- 
 gent of its complement, it follows that, when the change 
 in the angle is small, the change in the cotangent varies 
 approximately as the change in the angle ; except when 
 the angle is small, and then the change in the cotangent 
 becomes irregular ; it is never, however, insensible. 
 
 These results may also be deduced from the equation 
 cot - cot(0 + ^) = <5 cosec^e - S^cot 6 cosec^^, 
 approximately. 
 
 Secant— ^We have 
 sec(0 +S)-sece = S tan 6 sec ^ + ^( J + tan20)sec 0, 
 approximately, the third and succeeding terms being 
 small in comparison with those that are retained. 
 
THEORY OF PROPORTIONAL PARTS. 385 
 
 The ratio of the second term to the first is (5(JcotO+tan 6). 
 Hence, when the change in the angle is small, the change 
 in the secant varies approximately as the change in the 
 angle, except when the angle is small or nearly a right 
 angle. If the angle he small, the change in the secant is 
 irregular and insensible; if the angle be nearly a right 
 angle, the change in the secant is irregular but not 
 insensible. 
 
 Cosecant. — The cosecant of an angle being the secant of 
 its complement, it follows that, when the change in the 
 angle is small, the change in the cosecant varies approxi- 
 mately as the change in the angle, except when the angle 
 is small or nearly a right angle. If the angle be small, 
 the change in the cosecant is irregular ; if the angle be 
 nearly a right angle, the change is irregular and insensible. 
 
 These results may also be deduced from the equation 
 
 cosec — cosec(0 + (5) = ^ cot cosec Q — 6\^ + cot20)cosec Q, 
 approximately. 
 
 246. Logarithms of the Circular Functions.— >Sfme. — 
 We have 
 
 log sm(0 + (5) - log sm = log — ^T^^-g— 
 
 = log(cos ^ + cot sin S) 
 = log(l+^cot0-i^2_ __) 
 
 = fi{S cot e - i^2cosec20 +...), 
 
 fx being the modulus of the common system of logarithms, 
 
 and the third and succeeding terms being neglected in 
 
 comparison with those that are retained. 
 
 The ratio of the second term to the first is 
 
 — ^S cosec^^ tan 6 or —S cosec 20, which is small unless 
 
 Q be small or nearly a right angle. 
 
 2 b 
 
386 THEORY OF PROPORTIONAL PARTS. 
 
 Hence, when the change in the angle is small, the 
 change in the logarithm of the sine varies approximately 
 as the change in the angle, except when the angle is 
 small or nearly a right angle. If the angle be small, the 
 change in the logarithm of the sine is irregular but not 
 insensible ; if the angle be nearly a right angle, the 
 change is irregular and insensible. 
 
 Cosine. — When the change in the angle is small, the 
 change in the logarithm of the cosine varies as the change 
 in the angle, except when the angle is small or nearly a 
 right angle. If the angle be small, the change in the 
 logarithm of the cosine is irregular and insensible ; if the 
 angle be nearly a right angle the change is irregular. 
 
 These results may also be deduced from the equation 
 log cos e - log cos(0 +S) = jjl8 tan 6 + iimShec^e, 
 approximately. 
 
 Tangent — From the two preceding results we have 
 
 logtan(O + ^)-logtan0 = logsin(0 + ^)-logcos(^ + ^) 
 — (log sin 6 — log cos 6) 
 = juiS(cot e + tan 0) - ifxS^cosec^O - sec^^) + . . . 
 = 2jj.S cosec 20 - 2iuLS^cosec 26 cot 20+.... 
 
 Now, cot 20 being great when is small or nearly 
 a right angle, it follows that, when the change in the 
 angle is small, the change in the logarithm of the tangent 
 varies approximately as the change in the angle, except 
 when the angle is small or nearly a right angle. In both 
 these cases the change in the logarithm of the tangent is 
 irregular but not insensible. 
 
 Cotangent. — The same results are true for the loga- 
 rithm of the cotangent. 
 
 Also, since 
 log cot - log coi(0 -\-S) = log tan {0 + S)- log tan 0, 
 
THEORY OF PROPORTIONAL PARTS. 387 
 
 it follows that the diifereDces for any small change in 
 the angle are numerically the same in the logarithms of 
 the tangent and cotangent of the angle. 
 Secant and Cosecant. — Lastly, since 
 log sec(0 -\-S) — log sec Q = log cos 0— log cos(0 + S), 
 and log cosec(0 + ^) — log cosec d = log sin Q — log sin(0 + (5), 
 it follows that the results for the logarithms of the secant 
 and cosecant are the same as for those of the cosine and 
 sine, respectively ; also that the differences for any small 
 change in the angle are numerically the same for the 
 logarithms of the cosine and secant and for those of the 
 sine and cosecant. 
 
 247. If f{0) denote any circular function of an angle 6^ or its 
 logarithm, we have seen that, in every case, 
 
 f{d+^)-f{e)=A^+B^^+..., 
 
 where A, B, etc., ar^ functions of 6 but not of 8. 
 
 If B8^ be not small compared with A8, the change in the circular 
 function or its logarithm is irregular. If A8 be small compared 
 with 8, the change is insensible. 
 
 248. When the Angle is given. — In determining the value of any 
 circular function of an angle or its logarithm, by means of the prin- 
 ciple of proportional parts, we have to take into account only the 
 irregularities in its change. 
 
 The largest difference-angle with which we have to deal is just 
 less than one minute, and therefore the irregularity in the change 
 of the function may affect the seventh place of decimals if B8^ be 
 not less than '0000001, where 8 is tlie number of radians in one 
 minute. The limiting value of the angle 6 is given by the equation 
 
 B=-ooooooix(15§ooy. 
 
 249. When the Circular Function or its Logarithm is given. — The 
 accuracy of the calculated value of the corresponding angle to the 
 nearest second depends both on the insensibility and the irregularity 
 in the change of the function. 
 
388 THEORY OF PROPORTIONAL PARTS. 
 
 The change in the function will be insensible for two angles 
 differing by one second, if Ah be less than '000000 1, where S is the 
 number of radians in one second. The limiting value of the angle 
 6 is given by the equation 
 
 ^ = •0000001x548000 
 
 IT 
 
 The calculated value of the angle may, on account of the irregu- 
 larity in the change of the function, differ from the true value, if 
 ji?82 be not less than -^ of ^8, where S is the number of radians in 
 one minute, this being just greater than the largest difference- 
 angle with which we have to deal. The limiting value of the 
 angle 6 is given by the equation 
 
 i?_ 10800 
 
 A GOtt* 
 
 Miscellaneous Examples. III. 
 
 a. 
 
 1. If ^+5+ C=7r, prove that 
 
 sin3^ sin(5 - C) + sin35 sin((7- ^) + sin3(7 sin(^ - 5) = 0. 
 
 2. If^+5+a=f, prove that 
 
 cosec A cosec B cosec G—cotB tan G-cotC tan B 
 — cot G tan ^ — cot ^ tan G— cot A tan 5— cot B tan A = 2. 
 
 3. If.4 + 5+a+i) = 2'7r, prove that 
 
 cos J^ cos JD sin J5sinJC— cos JJ5cos JCsin hA sin JD 
 = sin i(A-{-B)sin ^{A + C)cos i(A +Dy 
 
 4. Prove that 
 
 S cos 2a sin(/3 - y) + 2 sin(;8 - y) . S cos(/3 + y) = 0. 
 
 5. Prove that 
 
 S cos 2a cot J(y - a)cot J(a - /3) = 2 cos 2a + 22 cos(/3 + y). 
 
 6. Prove that 
 
 2 sin ^ sin y sin(/? — y)sin(3a+/3 + y) 
 +sin(a+/3 + y) .nsin(^-y) = 0. 
 
MISCELLANEOUS EXAMPLES. 389 
 
 TT 1 1 
 
 1. Prove that ^ = 2tan-^^ + tan-ij=. 
 
 2. Prove that •-- = tan-i^ + tan-i-+tan~i-. 
 
 3. Prove that :r = 2 sin" ^—7-rr — sin ~i- 
 
 4. Provethat2 tan~i- — tan-i^ = j,2tan-ij, + tan-^^ = 2-, 
 
 5 1 TT 
 
 2tan-i— — tan-i^^=— , and, generally, that 
 
 2tan-i^"-' + (-irtan-i— = ^, where l,^\-^2 
 are successive convergents to ^2. 
 
 5. Prove that 
 
 tan-^=,=tan"i7i-tan-^K.tan-^-7Tr=tan-^— r-tan-^Ts, 
 7 2 3' 41 12 17 
 
 tan"^^^ = tan-^j=^ — tan-i^, and, generally, that 
 
 tan-i — = tan-^ tan~^ , where 1, —,—,... 
 
 are successive convergents to ^2. 
 
 6. Prove that tanh-^-^ = tanh-i^ — tanh-i->,, tanh'^Tr;^ 
 
 17 12' 99 
 
 = tanh-i„ — tanh-i— , and, generally, that 
 tanh-i = tanh-^ tanh"^ — , where 1, — , 
 
 P-2n+l q2n 'p2n ?i 
 
 — ,... are successive convergents to ^2. 
 
390 MISCELLANEOUS EXAMPLES. 
 
 y- 
 
 1. If in a triangle the median which bisects the base c 
 
 is perpendicular to the side 6, then 
 2tan^ + tana=a 
 
 2. In any triangle S- + 6i2=2i2. 2—. 
 
 T-^ Ob 
 
 3. If a, P, y be the lengths of the lines joining the feet 
 
 of the altitudes of a triangle, then 
 
 a- 62+^2- 2a6c * 
 
 4. Shew that the line joining the middle point of BG to 
 
 the middle point of the perpendicular from A on 
 BG, makes with BG an angle whose cotangent is 
 cot B <- cot G. 
 
 5. Find the inclination of the line joining the centres of 
 
 the inscribed and circumscribed circles of a triangle 
 to its base. 
 
 6. If straight lines be drawn through the vertices of a 
 
 triangle, bisecting the exterior angles, and if A be 
 the area and P the perimeter of the original 
 triangle, and A', P' the corresponding quantities 
 in the new triangle, shew that 
 4AA' = Pa6c, 
 
 and PP' = 4AY cos ^ + cos ^ + cos ^ j. 
 
 8. 
 1. In any quadrilateral figure whose diagonals intersect 
 at right angles, if S and D be respectively the 
 sum and difference of two opposite sides, and S\ 
 ly the sum and difl'erence of the other two oppo- 
 site sides, then 
 
MISCELLANEOUS EXAMPLES. ^91 
 
 2. In any triangle, whose perimeter 2s is given, the value 
 
 of Rr^^^ is greatest when the triangle is equi- 
 
 lateral, and is then equal to ^. 
 
 3. A polygon of Zn sides, which are a, b, c successively, 
 
 repeated n times, is inscribed in a circle; if the 
 angular points be A, B, G, D, E, etc., and the 
 radius of the circle be denoted by r, prove that 
 
 AC'^ = I ac + 2hr Hm-]U}C + 2ar sin -]-r-(ab-\- 2cr sin- j, 
 
 with similar expressions for BD and CB. 
 
 4. The circumference of a circle is divided into twelve 
 
 arcs, whose lengths, taken in order, are in arith- 
 metical progression, and the first six together form 
 a quadrant. Shew that the area of the polygon 
 bounded by the chords of the twelve arcs is 
 
 — cosec fz-x, where a is the radius of the circle. 
 
 o 72 
 
 5. If the diameters AA\ BB\ CC of the circumcircle of 
 
 a triangle ABC cut the sides in D, E, F respec- 
 tively, then 
 
 ad^be^cf~b: 
 
 ^^^ 11^^ WE'^ GT^ ¥R ^^^^ '^ ^^^ ^ ^^^ ^^ ^^' 
 
 6. A flagstafi" on the top of a tower is observed to subtend 
 
 the same angle (a) at two points in a horizontal 
 plane on the same line through the centre of the 
 base, whose distance from each other is 2a, and 
 an angle /3 at a point half way between them. 
 Find the height of the flagstaff. 
 
MISCELLANEOUS EXAMPLES. 
 
 1. One of the aogles of a plane triangle is 60°, and the 
 
 sides including it are in the ratio of 3 to 5 ; find 
 the tangents of the other angles. 
 
 2. A, B, G are three points in a straight line such that 
 
 AB and BG each subtend an angle of 80° at a 
 point P. If AB = a, BG=c, shew that the differ- 
 
 ence between AP and GP is , - > 
 
 3. From a point P in the side AG of a, triangle ABG, a 
 
 line is drawn bisecting the triangle and making 
 an angle 6 with AG, shew that 
 2AP^ 
 
 cot = -Tn Tri COSCC A—COtA. 
 
 AB .AG 
 
 4. The alternate angular points of a regular pentagon are 
 
 joined by straight lines. Find the length of a side 
 of the pentagon formed by these lines, and shew 
 that the radius of the circle circumscribing it is 
 
 as/^ — ^ls/^f where a is the side of the original 
 pentagon. 
 
 5. Through the angular point C of a triangle ABG a 
 
 straight line GPQ is drawn, on which are let fall 
 the perpendiculars AP, BQ ; prove that 
 PQ = AP cot B~BQ cot A. 
 
 6. A regular pentagon and a regular hexagon are in- 
 
 scribed in a circle, so as to have an angular point 
 in common, and the other adjacent angular points 
 are joined ; shew that the perimeter of the figure 
 so formed is 4rsin 18° sin 15°cosec3°, where r is 
 the radius of the circle. 
 
MISCELLANEOUS EXAMPLES. 393 
 
 t 
 
 1. A circle whose centre is I and radius r is inscribed 
 in the triangle ABC, and touches the sides in 
 D, E, F. Circles whose radii are r^^ r^, r^, are 
 inscribed in the quadrilaterals AEIF, BFID^ 
 CD IE; shew that 
 
 2. In an isosceles triangle a series of circles is inscribed, 
 
 the first of which is the inscribed circle of the 
 triangle, and the others touch the preceding one 
 and the two equal sides of the triangle. If the 
 sum of the areas of the circles is equal to four- 
 thirds of the first, find the ratio of the sides of the 
 triangle. 
 
 3. A triangle is divided into two parts by a line through 
 
 one of the angular points A, such that the circles 
 inscribed in the two parts touch the dividing line 
 in the same point. Shew that 0, the inclination of 
 the dividing line to the opposite side, is given by 
 the equation 
 
 cot (p = jf tan ^ ~ tan ^^j. 
 
 4. A plane polygon whose sides are a^, a^, a^..., and area 
 
 A, is divided into triangles by joining its angular 
 points with a point, at which the sides subtend 
 
 angles 0-^, 0^, 0^ If the centres of the circum- 
 
 circles of these triangles be joined in order, the 
 area of the polygon thus formed is 
 
 i{A-i^{a^cote)}. 
 
894 MISCELLANEOUS EXAMPLES. 
 
 6. From a point A outside a circle two lines of equal 
 length AB, AC, drawn to the ends of a diameter 
 BG, and the circumference m D, E -, if BE, CD 
 meet in 0, the area of the quadrilateral 
 ABOC= ^BC^cot BA C. Compare that part of the 
 area of the circle which is outside the triangle 
 ABC with that which is included within it. 
 
 6. PX and PY are two fixed right lines meeting at an 
 acute angle P. On PX two fixed points B and C 
 are taken. Shew that, if A be the point on PF at 
 which BC subtends the greatest angle possible, 
 
 and that when P is a right angle, this reduces to 
 PC-PB 
 
 sin B AC: 
 
 PC+PB' 
 
 1. If a = -y- and x = cos a, shew that 8x^ + 4!X^ — 4a3 — 1 = 0, 
 
 and that the other roots of the equation are cos 2a 
 and cos 3a. 
 
 2. If a = ~ and x = sma, shew that a^-'^x^ + ^ = 0, 
 
 and that the other roots of the equation are sin 2a 
 and sin 4a. 
 
 3. If a = -i^, then cos a + cos 2a + cos 3a = — J, 
 
 cos a cos 2a + cos a cos 3a + cos 2a cos 3a = — J, 
 cos a cos 2a cos 8a = J. 
 
MISCELLANEOUS EXAMPLES. 395 
 
 27r 
 
 4. If a = ^, then sin a + sin 2a + sin 40 = 1^^7, 
 
 sin a sin 2a -f- sin a sin 4a + sin 2a sin 4a = 0, 
 
 sin a sin 2a sin 4a = — Jx/'^- 
 
 27J- 
 
 5. If a = -=-, then cos^a + cos22a + cos^Sa = f , 
 
 sin^a + sin22a + sin24a = |. 
 
 6. If ABGDEFG be a regular heptagon inscribed in a 
 
 circle of unit radius, then will 
 
 e. 
 
 1. Prove that one solution of the equations 
 
 x^ = a^ + ay, y^ = a^-\-xy 
 X y a 
 
 sin — sm -y sm y 
 
 and find the other solutions. 
 
 2. Prove that the equations 
 
 ay+a^ = x^, xz-\-a^ = y^, yz + a^ = z^, 
 
 are satishea by -. — rr- = -r-^^ = — — t- — - — , 
 '' sm 2a sin 3a sm 4a sm a 
 
 when c(=q, and solve the equations. 
 
 8. Shew that one root of the equation ^x^ — 4a;2 — 4;^? + 1 = 
 
 is cos -=, and find the other roots. 
 
 4. Shew that sin = is a root of th'e equation 
 
 x^ + ^x''-^^ = 0, 
 and find the other roots. 
 
396 MISCELLANEOUS EXAMPLES. 
 
 6. If o. = jxy then will sin a + sin 13a = —J, 
 
 and sin a sin 13a = — J. 
 
 6. Prove that 
 
 (a: — 2 cos -R- )( ^ — 2 cos -r- )( !» — 2 cos -v )( ^ — 2 cos -^j 
 
 =a;H2a;8-a;2_2a;+l. 
 
 1 . Prove that cc = tanh X'\-\ tanh^a? + ^ tan \\^x + ...ad. inf. 
 
 2. Prove that 2(cos ^ + J cos^^ + \ cosM + . . .) 
 
 = cos2— - sin2^ +-(^cos*2 - sill 9 y 
 
 ■^iG^^'2 "^'"'2 ) + •••• 
 
 3. Prove that 2«cos Q cos 20 cos 2^0 .. . cos 2*^0 
 
 = cos + cos 30 + cos 50+ ... +cos(2'^+i - 1)0. 
 
 4. Find the sum of n terms of the series 
 
 log(l + cos 0) + \og\l + cos 2) + \og{\ + cos^aj + • . • • 
 
 5. Prove that 
 
 ( j = 1- sin^Jic-cos^Ja? sin^Jaj-cos^liccos^Jicsin'^Ja;-.... 
 
 6. Sum to infinity the series 
 
 1. Tr.l, TT.l, TT, 
 
 22^^^22'^23 23"^2^ 2'*"^*'" 
 
 X. 
 
 1. If /i be positive, and < < ^, then 
 
 sin0 sin(0 + ^) 
 ~0~ ~0+r"' 
 
MISCELLANEOUS EXAMPLES. 897 
 
 . . x^ 
 
 2. Find the limit of , when x = 0. 
 
 1 — cos mx 
 
 3. Find the limit of ^^ , when Q = 0. 
 
 4. If the unit of measurement be a right angle, find the 
 
 limit of ^3 , when = 0. 
 
 5. Find the limit of ^^ -, when x = 0. 
 
 SlW^X 
 
 6. Three mountain peaks A, B, G appear to an observer 
 
 to be in a straight line when he stands at each 
 of two places P and Q in the same horizontal line ; 
 the angle subtended hy AB and BG at each place 
 is a, and the angles A QP, GPQ are (p and \[r, 
 respectively. Prove that the lieights of the 
 mountains are as 
 cot 2a+cot i/r : J(cota+coti/r)(cota+cot0)tana : cot2a+cot0, 
 and that, if QB cut AG in D, 
 
 AG= GD . sin 2a(cot a + cot i/r). 
 
PART III. 
 COMPLEX QUANTITY. 
 
 "Every combination of symbols can be explained, and everything 
 explicable is a line of definite le ngth and direction, and every such 
 line can be represented byp + gV-1." — De Morgan. 
 
 CHAPTER XVI. 
 COMPLEX NUMBERS. 
 
 250. On the Representation of Positive and Negative 
 Numbers by Straight Lines. — We know that if lengths 
 measured along a straight line, from a point in the line as 
 an origin, be denoted by positive numbers, then lengths 
 measured from the origin along the line in the opposite 
 direction will be denoted by negative numbers. 
 
 Conversely, all positive numbers may be represented 
 by lengths measured in one direction along a straight line, 
 and all negative numbers by lengths measured from the 
 same origin in the opposite direction. « 
 
 In the same manner, quantity of any kind may be 
 represented by lengths measured along a straight line by 
 taking a unit of length to represent a unit of the quan- 
 tity considered. Thus, if angles are the quantities con- 
 sidered, a radian may be represented by an inch, and 
 
 398 
 
COMPLEX NUMBERS. 399 
 
 radians may then be added or subtracted by operating 
 upon their representative lengths. 
 
 All real number or quantity may accordingly be repre- 
 sented by length measured along a straight line ; and, in 
 the case of quantity which can be conceived as existing 
 in opposite conditions, if one of these conditions be repre- 
 sented by lengths along the line in one direction, the 
 other condition will be represented by lengths measured 
 in the opposite direction. 
 
 In some cases these opposite conditions of quantit}?- are 
 inconceivable, for example, a negative number of lbs. of 
 matter, or a negative number of ergs of energy are in the 
 fullest sense of the word ' impossible ' quantities. 
 
 251. On the Direction of Number. — Hitherto, a nega- 
 tive number has been obtained by measuring a length 
 from the origin in the negative sense of the straight line 
 along which numbers are represented. 
 
 The same result may be obtained by measuring the 
 same length from the origin in the positive sense, and then 
 rotating the length about the origin through two right 
 angles in a plane containing the line ; thus, a negative 
 number is the positive number equal to it in magnitude 
 turned through an angle of two right angles. We may 
 accordingly substitute for the symbol — a the equivalent 
 symbol {a, tt), where a denotes the magnitude only of the 
 number and tt the number of radians through which 
 rotation has taken place. 
 
 This mode of representing negative numbers suggests 
 
 an extension of the idea of number. From the origin we 
 
 may take a length along the line of positive number, and 
 
 * then rotate the length about the origin through any given 
 
400 COMPLEX NUMBERS. 
 
 angle. The line thus obtained is called a vector, the 
 number it represents a complex number ; the magnitude 
 of the line is called the modulus of the number, and the 
 angle through which rotation has taken place the ampli- 
 tude of tlie number. The vector or number is fully 
 denoted by the double symbol (a, a), where a is the 
 modulus, and a the amplitude of the number. 
 
 The modulus of a given complex number a; is a one- 
 valued quantity and is denoted by mod(a;). The ampli- 
 tude is many-valued, its values forming a series of angles 
 of constant difference 27r, and is denoted by Amp(.'z;). 
 The principal value of the amplitude is that value which 
 is greater than — tt and not greater than tt, and is denoted 
 by amp(aj). 
 
 It will be observed that 
 
 Amp(aj) = 2n7r + amp(aj), 
 
 where n is any integer, positive or negative. 
 
 The successive values of the amplitude obtained by 
 assigning to n in the equation 
 
 Amp(cc) = 2n7r + amp(a;) 
 
 the values 1, 2, 3,... are called the 1st, 2nd, 3rd, ... positive 
 values, while the values obtained by putting n— —1, 
 — 2, —3,... are called the 1st, 2nd, 3rd, ... negative values. 
 Positive and negative numbers are special cases of com- 
 plex numbers, e.g., -|-3 = (3, 0) or (3, 27r) or, generally, 
 (3, 2n'7r) where n is zero or any positive or negative 
 integer ; and — 5 = (5, tt) or (5, 2n+l . tt) where n is zero 
 or any positive or negative integer. Thus, 
 
 mod( + 3) = 3, Amp(-|-3) = 27i'7r, amp(-f-3) = 0, 
 
 and 
 
 mod( — 5) = 5, Amp{ — 5)={2n + l)7r, arap( — 5)=^. 
 
COMPLEX NUMBERS. 401 
 
 252. The Addition of Complex Numbers. 
 
 OZ, OB represent any complex numbers 
 {a, a), (h, P) respectively, and if from 
 A a line AC he drawn equal to OB, 
 and inclined to OX the line of positive 
 number (called the primary axis) at 
 the same angle as OB, then the com- 
 plex number represented by 00 is 6 X 
 called the sum of the complex numbers (a, a), (b, ^) ; or, 
 if (c, y) be the number represented by 00, then 
 
 {a,a)+(6,/3) = (c,y). 
 
 By making a and /3 equal to or tt we obtain the sum 
 of two positive or negative numbers as defined for arith- 
 metical or real algebraical quantity. 
 
 253. The order of addition of two complex numbers 
 is indifferent. 
 
 Complete the parallelogram OAGB (see fig., art. 252), 
 then, by Eucl. I. 34, BC is equal and parallel to OA, there- 
 fore, by the definition of addition, the sum of OB and OA 
 is also represented by 00 ; hence 
 
 OA + OB^OB + OA, 
 i.e., the Commutative Law in Addition holds for complex 
 numbers. 
 
 The addition of two complex numbers may be repre- 
 sented in three ways : — 
 
 (1.) 0^ + 0^ = the third side of a triangle whose other 
 sides taken in order are equal to OA 
 and OB in magnitude and direction ; 
 (2.) OA + 0^ = the diagonal through of the parallelo- 
 gram whose sides are OA, OB. 
 
402 COMPLEX NUMBERS. 
 
 (3.) Oil + Oi? = twice the complex number represented 
 by the median through of the 
 triangle OAB. 
 
 254. The Multiplication of Complex Numbers.— In 
 
 considering the addition of complex numbers, it was 
 necessary to extend the definition of the operation of 
 addition ; for multiplication no such extension is required. 
 We employ the following definition used in arithmetic and 
 real algebra : 
 
 Def.— To multiply one number by a second, we do to 
 the first what is done to unity to obtain the second. 
 
 255. With this definition we can prove that The 'pro- 
 duct of two complex numbers is a complex number whose 
 modulus is the product of the moduli of the factors, and 
 whose amplitude is the sum of the amplitudes of the factors. 
 
 Let (a, a) and (b, /3) be two complex numbers, then 
 shall (a, a)x(6, /3) = (a6, ^+;g). 
 
 To multiply (a, a) by (6, /3) we do to (a, a) what is done 
 to unity to obtain (b, /B). 
 
 Now, to obtain (b, (3) from unity, we multiply the unit 
 by the number 6, and rotate the resulting length through 
 an angle ^. 
 
 Hence, to multiply (a, a) by (b, /3) we multiply the 
 length of (a, a) by the number b, thus obtaining a length 
 ab, and then rotate the length ab through an angle (3 
 from its former direction, thus finally obtaining a length 
 ab making an angle a + /3 with the primary axis, i.e. a 
 complex number whose modulus is ab, and whose ampli- 
 tude is a + /3. 
 
 Hence, (a, a) X (b, /3) = (ab, a + ^). 
 
 Cor. (1, a)xa = (a, a). 
 
COMPLEX NUMBERS. 
 
 403 
 
 256. The order of multiplication of two complex 
 numbers is indifferent. 
 
 We have shewn that 
 
 (a, a) X (h, /3) = (ah, ^+2) (art. 255) 
 
 and that (6, /5)x (a, a) = {ba, 13 + a) „ 
 
 Now, ah = ha SLYid a+/3 = ^-ha, 
 
 therefore (ah, a + /3) and (ha, /3 + a) are one and the same 
 complex number, and therefore 
 
 (a, a) X (h, 0) = (b, /3) X (a, a), 
 i.e., the Commutative Law in Multiplication holds for 
 complex numbers. 
 
 257. The Distributive Law in Multiplication. — To 
 
 prove tha^t 
 
 (u-\-v)xw = uxw+vxw, 
 where u, v, lo are complex numbers. 
 
 Let OA=u, AB = v, 
 
 then OB = u + v. 
 
 Let m = mod(w), = amp(^t;). 
 
 In OA take a length Oa equal to 
 m. OA, and from a draw ah parallel 
 to AB and equal to m.AB; then, by 
 similar triangles OAB, Oab, Ob is 
 in a straight line with OB, and the 
 length Ob = m. OB. Euel. YI. 6. 
 
 Next, rotate the triangle Oab' about without change 
 of size or shape through an angle to the position OA'B\ 
 then, by the definition of multiplication, we have 
 OF'='OBx(m,e) = OBxw, 
 UT = qA X (m, 0) = OA X w, 
 and A7B' = ABx(m,e) = ABxw. 
 
 But 6B = 0A'+A'B\ (Def. of Addition) 
 
404 
 
 COMPLEX NUMBERS. 
 
 OBxw=OAxw+ABxw, 
 or (u+v)xw=uxw-{-vxw, 
 
 i.e., the Distributive Law in Multiplication holds for 
 complex numbers. 
 
 258. The Grouping of Complex Numbers in Addition 
 iC and Multiplication. — Let u, v, %v 
 be complex numbers ; then shall 
 
 {ii+v)-\-w = u-\-{v+iu) 
 and {uxv)xw = ux{vxw). 
 
 (1) Let OZ, AB, BG represent 
 the numbers u, v, w respectively, 
 then, 
 
 {u+v)+w = (OA-\-AB)-^BG 
 (Def. of Addition) 
 
 O X 
 
 = OB+BG 
 
 and u-h(v-hiu) = OA-{-(AB-\-BG) 
 = qA+AG 
 = 00; 
 (u-\-v)+w = u+(v+w). 
 
 (2) Let a, a be the modulus and amplitude of u; b, ^ 
 of V ; and c, y of w. 
 Then (uxv)xw = [(a, a) X (b, /5)] X (c, y) 
 
 = (ab, ^+^) X (c , y) (art. 255) 
 
 = (abc,a + /3 + y), 
 and ux{vxw) = (a, a) x [{b, /3) x (c, y)] 
 = (a, a) x(bc,/3+ y) 
 = (a6c, a + /3+y). 
 Therefore {uxv)xw = ux(vxw). 
 
 Thus, the Associative Law in Addition and Multipli- 
 cation holds for complex numbers. 
 
COMPLEX NUMBERS. 405 
 
 259.' Conjugate Complex Numbers. — Def.— If two 
 
 complex numbers have equal moduli, and amplitudes 
 of equal magnitude and contrary sense, the complex 
 numbers are said to be conjugate to one another. 
 
 Or, in symbols, (a, a) and {a, — a) are conjugate com- 
 plex numbers. 
 
 Addition of Conjugate Numbers.— Let OA, OA' re- 
 present two conjugate complex num- ' ^^a 
 bers {a, a), (a, — a); join AA', and 
 let AA' meet the primary axis in N, 
 then, by elementary geometry, ON ^ 
 bisects AA' at right angles. 
 
 Since N is the mid-point of AA'^ a! 
 
 we have '0A + 0A'=20N, (art. 253) 
 
 (a, a) -f (a, — a) = (2(x cos a, 0) = 2a cos a. 
 
 In like manner, it may be shewn that 
 
 (a, a) — {a, —a) — {a, a) -|- (a, tt — a) = 2la sin a, |^J. 
 
 Multiplication of Conjugate Numbers. — We have 
 (a, a) X (a, — a) = (a^, a — a) = (a^, 0) = a^, 
 or, the product of two conjugate numbers is a number 
 whose modulus is the square of the modulus of each of 
 the numbers, and whose amplitude is zero. 
 If a = l, we have 
 
 (1, a)x(l,-a) = (l,0) = l, 
 or conjugate complex numbers of unit modulus are re- 
 ciprocal to one another. 
 
 260. Powers and Roots of Complex Numbers.— 
 Powers. — By art. 255 we have 
 
 (a, a)x(6, ^) = (a6, ^+^), 
 hence, by repeated application of the rule, we get 
 (ttp ai) X (a^, ag) X . . . X (a„, an) = (a^a^ ...an, ai + a2+...+a„). 
 
406 COMPLEX N UMBERS. 
 
 Let the moduli a^ a^,-.. an be each equal to a, and the 
 amplitudes aj, 02... an be each equal to a, then we get 
 
 (a, a)" = (a^'r^a), 
 i.e. the nih. power of a complex number is a complex 
 number whose modulus is the nth. power of the modulus 
 of the original number, and whose amplitude is n times 
 the amplitude of the original number. 
 
 Roots.— Since (a, aY^ia"^, no) it follows that (a, a) 
 is an nth. root of (a**, no), or, putting r for ct", and for 
 na, and denoting the arithmetical nth root of r by Jijr, 
 we see that 
 
 (^r, - j is an nth. root of (r, 0). 
 
 Again, since (r, 0) = (r, 0+2m7r), where m is any in- 
 teger, we see, further, that 
 
 (jijr, -) is an nth root of (r, 0). 
 
 By giving m the series of values 0, 1, 2, ...n — 1 we get 
 as nth roots of (r, 0) n complex numbers no two of which 
 have the same amplitude, hence n distinct nth roots of 
 any complex number exist. 
 
 We may assign to m integral values other than those 
 of the series 0, 1, 2 ...n—1, but we shall not in this way 
 obtain any additional roots. For, if we put m = rn+8 
 where r is any integer and s an integer of the series 
 0, 1, 2...n-l then 
 
 [:^r, —^-) = [^r, -^^+ 2rxj = [^r, -^^), 
 
 Thus the nth root of a given complex number x is an 
 -^-valued quantity. The many-valued nature of the root 
 
 is indicated by using double brackets, thus ((x)y. 
 
COMPLEX NUMBERS. 407 
 
 We define as the 'principal nih root of x that root 
 whose amplitude is the nth. part of the principal value 
 of the amplitude of x, and we denote this principal 
 
 root either by {xY or a;" or ^cc. 
 
 Fractional Indices. — Let n = -, where p and a are 
 positive integers prime to each other. 
 
 Then, defining the -th power of a number as the pth 
 
 power of the ^th root of the number, and using double 
 brackets as before to denote the many-valued-ness of the 
 fractional power, we have 
 
 {{r,6)Y = {{{r,e)ff 
 
 If be the principal value of the amplitude of (r, 0) we 
 
 p 
 get the principal value of ((r, 0))q by putting m = 0, and 
 
 the 1st, 2nd, 3rd,... positive and negative values by 
 putting m = l, 2, 3, ... or —1, —2, —3, ... respectively. 
 
 We shall obtain the same values of ((r, 0))q, but, in 
 general, in a different order, by giving to m the series of 
 
 values 0, 1, 2,...q — l in the expression l^^^,— —), 
 
 provided that - is in its lowest terms. 
 
 Square Roots of - 1. — The two square roots of (1, tt) 
 
 are (l, |) and (l, -g. 
 
 Now, (1, 7r)=-l, 
 
 therefore the square roots of — 1 are the complex numbers 
 
408 COMPLEX NUMBERS. 
 
 f 1, ~j and f 1, — r), tlie former being the principal value 
 of the square root, aijd in consequence that denoted by 
 
 The symbol i is used as an abbreviation for the com- 
 plex number f 1, ^j or \/—l. 
 
 Since f 1, ^j and f 1, —-) have equal moduli and 
 opposite directions, 
 
 0'-|)=-(^-|> 
 
 thus, the two square roots of —1 are i and —i where 
 Cube Roots of 1. — The three cube roots of (1, 2?i7r) 
 
 -(i.o).(i,^)(i,-|). 
 
 Therefore the cube roots of 1 are 1, (l,-^) and 
 / 27r\ . . . \ ^/ 
 
 (l, — S-), the first being the principal value of the cube 
 
 root, the second the first positive value, and the third 
 the first negative or second positive value of the root. 
 
 If we use the symbol o) as an abbreviation for the com- 
 plex number f 1, -^ V then f 1, -^j =0)^, and the cube roots 
 of unity are 1, w, w^. 
 
 It will be observed that (a))22 = (l, ^) = (l, ^) = co; 
 thus either of the roots co, or' is the square of the other. 
 
COMPLEX NUMBERS. 409 
 
 Negative Indices. — Let n= —m where m is positive. 
 Then, defining a negative power of a number as the recip- 
 rocal of the corresponding positive power of the number, 
 we have 
 
 (a, aY = (a, «)""*= --i^=^ — ^ — -, 
 ^ ^ ^ • ^ {a,aT' (a"*, ma) 
 
 = (^^>-^a) (art. 259) 
 
 = (a", na). 
 
 261. The Resolution of Complex Numbers. — Jl ^2/ 
 
 complex number may he expressed as the sum of two 
 complex numbers having given amplitudes not differing 
 by a multitude of x. 
 
 Let OP = any complex number B^ 
 
 (r, 0), and let OA, OB make any 
 given angles a, /3 with the primary 
 axis OX. 
 
 Draw PN parallel to OB to meet q 
 OA in iV, then 
 
 OP=ON+NP, 
 i.e., (r,e) = m',a) + (NP,fi}. 
 
 It may be shewn that ON or mod (OiY)= . }^ — -^, 
 
 •^ ^ ^ sm(/3-a) 
 
 and that NP or mod(:^)=:^^l'^^'^. 
 
 ^ ^ sin(/3 — a) 
 
 Let a = 0, /3=Q, so that 0^ coincides with OX the 
 primary axis, and OB coincides with a line OY, making 
 an angle ^ in the positive sense with OX (called the. 
 secondary axis), and let x, y denote the lengths ON, NP ; 
 
410 COMPLEX NUMBERS. 
 
 then OP=ON+NP, 
 
 or . (r, e) = (x, 0)-\-[y, '^) = x+iy 
 
 = rcos 6 + i. rsin 
 = r{cofi 0+i sin 0). 
 
 Def. — If a complex number (r, 0) is expressed in the 
 form x-\-iy, the term x is called the real part, and the 
 term iy the imaginary part of the complex number (?', 6). 
 
 The expressions " real part " and " imaginary part " are 
 to be regarded as conventional expressions only. Con- 
 sidered as abstractions, all numbers, positive, negative, 
 and complex alike, which obey their definitions and laws 
 of combination, are equally real, or, if we will, equally 
 imaginary. 
 
 Considered as applied to things that can be counted, 
 or to quantity that can be divided into parts that may 
 be counted, positive numbers are always real ; when the 
 quantity is of a nature such that it can be conceived as 
 existing in opposite conditions, negative numbers are 
 real ; and, when the quantity has the attribute of direc- 
 tion, complex numbers are real. Thus (3, ^] feet-per- 
 
 second, 5^ poundals, or an impulse denoted by a + ib 
 units of impulse, are as real expressions as 10 shillings, 
 or 30 miles-per-hour. 
 
 Further, when applied to quantity having direction, 
 such as distance, velocity, acceleration, force or mo- 
 mentum, complex numbers furnish the direct and 
 complete representation of the object, and the results 
 obtained by the study of complex numbers regarded as 
 abstractions may be transferred to concrete quantity 
 having direction in the same manner, and with the same 
 
COMPLEX NUMBERS. 411 
 
 degree of confidence, as is the case when the properties 
 of arithmetical numbers are applied to examples dealing 
 with concrete magnitude. 
 
 The term "complex number" is often used to denote 
 a mixed quantity a-\-h\/ — \ partly real and partly 
 imaginary, the sign x/ — 1 and the sign of addition being 
 unexplained .symbols subject to the laws of real algebra. 
 In the treatment here adopted the word "complex" refers 
 to the double nature of the number, i.e., its modulus or 
 arithmetical magnitude and its amplitude or directional 
 magnitude. The mode of representing a complex number 
 as the sum of two parts, one along the primary axis (the 
 real part), the other along the secondary axis (the imagi- 
 nary part), is for many purposes of the greatest value ; 
 but it is to be regarded as one among an infinite number 
 of similar modes of resolving a complex number into 
 component parts. 
 
 Equivalent Forms of Results. — The following pairs of 
 equivalent forms of results already established are to be 
 noted : — 
 (l,a)x(l,/3) = (l,i+;8), 
 
 (cosa + isina)(cos^ + ^sin/5) = cos(a + /3) + ^sin(a+/3); 
 (1, a)x(l, -a) = l, 
 
 (cos a + i sin a)(cos a — -i sin a) = 1 ; 
 (1, a)« = (l,^a), 
 
 (cos a + '^ sin a)'* = cos na + i sin iia (n integral) ; 
 ct + 2r7r^ 
 
 ((1, «))"- = (l, 
 
 n 
 
 ((cos a + ^ sm a)r = cos h i sin , 
 
 where '^^ is a positive integer, and r = one of the numbers 
 0, 1, 2...^^i^l; 
 
412 COMPLEX NUMBERS. 
 
 ((l,a))-(l,^^±f^> 
 
 ((cosa + isina))' = cos ^«+^-^"" +isin ^"+^-^"" , 
 
 where 7' = any one of the numbers 0, 1, 2, ... g^ — 1. 
 
 Demoivre's Theorem. — The statement that, when n is 
 any real number, positive or negative, integral or frac- 
 tional, cos 710+'?^ sin 710 is a value of (cos + -^ sin 0)**, is 
 known as "Demoivre's Theorem." The theorem was 
 given by Demoivre in the form 
 
 where a? = cos 0, Z = cos nO. 
 
 262. Trigonometrical Formulae derived from the 
 properties of Complex Numbers. — In the demonstra- 
 tions of the preceding articles we have assumed the Com- 
 mutative and Associative Laws for the addition and 
 multiplication of real quantities (ab = ba, a + l3 = P+a, 
 art. 256, and ab.c = a .be, a + /3 + y = a+j8 + y, art. 258), 
 the theory of parallel straight lines, and of similar 
 triangles. It should be noticed that Eucl. I. 47 has not 
 been used. The theorem that the sides of a triangle are 
 proportional to the sines of the opposite angles, employed 
 in art. 20 1, is an immediate consequence of the definition 
 of the sine of an angle. 
 
 Without further assumption we can derive all the 
 fundamental properties of the circular functions. From 
 the relation (1, a) x(l, — a) = l we get 
 
 (cos a -h "^ sin a) (cos a - i sin a) = 1, 
 and, therefore, distributing the product on the left hand 
 side, cos^a -|- sin^a = 1. 
 
COMPLEX NUMBERS. 413 
 
 From (1, a)x(l, /3) = (1, a + ^) we get 
 
 cos a cos j8 — sin asin/5 + i(sin a cos /5 + cos a sin ^) 
 = cos(a + )5) + ^sin(a+)8), 
 and, therefore, since a complex number can be resolved in 
 two given directions in one way only, we have 
 cos(a + jS) = cos a cos /3 — sin a sin /3, 
 and sin(a + /3) = sin acoS|8 + cosasiny8. 
 
 Similarly, from the continued product 
 (l,a)x(l,^)x(l,y)X...X(l,X)=(l,a-|^+7+TrTX), 
 we get the general formulae 
 
 COS(.ll + ^2+---+^„)=0„-2a_2^2+---> 
 sin(^ + ^2-f...+^n) = 2>SfiOn-l-2lSf3C«-3+.... 
 
 And from the equation 
 
 (cos a-{-i sin a)'^ = cos na + i sin na, 
 n being a positive integer, we have 
 
 cos?la = cos*^a ^j — ^^— cos'*~2asin2a+..., 
 
 „ 1 n.n — l.n — 2 . „ « o , 
 sm na = n sm a cos'^-^a .. ^ ^ sm^a cos^~^a+ . . . . 
 
 Since the fundamental laws of algebra have been 
 shewn to hold for complex numbers, we may substitute 
 such numbers in any algebraical identity dependent on 
 these laws. Then, resolving the complex numbers along 
 the primary and secondary axes, we derive (in the case in 
 which the modulus of each number is unity) two trigo- 
 nometrical relations connecting the angles which represent 
 the amplitudes of the numbers. 
 
 For example, in the identity a^—¥ = (a — h) (a^ +ah + 6^) 
 let a = cosa-|-isina, 6 = cos/5 + ^sin/5; 
 
414 
 
 COMPLEX NUMBERS. 
 
 then we obtain the trigonometrical identities 
 cos 3a — cos 3/3 
 
 = (cos a — cos ^)(cos 2a + cos a + /S + cos 2/3) 
 
 — (sin a -- sin ^) (sin 2a + sin a + )8 + sin 2^8), 
 and sin 3a — sin 3/3 
 
 = (cos a — cos /3)(sin 2a + sin a + /8 + sin 28) 
 
 + (sin a — sin ^)(co3 2a + cos a + ^8 + cos 2/3). 
 
 263. The following examples will further illustrate the 
 use of complex numbers : — 
 
 Example 1. — Shew that the sum of the nth powers of the five 
 fifth roots of unity is either 5 or 0. 
 
 The fifth roots of unity are the five complex numbers of unit 
 modulus, and having amplitudes 0, a, 2a, 3a, 4a, where 5a = 27r. 
 
 The ?ith powers of these have unit modulus, and, rejecting 
 multiples of Stt, the amplitudes 
 
 0, 
 
 0, 0, when 
 a, 2a, 3a, 4a 
 
 is of the form 5m, 
 „ „ 5m + 1, 
 
 „ „ 5m + 2, 
 
 „ „ 5m + 3, 
 
 5m +4. 
 
 2a, 4a, a, 3a „ „ 
 
 3a, a, 4a, 2a „ „ , 
 
 4a, 3a, 2a, a „ „ , 
 
 In the first case the sum of the ?ith powers = 5. 
 In the other cases the sum is zero, since it is in each case the sum 
 of five complex numbers of unit modulus whose directions are 
 symmetrically distributed about the origin. 
 
 Example 2. — Demoivre's pro- 
 perty of the circle. See art. 
 226. 
 
 Take OP as primary axis. 
 
 Let LPOAi = e, 
 
 LPOAr=e+(r-i)^-^; 
 n 
 
 let OP=r, and a = radius of the 
 
 circle. 
 
 We have 0A^-\- A^=OP= r, 
 
COMPLEX NUMBERS. 
 
 415 
 
 .-. {ArP-rr={-moArr={-\r{a, e+{r-\f^y 
 
 i.e, ArP is a root of the equation 
 
 (^-r)"-(-l)'V,^^)=0- 
 The absohite term of this equation is (-l)"[r"-(a", n6)\ 
 
 hence A^ . A^ . . . I^= f" - (a'^ n 6). 
 
 Let ar be the image oi Ar with respect to the primary axis, i.e. a 
 
 point such that Ara^ is bisected at right angles by OP^ then by 
 
 changing the sign of 6 we get 
 
 a[P .^ ...aj'=f'-{a\ -nO). 
 
 Multiplying, and putting A^P .arP^p?., where p^ = mod(Jr^) 
 = mod(a^/^) we get 
 
 ^,,2n_2aV^cosw6'+a^". 
 
 PiW 
 
 Example 3, — If ^i, ^2» ^3) ■•• -4„ be 7i points at equal distances on 
 
 the circumference of a circle of centre and radius a, and if P be 
 
 any point and OP=r, and A^OP=$; prove that the sum of the 
 
 angles which A^P, A^P, A^P ... A^P make with OAi is an angle 
 
 1 , J. • r'^sinw^ 
 
 whose tangent is y, 
 
 r"cos nO — oT' 
 
 Take OAx as the primary axis. 
 
 We have OA^ + A^P = 0P_ = (r, 6), 
 
 A^P-{r,e)=-OAr,__ 
 
 .'. [ArP-{r,e)f = {-mOA;)- 
 
 =(-l)V, 
 
 .*. ArP is Si root of the equation 
 
 [.r-(r, (9)]'^-(-l)^a" = 0. 
 The absolute term of this equation is (-1)"[(^', Oy^-a^], 
 hence A^.A^\..A^=:(r, 6y-a"=r''(cosnd + ismn0)-a''' 
 
 Let pr=mod{ArP), ^^=amp(J^/^), 
 
 th en P1P2 '■• Pn(cos 2^ + ?' sin 2^) = r'*(cos w ^ + 1 sin w ^) - a'*. 
 
416 COMPLEX NUMBERS. 
 
 pip2 . . . p„cos 2<^ = r"cos n6 - a**, 
 and P)P2'-' PiM^ 2^ = r**sin nO, 
 
 r**sin nO 
 
 and therefore tan 2^ = 
 
 Example 4. — A man walks on a plane in such a manner that, 
 when he has passed over a distance a in a straight line, he always 
 changes his path through the same angle a in the same direction. 
 Shew that when he has done this n times, his distance from his 
 
 starting point is a sin ^ /sin ~, and that this distance makes an 
 angle (n—l)^ with his first path. 
 
 Let OAxA<:i... An be the path. Take 
 OAx as the primary axis. We have 
 OZn =(«, 0) + (a, a) + (a, 2a) + ... H-(a,^L a) 
 = a(l + cosa + cos 2a+... + cosM- la) 
 + ia(sin a + sin 2a + . . . + sin n - la) 
 
 cos(w - \)% sin ^ sin '^-^ sin(w - 1)^ 
 
 = a +za 
 
 • a -a 
 
 Sin — sm - 
 
 2 2 
 
 or modi{OAn)=a -. Amp(OJ„)=(w- 1)-. 
 
 |cos a + 0) cos^ a + ^) + w^cos^ a + ?^) V 
 
 = (- j Icos na-\-bi cos( wa + ^) + w^cos(^7ia + -r ))■ 
 
 Wehavea,= -1, (l, |) or (l,-|). 
 
 If 0)= - 1, each side of the equation vanishes. 
 
COMPLEX NUMBERS. 417 
 
 If o) = (l, I), take Oi;;i=(l, p, ar2=(l, ^), 02 = (1, a), 
 
 a5=(l,-a). 
 
 Draw the chords ALB, BMP, BNQ 
 at right angles to OX, 0(0i, Oin^ re- -^ 
 spectively. 
 
 Then, 
 
 cos a + (0 cosf a + ^ J -f- w^eosf a + -^ j 
 
 ='OL^OM+ON 
 
 '=^{OB+OA + OB + OP+OB+OQ) 
 =1 OB, since ^+0^+0^=0 
 =Kl,-a). 
 In like manner, writing na for a, we have 
 
 cos ?ia + w cos(^ wa +^) + w'-^cosfwa-l- ^)=f (1,- wa). 
 
 But [|(l,-a)]« = (|r(l,-^a), 
 
 |cos a + 0) cos^ a + ^ j + w2cos( a + — H 
 
 = (f r"^| cos wa + 0) cos(7^a + - ") + a)2cos(wa + ^ ) I . 
 If (o=M,-^ j, each side of the equation = (§)"(], na). 
 
 Example 6.— If 
 
 ^ = cos a + 1 sin a, B=cos/3 + i sin /?, (7= cos y + 1 sin y, 
 
 express ^^^4±^ in the form P+^i, and prove that 
 
 ^ _ _ 4 sin ^(/? + y - 2a)sin |(y + a - 2/3)sin K« + jS - 2y) 
 1 + 8 cos(/3 - y)cos(y - a)cos(a - ;8) 
 We have 
 
 ^(7+(7^ + ^^ _^ cos(;8+y)+...+^•{sin(^+y) + ..■} 
 A^ + B'+C'^ cos2a+...+^{sin2a+...} 
 
 ^ [cos(jg + y) + . ■ ■ + 1 {sin(^ + y) + . . . }][cos 2a + . , . - z{sin 2a + . . .}] 
 
 , (cos2a+...)H(sin2a+...)^ 
 Denoting this expression by P+ Qi, we have 
 2d 
 
418 COMPLEX NUMBERS. 
 
 ^ _ [8in(^ + y) + ■ . ^[cos 2a +...] - [co8(/3 + y) + ■ ■ .][8in 2a -t- . . .] 
 
 ^ (cos2a+...)H(sin2a+...)^ 
 
 _ 8in(;8+ y -2a)+two similar terms ^ _4n{sin ^(/? + y-2a)} 
 ~ 3 + 2 cos 2( j8 - y) + two similar terms 1 + 8n{cos(^ ^ )} 
 
 Example 7. — To prove that 
 
 cos 2a/sin ^~ i^ sin ^!-l2 sin ^5lZ_ -f. three similar terms 
 
 '2 2 2 
 
 =88m«+%r_+S. 
 
 Supposing the trigonometrical identity to be deduced from an 
 algebraical one, we may determine the latter as follows : — 
 Let a, h, c, d stand for complex numbers cos a + i sin a, etc., ' 
 
 then a-6 = 2isin^(cos^+tsin?±i?| 
 
 (a - 6)(a - c)(a - c^; = - 8i sin ^^ sin ^5^ sin ^(1, 1+^), 
 
 2i Ji Jt 
 
 where «=^(a + /? + y + 5) ; 
 
 also, Va6c5=(l, s\ 
 
 \/ara ^ (1, -2a) 
 
 a{a-h\a-c\a-d) .g^'sin ^"/^sineir^rsin^ 
 
 2 2 2 
 
 _ • ^' cos 2a + sin 2a 
 
 "8sin«-:;^sin±Z2sin^* 
 
 2 2 2 ^ 
 
 But 1 =(l,-,)=co8^±to±i-isin^±^+Z±S. 
 
 Hence the trigonometrical identity will be true if we can shew 
 
 that —. ,, , ^ ^ w T. + 3 similar terms = - . , 
 
 a(a-6)(a-cXa-o?) ^ahcd 
 
 i.e. that 2 ,. ■ ^ ■ ■ -j.-\- \-.=^. 
 
 a{a - o){a - c){a - a) abed 
 
 This algebraical identity is readily obtained by assuming 
 
 :v{x-a){x-b){a;-c)(x — d) a: — a x-h x~c x-d x 
 determining A^ B, C, 7), E^ and substituting their values in 
 A-\-B^C+D^-E=0, 
 
COMPLEX NUMBERS. 419 
 
 Examples XXVI. 
 
 1. Prove that amp((X + 6i) is equal to tan"^-, tan"i-4-7r, 
 
 or tan"^ — tt, according as a is positive, a negative 
 
 and b positive, or a negative and h negative, 
 respectively. 
 
 2. If a = cos A+i sin A, b = cos B+i sinB, c = cos C+isinC, 
 
 where A, B, G are the angles of a triangle, then 
 abc= —1. 
 
 8. If a = y^, then 
 
 [(1. a) + (l, 2a) + (l,4a) + ip 
 = 2[(1, a) + (l, 2a) + (l, 8a) + (l, 4a) + (l, 5a) + (l, 6a)] + i. 
 
 4. If (X = (l, a), then 2cosa = a+-, 2isin a = a , and 
 
 a^ — 1 
 
 i tan a = o , -, - 
 
 5. If a = (1, a), then 2 cos 2a = a^ +-^, and 2^ sin 2a = a^ — s. 
 
 6. If a = (l, a) and n be any positive integer, then will 
 
 1 . 1 
 
 2cos7ia = a^+— :, and 2isinna = a'^ — — . 
 
 a a 
 
 7. Find the simplest form of (<^o^^+^-}^'''^)\ 
 
 (cos 2/ — V — 1 sin yy 
 
 8. Find the real and imaginary parts of the expression 
 
 (cosa+x/— 1 sina)(cos/3 + /v/ — 1 sin^) 
 (cosy + x/ — 1 siny)(cos(5+/v/ — 1 sin^) 
 
 9. Find the simplest form of the expression 
 
 (cosg + V—lsinay 
 (sin;8 + v/-lcos/3/ 
 
420 COMPLEX NUMBERS. 
 
 10. Determine the simplest form of (cos - V --1 sin Oyo 
 
 (cos a + V — 1 sin a)^^ 
 
 11. Find the value of (-l + V^)H(-l->v/^)^ 
 
 12. Apply De Moivre's theorem to express the real and 
 
 imaginary parts of (a+6/v/ — 1)**, when n is any 
 integer. Find the value of 
 
 (1 + ^33)10 4. (i_ ^113)10, 
 
 13. Shew that 
 
 p 
 
 {m+nj -lY+(m-nsJ -\y = 2(m2 + n^f cos pO. 
 
 where 0=amp(m+'>ix/— 1). 
 
 (^cos|^-x/^sin^j 
 
 14. Simplify the expression -. 
 
 (^cos^ + x/-lsin^J 
 
 15. Find the cube roots of V— 1. 
 
 1 6. Find the four values of (( - 1 + J'^J^)% 
 
 17. Exhibit the four fourth roots of 1 + ^ — 3. 
 
 18. Find the three values of ((l + x/-'T))i 
 
 19. Find the three cube roots of >^3+i. 
 
 20. Find the five values of the expression (( '^ ~ )]"• 
 
 21. Express by De Moivre's theorem all the values of 
 
 ((-1))^^ 
 
 22. Reduce ^^^ to the form A+Bs/~^. where 
 
 a+o 
 
 a = cos a + /v/ — 1 sin a, and 6 = cos ^8+ V — 1 sin /3. 
 
 23. If cCy = cos^+ V— Isin J, prove that, the product 
 
 being continued to infinity, 
 
 X^^^^ ... = cos TT. 
 
COMPLEX NUMBERS. 421 
 
 24. Prove that 
 
 (sina; + V— 1 co^ xY = Qo^ ni-^ — x) + s/ — 1 &mn(^—x\ 
 
 25. Calculate in a form free from imaginary quantities the 
 
 value of 
 
 [cos — cos ^ + x/ — l(si n — sin 0)]** 
 
 + [cos d — cos (p — \/ — l(sin 6 — sin 0)J\ 
 
 26. Prove that (^ + ^y"' is reducible to the form 
 
 p{cos6-\-iiiin0), and find the values of p and 0. 
 
 27. Apply De Moivre's theorem to shew that, if 
 
 l/{a+bs/-l)-\-'^(a-hs/-l) = 2l/{a^ + ¥).cosie. 
 
 28. Prove that the expression (a-\-ihy'^+{a+ih'y' is 
 
 reducible to the form JS(cos0 + isin0), and find 
 the values of M and </>. 
 
 3 1 1 1 
 
 29. Shew that —---„ = -—; f-^; h-, 77-, where a, 8 
 
 1+x^ l-^x 1 — ax 1 — px '^ 
 
 are the imaginary values of (( - 1))^, and deduce, by 
 writing .t = cos 20 + V— 1 sin 20, that 
 
 3 tan SO = tan e - cot^O + ^) - cot^O - ^). 
 
 30. If ^, 5, be the angles of a triangle, then 
 
 2cos3^ + 3 = 2cos^.2(cos2^+cos^) 
 — S sin J. . 2(sin 2^ — sin ^), 
 and 2 sin 3 J. = 2 cos ^ . S(sin 2A — sin A) 
 + 2 sin ^ . 2(cos 2^ + cos ^). 
 
422 COMPLEX NUMBERS. 
 
 31. If 0) be an imaginary cube root of —1, prove that 
 i cos a + ft) cosf a + ^ j + w^cosf a + -^) f 
 
 X |cos^+ft,cos(/3+|) + a)2cos(^4-^)} 
 
 = f |cos(a + i8) + ft) cos(a + /3 + 1) + a)2cos(a + )^ + ^)}, 
 and deduce the value of 
 
 jcosa + ft)COsfa + ^j + ft)2cosfa+-i^H . 
 
 32. Shew that the roots of the equation 
 
 {{a-\-h)x-{a-h)y'^{a + h-{a-h)xY 
 
 Tit ... Ttt 
 a cos %b sill — 
 
 are the values of , where r has 
 
 a cos f-it)sm — 
 
 n n 
 
 any positive integral value between and n — 1 
 inclusive. 
 
 33. Determine the values of x from the equation 
 
 (cos A +x sin J.)(cos B+ x sin S) =cos(J. -{-B) + x sin(^ + 5). 
 
 34. Find m in order that (cos + msin 6)^ may be equal 
 
 to cos nO+m sin nO for all integral values of n. 
 
 35. If, in the identity 
 
 1 ^ 1 1 
 
 (x — a){x — h) {a—b){x—a) (a—b)(x — by 
 
 cos 26 + s/^ sin 20, cos 2a + \/^l sin 2a, and 
 cos2/3 + x/ — lsin2^ be written for x, a, b, re- 
 spectively, obtain the trigonometrical identity 
 resulting from equating the real parts of the two 
 expressions which are identical. 
 
 4 
 
COMPLEX NUMBERS. 423 
 
 36. From the identity 
 
 obtain by writing for a, cosa + isin a, and similar 
 substitutions, 
 sin(a-/3)sin(y-^) = sin(a-^)sin(y-^) + sin(a-y)sin(/5-^;. 
 
 37. If (r, 0) = (1, a) + (l, /3l prove that r = 2cos'^'^, 
 
 = ^ J^ . Hence shew that 
 
 cos a + cos /3 = 2 cos ^' cos — ^- , 
 and that 
 
 sin a + sin /3 = 2 sin ^^ cos ^^. 
 
 38. If a = -S-, then cos a + cos 2a + cos 4a = — J, 
 
 and sin a + sin 2a + sin 4a = J v^''- ' 
 
CHAPTER XVII. 
 
 SERIES OF COMPLEX NUMBERS. 
 
 264. Finite Series. — Let Uj^+U2+u^+...-\-Un be any 
 series of complex numbers, and let Sn denote the sum of 
 n terms of the series. 
 
 If all the terms of the series have the same amplitude, 
 the vectors representing them form a straight line, of 
 length equal to the sum of the moduli of the terms, and 
 inclined to the primary axis at an angle equal to the 
 common amplitude of the terms ; 
 or, if /Sfn=(r, 0), 
 
 then r = the sum of the moduli of the terms, 
 and = the common amplitude of the terms. 
 
 If the terms have not the same amplitude, the vectors 
 representing them do not form a straight line, and it 
 follows, by Eucl. I. 20, that the modulus of Sn is less than 
 the sum of the moduli of the terms. 
 
 In this case, if /Si„ = (r, 6), then r has a value less than 
 the sum of the moduli, and dependent on the values of the 
 moduli and amplitudes of the terms, and has a value 
 also dependent on these moduli and amplitudes. 
 
 265. Definitions. — If the sum of the first n terms of 
 
 a series of complex numbers tends to a limit S of finite 
 
 424 
 
SERIES OF COMPLEX NUMBERS 425 
 
 modulus and fixed amplitude, when the number n is 
 indefinitely increased, the series is said to be convergent, 
 and 8 is called its sum. 
 
 If the modulus of the sum increases without limit as n 
 is indefinitely increased, the series is said to be divergent. 
 
 If each term of a series of*complex numbers is expressed 
 in the form x-{-iy, the conditions of the definition of con- 
 vergency will be satisfied if the real series 'Zx and Zy are 
 each convergent ; and if one or both of the real series 2cc 
 and 22/ be divergent, the series of complex numbers is 
 divergent. 
 
 If the series whose terms are the moduli of the 
 terms of the original series is convergent, the original 
 series is said to be absolutely convergent. (See art. 266.) 
 
 If the series of complex numbers is convergent, and 
 the series of moduli of its terms divergent, the original 
 series is said to be semi-convergent. 
 
 If the modulus of the sum does not increase without 
 limit as n is increased indefinitely, and the sum does not 
 tend to a limit of finite modulus and fixed amplitude, the 
 series is said to oscillate. 
 
 The following are examples of oscillating series. 
 
 Example 1.— Consider the series whose wth term is +(l,~\. 
 Let 0AiA2A^A^A^ be a regular hex- 
 agon, then if OAi make an angle ^ with 
 
 the primary axis, the sides of the hex- 
 agon taken in succession, and repeated A^^ 
 continually, represent the terms of the 
 series, and the sum of n terms is 
 
 zero, all, ^> ^^ ^' or OA^, ^^ ^^ — ^ 
 
 according as n is of the form '^ 
 
 6m, 6m+l, 6m + 2, 6m + 3, 6m + 4, or 6m + 5, respectively. 
 
426 SERIES OF COMPLEX NUMBERS. 
 
 The series consequently oscillates^ and has any one of six distinct 
 values, each of finite modulus and fixed amplitude. 
 
 Example 2.— Consider the series obtained by placing in a circle 
 of unit radius a succession of chords A1A2!, 
 A2A3, ^3^4, ... of lengths hhh "" 
 We have 
 
 OA„=OA, + A^A2+A2A3+...+An-iAn. 
 Now the modulus of the sum OAn is unity 
 for all values of ?i, but in consequence of 
 the divergency of the series 1+^ + ^+... 
 the point An does not tend to any fixed 
 point on the circumference when n is indefinitely increased. 
 
 The series consequently oscillates, and has any one of an infinite 
 number of values, each of unit modulus. 
 
 266. A series of complex numbers is convergent when 
 the semes of moduli of its terms is convergent. 
 
 Let OAn represent the sum of "the first n terms of the 
 series of the complex numbers. 
 
 Then, since the series of moduli of the 
 terms is convergent, and since the modulus 
 of the sum is less than the sum of the moduli 
 of the terms, therefore mod (OAn) is finite, 
 however great n may be. Next, let AnA^ 
 represent the sum of the m terms immed- 
 iately following the first n terms. Since 
 the series of moduli is convergent, it follows 
 the modulus of AnA^ can, by sufficiently in- 
 creasing n, be made as small as we please, and this 
 however great m may be; and therefore OAm can be 
 made to differ in modulus and amplitude from OAn by 
 as little as we please. Hence, the series is convergent. 
 
SERIES OF COMPLEX NUMBERS. 
 
 427 
 
 267. Example 1. — Consider the 
 series 
 
 ^-^+^-"-<^dinf., 
 
 where ^ is a complex number. 
 The test ratio of the series of 
 
 moduli =i?^i^^^, and this can 
 
 be made as small as we please 
 by increasing n, hence the series 
 is absolutely convergent. 
 
 The diagram represents the 
 first four terms of the series, 
 when x^^sl^-'ri or (2,^), and 
 
 shews the rapid convergence of 
 the series after the second term, 
 the vector OAq representing the 
 sum of four terms, and approxi- 
 mately the sum to infinity. 
 
 Example 2.— The diagram re- 
 presents the series 
 
 -V.2 /»,4 ^6 
 H- — -U- 4- — -4- 
 
 (2+|4+^_ + 7 
 
 when x=J'i-^i, the vector OA 
 giving the sum of four terms, 
 and approximately the sum to 
 infinity. 
 
 268. If %, a^, a^... an he a series of constantly de- 
 creasing ^positive quantities, and if Lt. an = 0, and if /3 
 
 be not equal to zero or a multiple of 27r, then ivill the series 
 
 a,{l, a) + ai(l, a + ^) + a^{l, a + 2/5)-f... 
 be convergent. 
 
 It has been shewn in art. 210 that with the given 
 conditions each of the series 
 
 a^cos a i- aiCos(a + /3) + a^cos(a + 2/3) + . . . 
 
428 SERIES OF COMPLEX NUMBERS 
 
 and aQsina + ai8in(a + /8)+a2sin(a + 2^)+... 
 
 is convergent, hence the series 
 
 ao(l> a) + «i(l, ^Hh8) + a2(l, a + 2/3) + ... 
 is also convergent (art. 265). 
 
 269. If the series aQ-\-a^X'\-a.fy^-\- ...ad inf., where 
 aQ, dj, a2, ... are real quantities and x a complex number, 
 be absolutely convergent when mod{x) = R, it will be a con- 
 tinuous function of x for all values of x such that 
 mod(x) < R 
 
 Let X = (r, 6) where r < R. 
 
 Then aQ-{- a^x + a^'^ -\- . . . 
 
 — aQ+a^r cos 6 + a^r^cos 20+... 
 + i (a^r sin Q + a^r'^sin W+...) 
 = G+iS, say. 
 First, let 6 remain constant while r changes. Then 
 each term of the series G is numerically not greater than 
 the corresponding term of the series 
 
 and, by hypothesis, this series is absolutely convergent, 
 therefore G is absolutely convergent, and consequently G 
 is a continuous function of r so long as r<R (art. 211). 
 
 Similarly, the series /Sf is a continuous function of r ; 
 therefore G + iS, or a^ + a^x + a^^ + . . . 
 
 is a continuous function of mod(ic) so long as iaod(x) < R. 
 
 Next, let r retain a constant value less than R, and let 
 6 change from 0^ to 0^, where 6^ < 62- 
 
 Let G^=aQ + a{r cos 0^ + a2r2cos 20^+..., 
 
 (72=^0+ a^r cos O2 + a^r^cos 2^2 + • • • > 
 then 
 
 G^ - C2 = air(cos Oi - cos ^2) + a2^2(cos20i - cos 20^) +.... 
 
 Suppose that a^, a^, a^,... are all positive, then since 
 
 cos ~ cos (/)' is numerically less than 0-0' (art. 89), we 
 
SERIES OF COMPLEX NUMBERS. 
 
 429 
 
 see that 6\ ~ G^<{e^ - e^){a{t^-\-'ia^T'^+2,a^r^+ . . .), 
 
 and, as in art. 211, it may be shewn that for any fixed 
 
 value of r < -R, the series 
 
 a^t' + ^a^f^ + SagT^ + . . . 
 is convergent and therefore finite. 
 
 Therefore G^ ~ G^ diminishes indefinitely with Q^ - 0^ 
 or is a continuous function of 6. 
 
 The result follows, a fortiori, if the coefficients a^, a^, otg, 
 etc., are not all of the same sign. 
 
 Similarly, the series >S' is a continuous function of 0, 
 therefore 0+ iS, or a^ + a-^x + a^p^ + . . . 
 
 is a continuous function of amp(ic) so long as mod(aj) < R. 
 
 Combining these results, we see that 
 
 is a continuous function of x for all values of x such that 
 mod (a?) < R. 
 
 Geometrical Illustration. 
 
 Let OP-=x, 
 
 CE=aQ+ap-^acfc^+.... 
 Then if P move continuously from P to P' within a 
 circle of radius P, 8 will move continuously to a new 
 position S' ; and if P and P' are indefinitely near to one 
 another, so also are 8 and 8'. 
 
430 SERIES OF COMPLEX NUMBERS 
 
 270. If the series aQ-\-a-^x + a^^-\- ... ad inf., where 
 a^, a^, ttg . . . are real quantities and x a complex number, 
 he convergent when x has a value (1, a), then the limit of 
 aQ+ai(r, a) + aj^r, a)^+... ad inf. as r increases up to 1 
 ivill be 
 
 aQ+a^(l, aJ+a^O-, a)^+... ad inf. 
 By hypothesis, the series 
 
 ao+ai(l, a) + a2(l, af-\-... 
 is convergent ; therefore each of the real series 
 
 aQ + (Xjcos a + a^coB 2a + ... 
 and a^sin a + agsin 2a+... 
 
 is convergent. 
 Now the series 
 
 <^o + ^iO'' a) + a2(?', a)H... 
 = (Xq + a^r cos a + cv'^cos 2a + . . . 
 + i{a^r sin a + agT^sin 2a+...}. 
 But, by art. 213, the limits of 
 
 ^0 + <^i^' ^^^ " + agT^cos 2a + . . • 
 and a^r sin a + a^r^sin 2a + • . • , 
 
 as T increases up to unity, are 
 
 % + ^iCOS a + dgcos 2a + . . . 
 rind a^sin a + dg^in 2a + . . . 
 
 respectively ; therefore the limit of 
 
 % + ^i(^'' «) + «2(^'> a)2 + . . . , 
 as r increases up to unity, is 
 
 ao+ai(l, a) + a2(l» a)2+.... 
 Cor. — If the limiting value of x, for which the series is 
 convergent, be (R, a) where R is any fixed modulus, the 
 limit of aQ + a^{r, a) + a^ir, a)^ + . . . , 
 
 as r increases up to i^, will be 
 
 aQ + a^(R, a) + ttgCi^, a)^ + . . . . 
 
SERIES OF COMPLEX NUMBERS 
 
 431 
 
 For if we put hn for anR^, and p for rjR, we may write 
 the series in the forms 
 
 and 60 + hip^ a) + \{p, a)^ + . . . , 
 
 and apply the theorem of the present article. 
 
 Geometrical Illustration. 
 
 Let 
 
 0^=(1, a), 
 
 OP = (r,a\r<l, 
 
 CS=aQ + a^(r, a) + a^{r, af+..., 
 then as P moves up to _p, 8 moves up to s. 
 
 271. Series involving the cosines and sines of angles in 
 arithmetical progression may be reduced to an algebraical 
 form by the use of complex numbers. 
 
 Thus, if (7=aoCosa + aiCOs(a + /3) + a2COs(a + 2/5)+... 
 and S = a^sin a + aisin(a + /3) + a2sin(a + 2^) + . . . , 
 
 and if i:c = cos a + isin a, 2/ = cos iQ+'i sin /5, 
 
 then G+iS = a^ + a^xy ■^■a^x'ip'-^- ... . 
 
 If the sum of the algebraical series 
 
 a^x-\-a^xy-\-a^x\p'-\- ... 
 is known, the values of C and B may be found by resolv- 
 ing the sum of the series of complex numbers into com- 
 ponents along the primary and secondary axes. 
 
432 
 
 SERIES OF COMPLEX NUMBERS 
 
 As simple examples of this method we may take series 
 leading to geometrical progres- 
 sions. Examples of binomial, 
 exponential, and logarithmic 
 series will be found in succeed- 
 ing chapters. 
 
 The accompanying diagram 
 shews geometrically the nature 
 of the process. 
 
 Here 
 
 Mo m;m, c s=on,+n^n, + n^n^+,.., 
 
 272. Example 1. — Find the sum of the series 
 
 cos a + cos(a + j8) + cos(a + 2^) + . . . ton terms, 
 and sin a + sin(a + /8) + sin (a + 2^) + . . . to n terms. 
 
 (See arts. 205, 206.) 
 Let C and S denote the sums of the series, and let « = (1, a), 
 6 = (1, ^). Then 
 
 C+iS=a+ab-\-ab^+... + ab''-'^ 
 
 _ a(l-6«) 
 1-6 • 
 
 This expression=^"^^°^+^'"^^"Xl-cosn/3-zsin7i/?) 
 l-cos/5-^sin^ 
 
 (cosa+isina)(sn !^-^cos'^) . 2sin^ 
 
 (sin^-icos^V 2sin^ 
 V 2 2/ 2 
 
 (cos a + z sin a)( cos ^ + 1 sin ^ )sin -^ 
 Tcos^+isin^ jsin^ 
 =^cos(a+7^^|) + ^sin(a+w^|)Jsin^^sin|; 
 
SERIES OF COMPLEX NUMBERS. 433 
 
 and >S'=sin(a + 7^^|)sin^/sin|. 
 
 Example 2. — Sum to infinity the series 
 
 cosa + ^cos(a + /3)+^2cos(a + 2^)+... 
 and sin a + .r sin(a + j8) + ^2sin(a + 2^) + . . . , 
 
 when a^ is less than unity. 
 
 Denoting the series by C and S, and the complex numbers (1, a) 
 and (1, /3) by a and 6, we have 
 
 C+iS=a+x . ah+x"^ . ab^+... . 
 This series is absolutely convergent, since mod(6.r) < 1, 
 C+iS-- 
 
 Hence, C+iS=-. 
 
 l-bx 
 cosa + isina 
 
 1—.V cos y8 — ix sin (3 
 _ (cos a + ^ sin a)(l - .a; cos /3 + ix sin /3) 
 
 1-2X008/3 + x^ 
 
 _ COS g - :r cos(a - (3)-\-i sin a - 2^ sin(a - ft) . 
 \—2x cos ft + x^ 
 jy_ cosa-^cos(a-^) 
 ~ l-2.rcos/3 + ^2~' 
 
 and ^_ sina-.ysin(a-/3) ^ 
 
 1 -2^cos^ + .<:^^ 
 Example 3. — Sum to infinity the series 
 
 cos 2^ + cos ^ cos 3^4- cos^^ cos 4^+... 
 
 and sin 2^ + cos ^sin3^+cos2^sin4^+ 
 
 Let C and S denote the series, and let ^=(1, ^)=cos ^+isin ^. 
 Then C+i'S'=^2+^cos^+^*cos2^+... . 
 
 If cos ^ is numerically less than 1, i.e. if 6^=mr, this series is 
 convergent, and we have 
 
 C+ iS= ^^ = — 
 
 1-^cos^ l-cos^^-icos^sin^ 
 
 ^ x'^ _ (1, 26>) 
 
 8in(9(sin6>-^■cos<9) gin ^(1,^3^) 
 
 ihZ+e) 
 
 TT , 
 
 2 ^ -sin 6+1 coa 9 , . , 
 1+ 1 cot 
 
 sin 6 sin 
 
 2e 
 
434 SERIES OF COMPLEX NUMBERS 
 
 Hence, if d^nir, C= - 1, and >S'=cot $. 
 
 If d = mry we have C=H-l + l + ... = oo, and >S'=0 ; thus in each 
 case there is discontinuity when ^=0, tt, Stt..., or — tt, -27r — 
 
 The curve of the first series is a straight line parallel to the axis 
 along which $ is measured, with a series of isolated points at 
 infinity ; that of the second series is the cotangent curve with a 
 series of isolated points on the axis of 6 corresponding to $=0, 
 e=Tr, ^ = 27r, ..., (9=-7r, ^=-27r, .... 
 
 The fact that cos20+cos ^ cos 3^+ cos^^ cos 4^+... ac/m/l is equal 
 to - 1 for any very small value of 0, say one-millionth of a second 
 of angular measurement, while when ^=0 the sum of the series 
 is infinite, may serve to shew that theorems such as those of arts. 
 211 and 213 are not self-evident truths. 
 
 Examples XXVII. 
 
 1. Shew that, if x be any complex number, each of the 
 
 series 
 
 '^+^ + i5 + - 
 is absolutely convergent. 
 
 2. If ic be a complex number, the binomial series 
 
 \-\-nx-\-^-—^ — x^+. . . ad mf. 
 
 is convergent for all real values of n, when 
 mod(a;) < 1. 
 
 3. If a? be a complex number, and oi real but not a posi- 
 
 tive integer, the series 
 
 l + nx+ ^'^~^ x^+...adinf. 
 is divergent when mod(a:;) > 1. 
 
SERIES OF COMPLEX NU3IBERS 435 
 
 4. Find the values of the oscillating series 
 
 5. Find the limit of 
 
 where x — ii\ ~), when r increases up to \m\ty. 
 
 r. n l^ . sina , sin 2a , sin 3a , j.^ m +««^o 
 
 6. Sum the series --^ — I — ^ — I — ^ — h... to 10 terms. 
 
 7. Sum the series 
 
 cosa + a;cos2a + i:c'^cos3a+... ad inf., when x<l. 
 
 8. Sum the series 
 
 ccsin a — cc^sin2a + a3^sin3a — ... ad inf., when x < 1. 
 
 9. Find the sum to n terms of 
 
 cos sin 2^ + cos20 sin W + cos^O sin 40+ ... . 
 
 10. Find the sum to n terms of 
 
 cos a + x cos(a + P)+ x^cos(a + 2/5) + • . . 
 and sin a 4- aJ sin(a + ^) + aj^sin(a + 2^) + . . . . 
 
 11. Given the sum to n terms of the series 
 
 cos a + cc cos(a + j8) + cc^cos (a + 2/3) + . . . , 
 deduce the sum to infinity when x < 1. 
 
 12. Given the sum to infinity of tlie series 
 
 sin a + a? sin(a + 1^) + a;2sin(a + 2^) + . . ., 
 deduce the sum to n terms. 
 
 13. Prove that the sum ofn terms of the series 
 
 ^^cosa^cos|a^cos3«_^_ is equal to 0, 
 cos a cos^a cos'^a 
 if na = 7r. 
 
 14. Sum to infinity 
 
 sin cos + sin20 cos 20 + sin^O cos 80 + . . . 
 and sin sin + sin^O sin 20 + sin^O sin 30 + ... . 
 
 15. Sum the series 
 
 -L sin e + l sin 20 + 2-^ sin 30+ . . . ad inf 
 
436 SERIES OF COMPLEX NUMBERS 
 
 16. Sum the series 
 
 sin 45°sin 0+sin245''sin 20+siii345°sin 30+ . . 
 to n terms. 
 
 17. Prove that, if A, <1, 
 
 l^h? 
 
 (l-/i)2cos2|-|-(l+^)2sin2| 
 
 = l + 2^cosaj+2^2cQg2a;+... ad inf. 
 
 18. Sum to infinity 
 
 cos cos + cos20 cos 20 + cos^^ cos 30 + . . . 
 and cos sin + cos^^ sin 20 + cos^^ sin 30+ ... . 
 
 % 
 
CHAPTER XVIII. 
 THE BINOMIAL THEOKEM. 
 
 273. If ii be a positive integer, x any complex number, 
 then 
 
 1.2 \r\n — r 
 
 The proofs usually given for real values of x depend 
 only on the fundamental laws of algebra, and therefore 
 hold for complex values of x. 
 
 274. If m, n be any real numbers, x any complex 
 number such that ■mod(a;)< I, and if the infinite series 
 
 - , , m(m — 1) 9 , 
 l+maj+-Y-2— ^ +... 
 
 be denoted by /(m), then 
 
 f{m)xf{n)=f{m+n). 
 
 The proof given for real values of x, by aid of Vander- 
 monde's theorem,* depends only on the fundamental laws 
 of algebra and on the absolute convergency of the series 
 /(m) and f{n), and therefore holds for complex values of x 
 if mod(£c) < 1. 
 
 * C. Smith's Treatise on Algebra, art. 279. 
 437 
 
438 THE BINOMIAL THEOREM. 
 
 275. Positive Fractional Index. — From the equation 
 /(m) xf{n)=f{m + n), we deduce, as in the case when x is 
 real, 
 
 {/(p)*-<'«)' ■ 
 
 p and q being positive integers, and mod(ic) < 1. 
 
 Hence, it follows thaty M j is equal to one or other of 
 
 the values of the g- valued quantity ({l-\-x)y. We shall 
 shew that this value is the principal value, or, if 
 1 +x=(p, (p) where — tt < < tt, then 
 
 For any given value of-, provided only that mod(a;)< 1, 
 
 we know that the series denoted by^f -j is absolutely 
 
 convergent, and that it is a one-valued, continuous fiinc- U 
 
 tion of X ; or, geometrically, if OF=yy^\ then ^ is at a fl 
 
 finite distance from 0, has a single fixed position when x 
 is given, and moves continuously for any continuous 
 change in x. 
 
 Let OG=\,GP = x, 
 
 then 0P = l+x, length OP = p, lGOP = 0. 
 
 Since mod(a3) < 1, P cannot lie outside a certain circle 
 with centre G and radius less than 1, and since 0C=1, 
 is evidently outside this circle. 
 
 Hence, p or OP never vanishes, and (f> or lGOP lies 
 between limits greater than — ^ on the one side and less 
 than ^ on the other. 
 
THE BINOMIAL THEOREM. 439 
 
 Make the angle COQq = --^, and the length OQ^ — pi, and 
 
 let OQq, OQy OQ2, ... OQq-i be a system of lines of equal 
 lengths symmetrically distributed about 0. Xhen the 
 
 vectors OQq, OQ^, OQ^, ... OQ^.i represent the q values of 
 
 p 
 
 ((l + a?))*^, OQq representing the principal value. 
 
 The statement thaty f- j is equal to one of the values 
 
 of ((1 -{■x))q may now be expressed geometrically by saying 
 
 that i^ coincides with one of the points Q^, Qi, ... Qg-i; and 
 
 we have further to shew that F coincides in all cases 
 
 with Qq. 
 
 As P moves continuously, the length OP or p changes 
 
 p 
 continuously, and therefore pi, or each of the lengths 
 
 OQq, OQ^, ... OQq-i changes continuously, and, siuce OP 
 
 never vanishes, therefore OQq, OQ^ ... OQq^i never vanish, 
 
 and therefore F cannot pass through from one of the 
 
 points Qq, Qi, ... Qq-i to another. 
 
440 THE BINOMIAL THEOREM. 
 
 Again, as P moves continuously, the angle COP or 
 changes continuously, and therefore — or lCOQq 
 
 changes continuously. 
 
 Hence, each of the points Qq, Qi,...Qq-i moves con- 
 tinuously, and therefore in no way can F pass from one 
 of these points to another. 
 
 Hence, if we can shew that for any one position of the 
 point P, F coincides wnth a particular point of the system 
 Qo' Qv ^2' ••• Qq-if ^^^^ fo>' ctll positions of P within the 
 limiting circle, F will remain coincident with that 
 particular point of the system. 
 
 Let P be at G, then p = l, and ^ = 0, 
 
 length OQo = 1, and lGOQ^ =^ = 0, 
 
 i.e., Qq is at G. 
 
 Also, since, when P is at G, x = 0, thereforeyf-j = 1, 
 
 and therefore 0F= 1, i.e. F is also at G. 
 
 Thus F and Qq are together when P is at G, and there- 
 fore F coincides with Qq for all positions of P ; i.e., in all 
 cases for which mod(£c) < 1, we have 
 
 OF=OQq, 
 /(p = ;(cos^^H.isin^^). 
 
 276. Negative Index. — Let n he a, real negative 
 number commensurable with unity, and equal to — m, say. 
 
 Then/(n)=/(-m) = J^^, 
 since /(-m)x/(m)=/(-m+m)=/(0) = l; 
 
THE BINOMIAL THEOREM. 441 
 
 f(n) = -- ^—r-. ^ (art. -273 or 275) 
 
 = p- "^(cos m</) — i sin m^) 
 = p^(co9 n<l) + i sin n(f)). 
 
 277. Example. — If n be any real number, and - - < 6 <^. then 
 
 4 4 
 
 will"-gi^=l- ^(f-J) tan^^+ ^^^^-^X^-^)i^-^ Jtan4^-...,and 
 cos^^ 1.2 1.2.3.4 ' 
 
 «^^^^=^tan^-^(V ^XV^>tan3^+.... 
 cos"^ 1.2.3 
 
 The principal value of ((cos ^+tsin ^))" is cosw^+isin?i^, since 
 
 6 lies between - tt and tt ; therefore the principal value of 
 
 ((l+*tan^))"=^+*^. 
 cos"^ cos"^ 
 
 But, since - 1< tan ^ < 1 , we have, by the binomial theorem, 
 
 the principal value of 
 
 ((1+2 tan d)f = 1 + n(i tan 6) + ^^^~}\ i tan Ofi-.... 
 Hence, costi^ -sin 
 
 cos'*^ cos''^ 
 
 Equating real and imaginary parts, we have 
 
 cosnd^ 1 _ n{n-l) ^^2^ + n{n - l)(n - 2Xn - S) ^^^^^ _ 
 cos**^ 1.2 1.2.3.4 ■■■' 
 
 and E£i^^n tan ^-^<^-^)^V^) tan3^+... . 
 
 cos^t* 1.2.3 
 
 Examples XXVIII. 
 1. Find the modulus and amplitude of the sum of the 
 series l+nx+ '^y"^ x^ -{-... ad inf., when n = }- 
 
 and x= — . 
 4 
 
442 THE BINOMIAL THEOREM. 
 
 2. Draw the vectors representing the first three terms of 
 
 the series l + j(|:) + ilri)(|:y + ..., and the 
 
 vector representing the sum to infinity. 
 
 3. Sum the series 
 
 cosa + ^cos2aH — ^. ^ ^ cos3a+... 1 0(71+1) terms, 
 and 
 
 71/(71/ '~' 1^ 
 
 sin a +71 sin 2a + \ sin3a+... to (ti + I) terms. 
 
 4. Shew that 
 
 7l/(7h ~~" X ^ 
 
 sin2a+'nsin5a + --Y-2--sin8a+...to('n + X)terms 
 
 . 371 + 4 /_ 3a V 
 = sin-^-a.(2cos^j. 
 
 5. Shew that, if tan ^ = a; sin a/(l + ic cos a), then 
 
 77 (7) —" X I 
 l-\-nx cos a + -y-^-^ cos 2a+ . . . to (71 + X) terms 
 
 I 
 
 fx sin aV ^ , 
 
 ( ^ — TT ) cos 7?.^, and 
 \ sin ^ / ' 
 
 ^sin2a+... i 
 
 (x sin a\** • /. 
 = ( ---^) sin7ia 
 \ sin / 
 
 7?a;8ina+ ^.j ^ ^ sin2a+... to 71 terms 
 
 6. Sum the series 
 
 cos 710 + 7^ cos (71 — X )0 cos 
 71(^71 — X ^ 
 
 + \ ^ C0S(71 — 2)0 cos 20 + ... + COS 710. 
 
 7. Prove that in a triangle, where a is less than c, 
 
 cosji^ If, , a „,«-(«+ l)a^ oD 
 
 i 
 
 
THE BINOMIAL THEOREM. 443 
 
 8. If < ^, shew that 
 
 4 
 
 r==o l??4-2r 1 
 
 cos*^0cos^0= S (-1)^- ^ ^ - tan^^a 
 
 r=o ['yi-1 [2r 
 
 9. Shew that, if < t. then 
 
 4 
 
 cos^e sin '^^0 = 7Z tan e - ' ^('^ + 1X^ + ^) tan^O 
 
 + 1.2.3.4.5 tan0-.... 
 
 10. Shew that 
 
 mo fi/3/1 ,1 5.8 1 5.8.11.14 1 \ 
 
 cos 10 =VJ|1 + 33-3;-^ -3-0+ 3.4.5.6 •39--I 
 
 and hence calculate the value of cos 10° to three 
 places of decimals. 
 
CHAPTER XIX. 
 
 THE EXPONENTIAL SERIES. 
 
 278. For all values of x, whether real or complex, the 
 series 
 
 is absolutely convergent, and therefore also a continuous 
 function of x. The series is also one-valued, since each 
 of its terms has one, and only one, value for a given value 
 of ic. 
 
 This absolutely convergent, one-valued, continuous 
 series is called the exjponential series, and is denoted by 
 the symbol exp(a;). 
 
 279. If X and y he any nv/nihers. real or complex, then 
 
 exp(a?) X exp(2/) = exp(aj + y). 
 In the product of the absolutely convergent series 
 
 11 I? \L 
 
 the term of the (m + ny^ degree is 
 
 ^m+n ^m+n-1 y ^m+n-2 y2 ym+n 
 
 [m+n [m+n — l '\l \m-{-n — 2 ' [2 '" \m+n 
 
 444 
 
THE EXPONENTIAL SERIES. 445 
 
 or 
 
 1 L«+n+2!^a;«.+n-w("^+^X;^+^-l)^^+n-y^., 
 
 \m+n \, U: I? 
 
 +«|, 
 
 and, by the binomial theorem, this is equal to ^ — ^ — , 
 
 ' -^ ' ^ \m-\-n ' 
 
 which is the general term of exp(a; + 2/); we have therefore 
 exp(ir) X exp(2/) = exp(a;+2/). 
 Cor. — By repeated applications of this theorem we have 
 exp(aJi) X exp(aj2) x . . . x exp(fl?„) = q^^{x^+x^+ . . . +Xn). 
 
 280. If X he a real number, then 
 
 exp(a;^) = cosa) + '^sinaj. 'i^ -l*^^-:' <^''*-'^ i"' 
 
 We have, for real values of x, 
 
 x^ x^ 
 cosa; = l-r- + ^--... (art. 215) 
 
 [2 [4 
 
 and smx = x-^+%-..., (art. 216) 
 
 therefore cobx = 1 + ^j^+^^+ ... 
 
 and i8mx = (xi) + i^+^^^X-^"-> 
 
 [3 [5 
 
 therefore, by addition, 
 
 [2 ^ 
 
 cosx + i8mx=l + (xi) + ^-^+^^+..., 
 
 and therefore 
 
 exp(a;'i) = cos ii! + ^ sin x. 
 Cor. — Since 
 
 exp(a;i) = cos x + i sin x 
 and exp( — a5i) = cos( — fl?)+isin( — a?) 
 
 = cosx — ismx, 
 
446 
 
 THE EXPONENTIAL SERIES. 
 
 and 
 
 cos a; = -{exp(a;i) + exp( — ici)} 
 sin X = --.{exp(a;i) — exp( — a;i)}. 
 
 281. Tfx+yi be any comj^lex number, then 
 exp(a; + yi) = exp(£c)(cos y+isiu y). 
 
 For by art. 279, exp(aj+2/i) =exp(a;) x exp(2/i), 
 and by art. 280, exTp(yi) = cos 2/ + i sin y, 
 
 therefore exp(aj + yi) = exp(a;) (cos y-\-ismy). 
 
 From this theorem, we see that exp(x + yi) is a complex 
 number whose modulus = exp(aj) and whose amplitude is y. 
 
 Since cosy = co8(y-{-2n7r) and smy = s'm(y-\-2n7r), it 
 follows that, if z = x + yi, exp(0) is a periodic function of 
 s, and that its period is the imaginary quantity 27ri. 
 
 282. Example 1. — Represent geometrically the value of 
 e.p(2+|.). 
 
 We have exp(2) = 7"4 ajDproximately, 
 
 e.p(2 + |i) = (7-4, I). 
 
 Make ^vOP=1, OP =7% 
 
 o 
 
 )■ 
 
 then 
 
 0F=exp\2 + '^t 
 
 Example 2. — Sum the series 
 
 1+- cos ^+ A cos 2 
 
 } \1 
 
 and 
 
 ^ + — coa36+...adinf., 
 ^ sin ^+ i sin 20 + ,4 sin 3^+ . .. ao? m/ 
 
 li (2 L^ 
 
 Let (7 and >S' denote the sums of the series, 
 and let a: = cos d+isin 6, 
 
THE EXPONENTIAL SERIES. 447 
 
 then (7+^•>S=l+-^+^+^ + ... 
 
 lJi \1 If 
 
 =exp(^) 
 
 = exp(cos d + i sin ^) 
 = exp(cos ^)[cos(sin 0) + ? sin(sin d)\ 
 C=exp(cos ^) . cos(sin ^), 
 and /S'= exp(cos d) . sin(sin B). 
 
 Examples XXIX. 
 
 1. If cc be a real number, prove that 
 
 exp(a;) = cosh X + sinh x. 
 
 2. Shew that 
 
 exp(a; + yi) = (cosh x + sinh aj)(cos y + i sin y). 
 
 3. Calculate the values of exp(l) and exp(2) each to four 
 
 places of decimals. Verify the result by substitut- 
 ing the values in the equation 
 
 exp(2) = exp(l) x exp(l). 
 
 4. Calculate, to three places of decimals, the values of 
 
 exp(J) and exp(f). Verify by substituting in the 
 equation exp(f)-i-exp(J) = exp(l). 
 
 5. Represent by vectors the values of 
 
 exp(^^-|-ij and exp(^2~|)' 
 
 6. Sum the series 
 
 II n /I . sec^O sec^O ..^ , , . , 
 
 l + sec0cos04- |t) cos 20+ cos 30+... adinf. 
 
 and 
 
 secOsin^H — ^r— sin20H — r^— sin S0+... ad inf. 
 il If 
 
 7. Find the sum of the series 
 
 - , X cos , a;^cos 20 a?^cos 30 7 . /. 
 
 IH ^ 1 — Y~2 — 12 3 "^•" ^'^•^* 
 
448 THE EXPONENTIAL SERIES. 
 
 8. Prove that 
 
 csm 0+ ^-^+y-2-gH- ... ac« in/ 
 
 = exp(c cos 0)sin(c sin 0). 
 
 9. Sum to infinity the series 
 
 , csin(a + i8) , c2sin(a + 2^) , 
 sinaH ii — ^— H yy 1-.... 
 
 10. Find the sum of 
 
 sm A + sin(^ + J5)cot Q H — ^ .^ ' -\-...ad inf. 
 
 l± 
 
 11. Find the sum of 
 
 , . e sin20 . 02 sin30 ^e^^m^B , , . . 
 
 l + (l-^i^ + g^I^ + 5 sin^e +...acZtn/ 
 
 12. Sum to infinity the series 
 
 cosaj+acos2ic+ + 
 
 1 . ^ 
 
 13. Shew that 
 
 1 , zi n , cos20cos2O , cos^0cos30 , ^7 ._/. 
 1 +COS 0co:4 e + ^— ^ + ""i~2T~ •^* 
 
 sin 20^ 
 
 = exp(cos20)cosf — - — j 
 
 
 
 ~''^'- 
 
CHAPTER XX. 
 
 LOGAEITHMS OF COMPLEX NUMBERS. 
 
 283. Def. — If y = exip(x), where x is any number, real 
 or complex, then x is called the logarithm of y. The 
 logarithm of y, thus defined, is denoted by Log 2/, the 
 capital letter indicating that the logarithm is many- 
 valued (see art. 284). 
 
 When X is & real commensurable number the above 
 definition is equivalent to that obtained by substituting 
 e^ for exp(a;), provided that we restrict e'^, when x is not 
 integral, to its arithmetical value. These real logarithms 
 to base e of arithmetical numbers are called Napierian 
 logarithms, being closely connected with the logarithms 
 of sines calculated by Napier. It will be assumed that 
 the fundamental properties of Napierian logarithms are 
 known. 
 
 When X is Si complex number, or when the value of e* 
 is unrestricted, the definitions are not equivalent, for (1) 
 when cc is a complex number e'^ is an undefined and 
 therefore a meaningless expression, and (2) when i:c is a 
 
 fraction —, where p and q are integers, e* has q values, 
 
 while ex-p(x) has one value only. 
 
 For example, if ^e denote the positive value of the 
 2f 449 
 
450 LOGARITHMS OF COMPLEX NUMBERS. 
 
 square root of e, one value of ((e))^ is — ^e, and accord- 
 ingly, with a base-index definition of logarithms, J is a 
 logarithm of — mJc to the base e ; while with the inverse- 
 exponential definition given above it will appear that 
 - /^e has an infinite number of logarithms, and that no 
 one of these is J. 
 
 284. To 'prove that, if (r, 6) he any complex number, 
 
 Log (r, e) = logr+(e+2n'7r)i, 
 ivhere log r denotes the Napierian logarithm of the 
 positive number r* 
 
 Let Log(r, 0) = x-\-yi, 
 
 then, by definition, (r, 6) = exp(£C + yi). 
 
 But exp(aj + yi) = exp(a^)[cos y+i sin y], 
 
 (r, 0) = exp(fl?)[cos y + i sin y], 
 hence, r = exp(a;) or a; = log r, 
 
 and y = 0+2n7r, 
 
 Log(r, 6) = log r + (0 + 2mr)i. 
 
 285. Def. — If r be the modulus and 6 the principal 
 value of the amplitude of any complex number x, then 
 \ogr+6i is called the principal value of Logo; and is 
 denoted by logic. 
 
 Thus we may write 
 
 log(a +bi) = i log(a2 + 62) + Qi 
 where 6 = amp(a + bi) ; 
 
 or, if a is positive, 
 
 log(a + 6i) = I log(a2+?)2) +^ tan'^- ; 
 
 a 
 
 * For a geometrical discussion of the exponential function and the 
 inverse exponential, or logarithmic, function, see Chrystal's Algebra, 
 chap, xxix., § 19. 
 
LOGARITHMS OF COMPLEX NUMBERS. 451 
 
 if a is negative and h positive, 
 
 log(a -[■hi) = l log(a2 + 6^) + i tan " ^—\-iri; 
 and if a and h are both negative, 
 
 log(a + 6i) = J log(a^ -\-h^)-\-i tan - ^- — 7ri. 
 
 The following particular cases may be noticed : 
 logl = 0, log(-l) = 7ri 
 log^ = |i, log(-^)=-|^. 
 
 286. If a he any number real or complex, and x any 
 real number commensurable with unity, then will 
 
 {{a)Y and exp(a; Log a) 
 
 have the same values ; and the principal value of {{a)y 
 will be equal to exp(cc log a). 
 
 Let a = {r, Q), where — tt < :!> tt, then 
 
 Log a = log r + (0 + 2ii7r)i, (art. 284) 
 
 and therefore 
 
 exp {x Log a) = exp {a?[log r + (0 + 2n7r)i] } 
 
 = exY){xlogr){cosx(0-}-2n'7r) + i8mx{0-\-2n'7r)]. 
 But, by the properties of Napierian logarithms, 
 exp(a; log r) = exp (log r^) = r^, 
 .*. exp(a; Log a) = r*{cos x{6-\-2n7r) + ism x{0-{-2n'7r)}. 
 Again, we have 
 
 {(a)f = r'^{cosx(e + 2n'7r) + is'mx(e-h2n7r)}. (art. 260.) 
 Hence, ((«'))* = exp (a? Log a). 
 
 Putting n = 0, we see that the principal value of 
 ((a))*, or a^, =exp(£cloga). 
 
462 LOGARITHMS OF COMPLEX NUMBERS. 
 
 Example. — In illustration of the method of the foregoing proof 
 we will shew that ((9))* and exp(^ Log 9) have the same values. 
 The values of ((9))* are +3 and -3. 
 Also Log 9 = log 9 + 2w7ri, 
 
 exp(^ Log 9) = exp{^(log 9 + 2w7n)} 
 
 = exp(^ log 9) (cos nir + i sin rnr) 
 = 3(cos nir + i sin mr). 
 Assigning to n the values 0, 1, 2, 3, ... - 1, -2, -3, ... we get as 
 the only values of exp(| Log 9) the real numbers 3 and - 3. 
 Hence, ((9))* and exp(^ Log 9) are equivalent symbols. 
 
 287. If z he any number, real or complex, such that 
 mod(z) < 1, then will 
 
 2 3 
 
 log(l+«) = 2;~|+|-...acZm/ 
 
 By the Binomial Theorem, if ic be a real number com- 
 mensurable with unity, we know that the principal value 
 
 of ((1+2))- 
 
 = l+a,.+^^)«H^<^^^t|^^ + .... (1) 
 
 Also, by the last article, the principal value of ((1+0))* 
 
 = l+a;log(l+0) + -^^'-{log(l+«)F+ (2) 
 
 Therefore the series (1) and (2) are equal to one another. 
 The series (1) is convergent for all real values of x, 
 provided that mod(2;) < 1, and the series (2) for all values 
 of X, provided that log(l+0) is finite, which is always 
 the case when mod(0) < 1. Hence, we may equate co- 
 efficients of the powers of x in (1)* and (2). 
 
 * The argument is incomplete. It has not been shewn that the series 
 (1) remains convergent when its terms are re-arranged according to 
 powers of x ; or, if convergent, that it converges to the same limit as 
 before. For a complete treatment see Chrystal's Algehra, chap. xxvi. 
 §§ 32-.S5, chap, xxviii. § 9, chap. xxix. § 22. 
 
LOGARITHMS OF COMPLEX NUMBERS. 453 
 
 Equating coefficients of x, we have for all values of z 
 such that mod(2;) < 1, 
 
 \og{l + z) = z-t^t- ...adinf. 
 
 Cor. — Since Log(l+0) = log(l -\-z) + 2n7ri, we have 
 
 z'^ z^ 
 Log(l+z) = 2n7ri-{-z — —+— — ... ad inf. 
 
 when niod(2;) < 1. 
 
 288. If mod{z) — \, and Amy{z)=\=^{2n + l)Tr, then will 
 
 log(l + ^)=^-|'+|'- ... ad inf 
 
 Let a = Amp(0). 
 
 We know, by art. 2G8, that the series 
 
 (1, a) + J(l. ^Hh§) + W . ^+2^) + . . . ac^ inf 
 is convergent, provided that P=\=2n7r. 
 Let /3==a + 7r, then the series 
 
 (1, a) - i(l. 2a) + 4(1, 3a) - . . . acZ inf, 
 i.e., the series z — ^z'^+^z^— ... ad inf, 
 
 is convergent, provided that a=\=(2n + l)Tr. 
 Hence, by art. 270, the series 
 
 z^ , Z^ 7 . /. 
 
 z--^+--...adinf 
 
 is a continuous function of z up to the limit when 
 mod(0) = l, provided that Am-p{z)=\=(2n-\-l)7r. 
 
 Also log(l+2;) remains finite, and is a continuous 
 function of z under the same conditions. 
 
 Therefore, since the equality of log(l+0) and the series 
 
 s;— -^+^— ... holds, however nearly mod(2;) approaches 
 
 to unity, it will also hold when mod(:5) = l, provided that 
 Amp(2;)=i=(27i + l)'7r. 
 
454 LOGARITHMS OF COMPLEX NUMBERS. 
 
 289. The geometrical interpretation of some of the 
 symbols used in arts. 287 and 288 may be noticed. 
 
 Let z={r, 0), l+z = (p, <f>\ then, since log(l+;$;) is 
 
 the principal value of 
 j^ N^ Log(l+2;), cannot 
 
 be numerically greater 
 than TT. 
 
 Take^Oa=l, GP = z, 
 then 0P = 1+^. By 
 hypothesis, r or mod(;2) 
 < 1, therefore lies outside a circle whose centre is C 
 and radiusr. If P move round the circle from A to A' 
 in the positive sense, increases from to tt, or, gene- 
 rally, from 2mr to {2,n + V)ir, and ^ increases from to 
 a maximum value sin-V, which it attains when OP 
 touches the circle, and then decreases to 0. 
 
 If P move round the circle from A' io A in the posi- 
 tive sense, 6 increases from tt to 27r, or from — tt to 0, 
 or, generally, from (27i — l)7r to 2mr, and ^ is negative 
 and increases numerically from to a maximum value 
 sin~V, and then decreases numerically to 0. 
 
 It will be seen also that p decreases continuously from 
 OA to 0A\ i.e. from l-\-r to 1 — r, and then increases 
 from 0^' to OA.i.e. from 1 — r to l-j-r; hence log/) is 
 always finite. 
 
 The limiting case, in which r=\ should be carefully 
 considered. A' then moves up to, and ultimately coin- 
 cides with, 0. 
 
 As P moves from A to 0, 6 increases from ^nir to 
 
 (27i-|-l)7r, and from to ^. As P passes through 0, 
 
r 
 
 LOGARITHMS OF COMPLEX NUMBERS. 455 
 
 TT 
 
 (t) changes suddenly from +-^ to — — , and then increases 
 
 from — ^ to as P moves onward from to A. During 
 
 the same period p changes from 
 2 to 0, and then from to 2. 
 
 The excepted case of art. 288, 
 in which Amp (2;) = {^n + Ijtt, q\ 
 is that for which P coincides 
 with 0. 
 
 The discontinuity of the am- 
 plitude as P passes through gives rise to some inter- 
 esting results in the values of certain infinite series derived 
 from the logarithmic series. See art. 293, Examples 1, 2, 3. 
 
 290. If z he any number, real or complex, such that 
 mod 2; < 1, then will 
 
 Z^ 03 
 
 log(l-0)= -^- 2~ 3~--* ^^ ^''^•^• 
 Changing z into —0 in the equation 
 
 log(i+«)=^-i'+J-..., 
 
 we obtain the required result. 
 
 Co7\ — If mod(0) = l, and Amp{z)=\=2mr, 
 
 then log(l -2;)= -0-- ---.... 
 
 If, as in art. 289, CP = z, 
 and if PC meet the circle 
 again in Q, then OQ = l—z 
 or (p, (p), and the changes 
 in and p may be dis- 
 cussed as before. 
 
 The geometrical interpre- 
 tation of the limiting case when mod(0) = l should be 
 considered. 
 
456 LOGARITHMS OF COMPLEX NUMBERS. 
 
 291. Gregory's Series.— //a; he a real number behveen 
 — I and +1, both limits included, then will 
 
 /y.3 ^6 
 
 t8in-^x = x— — +——... ad inf. 
 o 5 
 
 Under the given conditions with respect to x, we have 
 \og{l+xi) = xi-i(xiy+i(xi)^- ..., 
 and log(l —xi)=—xi — i(xi)^ — i{xif - ..., 
 
 log(l + xi) — log(l — xi) = 2i(x — ix^+\x^— ...). 
 . Now 1+xi and 1—xi are conjugate complex numbers, 
 having the common moduhis +s/T+x^, and the ampli- 
 tudes tan"^a; and — tan"^a; respectively; hence 
 
 log( 1 + ici) — log(l — xi) = \og\/l-\-x^-\- i tan - ^x 
 - logVl +x^+i tan " ^x 
 = 2iiaii~^x, 
 tan -^x = x — Jcp^ + \x^ — ...ad inf. 
 Cor. — Since Ta,Ti-^x = n7r + tsLn-^x, we have for values 
 of a; between —1 and +1, both limits included, 
 Tei,n~^x=n'7r-\-x — ix^+^x^—..., 
 where n is to be so chosen that T&n'^x — nTr may lie 
 between — ^ and + ^. 
 
 292. Numerical Value of tt.— By aid of Gregory's 
 Series the numerical value of tt may be obtained to any 
 required degree of accuracy. The following methods 
 may be noticed : — 
 
 (1) In Gregory's Series, let x = l, then 
 
 4 3^5 7^-* 
 
 (2) Let a? = — 7H,then 
 
LOGARITHMS OF COMPLEX NUMBERS. 457 
 
 6~V3\ 8 3^5 32 7 33^" 
 and therefore 7r = 2^3(^l-^- 3 + 5 '32-7 '33 + 
 
 (3) -^ = tan-i^ + tan-ii (Euler's formula). 
 
 4 2 3"23"^5'25 ••• 
 ^3 3 33^5 3^ •••• 
 
 TT 1 1 . 
 
 (4) ~ = ^tsji-^~ — iQXi-^^^ (Machin's formula). 
 4 
 
 -4^1-1.1 + 1.1- 
 
 ~A5 3 53^5 55 ••• 
 
 \239 3 (239)3^ "V 
 
 (5) ^ = 4tan-i--tan-y^-+tan-i^ 
 
 (Rutherford's formula). 
 
 293. Example 1. — Sum the series 
 
 cos ^ + ^ cos W+^ cos 3^ + . . . ad inf.^ 
 
 and sin ^ + ^ sin 2^ + ^ sin Z6+ ... ad inf. 
 
 Let C and S denote the sums of the series, and suppose that Q 
 is not an even multiple of tt. 
 
 Let .r=cos ^+^sin ^=(1, 0. 
 
 Then (7+ iS=x-\- \x^ + |a^ + . . . ad inf. 
 
 = -log(l -x\ since O^^nir. 
 Now, l-^=l-cos^-isin^, 
 
 mod(l - x) = v^(l - cos ^)2 + sin2(9 = ^2 - 2 cos ^ = 2/y^sin2?, 
 
458 LOGARITHMS OF COMPLEX NUMBERS. 
 
 and amp(l -x)=^ tan~M ~^^"^ ), since 1 - cos ^ is positive, 
 
 \ J. — " COS \j/ 
 
 = -tan-i(cot|), 
 
 hence, log(l -a;)=logf 2\/sin2_ j -i tan-M cot ^\ 
 
 C-\-iS= -log(2.^^^) + ?-tan-i(cot|), 
 
 C= - log(2A/sin2| V and ^=tan-^(cot |). 
 
 Since tan~M cot - ) = tan-M tan^- ^) = n7r+J--, where n is so 
 chosen that n7r + ^-^ lies between -^ and ^, we may write 
 
 ^=W7r + ^--, with the above condition with respect to the vahie 
 
 of n. 
 
 If ^=2?i7r, we have immediately from the series, C=xf, S=0. 
 
 Example 2. — Sum the series 
 
 cos <^ - ^ cos 2<^ + ^ cos 3<f>- ... ad inf., 
 and sin </> - ^ sin 2<^ + ^ sin 3^ - . . . ad inf. 
 
 In the results of Example 1, put 6=7r-c{), then, provided that 
 ^>=|=(27i+ I)7r, we have 
 
 cos </) - ^ cos 2(^+^ cos 3<^ - . .. =log( 2A/cos2r V 
 and sin^-i^sin2<^+^sin3<^-... =tan-*(tan^j 
 
 where n is so chosen that - - < ^t + ^ < ^. 
 
 If <^=(27H-l)7r, then, immediately from the series, 
 cos ^-\ cos 2(/) + ^ cos 3i/) - . . . = - 00 , 
 and sin<^-^sin2^+^sin3^-... =0. 
 
LOGARITHMS OF COMPLEX NUMBERS. 
 
 459 
 
 The curves of the series in Examples 1, 2, 3 are represented in 
 following figures. 
 
 y = cos x+ f cos 2x-\-j cos 3x * — 
 
 y =cos x--fcos 2X + -^C0S JX 
 
 y =COS X + jCOS JX + jCOS JX+ — 
 
 There are no isolated points. 
 
 .'-'O 
 
 • y = sin X -t- -f-sin 2x ■*■ -j- sin jx* - — -■ 
 
 TTtere are isolated joints along Ox at -ztt, O, 2T, etc. 
 -y= sin X --j-stH 2x -t-j sinjx- 
 
 TTiere are isolated points along Ox at-tr, IT, Sir, eto. 
 
 -y = sin X ■t--i-sin jx *■ 4-sin jx^r. 
 
 There . are isolated Points along Ox at -2Tr, -it, 0, v, 2ir, etc. 
 
460 LOGARITHMS OF COMPLEX NUMBERS. 
 
 Example 3. — Sum the series 
 
 cos ^+^ cos 3^ + ^ cos 5^+..., 
 and sin ^+i sin 3^+^ sin 5^+.... 
 
 If ^=|=»7r, we have by examples 1 and 2, 
 
 cos ^ + ^ cos 2^+ J cos 3^+... =-log('2A/sin2|\ 
 
 and cos^-^cos2^+^cos3^-... =log(2A/cos2|V 
 
 .*. by addition, 
 
 cos ^ + ^cos3^+^ cos 5^+ ... = -^ log(A/tan2|Y 
 
 If d=mr, the series =±(1+^ + 1 + ...) =±oo. 
 Again if ^=i=W7r, we have 
 
 sin6^ + ^sin2^+Jsin3^+... =tan-Mcot|V 
 
 and sin ^-^ sin 2^+^ sin 3^-... =tan-Mtan|j, 
 
 .'. by addition, 
 8in^+^sin3^+isin5^+... =^{tan-^(cot |) + tan-i(tan ^)]- 
 
 the upper or lower sign being taken according as tan - is positive 
 or negative ; thus from ^=0 to 9=ir^ the series = ^, from ^=7r to 
 d='2.Tr its value is-^* and so on. If d = mr, we have immediately 
 sin^ + ^sin3^+... =0. 
 
 Example 4. — Supi the series 
 
 7i sm a + -^sm 2a + — sin 3a + . . . aa inf.^ 
 2 3 
 
 where n<\. If ^ is the sum of the series, shew that 
 
 sin ^=wsin(^+a) and that taxi(e+^=\±^ tan % 
 
 V 2/ 1-n 2 
 
 ^2 
 
 Let ^=7isin a+ — sin2a+... 
 
LOGARITHMS OF COMPLEX NUMBERS. 461 
 
 ^=yiCOSa + — cos2a+ ..., 
 
 and let a = cos a + i sin a. 
 
 Then </) + 6'^ = wa + ^^-^ 4- — ^ + . . . 
 
 = -\og{l-na) 
 
 = -log{(l - 71 COS a)2+(n sin a)2}* -itan-^f -?isina \ 
 °^ ■^ Vl-7iC0sa/ 
 
 = - log\/l-27^eosa+w2 + ^ tan-^f _^i^HL^\ 
 
 \l-wcosa/ 
 
 and.-. • ^=tan-^f-Ji!HL^y 
 
 From this result we see that 
 
 sin ^ . (1 - ?i cos a) = cos ^ . 71 sin a, 
 and therefore sin ^ = n sin( ^ + a). 
 
 We have, further, 
 
 sin(^ + a) + sin^ _ l+7i 
 sin(^ + a)-sin ^ 1 — n' 
 
 whence tan((9+^Hl±^tan^. 
 
 V 2J 1-71 2 
 
 Example 5. — If sin ^=72'Sin(^+a) where n<l, find an expansion 
 for ^ in a series of ascending powers of n. 
 
 Since the equation is unaltered by the addition of any multiple 
 
 of TT to 6, we may consider 6 to lie between — ^ and ^. 
 
 2 A 
 
 Let ^ = (1) 0), a = {lj a), and /. a:r=(l, ^+a), 
 
 then 2^ sin ^ = .r - - and 2^ sin( 0-\-a) = a.v- — 
 
 a: ax 
 
 Substituting in the given equation, we have 
 
 1 / 1\ 
 
 x--=n{ax- — I, 
 X \ ax) 
 
 1-- 
 
 and therefore x'^= 
 
 1—na 
 Taking logarithms we have 
 
 26>i + 2r7r?'=log(l --) -log(l 
 
462 LOGARITHMS OF COMPLEX NUMBERS. 
 
 In this equation r must be zero; for, if ainp(l--J = ^, and 
 
 therefore amp(l - wa)= - ^, the right-hand member of the equation 
 reduces to 2<^i (art. 285) ; also, since 1 - ?i cos a is positive, <^ lies 
 
 between - ^ and ^, and consequently, with the restricted value of 
 2 2 
 
 0^ the value zero is the only admissible value of r. 
 Expanding the logarithms we obtain the equation 
 
 and .-. ^=7isina+^sin2a-|-^8in3a+... . 
 
 Example 6. — If tan a=7i tan ^, where w > 1 , then will 
 a+nr = ^ + wisin2/3+^sin4^-f-^sin6^+...a<]?Mi/-. 
 
 where m=^?l^I— . 
 
 71+1 
 
 Let a=(l, a), 6=(1, /?), 
 
 then itana=^-^- and ttan^= ,.,"~ . 
 
 a^+1 ^ 6H1 
 
 Substituting these values of tan a and tan /?, the given equation 
 
 may be written 
 
 a^-l _ 62-1 
 
 a2+l VTi* 
 
 1 -!?^ 
 
 whence a^=h^ -^. 
 
 1 -mo^ 
 
 Taking logarithms we have 
 
 2az + 2r7rj = 2/3^ + log^l - p) - log(l - w62) 
 
 2 
 
 or a + r7r = /?4-msin 2^4- — sin 4^+... adinf. 
 
LOOARITHifS OF COMPLEX NUMBERS. 
 
 463 
 
 294. Short Table of Napierian Logarithms, or Real 
 Logarithms to Base e. 
 
 No. 
 
 log. 
 
 No. 
 
 log. 
 
 No. 
 
 log. 
 
 1 
 
 0-00 
 
 16 
 
 2-77 
 
 
 
 2 
 
 0-69 
 
 17 
 
 2-83 
 
 200 
 
 5-30 
 
 3 
 
 110 
 
 18 
 
 2-89 
 
 300 
 
 5-70 
 
 4 
 
 1-39 
 
 19 
 
 2-94 
 
 400 
 
 5-99 
 
 5 
 
 1*61 
 
 20 
 
 3-00 
 
 500 
 
 6-21 
 
 6 
 
 1-79 
 
 
 
 600 
 
 6-40 
 
 7 
 
 1-95 
 
 
 
 700 
 
 6-55 
 
 8 
 
 2-08 
 
 30 
 
 3-40 
 
 800 
 
 6-68 
 
 9 
 
 2-20 
 
 40 
 
 3-69 
 
 900 
 
 6-80 
 
 10 
 
 2-30 
 
 50 
 
 3-91 
 
 1000 
 
 6-91 
 
 11 
 
 2-40 
 
 60 
 
 4-09 
 
 1100 
 
 7-00 
 
 12 
 
 2-48 
 
 70 
 
 4-25 
 
 1200 
 
 7-09 
 
 13 
 
 2-56 
 
 80 
 
 4-38 
 
 
 
 14 
 
 2-64 
 
 90 
 
 4-50 
 
 
 
 15 
 
 2-71 
 
 100 
 
 4-61 
 
 
 
 Examples XXX. 
 
 1. Find the values of Log(l + V — 1), and represent them 
 
 geometrically. 
 
 2. Find the values of Log( — 20), and represent them 
 
 geometrically. 
 
 3. If \og{l+iidMa) = A-\-Bi, prove that J. = log sec a, 
 
 7r TT 
 
 and find -B, having given ~ « < « < o* 
 
464 LOGARITHMS OF COMPLEX NUMBERS. 
 
 4. li\og{x-\-iy) = a-\-ip, prove that 
 
 x^ -\-y^ = e^a, and y=xt&u /3. 
 
 5. Prove that Log( — ^e) = J + (2'/^+l)'7^^, and represent 
 
 the logarithms by vectors. 
 
 6. Shew that the expressions 
 
 Log(a + hi) and J log(a2 + 62) + Tan " i- 
 are not equivalent. 
 
 7. Prove that Log(cos 0+i sin 0) = {0-\-2nTr)it where n is 
 
 any integer ; 
 
 and that log(cos 6+i8mO) = (0+2n'7r)i, where n is 
 
 an integer so chosen that —tt < (O + S-nTr) > ir. 
 
 8. Shew that Logxy = 'Logx-\-Logy-\-2mri, where n is 
 
 an integer. 
 
 9. If a and /3 are the principal values of the amplitudes 
 
 of two complex numbers x and y, then will 
 log xy = log X + log y + 2r'7ri, 
 
 where r= —1, when a-\-^ > tt, 
 
 r = 0, when — tt < a + ^S 4»7r, 
 r=l, when a + fi :^ — tt. . 
 
 10. From the series tt = 2^3(l -|- i+^ • i- ...),find the 
 
 value of TT to two places of decimals. 
 
 11. Find the value of tt to three places of decimals by 
 
 Euler's formula. 
 
 12. Find the value of tt to five places of decimals by 
 
 Ma chin's formula. 
 
 13. Find the value of tt to ten places of decimals by 
 
 Rutherford's formula. 
 
 14. Trace the curve 
 
 2/ = cosa7 — J cosS-Tj + ^cos 5rc— ... ad inf. 
 
LOGARITHMS OF COMPLEX NUMBERS. 465 
 
 15. Trace the curve 
 
 2/ = sin a; — J sin ^x + i sin 5aj — . . . ad inf. 
 10. If msin(m0 + O) = sinm^, where m < 1, then 
 
 = sin0 + imsin20 + Jm%in3O+.... 
 
 17. Sum to infinity, when x<\, 
 
 x^m e + lx^BmW-\-\xhmW-\-.., . 
 
 1 8. Shew that, \i x<l, 
 
 tan \— 7T = a:;sin0 — -^sin 26+-^smS0— .... 
 
 l-\-xcosO 2 3 
 
 19. Sura the series, when m<l, 
 
 m sin^a — ^mhm^2a + Jm^sin^Sa — ... ad inf. 
 
 20. In any triangle, if a > 6, shew that 
 
 h b^ b^ 
 
 log c = log a - - cos 0- ^2^08 20- — 3C0S 3C- . . . . 
 
 21. In any triangle, if & < c, shew that 
 
 j5 = -sin^ + Hsin2^ + J^sin3^ + .... 
 c c c 
 
 22. In any triangle, if 6 < c and a<c, shew that 
 
 , b bcosA — acosB, b'^cos 2A — a^cos 2B 
 
 '°Sa= c + 2^5 
 
 , fe^cos SA — a^cos 35 
 "^ 3c^ "^ 
 
 23. If cot y = cot X + cosec a cosec x, shew that 
 
 2/ = sin aj sin a — I sin 2a;sin^a + 1 sin Sa^sin^a — . . . . 
 
 24. Sum the series 
 
 cos ^ sin + J cos^O sin 20 + J cos^0 sin 80 + . . . ad inf. 
 
 25. Shew that cot-i(l + cot0 + cot2e) 
 
 . sin . ^^ sin20 , . ^^ sin^O . 
 
 = sm . — sm 20 . — ^- + sin 30 . — -^ ...ad %nf. 
 
 26. Sum to infinity 
 
 cos — J sin 20 — J cos 30 + J sin 40 
 
 + i cos 50 — ^ sin 60 — ... . 
 2g 
 
466 LOGARITHMS OF COMPLEX NUMBERS. 
 
 27. Prove that 
 
 \ log sec^a; = sin^ic — J sin^2fl? -h i sin^Sa; — . . . . 
 
 28. Sum the series 
 
 cos^e - i cos220 + J cos^se -...ad inf. 
 
 29. Sura the series 
 
 sin^flj — J sin^Sa; + \ sin^Sa? — ...ad inf. 
 
 30. If be an angle whose cosine is positive, expand 
 
 log cos in a series of cosines of multiples of Q. 
 
 31. Expand log(l — 2acos0+a2) in a series of cosines of 
 
 multiples of 0. 
 
CHAPTER XXL 
 
 COMPLEX INDICES. 
 
 295. The expressions {{A)y and A^ have been defined 
 for those cases in which A is any number, real or complex, 
 and 5 is a real number commensurable with unity. It 
 has been shewn that under these conditions the values of 
 ((A)y are identical with those of ex-p{B . Log A). We 
 now extend the meaning of ({A)y by the following 
 definition : — 
 
 Def. — For all values of A and B, real or complex, the 
 expression exp(5. Log^) is denoted by ({A)y. 
 
 296. To reduce ((A)y to the form x-\-yi. 
 LetA=(a,a),B=ib,l3)> 
 
 then, by definition, ((A)y =exp(5. Log^) 
 
 = exp[(6 cos ^ 4- ^^ sin /3)(log a-\-a-\- 'Im-rr . i)] 
 — exp[6 cos P log a — 6 sin /3 . a + 2m7r 
 
 + ^(6 sin ^ log a + 6 cos yS . a + 2m7r)] 
 = exp(6 cos /5 log a — 6 sin ^ . a + 2m7r) 
 X [cos(6 sin ^ log a + 6 cos |8 . a + ^mw) 
 
 + i sin(6 sin ^ log a 4- h cos /3 . a + 2m7r)], 
 
 an expression of the ^ovm x-\-yi. 
 
 467 
 
468 COMPLEX INDICES. 
 
 Hence, {{A)y is a many- valued function of A and B, 
 whose modulus = exp (b cos ^ log a — 6 sin /3 . a -f 2m7r), 
 and whose amplitude 
 
 = 6 sin ^ log a+6 cos /5 . a + 2m7r+ 2ti7r. 
 
 Def. — The value of {(A)y obtained by putting m = 
 in the above result is called the ^^rmcipa^ value of {{A)y, 
 and is denoted by {Ay or A^. 
 
 Thus, 7l^ = exp(6cos/81oga— 6sin/3.a) 
 X [cos(6 sin ^ log a+ 6 cos |8 . a) 
 
 + i sin(6 sin ^ log a -b ^ cos /3 . a)]. 
 
 297. The formula of the preceding article is cumbrous, 
 but the method used may readily be applied to any 
 particular case, as in the following examples. 
 
 Example I. —To find the values of ((\/-l)) ~^- 
 By definition, ((\/^))^^=exp(V^ LogV'^l). 
 Now, -vrri = (l, |), therefore LogV^=|*'+2^T2, 
 
 ... ((x/-^))^=exp[z(|i + 2.m)] 
 
 = exp(-|-2n7r). 
 
 Let n=0, then 
 
 the principal value of ((\/^))'^^ = expf -^j 
 
 = 1-000 -1-571 + 1-234 --646+ -254 --080+ -021 --005 + -001-... 
 = -208 approximately. 
 
 Example 2.— To find the values of ((e))^' where ^ is a real 
 number. 
 
 By definition, we have ((e))^* = exp(^/Loge) 
 = exp[^2J(log e + 2w7^^)] 
 =exp[-2^27r.^+(9i] 
 =exp( - 2^?7r . ^)(cos ^+?"sin 0). 
 
COMPLEX INDICES. 469 
 
 Let M = 0, then e^*=cos ^+*sin ^=exp(^^). 
 
 Since the principal value of ((e))^* is exp(^i), S^ is often used in 
 the place of the one- valued symbol exp((90, or in place of 
 cos ^+isin d. 
 
 Thus, we may write cos^= — — , 
 
 and sin 0=^ % . 
 
 2^ 
 
 298. The symbol {{A)y having been defined for complex 
 values of A and x, we may examine the consequences that 
 result from the identification of a logarithm of a number 
 with the index of the power to which a given base must 
 be raised to obtain the number. 
 
 We have the following definition : — 
 
 Def. — If A,B,xhQ any numbers, real or complex, such 
 that one value of {(A)Y is equal to B, then x is called a 
 logarithm of B to the base A. 
 
 We express this relation by the equation x = Log^i^. 
 
 It will be shewn in the following articles that a; is a 
 two-fold many-valued function of A and B; and that 
 Jjog^B is identical with Log B as previously defined when 
 A = {e, 0), so that Log 5 is a particular case of Log,, 5. 
 
 It has already been shewn that when 5 is a real posi- 
 tive number the Napierian logarithm of 5 is a value of 
 Log 5. 
 
 299. To express hog^B in the form x + yi. 
 If Log^B = x-\-yi, 
 
 then, by the definition of the last article, we have 
 ((A)Y+^^ = B. 
 Let A = (a, a) and B = (6, /3). 
 
470 COMPLEX INDICES. 
 
 By tho definition of art. 295, 
 
 = exp{(ic+2/t)Log^} 
 
 = exp{(a;+2/^)(log a+a + 2m7r . i)} 
 
 = QX^(x\oga—y.a + ^mir+i{y\oga + x. a + 2mx)}, 
 or {{A)y'^y^ is a complex number whose 
 
 modulus = exp(a; log a — y.a + 2mir), 
 and amplitude = y log a + a; . a + 2m7r. 
 
 But mod(5) = 6, and amp(5) = jS, 
 .-.since ((A)y+'J' = B, we h&ve 
 
 x\oga — y.a + 2m7r = log h, 
 and yloga-\-x.a-{-2m7r = ^ + 2n'7r. 
 
 „ log g . log 6 + (g + 2m7r)(^ + 2'y^7r) 
 
 nence, x- (ioga)2+(« + 2m7r)2 
 
 and (/3+27i7r)loga-(a + 2m7r)log6 
 
 (logtt)2+(a + 2m7r)2 
 Thus we see that Log^B is a two-fold many-valued 
 function of A and B. 
 
 The value of Log^5 obtained by putting m = 0, ?i = in 
 the above result is called the principal value of Log^5, 
 and is denoted by log^5. 
 Thus, we have 
 
 j^_\ogaAogb + a§.§\og analog b 
 ^^^^^~ (loga)2 + a2 "^^ (loga)2 + a2' ' 
 
 where a and 6 are the moduli, and a and /3 the principal 
 values of the amplitudes of A and B, 
 
 If a = and ^ = 0, so that A=a and B = b, we obtain 
 the known arithmetical formula 
 
 \ogab = ,^^- 
 ° los: a 
 
COMPLEX INDICES. 471 
 
 300. In the following examples the method of the last 
 article is applied in special cases. 
 
 Example 1. — Find the general value of Logee. 
 Let 0=(r, 6). 
 
 Assume Loge2=^+3/^ 
 
 then {{e)Y+y^=z={r, 6). 
 
 We have also ((e))*+^=exp{(.r+?/i)Loge} 
 
 = exp{(^ +3/0(1 + 2w7r . ^)} 
 = exp{.27 - y . '2,imT + i{y + x . ^tnir)}, 
 or {{e)y^y'' is a number of modulus exp(^-y . 2m7r) and of amplitude 
 
 1/ + X . '2.17177. 
 
 Hence, ^-y - 2wi7r = log r, 
 
 and ?/ + ^. 2m7r=^+2w7r. 
 
 Solving for x and y we obtain the result 
 
 -r ^_ logr+(^ + 2?i7r). 2w7r . 6'+2^7r-2m7rlogr 
 ^' l+(2m7r)2 "^^ l + (2m7r)2 " 
 
 If m=0, i.e., if the value of e be restricted to (e, 0), 
 we have Log(e, q)Z = log r + 1{ + 2w7r), 
 
 and . '. Log(e, o)Z = Log z. 
 
 If s=l, thenr=l, ^=0, 
 
 and we have LogJ = ^^"- ^7"+ ^^ 
 
 l+(2m7r)2 
 
 Example 2. — Find the general value of Log4( - 2), and shew that 
 one value is ^. 
 
 Let Log4( -2)=x+ yi, 
 
 then ((4))^+3'»= -2. 
 
 But ((4))*+2'» = exp{(ji' + yi) Log 4} 
 
 = exp{(.r + ?/0(log 4 + 2m7ri)} 
 = exp{x \og4-y. 2w7r + i{y log 4 + ^. 2w7r)}. 
 Also, -2 = (2, tt) 
 
 exp(.r log 4 - ?/ . 2m7r) = 2, or .^7 log 4 - ?/ , 2w7r = log 2, 
 and ?/log44-^. 2w7r = (2?i+l)7r. 
 
 Solving for x and y we obtain the result 
 T oo- r - 2^ - (log 2)^ + ^^7^(271 + l )7r . (2w + Dtt log 2 - mw log 2 
 ^''^^ '^ 2(log2)2 + 2(wi7r)2 ^^' 2(log2)2 + 2(?n7r)2 ' 
 If m = l, 71=0, we have the value ^. 
 
472 COMl'LKX IM>l('KS, 
 
 ExAMPLR 3.— Find the general value of Lo^xl-^-¥i^'\ aiid 
 shew that one vahic is \. 
 
 Let L<)g,( - H»y )=.<''+.y*, 
 
 then ((l))x+r*„_j + 4vV3 
 
 But ((l))*+«"-exp{(^'+yOLog 1} 
 
 -exp{(^+yt).2wm} 
 = ex jX - y . 2m7r + ix . 2«i7r) 
 
 oxi)( -y . SmTr)"! or -y . 2mrP'=0 
 
 and X. 2w7r - ??^ + 2M7r. 
 
 3 
 
 Now m cannot equal zero, for then - J+*^- = exp(()) = l,anim- 
 
 ]>o8Hible result ; 
 
 hence, .'/="0. 
 
 Weliavealso ' = -?-', 
 
 ^A-^-^':yt^' 
 
 a real quantity in all cases. 
 
 If m=»l, 71 = 0, we obtain the particular value i. 
 
 Examples XXXL 
 
 1. Find tho values of ((l)y. 
 
 2. Express ((l+i))' in the form x + y% and shew that 
 
 mod(l 4- if = '46 approximately, 
 and amp(l+'i)^=slog2. 
 
 '.\. Find tho ni\i roots of a*^-\ and shew that their sum 
 
 is zero. 
 4. Prove that the real part of (V - ly*'^^'^"'"^^ is 
 
 e 8 cos(i7r log 2). 
 
COMPLEX INDICES. 473 
 
 5. If {a - bs/^^iy"^^^ be exhibited in the form a + ps/^^, 
 
 where a and /3 are real quantities, find the values 
 of a and /3. 
 
 6. Shew that (a + biy+'i^ is entirely real if qlogm+2><l> 
 
 is a multiple of tt, where m = mod{a + hi) and 
 = amp(ct + 6i). 
 
 7. If -'^<0< J, shew that 
 
 (a+ia tan ^y''g(««ec0)-0f = exp{(log a sec 0)H^^}. 
 
 8. Reduce V- — r^!r^ — • to the form x-\-yi. 
 
 {a — hiy-'i' ^ 
 
 9. Find the values ofLog^^{\/—l). 
 
 10. Find the general value of Log^2, and shew that one 
 
 value is i log 2/2^. 
 
 11. Prove that log2,._ 1^-1} = -. 
 
 12. Prove that Log^^^^{S/-l} = (2n+^y(2m+^,), 
 
 where m, 7i are any integers. 
 
 13. Prove that 
 
 where m, n, p, (/ are any integers. 
 
 14. If af^ =a(cosa + isina) find the general value 
 
 of X. If a = 2, a = 0, shew that the result gives 
 x=±^2. 
 
CHAPTER XXII. 
 
 CIECULAR AND HYPEKBOLIC FUNCTIONS OF 
 COMPLEX NUMBEES. 
 
 301. The geometrical definitions of the circular and 
 hyperbolic functions imply that the variables are real 
 quantities. At the present stage sin(a + ^/3) and 
 co.sh(a + ^|8) are undefined and consequently meaningless. 
 Before extending the definitions, it will be well to 
 recapitulate briefly the properties of these functions so far 
 as real quantities are concerned. 
 
 Having defined the functions as the ratio of certain 
 lines connected with an angle, or with a hyperbolic sector, 
 we proved geometrically from a property of the circle 
 or rectangular hyperbola that 
 
 cos2i» + sin2a3=l, 
 and cosh^ic — sinh^aj = 1. 
 
 It was then shewn geometrically that certain addition 
 formulae were true, viz. : — 
 
 cos(a3 + y) = cos xcosy — sin x sin y, 
 cosh(ic + 2/) = cosh x cosh y + sinh cc sinh ?/, 
 
 sin(aj + y) = sin x cos y + cos x sin y, 
 sinh(i« + y) = sinh x cosh y + cosh aj sinh y. 
 From these formulae similar results for any number of 
 
 variables readily followed, and thence it was shewn that 
 
 474 
 
FUNCTIONS OF COMPLEX NUMBERS. 475 
 
 the functions might be exhibited as convergent series in 
 powers of the variables, thus 
 
 cosa; = l-|+g-..., 
 coshaj = l+-2 + |^ + ..., 
 
 [3 [5 _ 
 
 /y>3 /vi5 
 
 sinha; = aj + ^ + r5 + .... 
 
 It also followed from the definitions that each function 
 was one-valued, and that for real values of the variable 
 the circular functions were periodic, while the hyperbolic 
 functions were non-periodic, the hyperbolic sine, for 
 example, increasing continuously with the sector from 
 negative infinity through zero to positive infinity. 
 
 A division of the functions into the two classes of even 
 and odd functions was an obvious inference from the 
 geometrical definitions. 
 
 302. The series 
 
 [2 + 14 •••• 
 
 ■ - + I+I+-- 
 
 are convergent for complex as well as for real values of x, 
 and, when x is real, they are respectively equal to cos x, 
 cosh X, sin x, sinh x. 
 
476 CIRCULAR AND HYPERBOLIC 
 
 We may therefore extend the meaoings of cos a;, 
 cosh X, sin x, sinh x as in the following definition : — 
 
 Def. — When x is any quantity, real or complex, the 
 series 
 
 aj2 . a;4 
 
 .-!+-... 
 
 '2 • 14 
 
 aj2 . x^ 
 
 /yi3 /¥>5 
 
 are called the cosine of a?, the hyperbolic-cosine of a;, the 
 sine of a;, and the hyperbolic-sine of a?, respectively. 
 
 We further define sec x, sech a?, cosec a?, cosech a; as the 
 reciprocals of cos x, cosh x, sin a?, sinh x, respectively ; and 
 tan X, tanh x, cot a;, coth a; as the fractions 
 
 sin a; .sinh a; cos a; cosh a? ,. , 
 
 , ) — -. — > -. — ) -V-, — , respectively. 
 
 cos a? cosh a; sinaj sinh a; ^ '' 
 
 Since each of the defining series is a one-valued func- 
 tion of Xj it follows that each of the circular and hyper- 
 bolic functions of a complex number is a one-valued 
 function of the variable. 
 
 The defining series for cos x and cosh x involve even 
 powers only of x, hence cos x and cosh x are even func- 
 tions of the complex variable. In like manner, sin x and 
 sinh X are odd functions of the variable. Thus, for a com- 
 plex, as for a real, variable we have 
 
 cos X = cos( — x), cosh X = cosh( — a;), 
 sina;= —sin ( — a;), sinh a; = —sin (—a;), 
 and so on for the other functions. 
 
FUNCTIONS OF COMPLEX NUMBERS. ^^J'J 
 
 303. Exponential Values of the Circular and Hyper- 
 bolic Functions of a Complex Variable. — From the 
 definitions of art. 302 it follows, for all values of x, that 
 
 - , . , {xif , {xi)^ , 
 GO^x + i^mx^l+xi+Y^-^ ^ ^ g +... 
 
 = exp(a;i), 
 
 (xi^ ('T/h ^ 
 
 and that co^x—is>mx = l—xi+\ — o~r~oQ + "- 
 
 = ex-p{ — xi). 
 Hence, by addition and subtraction, we have 
 
 cosic = ^{exp(a;'^) + exp( — ^i)}, 
 
 and 8mx = -^.{ex-p{xi) — exYi( — xi)}. 
 
 In like manner we have, for all values of x, 
 
 T^ X 
 
 cosh aj + sinh cc = 1 + aj + —-+——-— + .. . 
 1 . Z 1. . Z . o 
 
 = exp{x), 
 
 X X 
 
 and cosha; — sinha;=l—i»+— - — -——- + ... 
 
 1 . Z i. . Z , o 
 
 = exp{ — x), 
 cosh cc = ^{exp(i:c) + exp( — a?)}, 
 
 and sinh a; = ^ { exp(ic) - exp( — x)}. 
 
 304. Periodicity . of the Circular and Hyperbolic 
 Functions of a Complex Variable. — If x be any 
 number, real or complex, and n any integer, positive 
 or negative, we have 
 
 exp(a:!i) = exp (xi + 2n7ri), 
 and exip( — xi) = exY)( — xi — 2n'7ri), 
 
 cosx = cos(x+2n7r), and sinfl? = sin(aj + 27?7r), 
 
478 CIRCULAR AND HYPERBOLIC 
 
 i.e., cos a; and sin a; are periodic functions of the variables, 
 having for the period the number 27r. 
 
 Again, since 
 
 exp(7^^) = cos tt + i sin tt = — 1, 
 exp(iri) = — exp(a:;i -\-'ln-\-l . iri), 
 and exip{ — xi)=—ex'p{ — xi — 2n + l.7ri), 
 
 cos ic = — cos(a; +2n-\- Itt), 
 and sin a; = — sin(ic + 27i + Itt). 
 
 From the above results we see that tan x and cot x are 
 periodic functions of x, and that the period is the number 
 TT. So also secaj and cosecaj are periodic, the period 
 being 27r. 
 
 We have also 
 
 exp X = exp(fl3 + 2n7ri), 
 and exp( — fl3) = exp( — a? — 27i7ri), 
 
 cosh X = cosh (x + 2mri), 
 and sinh x = sinh(a; + ^niri), 
 
 i.e., the hyperbolic cosine and sine of x are periodic func- 
 tions of X having the imaginary period 2^1. It readily 
 follows that the hyperbolic tangent and cotangent have 
 the imaginary period iri. 
 
 Thus, we see that when complex variables are considered, 
 the analogy between the two classes of functions is com- 
 pleted by the establishment of the periodicity of the 
 hyperbolic as well as of the circular functions. 
 
 305. By art. 303, we have 
 
 cos a? -}- i sin a; = exp(a?i), 
 and cosa? — 'isina! = exp( — a;i), 
 
 therefore, by multiplication, 
 
 cos^a: + sin^a? = exp(O) = 1 . 
 
FUNCTIONS OF COMPLEX NUMBERS. 479 
 
 In like manner, 
 
 cosh X + sinh x = exp(cc), 
 and coshfl? — sinha3 = exp( — cc), 
 
 cosh^i^c — sinh^^ = exp(O) = 1. 
 Again, 
 cos(ir + 2/) 
 = i{exp(aj + 2/-'^) + exp{-a) + 2/.^)} 
 = J {exp(fl3'i)exp(2/^) + exp( - a;7:)exp( - ?/i) } ; 
 and cos a; cos 2/ — sin i:c sin ;y 
 
 = i{exp(a;i) + exp(-a;i)}{exp(2/i) + exp(-2/^)} 
 
 -4p{expW-exp(-a;^)}{exp(2/i)-exp(-2/i)} 
 
 = i{exp(a3^)exp(2/^) + exp( - aji)exp( -yi)}; 
 cos(i:c + 2/) = cos x cos 2/ — sin a; sin y. 
 In like manner, we have 
 
 sin(a; + 2/) = sin aj cos 2/ + cos a? sin 2/, 
 cosh(aj + 2/) = cosh x cosh 2/ + sinh x sinh 2/, 
 and sinh(a; + y) = sinh aj cosh y + cosh a? sinh y. 
 
 Thus, the fundamental formulae, and therefore also 
 all the consequences of these formulae established in 
 the preceding chapters, hold for complex values of the 
 variables. 
 
 306. Formulae of Interchange of Circular and Hyper- 
 bolic Functions. 
 
 We have cos(aji) = J{exp(a;i. ^) + exp( — a;^.^)} 
 = J{exp( — ic) + exp x] 
 = cosh X. 
 In like manner, 
 
 sin xi = i sinh x, 
 cosh xi = cos X, 
 sinh xi = i sin x. 
 
480 CIRCULAR AND HYPERBOLIC 
 
 aud, consequently, 
 
 tan ict = i tanh x, 
 tanha;i = 'itanaj. 
 When ic is a real quantity we have defined gd ^^ as a 
 certain angle between — ^ and ^, and have shewn that if 
 
 = gd u, then tan - = tanh ^. We now extend the mean- 
 ing of gd u by the following definition. 
 
 Def. — If u be any number, real or complex, and ^ be a 
 number such that tan ^ = tanh-, then is called the 
 gudermannian of u. 
 
 With this definition, we have for all values of u 
 
 l + tan^l l+tanh2| 
 
 sec 6 = ^ = = cosh u, 
 
 1— tan^^ 1— tanh^^ 
 
 2 tan - 2 tanh ^ 
 
 tan 6 = ^ = = sinh u, 
 
 l-tan2? l-tanh^l 
 
 and exp (u) = cosh t6 + sinh u = sec + tan = tanf J+^j, 
 
 or u = LogtSin(~+^. 
 
 The following example is introduced to shew the equi- 
 valence of the relations 
 
 tan ^ = tanh ^ and u = Log ^^^^f r +^)- 
 
FUNCTIONS OF COMPLEX NUMBERS. 481 
 
 Example.— If w=Logtaii(^ + |) prove that tanli- = tan ^. 
 We have Log tan^| + |W log tan^| + 1) + ^mri, 
 
 | = ilogtan(J + |)+.^,^^•, 
 .-. tanh|=tanh{ilogtan(|+|)} 
 
 ^exp{ilogtan(|+^)}-exp(-^logtan(|+|)} 
 ~exp{ilogtan(| + |)}+exp{-ilogtan(^+|)} 
 
 exp(logtan(|+|)}-l 
 exp{logtan(^ + ^)} + ] 
 
 V4^2/ ^ ^ 
 
 = 7i = tan -. 
 
 307. Inverse Circular Functions. 
 
 Def. — If cos(a? + m) = a + |8i, then x-\-yi is called the 
 inverse-cosine of the number a + /3i. 
 
 Since cos(2n'jr±x-\-yi) = cos(x-\-yi), 
 
 it follows that the inverse-cosine of a given number 
 a + fii has an infinite number of values. The notation 
 for this many- valued function is Cos-\a + pi). 
 
 For any given value of the real number x, we may 
 evidently determine a positive or negative integer n such 
 that either 2n7r + x or 2n7r — x shall lie between and tt, 
 and this can be done in one way only. For example, 
 
 if ic lie between and tt we make n = in 2mr + x, 
 
 \i X „ TT „ Stt „ n = \m2nir—x, 
 
 iix 
 
 )) 
 
 „ -TT ,. 
 
 71 = in 2nir — x, 
 
 iix 
 
 5J 
 
 -TT „-27r ,. 
 
 71 = 1 in 2mr-\-x, 
 
 and so on. 
 
 
 
 
 2h 
 
482 CIRCULAR AND HYPERBOLIC 
 
 The value of the general expression 2n'n'±{x-{-yi), 
 whose cosine is a + ^i, for which Inir + x or ^nir — x 
 lies between and tt, is called the principal value of 
 Cos-^(a + /3i), and is denoted by QO^-\a + ^i), and we 
 'have Cos - \a + /3^) = 27i ± ttCOS - \a + /3i). 
 
 In like manner, if sec(a; + 2/'^) = a + /3i, 
 then iC + 2/'^ = Sec"^(a + ^i), 
 
 and the value of 2n'7r±(x-\-yi), for which 2mr-\-x, or 
 2mr—x, lies between and tt, is called the principal 
 value of Sec-^(a + /3i), and is denoted by sec-i(a + ^i). 
 
 We have also ^QC~\a-\- fii) = 'tnir±^QQ,~\a-\- ^i). 
 
 Similar definitions may be given of the other inverse 
 circular functions, the principal values of the inverse 
 sine, cosecant, tangent and cotangent, having their real 
 
 parts between — ^ and ^. With this convention, 
 
 Sin-i(a+)Si) = '^7r + (-l)^sin-i(a + /3^), 
 Cosec-i(a + ;8i)=7i7r + (-ircosec-i(a + j8i), 
 Tan-l(a+iS^) = '^^'7^ + tan-l(a4-/3^), 
 Cot-i(a + j8i) = ?i7r + cot-i(a + /3i). 
 
 308. Inverse Hyperbolic Functions. — Definitions simi- 
 lar to those of the last article may be given for the 
 inverse hyperbolic functions of a complex variable. 
 
 The periods of the hyperbolic functions are imaginary, 
 and we take as the principal value of Cosh-^(a + /5^) or 
 Sech"^(a + /3'i) that value for which the imaginary part 
 lies between and iri ; and for the principal value of each 
 of the remaining functions that value for which the 
 
 imaginary part lies between — ^ and + y* 
 
FUNCTIONS OF COMPLEX NUMBERS. 483 
 
 With this convention, 
 
 Cosh - i(a + /3i) = 27i,7ri ± cosh - i(a + /3i), 
 Sinh-Xa + /3^)=7^7^^ + (-l)^sinh-l(a + /3^), 
 Tanh " \a + ^i) = niri + tanh -\a + /Si). 
 309. We add a few examples in illustration of the sub- 
 ject of this chapter. 
 
 Example 1. — Eeduce cos{a + (3i) to the form .v+i/i. 
 We have cos(a + /3t) = cos a cos f3t — sin a sin (Si 
 
 = cos a cosh ^ - sin a . i sinh /3 
 = cos a cosh j3-i sin a sinh /?. 
 Example 2. — Eeduce tanh(a+/5i) to the form x+yi. 
 We have tanh(a + ^*):='J"MM^^)^ 
 
 cosh(a + ^^) 
 _ sinh g cos /? + 1 cosh a sin j8 
 cosh a cos /? + z sinh a sin ^ 
 _ (sinh g cos ^ + ^' cosh a sin ^)(cosh a cos ^ - 1 sinh a sin ^) 
 
 cosh'-^g cos^/^+sinh^g sin^/? 
 _ sinh g cosh g(cos2/5 + sin^^) + i sin /3 cos /^(cosh^a - sinh^g) 
 
 cosh^a cos^;8 + sinh% sin^jS 
 _ sinh g cosh g + 1 sin ^ cos ^ 
 cosh^g cos'^/? + sinh^g sin^yS 
 _ sinh 2g + 1 sin 2/5 
 cosh 2g + cos 2j8 ' 
 Example 3. — Express sin~Ya + /30 in the form x-\-yi. 
 We have g+/3^ = sin(^+^^) = sin;rcoshy + ^■cosa?sinhy, 
 
 sin ^ cosh y = g, (1) 
 
 and cos ^ sinh y=)S. (2) 
 
 From (2), 
 
 (l-sin2^)(coshV-l)=/3^ 
 and . •. by ( 1 ), cosh^ + sin% ^d^+l + p^, (3) . 
 
 and consequently from (1) and (3), remembering that 
 cosh 3^ > 1 > sin ^, we have 
 
 cosh j/ + sin .r = */(g + 1 f + ^^ 
 and coshy-sin^=\/(g-l)''^4-^^ 
 
 cosh.v=i{v/(or+T)2+F+V(g- l)2 + iS-'}, 
 and sin.r = i{V(g+l)2 + /?'-v'(g-l)2 + y82}. 
 
484 CIRCULAR AND HYPERBOLIC 
 
 Now, since sin~Xa+/8i) is the principal value of the function, x 
 must lie between -J and J, therefore cos^ is positive, and there- 
 
 fore by (2) y and fS have the same sign. 
 Hence, finally, 
 
 ±zcosh-^J{V(T+T7+^+ sl{a-\f-\-p% 
 the upper or lower sign being taken according as /? is positive or 
 negative. 
 
 Example 4. — To express ta.n~\a + fii) in the form x+yi. 
 
 Since tan~\a+^0 is the principal value of the function, its 
 
 real part lies between -^ and +^, i.e., -!^ < ;r<^' 
 
 We have a+^t=tan(^4-yi), 
 
 and, by an easy reduction, similar to that of Example 2, 
 . / , •v_ sin 2a7+^sinh 2y . 
 cos 2.27+ cosh 2y 
 thus we have 
 
 ^^ s in 2.g ,.. 
 
 cos 2jp+ cosh 2y 
 
 B= ^^"^^3/ . (ii) 
 
 cos 2.r + cosh 2^^ 
 
 From (i), a and x have the same sign, since -~ <cx<'^\ and from 
 
 (ii), yS and y have the same sign. 
 
 Affain g^+jg^— sin^2.y+8inh^2?/ cosh^ 2y - cos''^2^ 
 ^ ' '^ (cos 2.r+ cosh 2y)2~(cos2^+ cosh 2y)2 
 
 _cosh2y-cos2.r 
 ~ cosh 2y + cos 2.r' 
 
 ... l-a2-^2^ 2cos2^ (iji) 
 
 ' cosh2y + cos2^ 
 
 and l + a2 + /32= 2cosh2L^ ^i^^ 
 
 cosh 2y + cos 'zx 
 
 From (ii) and (iv), tanh 2y = ^^^f^^, , 
 
 and therefore y=itanh~^ /— tso* 
 
 •^ 2 l+a2+^ 
 
FUNCTIONS OF COMPLEX NUMBERS. 485 
 
 2a 
 Also from (i) and (iii), tan ^x= ^ — m> 
 
 and therefore 2^=r7r + tan~-^- n — p^, 
 
 where r is to be taken so that _^<^<^, and that x may 
 have the same sign as a ; hence if a^ + /3'^ < 1, 
 
 and, if a^ + yS^ > 1, 
 
 the upper or lower sign being taken according as a is positive 
 or negative. 
 
 Hence, finally, if a'^+(3^<l, 
 
 andif a^ 4-/52 >1, 
 
 tan-Xa-f^z}= ±|+itan-l^_^, + ^itanh-^^J^^ 
 
 the upper or lower sign being taken according as a is positive or 
 negative. 
 
 Example 5. — The roots of the cubic equation c(^ + ^IIx+G=0^ 
 where H is positive, may be obtained in the following manner. 
 
 Let x=—. the equation becomes 
 m 
 
 Comparing this equation with the equation 
 
 sinh^w + 1 sinh u — ^ sinh 3w = 0, 
 
 and taking m and u such that m^ = —j^ ainh.Zu= —4m^Gj we see 
 
 that 2= sinh w. 
 
 From the equations for m and u, we have sinh 32^= -—7-3, a 
 
 211^' 
 
 real quantity ; hence u can be determined from a table of hyper- 
 bolic sines. Since sinh 3w=sinh(3w±27ri), the three roots are 
 
 2\fH8mhu, 2x/iS^sinh(w + i^) and 2v/Fsinh(w-t!li:). 
 
486 CIRCULAR AND HYPERBOLIC 
 
 Example 6.— Given that 
 
 t*=logtanf^ + |j=a;+a3^ + a5^+a7;r^ + ..., 
 shew that x=u- a^u^ + a^u^ - a-jV? + . . . . 
 
 Silica ««==log tan(^+^j, we have x=gd. u, 
 
 and therefore tanh % = tan ^. 
 
 2 2 
 
 Now, tanh ^ = ^ tan ^*, and tan ^ = i tanh 5*, 
 
 ' 2 i 2' 2 i 2' 
 
 tan ^ = tanh ^, or ui = gd(ari). 
 2 2 
 
 Hence, ^' is related to ui in the same manner as u is related to 
 
 X, and therefore if u=x + asx^ + a^s^ + . . . , 
 
 we must have also 
 
 xi = ui + a^uif + a^{uiy + . . . , 
 
 or :i7=w-a32i3+a5W^-.... 
 
 Examples XXXII. 
 
 1. Shew that the difference of the squares of the moduix 
 
 of cos(a + j8i) and sin(a + ^i) is cos 2a. 
 
 2. If 2sin(a + )8V^T) = flj + 2/>v^^^, pi^ove that 
 
 aj2+2/^ = e2^+e-2^-2cos2a. 
 
 3. If sin(a + /3i)=a; + 2/^, prove that 
 
 X^ /Jj2 /jr.2 qj2 
 
 sin^a cos^a cosh^^ sinh^/?"" 
 
 4. Ifsin(O4-0x/ — l)=p(cosa + /v/ — lsina),when p,a,0,(p 
 
 are real quantities, then will 
 
 tan a = -7 jcot^. 
 
 e'^ + e""^ 
 
 r -n XT. i. X v/ . /D-\ tanhasec^/^ + isech^atan^ 
 
 5. Prove that tanh(a + m)= , , . ■■ . . — oo ^. 
 
 ^ ^ ^ l + tanh^atan^/S 
 
 6. Find the real and imaginary parts of sec{x-^ iy). 
 
FUNCTIONS OF COMPLEX NUMBERS. 487 
 
 7. Express sin(a + ^x/ ^) sin(a -/g^/^) -^ ^^^^^-^^^^ 
 ^ sin(a-|8x/-l) sin(a + /3%/-l) 
 form. 
 
 o T> . 1 . i. / . • N sin2a; + isinh2'?/ 
 
 S. rrove that tQ^mx-[-%y) = ^ — ; i-iv • 
 
 ^ ^^ cos zee + cosh 2^/ 
 
 9. If a; = (e^-{-e~^)cosa, 2/ = (e^ — e~^)sma, prove that 
 sec(a + 6 V - 1) + sec(a - 6/v/^) = ^alpT^a' 
 
 sec(a + 6V^-sec(a-6V^ = ^^^£2-- 
 
 10. Prove that 
 
 log sin (ot + /3^) = J log J (cosh 2/3 — cos 2a) + cjii, 
 where = amp (ein a cosh ^8 + i cos a sinh /5). 
 
 11. If a be a re^l number not numerically greater than 
 
 unity, prove that the equation ^\nx = a has no 
 imaginary root. 
 
 12. Eeduce cos"^(a + ^'i^) to the form x + yi. 
 
 13. If cos(0 + (pi) = cos a + i sin a, shew that sin^0 = sin^a. 
 
 14. If cos(a + /5i) = a + 6i, and sin(a + /3i) = c + <i'^, prove 
 
 that tan a = - = - - and hh " ^z? = (^2 4. d^)(b + c)2 
 
 15. Reduce tan-^(cos0 + isin 0) to the form A+Bi. 
 
 16. Assuming that sinu = ^. — and cosu = — ^ , 
 
 shew that, li x-\-iy = cid,T^{^-\-iti), then 
 tan2f=— ^^^ 
 
 c" — cc^ — 2/^ 
 
 17. coshfcc + ^j = '^sinha;. 
 
 18. sinhf a:; + ^y=icosha?. 
 
488 FUNCTIONS OF COMPLEX NUMBERS. 
 
 19. tanh(a;+^) = cotha;. 
 
 20. coth('a; + ^) = tanh£c. 
 
 21. ^Qch\x-^~\ = — i cosech x. 
 
 22. cosechraj + ^ j = — i sech x. 
 
 23. cosh('^— a;j= — isinhoj. 
 
 24. sinhf^ —xj=^icoshx. 
 
 25. ia.nh(l^-x) = -coth a;. 
 
 26. If and u are connected by the equation tan ^ = tanh ^, 
 
 find the general value of in terms of u, and, con- 
 versely, the general value of i6 in terms of 0. 
 
 27. If tan I = -^ tan |, then will log tan(| + 1) = ix. 
 
 28. If u = log tan(^^+|\ then will a;i = log tan('|+|^\ 
 
 on I?- J i-u rj.T- • 1 COS 20 , cos 40 
 
 29. Find the sum of the seiies 1 — z — ^-f- 
 
 1.2 ■ 1.2.3.4 
 
 30. Shew that 
 
 — sin a + |T sin 2a+...acZ inf. = sinhf a; cos ^jsinf a; sin ^ j. 
 
 31. Shew that the roots of the cubic equation 
 
 x^-\-SHx-\-G = 0, when H is negative, may be 
 determined by the aid of a table of hyperbolic or 
 circular cosines, according as 6^^4-4^^ is positive 
 or negative. 
 
MISCELLANEOUS EXAMPLES. 
 
 489 
 
 Miscellaneous Examples. IV, 
 
 1. Prove that 
 
 sin(^ — y) + sir«(y — a) + sin(a — ^) 
 
 + 4 sin i(^ - y)sin i(y - a)sin i(a - ^) = 0. 
 
 2. If no two of the angles a, /3, y are equal, or differ by 
 
 a multiple of four right angles, then Esin(/3 — y) 
 cannot vanish. 
 
 3. Prove that 
 2sin(^ + y) X E sin(/3 — y) = 2 sin /3 sin y, cos ^ cos y, 1 
 
 sin y sin a, cos y cos a, 1 
 sin a sin ^, cos a cos |8, 1 
 
 4. If a, /3, y are unequal, and if E cosec a(sec ^ — sec y) = 0, 
 
 then will 2sin(^ + y) = 0. 
 
 5. If acos^cosy + 6sin/5siny 
 
 = a cos y cos a + 6 sin y sin a 
 = a cos a cos ^8 + 6 sin a sin /? = c, 
 
 then will g + « Bi"(«+./3)+|i"(^+y) , 
 a 6 sin(a + y) 
 
 c c_ sin(a — /3) — sin(/3 — y) 
 a 6~ sin(a + y) 
 
 2sin(/5 + y) = 0, and -+1 + ^=0. 
 
 6. If (XC0SaC0S|8 + fesin asin^ = c, 
 
 and a cos /3 cos y + 6 sin /3 sin y = c, 
 then will 
 
 (i+c^-i)^"'"'"^'^+S+a^-p) 
 
 11 1 
 
 sin a sin y= -5+10 — o- 
 ' a^ Ir c^ 
 
490 MISCELLANEOUS EXAMPLES. 
 
 1. Having given 
 
 a^cos a cos /3 + a(sin a + sin j8) + 1 = 0, 
 a^cos a cos y + ci(sin a + sin y) + 1 = 0, 
 
 prove that 
 
 a^cos ^ cos y + a(sin ^ + sin y) + 1 = 0, 
 
 ^, y being unequal and less than x. 
 
 2. Having given 
 
 a^cos a cos j8 + a(sin a + sin ^) + 1 = 0, 
 a^cos a cos y + a(sin a + sin y) + 1 = 0, 
 
 prove that 
 
 cos a + cos ^ 4- cos y = cos(a + i^ + y), 
 
 j8, y being unequal and less than x. 
 
 3. If acos(0+^) + &cos(e-^)+c = O, 
 
 a cos{(f) + V^) + 6 cos(^ — \lr) + c = 0, 
 a cos(i/r + 0) + 6 cos( V^ - 0) + c = 0, 
 and, if 6, 0, yp- are all unequal, then 
 
 j^ a sm^6 + bsm^(l) _ h sin^^ + c sin^o^ _ c sm^O -\- a sin^cp 
 Fco^O + cco^ "" c cos^O + a coii^<t> a cos^O + h cos-^^' 
 then will a^ + 6^ + c^ = Zahc. 
 
 5. If 
 
 a cos a cos ^ + 6 sin a sin /3 = a cos /3 cos y + 6 sin ^ sin y 
 = a cos y cos ^ + 6 sin y sin (5 = a cos ^ cos e + & sin S sin e 
 
 = a cos e cos a + 6 sin e sin a = c, 
 
 then will 
 sin(a + /3) + sin(/3+y)+sin(y + ^)+sin(54-e) + sin(e+a) = 
 
 and 
 
 sin(a + y) + sin(y + e) + sin(e+/3) + sin(^+^) + siii(^ + a) = 0. 
 
MISCELLANEOUS EXAMPLES. 491 
 
 6. If 
 
 a cos a cos /3 + h sin a sin /5 = a cos /3 cos y + Z> sin p sin y 
 = (Xcosy cos(5 + ^siny sin ^ = acos ^cose+6sin ^sine 
 = a cos e cos a + 6 sin e sin a = c, 
 
 y- 
 
 1. Eliminate ^ and ^ from the equations 
 
 tan + tan (p = a, 
 
 tan tan ^(cosec 20 + cosec 2(p) = 6, 
 
 cos(0 + 0) = c cos(0 — 0). 
 J, j^ X cos 0^ , ysinO^ _^ x cos 0^ y sin 0^ _ 
 
 Z. XL 7 1. 7 i, 
 
 a ah ' 
 
 Oi — ^2 = 2«' ^^^^ will ""2 + 12= sec^a. 
 8. Prove that the elimination of 0, (p from the equations 
 r cos(20 —a) = 711. cos'^0, 
 r cos( 20 — a) = 7n cos^(p, 
 tan = tan + 2 sin /3, 
 
 m cos a 
 
 gives r = .j 2 — ^-g-^. 
 
 ^ 1 — cos^a srn^p 
 
 4. Eliminate from the equations 
 
 aj = asec(0 + a), y = htsiu((f) + /3). 
 
 5. Eliminate from the equations 
 
 cos^O + acosO = 6, 
 sin^O + a sin = c. 
 
 6. Eliminate from the equations 
 
 4(cos a cos 6 + cos 0)(cos a sin + sin 0) 
 = 4(cosacosO + cos'0-)(cosasin + sin \/r) 
 = (cos - cos ^)(sin — sin xj/-). 
 
492 MISCELLANEOUS EXAMPLES 
 
 s. 
 
 1. Shew that (l + sm0)(3siii^+4cos0+5) 
 
 is a perfect square. 
 
 2. Solve the equations 
 
 cosaj+cos2/ = a, 
 sin X + sin y = h. 
 
 3. If X sin^A cos B-y sm^B cos A + 0(cosM - cos^^) = 0, 
 
 and 
 
 z sin^Ccos A —xsin^A cos C+ y{cos^C— cos^J.) = 0, 
 where A, B, C are the angles of a triangle whose 
 sides are a, b, c, then ax -hy = cz. 
 
 4. Eliminate 6 from 
 
 sin 3(1+0) + 3 sin(|+ e) = 2a, 
 
 sin3g-O) + 3sing-0) = 26. 
 
 5. Eliminate from the equations 
 
 -cos 0-1 sin = cos 20, - sin 0+| cos 0=2 sin 20. 
 a o ah 
 
 6. Eliminate from the equations 
 
 (a + 6)(a3 + 2/) = cos + sin sin 20, 
 and 2(a2 ~ b^)(x^ - 2/^) = 3 sin 20 + sin320. 
 
 1. The sides of a triangle are a(l— m), a, and a(l+W/), 
 
 where m is small, shew that the mean angle 
 
 TT ,- 3^3 ,_ 
 
 = g -^ V 3 m^ — -^Y~ '^* — 3 V 3 m^. . . nearly. 
 
 2. Eliminate and from the equations 
 
 a cos(0 + a) sin .,,. \,7«/,, \ 
 r= — /A ; = ^-g, asm(0-a) + 6sm(0 + a) = c. 
 cos(0 — a). sm0 ^ _ ^^ 
 
MISCELLANEOUS EXAMPLES. 493 
 
 8. Eliminate Q and from the equations 
 
 cos 6 + cos = a, cot 6 + cot = 6, cosec 6 + cosec = c. 
 
 4. Establish the relation 
 
 tan((X + 6 + . . . + '^ + '>^) 
 _ 2 1 + cos 2a 2 l+cos2m 
 
 ~~ sin 2a— tan 6— sin 26— "* tan-T^ 
 
 5. Prove that if m be a positive integer, 
 
 cos(m + l)0 
 
 1 1 
 
 = cos 7YiO\ 2 cos Q • 
 
 2cos0- 2 cose 
 where 2 cos 6 is repeated m times. 
 6. Solve the equation 
 
 1111 
 
 — — cos 6) 
 
 2tane+ 
 
 tane+ 4tan0+ tane4- 4tane+*" 
 = 2x/3 cosec 20, 
 6 being a positive acute angle. 
 
 i- 
 
 1. Shew that if be a positive angle not greater than 
 
 a right angle, sin will be less than — -^. 
 
 2. Evaluate ^log(l+^) ^^^^^^ ^^^ 
 
 1 — cos fl? 
 
 3. Find the limit, when x is indefinitely diminished, of 
 
 e^-l + log(l-a;) 
 sin^cc 
 
 4. Find the limit, when x is indefinitely diminished, of 
 
 sinic + sinhfl? — 2a; 
 x^ * 
 
 5. Shew that, as x continually diminishes, 
 
 Sx~^tsinx + Sx-^ia.n~'^x — 6x-^ 
 continually approaches the value unity. 
 
494 MISCELLANEOUS EXAMPLES 
 
 6. If regular polygons of the same number of sides be 
 inscribed and circumscribed to the same circle, 
 then, when the number of sides is very large, the 
 difference between the perimeter of the circum- 
 scribed polygon and the perimeter of the circle 
 is double of the difference of the perimeters of 
 the inscribed polygon and circle. 
 
 V' 
 
 1. If tan = ic, tan nQ = a, we have 
 
 a = 7^ — 1^ » hence shew that 
 
 L-ms 
 
 tan + tan(e + -) + tan('e-l-— ) + ... to n ten 
 
 z= ^n cot n(j + e\ 
 
 2. Prove that 
 
 tan 6 tsLnfe + -\m (e-\-^\..ton factors 
 
 = (-1)1 or (-1)V tan 710, 
 according as n is even or odd. 
 
 3. Prove that 
 
 tan20 + tan2^0 + ^Vtan2(e +—) + ... to n terms 
 
 4. Find the value of 
 
 cot + cot^O + -) + cot^0 +-^'\-\-... ton terms. 
 
MISCELLANEOUS EXAMPLES. 495 
 
 5. Find the equation whose roots are 
 
 tan^—j -, tan^^, tan^—, tan^^j, tan^-^. 
 
 6. Find the value of 
 
 tan^^ + tan^l^+tan^^^ + tan^^ + tan^^. 
 
 a 
 
 1. Prove that ^-3!^+^^- J^-,...=^2. 
 
 2. Sum the series 
 
 1.22.3"^5.62.7^9.102.11"^13.142.15 
 
 to infinity. 
 
 3. Prove that, if J<^<7r, then 
 
 sin J. + J sin^ J. + i sinM + . . . 
 
 = 2(cot^ + icot3| + ...). 
 
 4. Prove that, if 0< ^ < J, then 
 
 \ sinM + \ sin* J. + \ sin^J. + . . . 
 
 = 2(tan2| + itan6^ + ...). 
 
 5. If _^<0<^, then will 
 
 2 sin^e + 1(2 sin20)2 + 1 (2 sin20)3 + . . . 
 
 = 2(tan20 + \ tan«0 + i tan^^^ +...). 
 
 6. Prove that 
 
 sin0 + isin3e + isin50+... 
 
 = 2(s;n0-isin3O + ism50-...). 
 
496 MISCELLANEOUS EXAMPLES. 
 
 K. 
 
 1. Obtain the expansion oi QOsnO in terms of powers of 
 
 cos Q by comparing the expansions of the equival- 
 ent expressions 
 
 log(l + 2a;cos0+a;2) and log(l+aJ0)+Wl + -\ 
 where = cos + i sin 0. 
 
 2. If cos(O-l-J) be positive, expand logcos(^+|^j in a 
 
 series of sines and cosines of multiples of 6. 
 
 3. Expand cos^"0 + sin*^0 in a series of cosines of multiples 
 
 of a 
 
 4. From the formula 2 sin cos = sin(0 + </>) + sin(0 — (fi), 
 
 deduce 
 
 [Tjlri |3 |2yi-2 |5 |27i-4 |2ti+ 1 [0 
 
 _ (a+l)2n+l4.(^_l)2n+l 
 
 2[2;yH-l 
 
 •^- ^^l+^^+^2 = ^o+%'^ + «2^'+--- prove that 
 a^ + ctiflj cos Q + agflJ^cos 20 + . . . ad. inf. 
 
 _ m + 'yipa;^ + {n + mp + 'nga?^)a; cos + mga;^cos 20 
 ~ 1 +p V + g2^4 -f 22Jiz;(l + ga;2)cos + 2g'a;2cos W ' 
 6. Sum the series 
 
 sin + 2c2sin 20 + Seisin 30 + . . . to ti terms. 
 
 X. 
 
 1. If Vn+\ = pv„ - gvn-i, and Vi =p, ^0=2^, shew by mathe- 
 matical induction that 
 
 , . -, y' yi(ti-r-l)...(?i-2r+l) n-2w 4. 
 ■^^ ^ 1.2.3...r ^- ^-T"-- 
 
MISCELLANEOUS EXAMPLES. 497 
 
 2. Find the sum of n terms of the series 
 
 sin0 + 2sin(0 + a) + 3sin(0+2a)+..... 
 
 3. Sum the series 
 
 l^sin + 22sin20+32sin3O+... to n terms. 
 
 4. If n be equal to 3m ± 1, prove that 
 
 {U^}-|— j2— + ^^ 1^ 
 
 f 7i(n-l)(n-2)(7i-3) _ ^(77.-l)(^-2)(7i-3)(7i-4) ) _ 
 
 + 1 ^ + ^ |3 -...-0. 
 
 5. If r and s be unequal inte£:ers, prove that 
 
 Ssecf-^- + e)sec(?^+0) = ^ or 0. 
 
 l-(-l)2cOS7l0 
 
 according as n is even or odd ; the summation 
 extending for all integer values of r and s from 
 to n — 1 inclusive. 
 
 6. Shew that the series whose first term is 
 
 _^ a2^uia + a^^m'2^a-\- ...-\-an^m{n — '\)a 
 % + ttgCOs a 4- (XgCOS 2a + . . . + a„cos(?i — l)a' 
 and 7'*^ term 
 
 _^ ar+isina + o^r+2sin2a+...+<Xr-isin('yi-l)a 
 tty + ttr+icos a + . . . + a^_iCOs('yi •- l)a 
 any term being formed from the preceding by a 
 cyclical interchange of letters, is an arithmetical 
 progression, whose common difference is a, pro- 
 vided that ua = ^ir. 
 
 2i 
 
MATHEMATICAL TABLES. 
 
 LOGARITHMS OF NUMBERS. 
 
 No. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 1004 
 
 0017337 
 
 7770 
 
 8202 
 
 8635 
 
 9067 
 
 9499 
 
 9932 
 
 0364 
 
 0796 
 
 1228 
 
 13 
 
 0056094 
 
 6523 
 
 6952 
 
 7380 
 
 7809 
 
 8238 
 
 8666 
 
 9094 
 
 9523 
 
 9951 
 
 1270 
 
 1038037 
 
 8379 
 
 8721 
 
 9063 
 
 9405 
 
 9747 
 
 0089 
 
 0430 
 
 0772 
 
 1114 
 
 1340 
 
 1271048 
 
 1372 
 
 1696 
 
 2020 
 
 2344 
 
 2668 
 
 2992 
 
 3316 
 
 3640 
 
 3964 
 
 1478 
 
 1696744 
 
 7038 
 
 7332 
 
 7626 
 
 7920 
 
 8213 
 
 8507 
 
 8801 
 
 9094 
 
 9388 
 
 1655 
 
 2187980 
 
 8242 
 
 8505 
 
 8767 
 
 9030 
 
 9293 
 
 9554 
 
 9816 
 
 0079 
 
 0341 
 
 1769 
 
 2477278 
 
 7524 
 
 7769 
 
 8015 
 
 8260 
 
 8506 
 
 8751 
 
 8997 
 
 9242 
 
 9487 
 
 1835 
 
 2636361 
 
 6597 
 
 6834 
 
 7071 
 
 7307 
 
 7544 
 
 7780 
 
 8017 
 
 8254 
 
 8490 
 
 2296 
 
 3609719 
 
 9908 
 
 0097 
 
 0286 
 
 0475 
 
 0664 
 
 0854 
 
 1043 
 
 1232 
 
 1421 
 
 2438 
 
 3870337 
 
 0515 
 
 0693 
 
 0871 
 
 1049 
 
 1228 
 
 1406 
 
 1584 
 
 1762 
 
 1940 
 
 39 
 
 2118 
 
 2296 
 
 2474 
 
 2652 
 
 2830 
 
 3008 
 
 3186 
 
 3364 
 
 3542 
 
 3720 
 
 2726 
 
 4355259 
 
 5418 
 
 5577 
 
 5736 
 
 5896 
 
 6055 
 
 6214 
 
 6374 
 
 6533 
 
 6692 
 
 30 
 
 4361626 
 
 1786 
 
 1945 
 
 2104 
 
 2263 
 
 2422 
 
 2581 
 
 2740 
 
 2899 
 
 3058 
 
 83 
 
 4445132 
 
 5288 
 
 5444 
 
 5600 
 
 5756 
 
 5912 
 
 6068 
 
 6224 
 
 6380 
 
 6536 
 
 2820 
 
 4502491 
 
 2645 
 
 2799 
 
 2953 
 
 3107 
 
 3261 
 
 3415 
 
 3569 
 
 3723 
 
 3877 
 
 3079 
 
 4884097 
 
 4238 
 
 4379 
 
 4520 
 
 4661 
 
 4802 
 
 4943 
 
 5084 
 
 5225 
 
 5366 
 
 3157 
 
 4992746 
 
 2883 
 
 3021 
 
 3158 
 
 3296 
 
 3434 
 
 3571 
 
 3709 
 
 3846 
 
 3984 
 
 58 
 
 4121 
 
 4259 
 
 4396 
 
 4534 
 
 4671 
 
 4809 
 
 4946 
 
 5084 
 
 5221 
 
 5359 
 
 3290 
 
 5171959 
 
 2091 
 
 2223 
 
 2355 
 
 2487 
 
 2619 
 
 2751 
 
 2883 
 
 3015 
 
 3147 
 
 3555 
 
 5508396 
 
 8518 
 
 8640 
 
 8763 
 
 8885 
 
 9007 
 
 9129 
 
 9251 
 
 9373 
 
 9495 
 
 61 
 
 5515720 
 
 5842 
 
 5964 
 
 6086 
 
 6208 
 
 6329 
 
 6451 
 
 6573 
 
 6695 
 
 6817 
 
 498 
 
MATHEMATICAL TABLES. 
 
 499 
 
 No. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 3570 
 
 5526682 
 
 6804 
 
 6925 
 
 7047 
 
 7169 
 
 7290 
 
 7412 
 
 7534 
 
 7655 
 
 7777 
 
 3667 
 
 5643109 
 
 3228 
 
 3346 
 
 3464 
 
 3583 
 
 3701 
 
 3820 
 
 3938 
 
 4056 
 
 4175 
 
 68 
 
 4293 
 
 4412 
 
 4530 
 
 4648 
 
 4767 
 
 4885 
 
 5004 
 
 5122 
 
 5240 
 
 5359 
 
 3701 
 
 5683191 
 
 3308 
 
 3426 
 
 3543 
 
 3660 
 
 3778 
 
 3895 
 
 4012 
 
 4130 
 
 4247 
 
 10 
 
 5693739 
 
 3856 
 
 3973 
 
 4090 
 
 4207 
 
 4324 
 
 4441 
 
 4558 
 
 4675 
 
 4793 
 
 40 
 
 5728716 
 
 8832 
 
 8948 
 
 9064 
 
 9180 
 
 9297 
 
 9413 
 
 9529 
 
 9645 
 
 9761 
 
 3920 
 
 5932861 
 
 2971 
 
 3082 
 
 3193 
 
 3304 
 
 3415 
 
 3525 
 
 3636 
 
 3747 
 
 3858 
 
 25 
 
 8397 
 
 8507 
 
 8618 
 
 8729 
 
 8839 
 
 8950 
 
 9060 
 
 9171 
 
 9282 
 
 9392 
 
 54 
 
 5970367 
 
 0476 
 
 0586 
 
 0696 
 
 0806 
 
 0916 
 
 1026 
 
 1135 
 
 1245 
 
 1355 
 
 4021 
 
 6043341 
 
 3449 
 
 3557 
 
 3665 
 
 3773 
 
 3881 
 
 3989 
 
 4097 
 
 4205 
 
 4313 
 
 4169 
 
 6200319 
 
 0423 
 
 0527 
 
 0631 
 
 0736 
 
 0840 
 
 0944 
 
 1048 
 
 1152 
 
 1256 
 
 4472 
 
 6505018 
 
 5115 
 
 5212 
 
 5309 
 
 5406 
 
 5503 
 
 5601 
 
 5698 
 
 5795 
 
 5892 
 
 4512 
 
 6543691 
 
 3787 
 
 3883 
 
 3980 
 
 4076 
 
 4172 
 
 4268 
 
 4365 
 
 4461 
 
 4557 
 
 4670 
 
 6693169 
 
 3262 
 
 3355 
 
 3448 
 
 3541 
 
 3634 
 
 3727 
 
 3820 
 
 3913 
 
 4006 
 
 4984 
 
 6975780 
 
 5867 
 
 5955 
 
 6042 
 
 6129 
 
 6216 
 
 6303 
 
 6390 
 
 6477 
 
 6565 
 
 5040 
 
 7024305 
 
 4392 
 
 4478 
 
 4564 
 
 4650 
 
 4736 
 
 4822 
 
 4909 
 
 4995 
 
 5081 
 
 5410 
 
 7331973 
 
 2053 
 
 2133 
 
 2213 
 
 2294 
 
 2374 
 
 2454 
 
 2535 
 
 2615 
 
 2695 
 
 5690 
 
 7551123 
 
 1199 
 
 1275 
 
 1352 
 
 1428 
 
 1504 
 
 1581 
 
 1657 
 
 1733 
 
 1810 
 
 6258 
 
 7964356 
 
 4425 
 
 4494 
 
 4564 
 
 4633 
 
 4703 
 
 4772 
 
 4841 
 
 4911 
 
 4980 
 
 7045 
 
 8478810 
 
 8872 
 
 8933 
 
 8995 
 
 9057 
 
 9118 
 
 9180 
 
 9241 
 
 9303 
 
 9365 
 
 8000 
 
 9030900 
 
 0954 
 
 1008 
 
 1063 
 
 1117 
 
 1171 
 
 1226 
 
 1280 
 
 1334 
 
 1388 
 
500 
 
 MATHEMATICAL TABLES. 
 
 LOGARITHMS OF CIRCULAR FUNCTIONS. 
 Cosines. Sines. — Continued. Secants. 
 
 Angle. 
 
 LCOB. 
 
 23° 17' 
 
 9-9631082 
 
 18 
 
 9-9630538 
 
 53 14 
 
 9-7771060 
 
 . 15 
 
 9-7769369 
 
 67 10 
 
 9-7341572 
 
 Sines. 
 
 Angle. 
 
 X^sin. 
 
 1° 0' 
 
 8-2418553 
 
 16 23 
 
 9-4503452 
 
 24 
 
 9-4507747 
 
 35 14 
 
 9-7611063 
 
 37 14 
 
 9-7818002 
 
 40 8 
 
 9-8092691 
 
 9 
 
 9-8094189 
 
 41 13 
 
 9-8188250 
 
 14 
 
 9-8189692 
 
 27 
 
 9-8208358 
 
 47 19 
 
 9-8663534 
 
 48 26 
 
 9-8740085 
 
 46 
 
 9-8762361 
 
 47 
 
 9-8763468 
 
 60 19 
 
 9-9389076 
 
 Angle. 
 
 jLsin. 
 
 60° 20' 
 
 9-9389796 
 
 67 35 
 
 9-9658764 
 
 36 
 
 9-9659285 
 
 74 18 
 
 "9-9834872 
 
 76 7 
 
 9-9871236 
 
 8 
 
 9-9871549 
 
 41 
 
 9-9881628 
 
 Tangents. 
 
 Angle. 
 
 i^tan. 
 
 4° 48' 
 
 8-9241363 
 
 49 
 
 8-9256487 
 
 19 11 
 
 9-5414678 
 
 12 
 
 9-5418747 
 
 21 34 
 
 9-5968776 
 
 34 
 
 9-8289874 
 
 1 
 
 9-8292599 
 
 36 21 
 
 9-8668291 
 
 22 
 
 9-8670937 
 
 38 5 
 
 9-8941114 
 
 6 
 
 9-8943715 
 
 49 13 
 
 10-0641556 
 
 Angle. 
 
 i/sin. 
 
 21° 34' 
 
 53 38 
 
 39 
 
 100315215 
 10-2269815 
 10-2271532 
 
 Cotangent. 
 
 Angle. 
 
 iy cot. 
 
 24° 13' 
 
 10-3470119 
 
ANSWERS TO THE EXAMPLES. 
 
 PART I. 
 
 La. 
 
 1. 206086", 33° 44' 35". 2. -7755, -64672. 
 6. 80°, 60°, 40°. 6. 150°, 25°, 5°. 
 
 3. 135°. 4. 16. 
 7. 30°, 66°, 108°, 156° 
 
 Lb. 
 
 1. 127127", 58° 28' 21". 3. -09176234567901, -9450661728395. 
 
 3. 156°. 4. 20. 6. 36°, 72°, 72°. 
 
 6. 45°, 75°, 105°, 135°. 7. 3, 6. 
 
 ,- 15 15 17 
 **• TT> "8") T5- 
 
 13. JL, 7, V74. 
 
 V74 5 7 
 le ct2-62 2a& 
 
 23. 30°. 24. 0° 
 
 28. 0°, 90°. 
 31. 60°, 90'. 
 
 14. 
 
 ILa. 
 
 _2 3^ 3 
 
 Vl3' Vi3' 2' 
 
 ^^ V(10 + 2V5) 
 4 
 
 25. 15°. 
 
 20. 30°, 60°. 
 
 33. a2(62_i)=l. 
 
 _^ 9 41 41 
 
 12. TIT) -9-> TTT- 
 
 15 V5 3 A 
 3 2 V5 
 
 22. V5-i 
 
 26. 15°. 27. 45° 
 
 30. 0°, 60°. 
 
 34. a2 + 62 = c2 + d2. 
 
 35. (cc' - aa'f = (ab' - hc'){a'h - b'c). 
 
 n.B. 
 
 ,, 60 60 11 
 
 **• ¥TJ H' bit* 
 
 -J, 21 21 29 
 
 !*• ZTJ 2^» 2 1- 
 
 17. 
 
 V5-l 
 
 V(10 + 2v5)* 
 
 _« 12 12 13 
 
 12. TS") -5-) TT- 
 
 15. 
 
 22. I 
 
 4^5 19 4v5 
 21 ' 4^/5' 19 ' 
 23. 0°. 
 501 
 
 _« 12 35 37 
 *«• ¥T> ^T> 3T- 
 
 16. 
 
 l-m2 
 
 -w3 
 
 l+m-*' 2m 
 24 15°. 25. 0° 
 
502 ANSWERS TO THE EXAMPLES. 
 
 26. 0°. 
 
 27. 0°, 60°. 
 
 28. 
 
 30°. 
 
 20. 60°. 
 
 30. 90°. 
 
 31. tan-12. 
 
 32. 
 
 eos-t- 
 
 
 33. a2 + 62 = a=&2. 
 
 
 34. 
 
 «2 + «2 = 
 
 32 + r3. 
 
 36 . {mm' -pp'f = {mn' + np'){m'n + n'p). 
 
 III. 
 
 1 - cos^a sin^a 
 
 1. ; ""V o ■ a. 1. 3. tan6^ + 3tan4^ + 3tan2^ + l. 
 2 + cos''asm-a 
 
 6 45° 30° 7 ^v^(^-^^) \/(l-^'^) 
 
 ' * * nsJ{m^-\)' ^/(m2-l)* 
 
 lO. a2 + 2c = l. 11. aM(a^ + 6^) = l. 12. 2a6. 
 
 IV. A. 
 
 _ 6 28 185 o 6 3 6« « 2 013 131 8 
 
 !• UffTJ TTT' 2. -g-g-, ^-5. a. -JXTTS* TTOT- 
 
 2. cos a cos jS cos 7 - cos a sin /3 sin 7 - sin a cos ^ sin 7 — sin a sin § cos 7. 
 _g tan g + tan /3 + tan 7 - tan a tan j3 tan 7 
 1 - tan /3 tan 7 - tan 7 tan a — tan a tan ^' 
 
 __ 7 49 17 __240 720 10 3 3 8 
 
 ao. 3V5, 2 ?V5. 21; JL, 24 22 ^ 23. 0, JL, IZ. 
 
 7 7 15 V2 7 \/2' 52 
 
 24. W(2 + \/2), W(2-\/2), v^-1. 30. tana. 31. sec^a. 
 
 120 
 34. sin a. 35. cos 2a. 40. tan"^-— . 
 
 41. 5cosa-20cos3a + 16cos5a. 40. tan 2a. 50. -cot—. 
 
 66. 45°. 67. 15°. 68. 18°. 60. 0°, 60°. 
 
 60. 6°, 45°. 61. 0°, 30°. 62. 20°, 45°, 90°. 
 
 63. = 30°, = 0°. 64. 2a2 - 6 = 1. 66. a* + 62 = 2a2. 
 
 66. 2a%{h + c) = {c'^ + d?-a^-W){c'^ + (P-h^). 67. 1ahc = h'^-a\ 
 
 IV. B. 
 , 989 1431 « 110 5 33 5 « 21 140 
 
 4. 1,|. 6. -mf. 7. tan 2a. 
 
 12. sin a cos j3 cos 7 + cos a sin /3 cos 7 + cos a cos /3 sin 7 - sin a sin /3 sin 7. 
 
ANSWERS TO THE EXAMPLES. 
 
 503 
 
 13. |. 
 
 17. 
 
 36 4m(m^-l) 
 ' 77' m4-6m2 + l' ' 
 
 7.3 4 7. 7. 
 
 TOlTJ 7 2' 8" 
 
 4 3 3 
 "SJ T' 4' 
 
 18. 
 
 24 5^/11 
 18 * 
 
 21. 
 
 22. ft 
 
 23. 
 
 25' 
 
 JL 2 
 
 V2' 5' 
 -Q 4tana(l - tan^a) 
 
 1-6 tan-a + tan^a 
 49. sin3a/sin5o. 
 66. 75°. 
 
 L ^' a 
 V2' 128' 
 
 41. 5 sin a. - 20 sin^a +16 sin^a. 
 
 60. 2 cos22a/cos 3a. 
 
 57. tan~^J. (The root tan'^S is inadmissible in Part 1, being greater 
 
 than 45°. ) 
 
 58. 0°, 45°. 59. 7° 30', 30°. 60. 0°, 9°. 
 
 61. 30°. 62. 0°, ir 15'. 63. 15°. 
 
 64. a2 + 6 = l. 65. {a^+¥-l){a" + b^-2) = 2{a-\-l). 
 
 66. (ah + cd)tsin{a-p) = bc-ad. 67. 2¥=3b-a. 
 
 3. sin(a + ^ + 7)+sin(a-/3-7) + sin(a-jS + 7) + sin(a + ^-7). 4. ^2, 
 
 7. c2=a2 + 62 + 2a6cos(a-^). 7=tan-i-^^^5^±^^H^. 
 
 a cos a + o cos /3 
 18. ^ + 20 = 90°. 
 
 VI. A. 
 
 1. 
 
 5-3967519. 
 
 2. 
 
 1 •8757579. 
 
 3. 
 
 7-5929813. 
 
 4. 
 
 3-9615188. 
 
 5. 
 
 2-7315929. 
 
 6. 
 
 176681. 
 
 7. 
 
 -00523945. 
 
 8. 
 
 16387 1. 
 
 9. 
 
 •0202245. 
 
 lO. 
 
 143-638. 
 
 11. 
 
 -00880535. 
 
 12. 
 
 26-7453. 
 
 13. 
 
 •0207381. 
 
 14. 
 
 1-4714612. 
 
 15. 
 
 1-5599642. 
 
 16. 
 
 1-3028511. 
 
 17. 
 
 1 9093311. 
 
 18. 
 
 1-5038295. 
 
 19. 
 
 1-9918678. 
 
 20. 
 
 19° 0' 10". 
 
 21. 
 
 16° 2' 34". 
 
 22. 
 
 15° 0' 4". 
 
 23. 
 
 42° 45' 48". 
 
 24. 
 
 80° 56' 34". 
 
 25. 
 
 28° 17' 51". 
 
 26. 
 
 •2088221. 
 
 27. 
 
 -3215359. 
 
 28. 
 
 •7620948. 
 
 29. 
 
 21° 48' 8". 
 
 30. 
 
 78° 27' 47". 
 VI. 
 
 B. 
 
 
 
 
 1. 
 
 5-8010626. 
 
 2. 
 
 2-7105548. 
 
 3. 
 
 1-8641900 
 
 4. 
 
 4-7251420. 
 
 5. 
 
 3 •8982874. 
 
 6. 
 
 1-65006. 
 
 7. 
 
 -235567 
 
 8. 
 
 20-8484. 
 
 9. 
 
 249525. 
 
 lO. 
 
 •000176042. 
 
 11. 
 
 -000235408. 
 
 12. 
 
 1 -19680. 
 
 13. 
 
 •191998. 
 
 14. 
 
 1-8980337. 
 
 15. 
 
 0-5478477. 
 
 16. 
 
 1 -9260104. 
 
 17. 
 
 1 -0087298. 
 
 18. 
 
 1-4072257. 
 
 19. 
 
 1-9892056. 
 
 20. 
 
 69° 4' 25". 
 
 21. 
 
 73° 6' 16". 
 
 22. 
 
 61° 7' 41". 
 
 23. 
 
 18° 29' 21". 
 
 24. 
 
 32° 56' 39". 
 
 25. 
 
 69° 56' 6". 
 
 26. 
 
 -3290360. 
 
 27. 
 
 -7283544. 
 
 28. 
 
 2-2213844. 
 
 
 
 29. 
 
 3.3° 23' 2". 
 
 30. 
 
 36° 37' 10". 
 
 
 
504 ANSWERS TO THE EXAMPLES, 
 
 Vil. A. 
 
 1. 
 
 5 = 54° 33', 
 
 a = 99-759, 
 
 6=140-115. 
 
 a. 
 
 ^=65U1', 
 
 a = 2771-21, 
 
 6 = 1252-22. 
 
 8. 
 
 ^=48° 43', 
 
 6 = 793-72, 
 
 c= 1203-00. 
 
 4. 
 
 ^=25° 57', 
 
 a = 19526-0, 
 
 c = 44622-0. 
 
 6. 
 
 ^=34° 28' 21", 
 
 B = 55° 31' 39", 
 
 6 = 586-971. 
 
 6. 
 
 A = 74° 52' 54", 
 
 5=15° 7' 6", 
 
 a = 2946-37. 
 
 7. 
 
 ^=39°19'48^ 
 
 B = 50° 40' 12", 
 
 c = 1295 -39. 
 
 8. 
 
 A = 54° 38' 26", 
 
 5 = 35° 21' 34", 
 
 c = 971-138. 
 
 9. 
 
 5 = 27° 44' 18", 
 
 6=130-941, 
 
 c = 281 -330. 
 
 lO. 
 
 ^=26° 36' 21", 
 
 5 = 63° 23' 39", 
 
 6=182521. 
 
 11. 
 
 A =1^° 45' 57", 
 
 a = 384-014, 
 
 6 = 104-580. 
 
 >3- 
 
 ^=38° 37' 5", 
 
 5 = 51° 22' 55", 
 
 c = 642-499. 
 
 
 ^ 
 
 VII. B. 
 
 
 1. 
 
 P=18°45', 
 
 a = 3800-03, 
 
 6 = 1289-94. 
 
 a. 
 
 ^=30° 12', 
 
 a = 2868 -22, 
 
 6 = 4928-10. 
 
 3. 
 
 ^=79° 48', 
 
 6 = 8-6366, 
 
 c = 48-7708. 
 
 4. 
 
 ^=64° 17', 
 
 a =1285-23, 
 
 c = 1426-53. 
 
 6. 
 
 A= 8° 29' 42", 
 
 5 = 81° 30' 18", 
 
 6 = 6864-90. 
 
 6. 
 
 .4=57° 4' 18", 
 
 5 = 32° 55' 42", 
 
 a = 47928-7. 
 
 7. 
 
 ^=28° 57' 36", 
 
 5 = 61° 2' 24", 
 
 c = 1059-48. 
 
 8. 
 
 ^=57° 18' 11", 
 
 5 = 32° 41' 49", 
 
 c = 9520-64. 
 
 9. 
 
 A=i 9° 45' 44", 
 
 6 = 4068-56, 
 
 c = 4128-34. 
 
 lO. 
 
 ^=38° 56' 45", 
 
 5 = 51° 3' 15", 
 
 c = 93 -8615. 
 
 11. 
 
 A = 13° 18' 45", 
 
 5 = 76° 41' 15", 
 
 a =119-967. 
 
 12. 
 
 ^=58° 45' 4", 
 
 a = 6159-71, 
 
 6 = 3737-64. 
 
 vni. A. 
 
 1. 449 ft. 8 ins. 2. 131 ft. 2 ins. 3. 34 ft. 2 ins. 
 
 4. 13-24 ins. 6. 7-899 miles ; 9-349 miles. 
 
 6. 9ft. 7 ins. ; 116ft. 5 ms. 7. 82 ft. 2 ins. 8. 501 ft. 
 
 9. 2 m. 342 yds. ; 42° 20' 52" E. of N. 
 
 lO. 485 yds. 1 ft. 11. 74 ft. 
 
 12. 7 m. 4413 ft. 13. 27° 36'. 14. 2° 9' 33". 
 
 16. 2 ft. 5-12 ins. 16. ils/2j\/5 - 1. 
 
 18. 351 ft.7 ins., 455 1ft. 7 ins. 19. acos^cosec(a + /3), 52 ft 11 ins. 
 
 ao. tB.n-^^^^-^L^^-, 2.A>-g! ^-^^'M miles. 
 ^(4 tan^a - tan^/S)' V \ 4 tan^a - tan2/3/ 
 
ANSWERS TO THE EXAMPLES. 505 
 
 VIII. B. 
 
 1. 101ft. 7 ins. 2. 12ft. Sins., 36ft. Sins. 3. 107 '07 ft. 
 
 4. 4° 32' 54". 5. r 37' 14". 6. 77 ft. 3 ins., 372 ft. 4 ins. 
 
 7. 110 ft. 3 ins. 8. 500 ft. 7 ins. 9. 124 ft. 7 ins. 
 
 lO. 190ft. Gins. 11. 38.39ft. 12. 5182ft. 
 
 13. 3° 13' 45". 14. 2M9'54". 15. 386 ft. 1 in. 
 
 17. a tan o sec ^. 19. 3978 -78 miles. 20. 103 ft. 9 ins. 
 
 Miscellaneous Examples. L 
 
 a. 
 I. 26° 35' 41" ; 150° 3. 45°, 90°. 
 
 • 1885J 18 8 5- ^' ^" 11 ^" • 
 
 1. 60. e. ^=43° 35' 42", 5=46° 24' 18", a=9762. 
 
 7. 85 ft. 11 ins. 
 
 7. 
 1. 33°, 60°, 87°. 3. 0°. 
 
 6. 64cos7a- 112cos^a + 56cos^a-7cosa. 6. sin5a/sin7a. 
 
 7. 11 ft. 6 ins. 
 
 5. 
 
 2. coBa = '2q{p + q)j{p^ + 2pq + q^), ta.na=p{p + 2q)/2q(p + q), etc. 
 
 3. 3, A «. -^ 
 7. 459 ft. 9 ins., 2029 ft. 
 
 3. 3, -^ 6. J[=22°2'49", 5=67° 57' H", c=2730-67. 
 
 1. 30°, 25°, 125°. 3. 0°, 45°. 4. ^, f. 
 
 8 8 
 
 e. ab^=a + 2h, 7. 5-38 ft. per sec. 
 
 f. 
 
 1. cosa=(27H-l)/(27i2 + 2n+l), sina = 2w(7i+l)/(2n2 + 2n+l), etc. 
 
 2. c2(a + 2) = a62. 6. 5 = 70° 24', a = 335-452, & = 942-057. 
 7. 1255 ft. 6 ins. 
 
 6. 5=80° 6', 6=406-812, c=412-961. 7. 12° 46' 46". 
 
 3. m, 1. 5. 15°, 30°; 0°, 30°. 
 
506 ANSWERS TO THE EXAMPLES. 
 
 e. 
 
 ». Hii' fi. 0°, 90°. 6. a62=l. 7. 2. 
 
 *• r-~^^)^^^^^■ «• 2co83a. 6. 0', 7°30', 90' 
 
 2. Vt- 3- coseca. 5. 45°. 6. 6^^=40^^^ + 62-02). 
 
 IX. A. 
 
 1. ^ of a radian. 2. 6-981 feet. 3. 3957 miles. 
 
 4. 3-142. 6. 6 m. 150 yds. 8. 7r/180. 
 
 0. •7rr2. lO. -01745. 11. 50m = 27/i. 
 12. 11° 13' 52^-932; 0°0'33"-048; 78° 18'4"-212. 
 
 18. 32^ 48' 61" i ; 84" 21' 4" '§38271604 ; 48^ 13' 58"' -024691358. 
 
 - ITSlTT . 470035 
 
 • 9000 ' 200000' 
 
 IX. B. 
 
 1. 1/2880. 2. 2094-4 miles. 8. 7 miles. 
 4. 3-14. 6. 847849 miles nearly. 8. 7r/180. 
 
 9. 7rr2. 10. -052. 11. 250s = 81<T. 
 
 12. 48° 25' 46" -668; 4°36'31"-716; 0°0'0"-972. 
 
 13. 5^^ 61' 54" -320987654; 93^^ 57' 39' '209876543 ; O" 0' 15" '432098765. 
 
 14. 210297r/43200 ; 228137r/250000. 
 
 1. 38'. 2. 200/7r. 8. 50°, 27°. 4. 30n7r/(l+60?i). 
 
 6. 56if , 60°, 63xV. «. 85^°, 4^°. 7. 150°, 135°. 
 
 8. 49f4:°, y radians. 
 
 9. 45°, 75°, 105°, 135°; jr/i, 57r/12, 77r/12, 37r/4. 
 
 10. 47r/35, 97r/35, 27r/5, 197r/35, 247r/35. 11. (47r-3)/37r. 
 
 12. 65°27'16iS:". 14. 15°, 45°, 120°. 
 
 XL A. 
 
 8. -(2 + V2 + \/3 + /v/6). 4. 2/i7r±^J, 2nir±^. 
 
 5 5 
 
 6. 2nir+~. 6. nw, 7i7r + tan"^2, nir --. 
 
 ~3 4 
 
ANSWERS TO THE EXAMPLES. 507 
 
 3 "6 6' 4' 4 6 
 
 _Q TT TT TT 5t TtT 27r 5x IItT 
 
 * l2' 6' 3' 12' 12' y T' l2 * 
 
 XLb. 
 
 2. 0. 3. V3-2. 4. W7r±J, W7r±J. 6. mr±l. 
 
 4 3 6 
 
 6. nir + {-\r-^, n-rr-i-lT-^. 7. (2?i±l)5, 2(3n±l)7r 
 ' ^ ' 2 3 9 
 
 8. 2«7r±^, (2/i + l)7r+cos-il. 9. 60°, 120°, 240°, 300°, 420°, 480° 
 o 3 
 
 lO. J |, |:, 1^. 11. (2n + l):r + .|. 12. 2n:r + ^. 
 
 XII. A. 
 
 6. cos(a + /3 + 7 + 5) + cos{a + /3-7-5) + cos(a-/3 + 7-5) + cos(a-^-7 + 5). 
 
 34. tana. 35. Umr + ^, wt±J. 36. nx+—, nx-¥^. 
 
 3\ 2/' 3 -10 -10 
 
 37. O''"^)"' ^^" + ^"^^i 38. n^, {2n + lf-. 
 
 r~s ' r+s 
 
 44. (2«-i)7r±|. 45. (27i-l)7r-a±/3. 
 
 46. 71 . 360° + 69° 12' 13", n . 360° - 32° 20' 1". 
 
 50. iUs + s/5 + ^j5-^5), J(\/4 + N/6 + v/2+V4-V6 + Ay2), 
 i(V4-v/6 + x/2 + V4 + V6-N/2), -^^• 
 
 51. 4H7r + ?^ and 47^ + 5". 65. -^ + ^ = 1. 
 
 2 2 a'* 62 
 
 _Q +secasecj3 
 
 * l-7(sec2a-l)(sec2p-r)" 
 
508 ANSWERS TO THE EXAMPLES. 
 
 XII. B. 
 86. nir + l, !^. 36. ""^^ ?^. 37. ^t + |, (6n±lW3(r-l). 
 
 38. (2n + l)5, ?If. 89. (4n-l)5, (471+1)^^. 40. !^. 
 
 41. «7r, 7i7r±^. 42. «7r + {-l)"^, n7r + (-ir^, n7r-(-l)«?^. 
 
 4 6 10 10 
 
 43. TO7r + (-l)"^-l 44. (w-i)7r + (-l)«sin-iA- 
 
 46. W7r + 5, TOT + ^. 46. Ji. 180° + ( - 1)«(12° 55' 15") -26°33'54". 
 4 2 
 
 «0. K\/4 + v/6-V2-V4-x/6-^/2), l-{V2(x/2 + l) + ^2(^2- 1)}, 
 
 t(V4 - V6 - ^2 - 74 + ^/6 + ^/2), i(-N/4+N/6-v2+>/4-x/6 + V2). 
 
 61. 4n7r+?!rand4w7r + ^. 66. ^ + ^=1. 
 
 2 2 a^ 62 
 
 66. -1. 67. 45°. 
 
 xni. 
 
 14. _+a. 
 
 16. ^ = (2w+ l)7r, = - (2w - J)7r ; 
 
 ^=2n7r-2tan-i>^, <^= - (27i-|y + 2tan-i^. 
 2 \ 3/ 2 
 
 16. (2n+l)J ^ + (-I)n^. 17. (47i-l)|, (47i + 3)|. 
 
 18. "^ + {1)%. 18. 7i7r + tan-i$^if^±^. 
 
 2 12 sinasin/3 
 
 20. ±1±V2. 21. 1, ^-^^. 
 
 22. 0, 0; 00, I; 0, |; -<«, ^; 0, ,r; oo, ^; 0, ^; -oo, ^^j 
 
 0, 27r. 
 24. 4a2-8a + l = 0ora2=l. 25. coseca = 2. 
 
 26. (ac - 6/)2 + (a/- 6c)2=(a2_ 62)2^ where/=a6ci/(cd-a2-62). 
 
 27. ^y2.tana. 34. -7, --gT- 69. a;2=sin2y. 
 60. 2^=(m + 2n)7r + (-ina + /3)±(7 + a), etc. 
 
 62. ^ = m7r±^, = 2w7r; d = 2mir, = 2n7r + ^. 
 
I 
 
 ANSWERS TO THE EXAMPLES. 509 
 
 63 
 64 
 
 (2.-i)„(2».,|)j 
 
 65. n7r + icos-^('sm'%osec^y 06. mr±a, W7r + 2a. 
 
 67. 2mr±a, 2w7r ±008-^-2 cos a). 68. !^ + ( - 1)'Y^- a\ 
 
 60. (2n + l)^. 71.r!r + itan-^ msin2a 
 
 10 2 ?i 4- m cos 2a 
 
 72. ^ = 2?i7r±^ + a, 0=-2w7r + ^ + a. 73. 2mr, 2w7r-^. 
 
 74. (2m + 2wH-l)^, (?i-m)7r; {m-n)ir, (2rn + 2?i-l)5; (6m + 2H+l)J, 
 6 6 8 
 
 (2wH-6n-l)5; (3«-w + l)-, (3m-w-l)^. 
 8 4 4 
 
 76. 13. 83. Jab{ab-^)^{a + h)U\\c. 
 
 84. (6c-a)(ac + 6)-0. 85. a^ + fts^c^sec^l 
 
 87. aa' sin (j3 - j8') + 66' sin (a - a') = a'h sin (a - /3') + a6' sin (/3 - a'). 
 
 88. a = 6(l+c). 92. 2cos(/3 — 7)cos(7-a)cos(a-/3). 
 93. cot a + e cos(a - j3)cosec a. 98, 3 + 2(cos5 + sin^). 
 
 99 1 1 102 ^' + ^'-2 «'-^' 
 3' V3' 2 ' ^M^' 
 
 111. a + ^=2mr + (t>+\p. 112. (a^ + c^- 8ac)(a2+c2)= ±4ac(a2-c2). 
 
 113. cos a- cos /3, -cosa-cos/3; -cosa + cos^, cos a + cos ^. 
 
 115. ±V3±1. 
 
 
 
 
 
 XIV. A. 
 
 
 1. 
 
 6. 
 
 FTj 2 6> "ST* 
 
 3^7 3 3 
 -%-' 4' 4 
 
 
 2. 
 7. 
 
 120°. 
 6 2 3 
 
 ^/85' V85' 5' 
 
 &. 6 = 16, c = 8(x/3 + l). 
 8. h h I 
 
 9. 
 
 56 10864 
 65> 125455 
 
 168 
 19 3- 
 
 lO. 
 
 16 5 6 24 
 
 "• if- 
 
 
 
 
 12. 
 
 4=90°, i5 = 60= 
 
 ', c = l. 
 
510 ANSWERS TO THE EXAMPLES. 
 
 XIV. B. 
 
 1. 135^ 30% 15". a. 120°. 6. ^3-1, ^3 + 1. 
 
 j,12664 „2 14 «1313 
 
 «. T^» TS^ S' ■'• 5;^. ;^. 5' 8. t, % 1^. 
 
 -» 4 2 4 496 --. 84 4 _80 _, 7 
 
 12. ^=75°, 5=15°, c:a = 2V2:^/3 + l. 
 
 XV. 
 
 16. \{a + h + c). 
 
 23. ^=90°, 5 = 30°, C=60°; AB:BG=^:2; AB:AC=l:2. 
 25. ,^/{6c(a + & + c)(i + c-a)}^(6 + c). 27. 4, 5, 6. 
 
 38. The sum of the cosines of any two angles greater than the cosine of 
 the third. 
 
 XVI. A. 
 
 1. 
 
 A = 
 
 18° 56' 20", 
 
 6 = 
 
 7275-78, 
 
 c = 5043-04. 
 
 2. 
 
 A = 
 
 36° 18' 13", 
 
 b = 
 
 8484-69, 
 
 c = 7282-61. 
 
 3. 
 
 B = 
 
 48° 30' 40", 
 
 a = 
 
 9485-25, 
 
 c = 4686-30. 
 
 4. 
 
 B = 
 
 20° 16' 23", 
 
 G = 
 
 117° 27' 59", 
 
 a = 0932-81. 
 
 5. 
 
 A = 
 
 18° 7'54", 
 
 G= 
 
 41° 36' 26", 
 
 b = 3902-34. 
 
 6. 
 
 A = 
 
 32° 42' 21", 
 
 B = 
 
 68° 58' 55", 
 
 c = 551 -992. 
 
 7. 
 
 B = 
 
 22° 45' 46", 
 
 C = 
 
 107° 59' 44", 
 
 c = 6744-14. 
 
 8. 
 
 B = 
 
 29° 6' 37", 
 
 = 
 
 130° 34' 13", 
 
 c = 9530-82; or 
 
 
 B = 
 
 150° 53' 23", 
 
 c= 
 
 go 4^, 27", 
 
 c= 1917-53. 
 
 9. 
 
 Triangle impossibh 
 
 3. 
 
 
 
 lO. 
 
 A = 
 
 65° 18' 40", 
 
 c= 
 
 73° 3' 20", 
 
 a = 1104-07; or 
 
 
 A = 
 
 31° 25' 20", 
 
 = 
 
 106° 56' 40", 
 
 a = 633-508. 
 
 11. 
 
 A = 
 
 69° 10' 20", 
 
 B = 
 
 46° 37' 20", 
 
 a = 9531 -82. 
 
 12. 
 
 A = 
 
 78° 35' 5", 
 
 B = 
 
 60° 36' 39", 
 
 C= 40° 48' 16". 
 
 13. 
 
 A = 
 
 31° 4' 50", 
 
 B = 
 
 134° 56' 5", 
 
 (7= 13° 59' 5". 
 
 14. 
 
 A = 
 
 41° 41' 16", 
 
 B = 
 
 49° 41' 19", 
 
 a =88° 37' 25". 
 
 
 
 
 XVI. B. 
 
 
 1. 
 
 A = 
 
 28° 44' 25", 
 
 b = 
 
 8208-28, 
 
 c = 4670-13. 
 
 2. 
 
 B = 
 
 9° 42' 10", 
 
 a = 
 
 8756-13, 
 
 c= 13061 -7. 
 
 8. 
 
 = 
 
 87° 47' 50", 
 
 a = 
 
 5980-80, 
 
 6 = 2808-01. 
 
 4. 
 
 B = 
 
 65° 7' 15", 
 
 = 
 
 46° 9' 45", 
 
 a = 514-791. 
 
 5. 
 
 A = 
 
 50° 52' 17", 
 
 (7 = 
 
 34° 50' 43", 
 
 6 = 87-9286. 
 
 6. 
 
 A = 
 
 54° 40' 34", 
 
 B = 
 
 21° 5' 26", 
 
 c = 464 -759. 
 
 7. 
 
 B = 
 
 62° 12' 40", 
 
 G = 
 
 90° 35' 10", 
 
 c = 1188-37; or 
 
 
 B = 
 
 117° 47' 20", 
 
 C = 
 
 35° 0'30", 
 
 c = 681 -797. 
 
ANSWERS TO THE EXAMPLES. 511 
 
 8. 
 
 B= 37° 17' 32", 
 
 C= 
 
 83° 1'48", 
 
 a = 11063-4. 
 
 9. 
 
 Triangle impossible 
 
 
 
 
 lO. 
 
 A= 44° 53' 17", 
 
 B = 
 
 48° 6' 39", 
 
 a = 677-158. 
 
 11. 
 
 5= 99° 33' 15", 
 
 G = 
 
 55° 25' 31", 
 
 6 = 4996-59; o 
 
 
 B= 30° 24' 17", 
 
 C = 
 
 124° 34' 29", 
 
 & = 2564-37. 
 
 12. 
 
 A= 38° 29' 2", 
 
 B = 
 
 66° 29' 21", 
 
 C = 75° 1'37". 
 
 13. 
 
 ^ = 142° 6' 10", 
 
 B = 
 
 21° 43' 16", 
 
 (7= 16° 10' 34". 
 
 14. ^ = 153° 28' 37", 5= 16° 50' 52", G= 9° 40' 31". 
 XVII. 
 
 2 126° 18°- ^('^ + ^5) 8 
 
 ' V(10 + 2>y5)' \/(10 + 2V5)' 
 
 3. a = 7306-72, & = 11545-35, c = 12545-35. 
 
 4. a =1192-64, 6 = 820-64. 7. 5 = 80° 46' 26", (7= 63° 48' 34". 
 
 XVIII. 
 
 1. 1960-95 yds. 2. 131 ft. 6 ins., nearly. 
 
 4. a sin(/3 - 7)sin a cosec(/3 - a) ; 86 ft. 2 ins. 
 
 5. 2072-49 yds. 6. 1352-07 yds. 7. 1113'92 yds., 1980-54 yds. 
 8. 272-14 yds. 9. 80 ft. 7 ins. 
 
 lO. asin(a + j8):(a + 6)sina. 11. 495-21 ft. 
 
 12. 699-95 ft.; 60° 19' 11". 16.264ft. 
 
 17. /isin(a + /3)cosec(a-i3); 780-28 ft. 18. 1574-18 ft. 
 
 19. tan-i (< + ^Osinasin^ ^ ^^ ig ^^^ g j^^^ 
 
 ^'cos a sin /3 - i sin a cos j3 
 
 21 ^^^''^ , "^^"'^ ; where a and /3 are the angles sub- 
 
 l-tan(a + j3)tani3' cos(a + 2/3) 
 
 tended by the flagstaff and the tower. 
 
 23. miles ; -^!-—-: miles. 26. cos"^(cosacos iS). 
 
 V(4-V3) 13 
 
 27. tan-i^ (a + 6)sin a s in^^ 3^^ 41° 2'in direction N. 33° 55' E. 
 
 6 cos a sin /3 - a sin a cos /3 
 
 34. tan = (tan a - tan j3 cos ^)cot j8 cosec ^ and tan 5 = tan a cos <p, where 
 
 5 is the inclination of the common section to the horizon and 
 the angle between its direction and that of the dip of the first 
 plane. 
 
 35. tan-i(a - 6)cota/c S. of W. ; 27° 38' 20" S. of W. 
 
 36. hcotaco^d. 37. 29-72 ft. 
 
512 ANSWERS TO THE EXAMPLES 
 
 39. Direction of wind is <t> E. of N. where 
 
 , , tan a sin 6 
 
 tan <b - , „ ; 
 
 tan a cos ^ - tan /3 
 
 velocity of wind = "^^^^"j"^ ft. per sec. 
 
 40. 5(V7-l)ft. 
 
 41. Length of path = 59 '55 miles, height at appearance = 93 '04 miles, 
 
 height at disappearance = 34 00 miles. 
 (p + g)7r 
 
 43. 
 
 acos^ 
 
 4(7i-l) 
 
 2 sin P"" sin ^^ 
 4(»-l) 4(7i-l) 
 
 46. I cos a cosec(a +/S), where tan /8 = tan 8 cos 6. 
 
 49. ha - 6)tan a cot^ j ha + 6)tan /3 tan X 
 2 2 2 2 
 
 XIX. 
 
 1. 270 sq. ft., 16296 sq. ft. 3. ^bhm2A, if A, a, b he given. 
 
 4. 6 sin J Va^ - fe^ siii'^!4, if ^, a, 6 be given. 
 
 6. ia2sinj5sinC/sin(^+(7). 16. 3. 17. 6. 18. 5 + ^^5:8. 
 
 19. 3(2 + ^3)a2. 28. 32^, 9, 14, 35, 90 inches. 
 
 29. 63J, 9^, 12, 60, 190 inches. 30. 2 + /^3 inches. 60. 2Jsq. ins. 
 
 l-sin:| 
 
 65. 8(^2- l)r2. 79. r ?, etc. 
 
 1 + sin^ 
 
 82. TT, J. 107. rsec^^. 
 
 2 2 
 
 121 
 
 I p-Pi-Ps , I p-Pz-Pi ^ l p-Pi~Pi - I p + Pj + Ps 
 
 Miscellaneous Examples. II. 
 a. 
 1. 3-1416. 3. VE; (2n + l)^, ^ + (-ir^. 
 
 ^ 31 5 
 
 /3. 
 
 3. a = (2/i + 1 )1 or iS - 7 = nw. 6. 86° 16' 47". 
 
XSWEB.^ TO THE EXAMPLES. 513 
 
 7. 
 
 3. mr, iiT±^ I --, — . 6. 8.Mt. 1 111. 
 
 2 3 
 
 6. 396 ; 6O5 ; 4 ; 41, 44, 198. 
 
 5. 
 
 4, ^^^" ~ ^^ l 5. 66° 29' 10", 4236-29. 
 
 2.-4 sin ^ sin B sin (7. 3. vir + ^ and ?/7r + ^. 
 
 4 4 
 
 5. 2a sin a, 2a cos 2a. 6. 
 
 ^ - sm' 
 
 71 
 
 3. -^-. 5. j5 = 75°, (7=90°, r = 2v/2; or 5 =105°, (7 = 60*, c = ^/6. 
 
 I 
 
 '7- 
 
 3. ^J - ^^f cos (8 - a) + ^'I = sin2(|8 - a). 5. .4 = 106° 53' 54", /? = 32° 42' 6" 
 0" ab O" 
 
 ^- 3' ¥' ^^^^' ^3' V3' ^^^' ^' ^- •'^' ^' *' ^^' "'• 
 
 4. ?i7r, »7r+tan-i^ . 5. 5926-61. 
 
 \. 
 
 1. cc^-px^+pqx-p = 0. 3. a+/S + 7 = ??-7r ; o + /3 + 7 = (2?? + 1)' 
 
 4. a sec A tan ^ tan (7 ; 230'966. 
 
 (I. 
 1. 6, S. 3. 'ij, (4»-l)J- ^ 
 
 2 8 Va(26-a) 
 
 1. k, cc. 3. ^\^ ; 7?7r-l-tan-^2, )?7r + tan-MN/5-2), 7i7r - tan " Mx/5 + 2). 
 
 4. (/)- + (7'-)cot a +yi9(sin a - 2 cosec a). 
 
 2k 
 
514 ANSWERS TO THE EXAMPLES. 
 
 XXII. 
 1. IT radians. 2. pfq. 3. ^. 4. 1. fi. 1. 
 
 8. 1, e 2 J or 0, according as m is <, = , or > 2. lO. 1. 
 
 XXIII. 
 1. cos ^(n + 1 )5 sin ^n^/sin i^. 2. sin^na/sina. 4. cos ^^ sin w^/sin i^. 
 
 ^ 
 
 6. |cos(a + 1^^ + ( - 1 )'»cos(a - ^^V) } / 2 cos ^ 
 
 6. i(sin 2"+!^ - sin 2^). 7. coti^-cot2"-i<?. 
 
 8. cot^-2"cot2M. 9. tan 2"^- tan ^. 
 
 11. ^-cos(?i+l)^sinn&/2sin^. 13 ,-(2 cos 5 - cos 3^ - cos 5^). 
 2 lb 
 
 16. — — + — 17 ^^1^^^"^ sing 21. 
 
 * w' w 4}i' ' 2"cos2"g + l 2co8g+i" 
 
 23. sin 2na/2 sin a. 25. sin ^(3/i + 1 )a sin f wa/sin fa, 
 
 26. ^ + cos(n+l)asin7ia/2sina. 
 
 27 
 
 . lcos{3a + (n-I)|}sin^^/sin|^ + ^^os{a + (,.-l)^}sin|/sin| 
 
 28. ^-cos{2a + (?i-l)i3}sinw^/2sin/S. 33. l cot j^ - cot a ; --—!-. 
 ^ Zi Jt n, tan ci 
 
 34. sin%S/sin^co8acos(a + w/3). 35. tan"^(l + 7i + »t2)_^. 
 
 4 
 
 36. cosec2a;--^(cot2^ + lY 87. sinwj8/4sini3-*icos{2a + (n- 1)^}. 
 
 38. w(r2 + \x-), r = rad. of incircle, x = dist. of point from the centre, 
 n = number of sides. 
 
 46. i(cosec g - cosec 3«g). 47. sin^g - 2"sin2^. 
 
 48. ^cosec-g-2"-icosec22«g. 49. ^t&n^n{n+l)e. 
 
 60. |(tan 3"a;/3" - tan ar). 61. —^{aeondaec{n+l)d-8&oe}. 
 
 2sin0 
 
 53. a sin na/2 sin^a 8in(7i + !)«, where a = dist. of from AB. 
 
 64. (-l)'»-i?^. 66. 2«cosh«^sinh— . 
 
 i2n 2 2 
 
 lUin^^4lBu4-n]. 69. cot%-(lcot "V-f^f] _1). 
 
 2\ 2/ 2 J \2« 2"/ 3\ 4V 
 
 61. 2''-2cosec2«+ig-Jcosec2g.62. ^(tan 3"^ - tan g). 
 
 •-•2 
 
ANSWERS TO THE EXAMPLES. 515 
 
 63. if cosec2^ - cosec2?^Y 
 SV 2 2 / 
 
 64. sin !| sin '^cos 'Lz\a + /3) /sin | sin | 
 
 XXIV. 
 14. w/2«-i. 18. 7r2/8; 7r796. 22. 7rV384. 23. tt^^. 
 
 25. ^(l-0 20. log(l-cos2^)-log(l-cos2^_^)-»log2. 
 
 43. (l)%even; (2) 7iodd. 
 
 XXV. 
 
 4. 2. 7. ^ = 14° 54' 13", B = 75° 5' 47", h = 5598-37. 
 
 8. 5 = 149° 53' 7" -5, C = 6'52"-5. 
 lO c = a + 6-_^^ 4- «^ ab(a-b)d '^ „_ &g _ ah{a-b) d^ 
 2(a + &)' a + 6 6(a + 6)3'' a + 6 6{a + 6)3 ' 
 
 11. dab cos a, where a, b are the sides, a the included angle, and 6 the 
 change in the angle. 
 
 13. 6= -(f>- x//, y = {c(f> + b\f/ COS A +xsinB)cosec A. 
 z = {c(j) cos A+b\p-\-x sin C)cosec A , 
 
 14. = (?/ sin (7 - 2 sin 5 - Z;^ cos C), 
 
 a 
 
 ^ = (z sin ^ - y sin G -cd cos ^), x = y cos C + s cos B + 6b sin (7. 
 a 
 
 15. = f^-^ + ^cot^Van5, xly = (^-%- dcotA - d cotB)t3inB, 
 
 \b a I \a b I 
 
 z-{x-y cos G-Qb sin (7)sec B. 
 
 16. ^ = (a;-ycos(7-scos5)/6sinC, etc. 
 
 18. ^ = - W^ • ^, = - iVL^f ;/, = ^v^Tj^ . therefore 5 can be deter- 
 70 70 ^ 70 
 
 mined with the greatest accuracy. 
 
 10. 3960 miles. 20. ^fl-^"\ % 23. '07 inch. 
 
 a \ 2 / a- 
 
 24. -0000952. 26. Decreased by cd sin A sec C. 
 
 Miscellaneous Examples. III. 
 7- 
 
 ., - _i cos 5 + cos (7- 
 6. tan ^ . p — ^-p^ 
 sin ZJ - sin G 
 
oiG AXSW/'JRS TO THE EXAMPLES. 
 
 5. 
 
 a tun a ^ /. 
 
 2tan/3 
 tan /3 - tan a 
 
 6. 
 
 1. 5v/3, Ln^^. 4. ~^a. 
 
 a. 2:3:3. 6. 7r + 4a-2sm2a : 7r-4a + 28in2a 
 
 1. 0, - a ; or . ^^ = -^- = -^ , where a = ^^ or ^. 
 sin 2a sin3o sin a 7 7 
 
 a. 0, -a, ^(tVS-l); ±a, 0, +a; 
 
 or ^A^ = J^. = . ^^ =^ , where a = Z^T or i^^. 
 sin2a 8in3a sin 4a sma 9 9 
 
 8. 008 1:, 008 1?. 4. xiu^'. sinlO^. 
 
 4. 21ogsin^-21ogsm- -7ilog2. 6. I/tt. 
 
 X. 
 2. 2/m-^. 3. 3. 4. 7r3/16. 6. - ^. 
 
 Examples XXVI. 
 7. C08(4a; + 5y) + i8in(4a; + 52/). 8. cos(a + /3-7- 5), isin(a + j8-7-5). 
 9. cos(5a-4/3) + isin(5a-4^). lO. cos(10^ + 12a)-isin(10^ + 12a). 
 11.-2-'. 12.-2^0. 14.-1. 15. -^, i(±^3 + 0. 
 
 16. ±K^2(V3 + ^), ±i</2(-l + iV3). 
 
 17. ±V2(co8^ + isin-^-), ±V2(sin^2-**^°«F2)' 
 V2(co8^^ + i8inj^), V2(co8^ + mn3i^), -;/2(8in^2 + *'^°«r^)- 
 ^2(co8^..8in^'^), ^2(cos^^.^8inl^-), 
 
 V2(cos^-Msin^). 
 
 18. 
 10. 
 
ANSWERS TO THE EXAMPLES. 517 
 
 20. cosC2r + ^y + isin^2r + ^'\|, whei'er = 0, 1,2, 3or4. 
 
 21. cos(2r+ 1)T +ism(2r + 1)^, where r = 0, 1, 2, 3 ... 9. 
 
 22. ^^^f« - ^) {cos i(a + /?) + / sin i(a + /3)}. 
 cos^(a-/3) 
 
 25. 2-+i(-lfsin'»^-rJ'cos'^J^-\ weven; 
 2n+i( _ 1 )^'sin" t^ sin ^^+1), 71 odd. 
 
 26. /) = r"7^'", d = 7yia-na', where (r, a) = a+ <6, (r', a') = a' + e&'. 
 28. ^ cos = r'^cos ma + r'"cos na', J? sin = r"'sin ma + r'"sin na', 
 
 where (r, a) = a + i6, (r', a') = a' + i6'. 
 
 31. i-\ < cos??.a + wcos[ wa + M +w-cos( ?ia4--^ W. 
 
 33. ±i. 34. ±i. 
 
 35. sin(a + /3)cos(2^ + a + /3) 
 
 = sin(^ - ^)cos(^ + 2a + /3) - sin(<^ - a)cos(^ + a + 2/3). 
 
 4. 1, 
 
 Examples XXVII 
 
 7r 
 
 (^'l)--«- »-(i'I 
 
 6. -y2iisina-2sinlla + sinl0a)/(5-4cosa). 
 
 7. (cosa-a;)/(l -2a;cosa + a:^). 8- a;sina/(l +2a;co&a + a;-). 
 9. {cos^^ sin d - cos"+i^ sin(ri + 2)^ + cos'*+2^ sin( w +J_ )^}/sin-^. 
 
 10. {cosa-a;cos(a-/3)-a;"cos(a + n^) + a;"+^cos(a + ?<-- lj8)}/(l-2a:cosj3 + x% 
 {sina-a;sin(a-j8) -a;"sin(a + n^) + a;"+isin(a + n- l/3)}/(l -2a: cos /3 + x^). 
 
 14. sin ^(cos ^ - sin ^)/( 1 - sin 2^ + sin^^), sin'-^/( 1 - sin 26 + sin^^). 
 
 11 6 = ^ , the series oscillate. 
 2 
 
 15. V2sin^/(3-2;^/2cos^). 
 
 ■«+l n 
 
 16. {sin?i0 - ^/2sin(n+ 1)^ + 2 ^ sin ^}/{2-'(3- 2^/2 cos ^)}. 
 18. 0, cot 6, if d=\=mr ; <x,0,ii d = nir. 
 
 Examples XXVIII. 
 
 1. M..., 18° 26'. 3. 2"cos""cos/^^* + iy, 2"cos""sin^^^ + l'\o. 
 
 6. 2"-'cos"^J"^cos"(<^ + 0) + 2»-icos»^±^cbs"(^-0). lO. -985. 
 
 2 ' 2 2 2 
 
518 ANSWERS TO THE EXAMPLES 
 
 Examples XXIX 
 
 3. 2-7183, 7-3891. 4. 1-649, 4-482. 6. ecos(tan0), e8m(tan('). 
 
 7. exp(a:cos^)cos(a;sin^). 9. exp(ccos^)sin(a + csin/3). 
 lO. exp(co8Scot^)8m(^+sin5cot^). 11. 2cos^exp(^cot^). 
 
 13. exp(a cos a:)cos(a; + a sin cc). 
 
 Examples XXX. 
 
 1. -35 + tY-+2wirY 2. 3-00 + i(7r + 2n7r). 17. tan-V ^^"^^ 
 
 \4 / l-a;co8 6^ 
 
 19. ilog(H-m)-ilog(l+2mcos2a + ?n=^). 24. tan-^(cot^). 
 
 26. taii-i{tan('|-|U. 28. ^ log(4 x/33i^). 
 
 29. Oor^, ifa;=|=(2n+l)^; ^ifa; = (27i + l)|. 
 
 30. -log2 + co82^-icos4^ + icos6^- .... 
 
 31. - 2{a cos + ia2cos2e + Ja'co8 30 +...}. 
 
 Examples XXXI. 
 
 /rt X « 61oga + 2rr , ,. . 61oga + 2r7r 
 1. exp(2w7r). 3. cos — ^ +ism 5 
 
 5. exp( - c0)cos(c log r), exp( - c^)sin(c logr), where {r, <p) = a-hs/ -1. 
 
 8. cos 2(;?0 + g log r) + i sin 2(p0 + g' logr), where (r, 0) = a + 6t. 
 
 9. (47i + 1 )/(4m + 1 ). lO. — - * |5^, where 7n=l=0. 
 
 14. If a; = (r, 6) then 
 
 (a + 27i7r)sina + logacosa g, _ (« + 2M7r)cos a - log a sin a 
 ~ a ' a 
 
 Examples XXXII. 
 
 6. 2cosa;coshy/(cos2« + cosh2y), 2isina;sinh?//(cos2a; + cosh2y). 
 
 gl- cos 2a cosh 2^ . 
 cosh 2/3 - cos 2a * 
 
 12. C08-H{ V(a + 1)2+^-' - s/(a - 1 )2 + j32} 
 
 + icosh-H{\/(a + l)- + ia^+ V(a-l)=^ + j8--^}, the upper or lower 
 sign being taken according as /3 is positive or negative. 
 
ANSWERS TO THE EXAMPLES 519 
 
 15. A - ± -> B = h tanh~^(sin ^), where A and cos 6 have the same sign. 
 4 
 
 26. ^ = 27i7r + 2tan-i(tanh^y ?i = 2?i7rj + 2tanh-i( tan ^V 
 29. cos(cos ^)cosh(sin 6). 
 
 3 
 
 31. If G^ + 4H^ is positive, let cosh 3m = numerical value of \G({- H)^^ 
 then the roots are ±2sJ - H co^Yiu, +'2sJ - Hcosh.iu + i''-~\ 
 
 +2 sj - jSTcoshf ?< - i — ], the upper or lower signs being taken 
 
 according as G is negative or positive. 
 
 If G^ + 4,H^ is negative, let cos 3^ = numerical value of 
 
 I G/{-Hf, then the roots are ± 2 \/^;^cos 6, ±2 V^cos ( ^ + ^) , 
 
 ±2\/- //cos( ^--;^ )i the upper or lower signs being taken 
 according as G is negative or positive. 
 
 Miscellaneous Examples IV. 
 7- 
 
 1. a = h{l + c). 4. ^-^-^^t^h-^=sinHa-p). 
 
 ^ xslb^ + y^ b- + y^ 
 
 6. Two equations in sin 26 may be deduced from the given equations. 
 6. cos^a = cos,m(f> -xp). 
 
 5. 
 
 2. a; = 2m7r + ^±cos"^-, y = 2mr + 6 + cos~'^^, where {r, d) = a + bi. 
 
 4. {a + b)^+{a-b)^ = 2. 5. (^ + UY+(^-KY = 2. 
 
 \a bj \a bj 
 
 6. {ax + by)^ + {ay + bxf = 2^. 
 
 2. tan2a = (a2-62)2/(2S6V-2a^). 3. a{ib^ + {b^ - c^)^} = Sbc. 6. -. 
 
 o 
 
 a. 2. 3. -J. 4. /^. 
 
 V- 
 4. ncotnd. 5. a;5_55a^4.330:e3_4(j2a:2 + l65x-Il = 0. 6. 2365. 
 
.21) .L\'>yirA7Av TO TIJl-J i:xAMrLi::s. 
 
 2. ""i 1 
 
 M 
 
 1} 
 
 2. - log2 -- sin 2d + i cos 4^ + J sin 66 - J cos 8d - ^sin 10^+ ... 
 
 6. {csind-c3sind - (»+ l)c"+^sin(w+ l)d + 2(n+ l)c"+^sinnd 
 - (n + 1 )c'»+38in(„ -1)6 + nc"+hin{n + 2)d - 27ic''+^sin{n + 1 )d 
 + nC+^sin n6}/{ 1 - 2c cos d + c2)2. 
 
 i{6 + h{n - 2)a}8in '-^ / 2 sin^^ - n cos{5 + ^(2??. - 1 )a}/2 sin " 
 Z ' z 2 
 
 3. (2m - 1 )sin n6l4. sin=^ - sin '^ sin ^^^^ A sin-^^-n2co8(2!'+i)^ /28in t 
 Z 2, 2t I Z 2/2 
 
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