UC-NRLF ^B SE7 flS3 Digitized by the Internet Archive in 2008 with funding from IVIicrosoft Corporation http://www.archive.org/details/elementsofplanetOOIeverich 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 + 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. =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 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 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 + - 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+)=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

sm2a, W ={h — a) cos sin ^ + A, cos 20, })=a sin^^ — 2/i cos sin 6+h coa^O, then a'cos\

?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 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-cSf/a6. As before, the expression under the radical sign is positive. Also, 2Slbc is a proper fraction, if As{s-a){s-bXs-c) 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, , 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\" 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) a6sin^, 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 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 ah. 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^«-«> - 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 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 ) = OC cot ^ cos((9 - ), PRACTICAL APPLICATIONS. tan /? cos (2) OD OC cot acoa 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 ^')iS/(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 -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 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-^,co8ee + ^- (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' 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^) + ^(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-<- — > 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/smGrSn-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,_3S^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)" - ^ +!-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 + ...+(- 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?. 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 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- + ...+(-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)2Tx\ 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 • „ 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)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 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, , 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 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 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\. Again, i^^ < e '^ ^n 2^ j r_tauh2^ + tanh2u to(r^-r )termsl .:, a fortiori, Frs^=(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

,-^; (3) given x, y, 0, to find z, ) = 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,

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 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\rS'=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 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 mtan3^+.... 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, \ 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- ... 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 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 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 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> 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 -~ 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'^+(3^1, 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 + 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+...+ "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 = 81/4-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 E. of N. where , , tan a sin 6 tan 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- 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 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 \ i^ 0LA3G0W : rUIXTHD AT TOE UXIVERSITY PRBSS, BY ROBKRT MACI.EHOSE. Sji h\ 14 DAY USE RETURN TO DESK FROM WHICH BORROWED LOAN OEPT. This bookis due on the last date stamped below, or on the date to which renewed. Renewed books are subject to immediate recall. 250ci57KK 4^ f«^C D Lu mz REC'D LD OCT 22 19S7 1^3^ £a41 jeMov57KK ctld REC'D Ln NOV 71957 JULlfe'65-lPM 4*159^1 I-., ft . i,-. LD. /.PR ::liab9 LD 21-100m-6,'56 (B9311sl0)476 General Library University of Californij Berkeley 14 THE UNIVERSITY OF CALIFORNIA LIBRARY .7.