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With numerous Examples. AN ELEMENTARY TREATISE ON ANALYTIC GEOMETRY, embracing Plane Geometry, and an Introduction to Geometry of Three Dimensions. AN ELEMENTARY TREATISE ON THE DIFFERENTIAL AND INTEGRAL CALCULUS. With numi;iouB Exam- ples. AN ELEMENTARY TREATISE ON ANALYTIC MECHANICS. With numerous Examples. AN ELEMENTARY TREATISE ON HYDROMECHANICS. With numerous Examples. * A TREATISE ON PLANE AND SPHERICAL TRiaON^OMETEY, AND ITS APPLICATIONS TO ASTROE^OMY AND GEODESY, WITH NUMEROUS EXAMPLES. BY EDWARD A. BOWSER, LL.D., Pkofkssob of Mathematics and Enginbeuing in Rutgers College. BOSTON, U.S.A. : PUBLISHED BY D. C. HEATH & CO. 1892. I COPTRIOHT, 1892, By E. A. BOWSER. TvrOGRAPHY BY J. S. CUSHING & Co., BoSTON, U.S.A. Presswork by Berwick & Smith, Boston, U.S.A. 1 i "-""--,- " niTi PEEFACE. The present treatise on Plane and Spherical Trigo- nometry is designed as a text-book for Colleges, Scien- tific Schools, and Institutes of Technology. The aim has been to present the subject in as concise a form as is consistent with clearness, to make it attractive and easily intelligible to the student, and at the same time to present the fullest course of Trigonometry which is usually given in the best Technological Schools. Considerable care has been taken to instruct the student in the theory and use of Logarithms, and their practical application to the solution of triangles. It is hoped that the work may commend itself, not only to those who wish to confine themselves to the numerical calculations which occur in Trigonometry, but also to those who intend to pursue the study of the higher mathematics. The examples are very numerous and are carefully selected. Many are placed in immediate connection with the subject-matter which they illustrate. The numerical solution of triangles has received much attention, each case being treated in detail. The iii IV PREFACE. examples at the ends of the chapters have been care- fully graded, beginning with those which are easy, and extending to Jiose which are more and more diifi- cult. These examples illustrate every part of the sub- ject, and are intended to test, not only the student's knowledge of the usual methods of computation, but his ability to grasp them in the many forms they may assume in practical applications. Among these exam- ples are some of the most elegant theorems in Plane and Spherical Trigonometry. The Chapters on De Moivre's Theorem, and Astron- omy, Geodesy, and Polyedrons, will serve to introduce the student to some of the higher applications of Trigonometry, rarely found in American textr-books. In writing this book, the best English and French authors have been consulted. I am indebted especially to the works of Todhunter, Casey, Lock, Hobson, Clarke, Eustis, Snowball, M'Clelland and Preston, Smith, and Serret. It remains for me to express my thanks to my col- leagues. Prof. R. W. Prentiss for reading the MS., and Mr. I. S. Upson for reading the proof-sheets. Any corrections or suggestions, either in the text or the examples, will be thankfully received. E. A. B. RuTGEUs College, New Brunswick, N. J., April, 1892. TABLE OF CONTENTS. -ooj»;c PART I. PLANE TRIGONOMETRY. CHAPTER I. Measurement op Angles. ART. TkQ^ 1. Trigonometry 1 2. The Measure of a Quantity 1 3. Angles 2 4. Positive and Negative Angles 3 5. The Measure of Angles 3 6. The Sexagesimal Method 5 7. The Centesimal or Decimal Method 6 8. The Circular Measure 9. Comparison of the Sexagesimal and Centesimal Measures . . 8 10. Comparison of the Sexagesimal and Circular Measures 9 11. General Measure of an Angle il 12. Complement and Supplement of an Angle 12 Examples 13 CHAPTER II. ' The Trigonometric Functions. 13. Definitions of the Trigonometric Functions 16 14. The Functions are always the Same for the Same Angle 18 15. Functions of Complemental Angles 20 16. Representation of the Functions by Straight Lines 20 17. Positive and Negative Lines 23 V VI CO J^ TENTS. ART. PAQK 18. Functions of Angles of Any Magnitude 23 19. Changes in Sine as tlie Angle increases from 0° to .'JOO^ 26 20. Changes in Cosine as the Angle increases from 0" to 360°, . . 26 21. Changes in Tangent as the Angle increases from 0° to 360°. 27 22. Table giving Changes of Functions in Four Quadrants 28 23. Relations between the Functions of the Same Angle 29 24. Use of the Preceding Fonnulae 30 25. Graphic Method of finding the Functions in Terms of One . . 30 26. To find the Trigonometric Functions of 45° 31 27. To find the Trigonometric Functions of 60° and 30° 31 28. Reduction of Functions to 1st Quadrant 33 29. Functions of Coraplemental Angles 34 30. Functions of Supplemental Angles 34 31. To prove sin (90° + A) = cos A, etc 35 32. To prove sin (180° + A) = - sin A, etc 35 33. To prove sin ( — A) =^ — sin A, etc 36 34. To prove sin (270° + A) = sin (270° - A) = - cos A, etc. ... 36 36. Table giving the Reduced Functions of Any Angle 37 36. Periodicity of the Trigonometric Functions 38 37. Angles corresponding to Given Functions 39 38. General Expression for All Angles with a Given Sine 40 39. An Expression for All Angles with a Given Cosine 41 40. An Expression for All Angles with a Given Tangent 41 41. Trigonometric Identities 43 Examples 44 CHAPTER III. Trigonometric Functions of Two Angles. 42. Fundamental Formulae 50 43. To find the Values of sin (x + y) and cos (x + y) 50 44. To find the Values of sin (x — y) and cos (x — y) 52 45. Formulae for transforming Sums into Products 65 46. Useful FormulsB 56 47. Tangent of Sum and Difference of Two Angles 57 48. Formulae for the Sum of Three or More Angles 58 49. Functions of Double Angles 60 50. Functions of 3 x in Terms of the Functions of a; 61 61. Functions of Half an Angle 63 62. Double Values of Sine and Cosine of Half an Angle 63 il PAGE 50 60 62 65 56 68 60 61 63 63 i = a, m cos (/» = ft 128 84. To solve a sin ^ + & cos = c 129 85. To solve sin (« -f a;) = m sin a; 131 80. To solve tan (a + x) = m tan x 132 87. To solve tan (a + x) tan x = ?m 133 88. To solve m sin (d + x)= a, m sin ((p + x)—. h 134 89. To solve x cos a + y sin a = «t, x sin u — y cos « = n 135 90. Adaptation to Logarithmic Computation ■ 135 91. To solve 7'C0S(/> cos^ = a, rcos0 sin = b, rsiui^ = c 137 92. Trigonometric Elimination 138 Examples 140 CHAPTER VI. Relations between the Sides of a Triangle and the Functions OF its Angles. 93. FormuliB 146 94. Right Triangles 140 95. Oblique Triangles — Law of Sines 147 90. Law of Cosines 148 97. Law of Tangents 149 98. To prove c = a cos B + ?> cos A 149 99. Functions of Half an Angle in Terms of the Sides 150 100. To express the Sine of an Angle in Terms of the Sides 152 101. Expressions for the Area of a Triangle 153 102. Inscribed Circle '. 154 103. Circumscribed Circle 154 104. Escribed Circle 165 105. Distance between the In-centre and the Circumcentre 155 106. To find the Area of a Cyclic Quadrilateral 157 Examples 169 CHAPTER VII. Solution op Triangles. 107. Definitions 165 108. Four Cases of Right Triangles 165 CONTENTS. IZ ART. 109. 110. 111. PAOK .... 12(5 112. .... 128 11.3. .... 129 114. .... 131 115. .... 132 no. .... 133 117. ... 1.34 118. ... 135 119. ... 135 120. ... 137 1 I'-il- ... 1.38 :':' 122. ... 140 123. 124. 1 125. 12(5. NCTIONS 127. ... 140 ... 146 ..147 .. 148 ( .. 149 128. ..149 ! 129. ..150 i 130. ..152 131. .. 153 ' 1.32. ..154 : . . 154 133. . . 155 l 134. ..155 135. ..157 136. . 159 137. 165 PAOI Case I. — Given a Side and the Ilypotenufie 1(5(5 Ca.se II. — (iiven an Acute Angle and the llypotenuHe 1(57 Case III. — Given a Side and an Acute Angle 168 Case IV. — Given the Two Sides 160 When a Side and the Hypotenuse are nearly Kcjual 1(59 Four Cases of Oblique Triangles 172 Case I. — Given a Side and Two Angles 172 Case II. — Given Two Sides and the Angle opposite Gneof them, 173 Case III. — Given Two Sides and the Included Angle 176 Case IV. — Given the Three Sides 177 Area of a Triangle 180 Heights and Distances — Definitions 181 Heights of an Accessible Object 182 Height and Distance of an Inaccessible Object 182 An Inaccessible Object above a Horizontal Plane 184 Object observed from Two Points in Same Vertical Line .... 185 Distance between Two Inaccessible Objects 186 The Dip of the Horizon 186 Problem of Pothenot or of Snellius 188 Examples 189 CHAPTER VIII. Construction of Looauithmic and Tkioonometric Tables. Logarithmic and Trigonometric Tables 204 Exponential Series 204 Logarithmic Series 206 Computation of Logarithms 207 Sind and tan 6 are in Ascending Order of Magnitude 208 The Limit of -— is Unity 209 e Limiting Values of sin 6 and cos d 209 To calculate the Sine and Cosine of 10" and of 1' 211 To construct a Table of Natural Sines and Cosines 213 Another Method 213 138. The Sines and Cosines from 30° to 60" 214 139. Sines of Angles Greater than 45° 216 140. Tables of Tangents and Secants 216 141. Formulae of Verification 216 142. Tables of Logarithmic Trigonometric Functions 217 143. The Principle of Proportional Parts 218 X CONTENTS. ART. PAGK 144. To prove the Rule for the Table of Common Logarithms . . . 218 145. To prove the Rule for the Table of Natural Sines 219 146. To prove the Rule for a Table of Natural Cosines 219 147. To prove the Rule for a Table of Natural Tangents 220 148. To prove the Rule for a Table of Logarithmic Sines. .' 221 149. To prove the Rule for a Table of Logarithmic Cosines 222 150. To prove the Rule for a Table of Logarithmic Tangents 222 151. Cases of Inapplicability of Rule of Proportional Parts 223 152. Three Methods to replace the Rule of Proportional Parts . . . 224 Examples 226 CHAPTER IX. De Moivre's Theorem. — Applications. 163. De Moivre's Theorem 229 p 164. To find all the Values of (cos e + V-l sin ey 231 155. To develop cos nd and sin 7i6 in Powers of sin 6 and cos 0. . . . 233 166. To develop sin d and cos 6 in Series of Powers ot d 234 167. Convergence of the Series 235 168. Expansion of cos" 6 in Terms of Cosines of Multiples of ^. . . 235 159. Expansion of Bin"0 in Terms of Cosines of Multiples of d. . . 236 160. Expansion of sin" 6 in Terms of Sines of Multiples of ^ 237 161. Exponential Values of Sine and Cosine 238 162. Gregory's Series 239 163. Euler's Series 240 164. Machin's Series 241 166. Given sin0 = x sin (^ + a) ; expand in Powers of a; 242 166. Given tan x = n tan e ; expand x in Powers of n 242 167. Resolve x" — 1 into Factors 243 168. Resolve x" + 1 into Factors 244 169. Resolve x"" - 2 r" cos 5 + 1 into Factors 245 170. De Moivre's Property of the Circle 247 171. Cote's Properties of the Circle 248 172. Resolve sin 6 into Factors 248 173. Resolve cos $ into Factors 260 174. Sum the Series sin .' + sin( a+ j8) + etc 261 175. Sum the Series ce ^, + cos(rt + $)+ etc 252 176. Sum the Series sin™ a + sin'"(rt + )9) + etc 252 177. Sum the Series sin « - sin(a + /3) + etc 254 178. Sum the Series cosec e + cosec 2d + cosec 4 fl + etc 254 i Tssfosm^sram COJf TENTS. XI PAGE 218 219 219 220 221 222 222 223 224 226 ABT. PACK Q 179. Sum the Series tan 9 + ^ tan - + \ tan - + etc 256 2 4 180. Sum the Series sin « + a; sin ( a+ 3) + etc — 255 181; Summation of Infinite Series 250 Examples 267 , 0, ^ The circular measure is employed in the various branches of Analytical Mathe- matics, in which the angle under consideration is almost always expressed by » letter. Note 2. — The student cannot too carefully notice that unless an angle is obvi- ously referred to, the letters a, /3, ..., 0, ' ■.,■;■■,:■ 180 TT ,;^ '■..'",:. ' .M) = l^and^ = ^. 7^ 180 : . . 10 PLANE TRIGONOMETRY. m I I EXAMPLES. 1. Find the number of degrees in the angle whose circu- 1 ' ' ■-■ lar measure is | Here e = 2 TT 2 TT 90x7 22 " = 28° 38' 10"iA where ^ is used for TT. 2. Find the circular measure of the angle 59° 52' 30". • Express the angle in degrees and decimals of a degree thus: : •• 60 )52.5 : 59.875 ... ^=5i^_p^ = (.333...) 7r=-« .0453..-. 3. Express, in degrees, the angles whose circular measures TT TT TT TT 2 2' 3' 4' 6' ^ NoTB 1.— The student should ospeoially accustom himself to express readily iu ciruular measure ud uugle which is given ia d'^grees. 4. Express in circular measure the following angles 60°, 22° 30', 11° 15', 270°. Ans. IT TT IT 3: 3' 8' 16' 2 5. Express in circular measure 3° 12', and find to seconds the angle whose circular measure is .8. ,. ; . 4: T /^TakeTT^—.' Ans. — , 45° 49' 5"fV. 225 ^^ 6. One angle of a triangle is 45°, and the circular measure of another is 1.5. Find the third angle in degrees. Ans. 49° 5' 27"^^. Note 2. — Questions in which angles are expressed in different systems of meas- urement are easily solved by expressing each angle in right angles. .-_^i GENERAL MEASURE OF AN ANGLE. 11 7. The sum of the measure of au angle in degrees and twice its measure in radians is 23^; find its measure in degrees (tt = ^y? ) . Let the angle contain x right angles. '' " ^" Then the measure of the angle in degrees = 90 x. it t( l( ic a (( a radians = X. .: 90a; + «■« = 23^; .-. 90 a: + v-^ 163 ~ 7 ' .-. 652a; = 163, .-. a;: _1 4 the angle is i of 90° = 221°. 8. The difference between two angles is -, and tlieir sum is 56° ; find the angles in degrees. Arts. 38°, 18°. 11. General Measure of an Angle. — In Euclidian geom- etry and in practical applications of trigonometry, angles are generally considered to be less than two right angles ; but in the theoretical parts of mathematics, angles are treated as quantities which may be of any magnitude what- ever. Thus, when we are told that an angle is in some particu- lar quadrant, say the second (Art. 5), we kuow that the position in which the revolving line stops is in the second quadrant. But there is an unlimited number of angles having the same final position, OP. The revolving line OP may pass from OA to OP, not only by describing the arc ABP, but by moving through a whole revolution plus the arc ABP, or through any number of revolutions plus the aro ABP. For example, the final position of OP may represent geometrically all the fol- lowing angles : 12 PLANE TRIGONOMETRY. Angle AOP = 130°, or 360° + 130°, or 720° + 130°, or - 360° + 130°, or - 720° + 130°, etc. Let A be an angle between and 90°, and let n be any lohole number, positive or negative. Then (1) 271 X 180° + A represents algebraically an angle in the first quadrant. (2) 2w X 180° — A represents algebraically an angle in the /owrf/i quadrant. " (3) (2n + 1) 180° — A represents algebraically an angle in the second quadrant. (4) (2 n + 1 ) 180° + A represents algebraically an angle in the f/iiVd quadrant. . . , In circular measure the corresponding expressions are (1) 2nir+d, (2) 2mr-e, (3) {2n + l)TT-6, (4) (2n + l)7r+^. EXAMPLES. .. ■!• • State in which quadrant the revolving line will be after d??scribing the following angles : (1) 120°, (2) 340°, (3) 490°, (4) - 100°, (5) _ 380°, (6) Itt, (7) IOtt + IT 12. Complement and Supplement of an Angle or Arc. — The complement of an angle or arc is the remainder obtained by subtracting it from a right angle or 90°. The supplement of an angle or arc is the remainder obtained by subtracting it from tivo right angles or 180°. Thus, the complement of A is (90° — A). The complement of 190° is (90° - 190°) = -100°. The supplement of A is (180° - A). The supplement of 200° is (180° - 200°) = - 20°. The complement of f tt is ( ^ — ^tt j = The supplement of \ir is {j — \Tr)=\ir. |t- m\ 1 ' EXAMPLES. 13 or EXAMPLES. 1. If 192 square inches be represented by the number 12, what is the unit of linear measurement ? Ans. 4 inches. 2. If 1000 square inches be represented by the number 40, what is the unit of linear measurement ? Ans. 5 inches. 3. If 2000 cubic inches be represented by the number 16, what is the unit of linear measurement ? Ans. 5 inches. 4. The length of an Atlantic cable is 2300 miles and the length of the cable from England to France is 21 miles. Express the length of the first in terms of the second as unit. Ans. lOO^. 5. Find the measure of a miles when b yards is the unit. . 1760a Ans. b 6. The ratio of the area of one field to that of another is 20 : 1, and the area of the first is half a square mile. Find the number of square yards in the second. Ans. 77440. 7. A certain weight is 3.125 tons. What is its measure in terms of 4 cwt.? Ans. 15.625. Express the following 12 angles in centesimal measure : ' 8. 42° 15' 18". Ans. m 95\ 9. 63° 19' 17". 70B35^70^\98.... ned 10. 103° 15' 45". 1148 73V g^vv 1 11. 19° 0'18". 218 UN 66^\6. ider 12. 143° 9' 0". 1598 5^55^5. V 13. 300° 15' 58". 3338 62^ 90^\l234567896. 14. 27° 41' 51". 308.775. 00°. 15. 67°.4325. 748.925. ', .. 16. 8° 15' 27". 98l7^50^ 20°. 17. 97° 5' 15". 1078 87^ 50^\ 18. 16° 14' 19". 188 4^29^^.... 19. 132° G'. 1468 7r77^\7. 14 PLANE TRIGONOMETRY Express the following 11 angles in degrees, minutes, and seconds : 20. 105K 52^ 75^\ 21. 82« 9^54^ 22. 70K 15^ 92^\ 23. 158 0^15^ 24. 1548 7N24^\ 25. 324«13^88^7. 26. lOe 42^ 50^\ 27. 208 77^50^ 28. 8«75\ 29. 170M5^35^ 30. 248 o^25^\ Ans. 94° 58' 29".l. 73° 53' 9".096. 63° 8'35".808. 13° 30' 4".86. 138° 39' 54".576. 291° 43' 29".9388. 9° 22' 57". 18° 41' 51". 7° 52' 30". ■. 153° 24' 29".34. ■ 21° 36' 8".l. ' Express in circular measure the following angles : • ' 14537r 31. 315°, 24° 13'. 32. 95° 20', 12° 5' 4". 33. 221°, r, 57°.295. 34. 120°, 45°, 270°. 35. 360°, 3^- rt. angles. Ans. ^TT, 10800 143 TT 2719 TT 270 ' 40500* TT TT 8' 180' 1 radian. ir 2.09439, |, f TT. 27r, Itt. Express in degrees, etc., the angles whose circular meas- ures are : ■ 36. f '^' I'r' 9* 07 1 1 2 ^^' 4' 6' 3' 5 38. -, .7854. 6 Ans. 112°.5, 120°, ?? degrees. 45, 30, 120, — degrees, — degrees, degrees. .TT TT TT 47°43'38"3^, 45°. il EXAMPLES. 15 39. ^, j\w, 2.504. A,is. 257° 49' 43".39, 15°, 143°.468. 40. .0234, 1.234, |. o 1°20'27", 70° 42' 11", 38° 11' 50" 41. Find the number of radians in an angle at the centre of a circle of radius 25 feet, which intercepts an arc of 37^ feet. Ans. 1^. 42. Find the number of degrees in an angle at the centre of a circle of radius 10 feet, which intercepts an arc of 57rfeet. ■ ■ ■ ' - ' ■ Ans. 90°. 43. Find the number of right angles in an angle at the centre of a circle of radius 3^ inches, which intercepts an arc of 2 feet. . , j^jig^ 4^^ 44. Find the length of the arc subtending an angle of i^ radians at the centre of a circle whose radius is 25 feet. , ^ Ans. 112|ft. 45. Find the length of an arc of 80° on a circle of 4 feet radius. Ans. 5f^ ft. 46. The angle subtended by the diameter of the Sun at the eye of an observer is 32' : find approximately the diameter of the Sun if its distance from the observer be 90 000 000 miles. Ans. 838 000 miles. 47. A railway train is travelling on a curve of half a mile radius at the rate of 20 miles an hour : through what angle has it turned in 10 seconds ? Ans. 6^ degrees. 48. If the radius of a circle be 4000 miles, find the length of an arc w^ich subtends an angle of 1" at the centre of the circle. Ans. About 34 yards. 49. On a circle of 80 feet radius it was found that an angle of 22° 30' at the centre was subtended by an arc 31 ft. 5 in. in length : hence calculate to four decimal places the numerical value of the ratio of the circumference of a circle to its diameter. Ans. 3.1416. 50. Find the number of radians in 10" correct to four sig- nificant figures (use ff^f for tt). Ans. .00004848. 16 PLANE TRIGONOMETRY. H CHAPTER 11. THE TEIGONOMETEIO FUNCTIONS. 13. Definitions of the Trigonometric Functio.ns. — Let EAD be an angle; in AD, one of the lines containing the angle, take any point B, and from B draw BC perpendicular to the other line AR, thus forming a right triangle ABC, right-angled at C. Then denot- ing the angles by the capital letters A, B, C, respectively, and the three sides opposite these angles by the corresponding small italics, a, h, c,* we have the following definitions : opposite side a_ c hypotenuse h __ adjace nt side c hypotenuse is called the sine of the angle A. is called the cosine of the angle A. - = ^f — ~ is called the tangent of the angle A. adjacent side h_ adjacent side a opposite side c _ hypotenu se b adjacent side c_ hypotenuse a opposite side is called the cotangent of the angle A. is called the secant of the angle A. is called the cosecant of the angle A. If the cosine of A be subtracted from unity, the remain- der is called the versed sine of A. If the sine of A be sub- * The letters a, b, c are numbers, being the number of times the lengths of the sides contain some chosoo unit of length. TRIGONOMETRIC FUNCTIONS. 17 1 A. - tracted from unity, the remainder is called the coversed sine of A ; the latter term is hardly ever used in practice. The words sine, cosine, etc., are abbreviated, and the func- tions of an angle A are written thus : sin A, cos A, tan A, cot A, sec A, cosec A, vers A, covers A. The following is the verbal enunciation of these defini- tions: The sine of an angle is the ratio of the oj^posite side to the hypotenuse; orsinA = -' c The cosine of an angle is the ratio of the adjacent side to the hypotenuse ; or cos A = — ■ /' c The tangent of an angle is the ratio of the opposite side to the adjacent side; or tan A = a The cotangent of an angle is the ratio 6f the adjacent side to b the opposite side; or cot A = a The secant of an angle is the ratio of the hypotenuse to the adjacent side ; orsecA = — ..,.,' .\v h -:'" ;; ■• ^ ' ■■ •■ ■■ ■" ' ■' .'■■ ' ■'• The cosecant of an angle is the ratio of the hypotenuse to the opposite side; or cosec A = a The versed sine of an angle is unity minus the cosine of the angle; or vers A = 1 — cos A = 1 • c I ii The coversed sine of an angle is unity mini, i the sine of the a angle; or covers A = 1 - sin A = 1 These ratios are called Trigonometric Functions. The student should carefully commit them to memory, as upon them is founded the whole theory of Trigonometry. These functions are, it will be observed, not lengths, but 18 PLANE TRIGONOMETRY. ratios of one length to anotlier ; that is, they are abstract numbers, simply numerical quantities ; and they remain unchanged so long as the angle remains unchanged, as will be proved in Art. 14. It is clear from the above definitions that :M .n cosec A X or sinA = sin A' sec A 1 cos a' or cosA = tan A 1 — . . > or cot A = cot a' cosec A' 1 sec a' tan A The powers of the Trigonometric functions are expressed as follows; (sin A)2 is written sin^ A, (cosA)^ is written cos^A, >^>; ' ' -> and so on. Note, — The student must notice that • sin A ' is a single symbol, the name of a number, or fraction belonging to the angle A. Also sin^A is an abbreviation for (sin A)', i.e., for (sin A) x (sin A). Such abbreviations are used for convenience. 14. The Trigonometric Functions are always the Same for the Same Angle. — Let BAD be any angle ; in AD take P, P', any two points, and draw PC, P'C per- pendicular to AB. Take P", any point in AB, and draw P"C" per- pendicular to AD. A C C P Then the three triangles PAC, P'AC, P"AC" are equi- angular, since they are right-angled, and have a common angle at A : therefore they are similar. " AP AP'"~ AP"* But each of these ratios is the sine of the angle A. Thus, sin A is the same whatever be the position of the point P on either of the lines containing the angle A. \ i: FUNCTIONS OF COMPLEMENTAL ANGLES. 19 Therefore sin A is always the same. A similar proof may be given for each of the other functions. In the right triangle of Art. 13, show that a = c sin A = c cos B = & tan A = b cot B, b — a cot A = tt tan B = c cos A = c sin B, c = a cosec A = a sec B = 6 sec A = 6 cosec B. Note. —These results should be carefully noticed, as they are of frequent use in the solution of right triangles and elsewhere. EXAMPLES. 1. Calculate the value of the functions, sine, cosine, etc., of the angle A in the right triangles whose sides a, b, c are respectively (1) 8, 15, 17; (2) 40, 9, 41 ; (3) 196, 315, 371 ; (4) 480, 31, 481 ; (5) 1700, 945, 1945. Ans. (1) sin A = -^-j, cos A = \^, tan A = -^-^f etc. ; , (2) sin A = 1^, cos A = ^\, etc. ; (3) sinA = ff, tanA = |f, etc.; ; (4) sin A = f ff , tan A = -\\% etc. ; (5) sin A = III, tan A = f|f, etc. In a right triangle, given : 2. a = Vm^+ n% b = V2 mn ; calculate sin A. Ans. 3. a — -\/'m?—mn, b = n\ calculate sec A. 4. a = Vw^+ mn, c=m+n; calculate tan A. -y/ 5. a = 2 miij b = m^—n^; calculate cos A. m + n m — n n mn +n^ m" 7J/ 6. sin A = 1^, c = 200.5 ; calculate a. 7. cos A = .44, c = 30.5 ; calculate b. 8. tan A = -y-, b = ^; calculate c. 120.3. 13.42. T^rViSO. vj f 20 PLANE TRIGONOMETRY. 15. Functions of Complemental Angles. — In the rt. A ABC we have siu A = -, and cosB = -'• (Art. 13.) c c \ .'. sin A = cosB. But B is the comphMuent of A, since their sum is aright angle, or 90°; i.e., A B = 90°-A. .-. sinA =cosB = cos (90°— A) = Also, cos A sinB = sin (90°- A) = ", a tan A =cotB = cot (90°- A) = -, cot A =tanB = tan (90°- A) =-, (a/ sec A =cosecB = cosec (90°— A) =^, cosec A = secB = sec (90°— A) = -, vers A = covers B = covers (90° — A) =r 1 — -, covers A = vers B = vers (90° — A) = 1 — Therefore the sine, tangent, secant, and versed sine of an angle are equal respectively to the cosine, cotangent, cosecant, and coversed sine of the comi^lement of the angle. * 16. Representation of the Trigonometric Functions by Straight Lines. — The Trigonometric functions were for- merly defined as being certain straight lines geometrically connected with the arc subtending the angle at the centre of a circle of given radius. Thus, let AP be the arc of a circle subtending the angle AOP at the centre. I REPRESENTATION OF FUNCTIONS BY LINES. 21 Draw the tangents AT, BT' meeting 01* produced to T', and draw PC, VD ± to OA, OB. Then « cotangent (( « secant (I (( - cosecant i( ft versed sine (( (I ■ , . CO versed sine (I PC was called the sine of the arc AP. DC , " cosine " AT « tangent " BT' OT OT' AC BD * ■ Since any arc is the measure of the angle at the centre which the arc subtends (Art. 5), the above functions of the arc AP are also functions of the angle AOP. It should be noticed that the old functions of the arc above given, when divided by the radius of the circle, become the modern functions of the angle which the arc subtends at the centre. If, therefore, the radius be taken as unity, the old functions of the arc AP become the modern functions of the angle AOP. Thus, representing the arc AP, or the angle AOP by 6, we have, when A = OP = 1, ;:fl '. 1' 22 PLANE THIOONOMETRY. sintf =^^ = ^-^ = PC, I m PC or PC 1 AT AT and similarly for the other functions. Therefore, in a circle whose radius is unity, the Trigono- metric functions of an arc, or of the angle at the centre meas- ured by that arc, may be defined as follows : The sine is the perpendicular let fall from one extremity of the arc upon the diameter passing through the other extremity. The cosine is the distance from the centre of the circle to the foot of the sine. TJie tangent is the line which touches one extremity of the arc and is terminated by the diameter produced passing through the other extremity. The secant is the portion of the diameter produced through one extremity of the arc ivhich is intercepted between the centre and the tangent at the other extremity. The versed sine is the part of the diameter intercepted between the beginning of the arc and the foot of the sine. Since the lines PD or OC, BT', OT', and BD are respect- ively the sine, tangent, secant, and versed sine of the arc BP, which (Art. 12) is the complement of AP, we see that the cosine, the cotangent, the cosecant, and the coversed sine of an arc are respectively the sine, the tangent, the secant, and the versed sine of its complement. EXAMPLES. 1. Prove tan A sin A + cos A = sec A. 2. 3. 4. 6. 6. (( B' will be deter- mined if we know the distance of the point from O, and if we know also upon which side of O the A— point lies. It is therefore con- venient to employ the algebraic signs + and — , so that if dis- tances measured along the fixed line OA or OB from O in one direction be considered positive, distances measured along OA' or OB' in the oppo- site direction from will be considered negative. This convention, as it is called, is extended to lines parallel to A A' and BB'; and it is customary to consider distances measured from BB' towards the right and from A A' upivards as positive, and consequently distances measured from BB' towards the left and from AA' doivmvards as negative. 18. Trigonometric Functions of Angles of Any Magni- tude. — In the definitions of the trigonometric functions given in Aru. 13 we considered only acute angles, i.e., angles in the first quadrant (Art. 5), since the angle was assumed to be one of the acute angles of a right triangle. We shall now show that these definitions apply to angles of any mag- nitude, and that the functions vary in sign according to the quadrant in which the angle happens to be. I 24 PLANE TRIGONOMETRY. Let AOP be an angle of any mag- nitude formed by OP revolving from an initial position OA. Draw PM ± to AA'. Consider OP as always positive. Let the angle AOP be denoted by A ; then whatever be the magnitud'! of the angle A, the defini- tions of the trigonometric functions are sin A sec A MP . OM , cosA= , OP' OP ' tan A = OP OM' cotA = OM MP' cosecA = I. When A lies in the 1st quadrant, MP is positive because measured from M iqnvards, <^^M is positive because measured from O towards the right (Art. 17), and OP is positive. Hence in the first quadrant all the func- tions are positive. II. When A lies in the 2d quadrant, as the oh; use angle AOP, MP is positive because meas"red from M u^nvards, OM is negative becf ase measured from towards the left (Art. 17), and OP is positive. Hence in the second quadrant sin A = — — is positive ; cos A OM --— - IS negative : OP ^ ' ■ . MP . tsin A =^~: IS negative ; OM •' ' and therefore sec A and cot A are negative, and cosec A is positive (Art. 13). FUNCTIONS OF ANGLFS. 25 III. When A lies in the 3J quadrant, as the reflex angle AOP, MP is negative because measured from M downwards, OM is negative, and OF is jjositive. Hence in the third quadrant the sine, cosine, secant, and cosecant, are negative, but the tangent and cotangent are positive. IV. When A lies in the 4th quadrant, as the reflex angle AOP, MP is negative, OM is positive, and OP is positive. Hence in the fourth quadrant the sine, tangent, cotangent, and cosecant are negative, but the cosine and secant are positive. The signs of the different functions are shown in the annexed table. Quadrant. I. II. • III. IV. Sin and cosec + + — Cos and sec- + — — + Tan and cot — + — NOTB. — It is apparout from tliis table tbat tbe 8ij;nB of all the functions in any quadrant are known wlien those of the sine am! cosine are known. The tangent and cotangent are + or — , according as the sine and cosine have like or different signs. 19. Changes in the Value of the Sine as the Angle in- creases from 0° to 380°. — Let A de- note the aiigle AOP described by the revolution of OP from its initial posi- tion OA through 3G0°. Then, PM being drawn perpendicular to AA', sin A = MI!, OP whatever be the magnitude of the angle A. 26 PL A NE TRIGONOMETR Y. \\ I • When the angle A is 0°, P coincides with A, and MP is zero; therefore sin 0° = 0. As A increases from 0° to 90°, MP increases from zero to OB or OP, and is positive; therefore sin 90°= 1. Hence in the 1st quadrant sin A is positive, and increases from to 1. As A increases from 90° to 180°, MP decreases from OP to zero, and is p)ositive; therefore sin 180° = 0. Hence in the 2d quadrant sin A is positive, and decreases from 1 to 0. , :; \< ■■ As A increases from 180° to 270°, MP increases from zero to OP, and is negative; therefore sin 270° = — 1. Hence in the 3d quadrant sin A is negative, and decreases algebraically from to — 1 . As A increases from 270° to 3G0°, MP decreases from OP to zero, and is nega4ive ; therefore sin 3G0° = 0. Hence in the 4tli quadrant sin A is negative, and increases algebraically from — 1 to 0. f 20. Changes in the Cosine as the Angle increases from 0° to 360°. — In the figure of Art. 19 cosA = OM op' When the angle A is 0°, P coincides with A, and OM = OP ; therefore cosO°= 1. As A increases from 0° to 90°, OM decreases from OP to zero and is j^ositive; therefore cos 90°= 0. Hence in the 1st quadrant cos A is positive, and decreases from 1 to 0. As A increases from 90° to 180°, OM increases^from zero to OP, and is negative; therefore cos 180° =— 1.' ' Hence in the 2d qunlrant cos A is negative, and decreases algebraically from to — 1. As A increases from 180° to 270°, OM decreases from OP to zero, and is negative ; therefore cos 270° = 0. t CHANGES IN THE TANGENT. 27 Hence in the 3d quadrant cos A is negative, and increases algebraically from —1 to 0. As A increases from 270° to 360°, OM increases from zero to OP, and is positive; therefore cos 360°= 1. Hence in the 4th quadrant cos A is positive, and increases from to 1. 21. Changes in the Tangent as the Angle increases from 0° to 360°. — In the figure of Art. 19 tan A = — -• OM When A is 0°, MP is zero, and OM = OP; therefore tan 0°=0. As A increases from 0° to 90°, MP increases from zero to OP, and OM decreases from OP to zero, so that on both accounts tan A increases numerically; therefore tan90° = QO. Hence in the 1st quadrant tan A is positive, and increases from to cc. As A increases from 90° to 180°, MP decreases from OP to zero, and is positive, OM becomes negative and decreases algebraically from zero to — 1 ; therefore tan 180°= 0. Hence in the 2d quadrant tan A is negative, and increases algebraically from -co to 0. When A passes into the 2d quadrant, and is only just greater than 90*^, tan A changes from +co to — oo. As A increases from 180° to 270°, MP increases from zero to OP, and is negative, OM decreases from OP to zero, and is negative; therefore tan 270°= oo. Hence in Liie 3d quadrant tan A is positive, and increases from to 00. As A increases from 270° to 360°, MP decreases from OP to zero, and is negative, OM increases from zero to OP, and is /)osrt*'»;e; therefore tan 360°= 0. Hence in the 4th quadrant tan A is negative, and increases algebraically from — oo to 0. 28 PLANE TRIGONOMETRY. w The student is recommended to trace in a manner similar to the above the changes in the other functions, i.e., the cotangent, secant, and cosecant, and to see that his results agree ith those given in the following table. 22. Table giving the Changes of the Trigonometric Functions in the Four Q,uadrants. Quadrant. I. II. III. IV. sill varies from + Oto 1 + 1 too Oto - 1 -1 too cos " " + 1 toO to - 1 - 1 too + Oto 1 tan " " + to CO -00 toO to 00 -00 toO cot " " + 00 toO oto 00 4- 00 to to — 00 sec " " + 1 to 00 — 00 to — 1 — 1 to — 00 00 to 1 cosec " " + 00 to 1 + 1 to 00 — 00 to — 1 — 1 to — 00 vers " " + oto 1 4 1 to 2 + 2tol 1 toO 'i Note 1. — The cosecant, secant, and cotangent of an angle A have the same sign as the sine, cosiiie, and tangent of A respectively. The sine and cosine vary from 1 to — 1, passing through the value 0. They are never greater than tinity. The secant and cosecant vary from 1 te — 1, passing through the value oo. They are never mimerically less than unity. The tangent and cotungent are unlimited In value. They have all values from — w to +00. The versed sine and coversed sine vary from to 2, and are always positive. The trigonometric functions change sign in passing through the values and », and through no other values. In the 1st quadrant the functions increase, and the coJ\tnctiona decrease. Note 2. — From the results given in the ahove table, it will be seen that, if the value of a trigonometric function be Riven, we cannot fix on one angle to which it belongs exclusively. Thus, if the given value of sin A be ' .ve know since sin A passes through all values from to 1 as A increases from i to 90°, that one value of A lies between 0° RELATIONS BETWEEN FUNCTIONS. 29 and 90°. But since we also know that the value of sin A paaaes through all values between 1 and aa A increases from 90"^ to 180°, it is evident that there is another ♦ value of A between 90° and 180° for which sin A = J, 23. Relations between the Trigonometric Functions of the Same Angle. — Let the radius start from the initial position OA, and revolve p^ in either direction, to the position OP. Let 6 denote the angle traced out, and let the lengths of the sides PM, MO, OP be denoted by the letters a, b, c* The following relations are evident from the definitions (Art. 13) : cosec = — — , sec 6 = , cot $ = tan 6 = sin^ sin cos^ cos^ ta,nO For tan^: a b a c l c sin » cos 6 IL sin2^ + cos2^ = L For sin'^ + cos^e IIL sec''^=l + tan2 ^ + ^ = ^' + &' For sec^^=^^ = *-^ = l + ^^=l + tan»^. IV. cosec''^ = l + cot2 6. For cosec*^=— = a^ + ft' a' a" = l+-T=l-fC0t2ft a* Formulae I., IL, IIL, IV. are very important, and must be remembered. * a,b,c are numbers, being the number of times the lengths of the sides contain some chosen uait of length. i ll If! I li 80 PLANE TRIGONOMETRY. 24. Use of the Preceding Formulee. 1. To express all the other functions in terms of the sine. Since sin^^ + cos^^ = 1, .-. cos6 = ±Vl sin^ . sin^ sin^d. tan 6 = cot^ = sec^: cosec 6 ■■ cos 6 1 tand 1 cos 6 1 sind = ± = ± VI- sin- ^ VI - sin2^ sin^ 1 Vl-sin^^ II. To express all the other functions in terms of the tan- gent. sin ^ Since tan B = cos ^ sine = tand cos 6 = ^^ = ± tan^ cose = ^ cote = cosec e = sec^ 1 tan^ 1 sin 6 = ± seed 1 Vl + tan^d Vl + tan^d seed = ± Vi +"tai?^. = ± Vl+tan^d tan $ Similarly, any one of the functions of an angle may be expressed in terms of any other function of that angle. The sign of the radical will in all cases depend upon the quadrant in which the angle lies. 25. Graphic Method of finding All the Functions in Terms of One of them. ^B To express all the other functions in tei'ms of the cosecant. Construct a right triangle ABC, hav- ing the side BC = 1. Then a V coaeca-l RELATIONS BETWEEN FUNCTIONS. 31 . AB AB .^ cosec A = - — - = -~ = AB. Now •. AC = ± Vcosec''* A — 1. smA = — — - = AB cosec a' A AC , Vcosec^A — 1 cos A = — = ± ~ , AB cosec A tanA = ?^^ ^ AC Vcosec'^A — 1 and similarly the other functions may be expressed in terms of cosec A. 26. To find the Trigonometric Functions of 46". — Let ABC be an isosceles right triangle in which g CA = CB. Then CAB = CBA = 45°. Let AC = m = CB. Then :2 T-7^2 AB' = AC 4- CB' = m" + m^ = 2 m\ .-. AB = mV2. sin45° = 5C = AB cos45° = 4^ = m m V2 V2 m 1 AB .„iV2 V2 tan45°=?5 =- = L cot45°=l. AC 771 sec45°=V2. cosec45°=V2. 27. To find the Trigonometric Func- tions of 60° and 30°. —Let AB be an equilateral triangle. Draw AD perpen- dicular to BC. Then AD bisects the angle BAC and the side BC. Therefore BAD = 30°, and ABD = 00°. B fn 1 • ■ < j 1 '^ 1 4[ i' Ill: 32 PLANE TRIGONOMETRY, Let BA = 2m. .-. BD = m. Then AD = V4 mi^ — m'^ = »?i V3. • • 3in60°-^^-^;/^-W3. .-. cosec60° ^ AB 2 m V3 cos60"=.^^ = ^. .-. sec60° = 2. BA 2 tan60°-^|^-^^^-V3. .-. cot60°_ 1. BD m yg vers«0°= 1 - cos 60°= 1-1 = 1. ^ ' V " ' Also sin30° = ^ = ^ = ^. .-. cosec30°=2. r AB 2m 2 cos30°=-— — = — ^ — — iv3. .-. sec30°= • AB 2 m ^3 tan30°— ^"^— '"*' = ^ . • oc^'\(\°^-k/^ DA ™ a/:^ a/'? ; EXAMPLES. 3 1. Given sin ^ = - ; find the other trigonometric functions. Let BAG be the angle, and BC be perpendicular to AC. Represent BC by 3, AB by 5, and consequently AC ^ by V26^^ = 4. AC 4 Then cos^ AB 5' *-^=Ic = 3 .e. = f-«. 5 = 4' cosec0 = || = 5 "3' REDUCTION OF FUNCTIONS. 33 3 3 5 2. Given sin 6 = -: find tan 6 and cosec 6. Aiis. -) -• 6 4 3 3. Given cos 9 = --, find sin 6 and cot 0. o 4. Given sec^ = 4; find cot ^ and sin ^. 5. Given tan 6 = V3 ; find sin (9 and cos 0. '"'^' 12 6. Given sin ^ = t^ ; find cos^. 7. Given cosec ^ = 5 ; find sec 6 and tan B. ' 2V2 1 Vlii VTs' 4 W3,| _5 13* 5 2V() 2VG 40 J)^ 4l' 41' V5 3 3 8. Given sec = — ; find sin 6 and cot 0. 2 9. Given cot 6 = — ; find sin 6 and sec 0. V5 3 10. Given sin = - ; find cos 6, tan 6, and cot ^. , ^ V7 3V7 V7 .:;,-\,.,.. .-^ X' "T"' T"' 6 11. Given sin = - ; find tan 6. c 12. Given sin0 = 2 mil ; find tan^. 2 m?i m-— 11- 28. Eeduction of Trigonometric Functions to the Ibt Quadrant. — All mathematical tables give the trigonometric functions of angles between 0° and 90° only, but in practice we constantly have to deal with angles greater than 90°. The object of the following six Articles is to show that the trigonometric functions of any angle, positive or negative, can be expressed in terms of the trigonometric functions of an angle less than 90°, so that, if a given angle is greater than 90°, we can find an angle in the 1st quadrant whose trigonometric function has the same absolute value. V -« 84 P,LANE TRiaONOMETliy. 1 t I'll iii \ 'a ■r,i , 29. Functions of Complemental Angles. — Let AA', HB' be two diaiuuters of a circle at right angles, and let OP and OP' be the positions of the radins for b any angle AOP = A, and its comple- ment A0P'=1)()°- A (Art. 12). Draw PM and P'M' at right angles J to OA. Angle OP'M'= B0P'= AOP = A. Also 0P= OP'. Hence the triangles 0PM and OP'M' are equal in all respects. P'M' ^ OM ~ op" P'M'= OM. 01 jf Also, sin (90° - A) = cos AOP = cos A. OM' ^ PM ~0P* OM' = PM. 01 Jf cos (90° - A) = sin AOl' = sin A. Similarly, tan (90°- A) = tan AOP'= :!-^ = 1±!^ = cot A. The other relations are obtained by inverting the above. 30. Functions of Supplemental An- ^ gles, — Let OP and OP' be the positions of the radius for any angle AOP = A, p and its supplement AOP'=180°-A ,/ (Art. 12). Since OP = OP', and POA = P'OA', the triangles POM and P'OM' are geometrically equal. P'M' PM .-. 8in(180°-A)=sinAOP'=-i-^=^ = sinA, OP cos (180° - A) = cos AOP'= OM' OP' OM OP = — cos A, REDUCTION OF FUNCTIONS. 35 tan (180°- A) = tan AOr'= ^-^ = ~ = - tan A. OM' OM Similarly the other relations may bo obtained. 31. To prove 8in(00°+A) = co8A, cos (00° + A) = — sin A, and tan (90° + A) = - cot A. Let OP and OP' be the positions of the radius for any angle AOP = A, and AOP'=90°+A. Since OP = OP', and AOP = P'OB = OP'M', the triangles POM and P'OM' are equal in all respects. .-. sin (90°+ A) = sin AOP'== -^ = ^^ = cos A, cos (90°+ A) = cos AOP'= ^' = — — = - sin A, ^ ^ OP' 01' . tan(90°+A) = tanA0P'=-^'=-^^^ = -cotA. ^ ^ OM' -PM 32. To prove sin ( 180° + A) = - sin A, cos (180°+ A) =- - cos A, and tan (180° + A) = tanA. Let the angle AOP = A ; then the angle AOP', measured in the positive direction, = (180°+ A). p The triangles POM and P'OM' are equal. .-. sin (180°+ A) = sin AOP' = P'M' OP' OM' cos (180°+ A) =cos AOP' = ^^ = tan (180°+ A)=tanAOP' = OP' P'M' OM' - OM 36 PL A NE TRIG ONOMETR T. l:,l > 1 33. To prove sin ( — A) = — sin A, cos ( — A) = cos A, tan ( — A) = — tan A. B Let OP and 01" be the positions of the radius for any tuiual angh>s AOP and AOP' measured from the initial line AO in opposite directions. Then if the angle AOP be denoted by A, the numerically equal angle AOP' will be denoted by — A (Art. 4). The triangles POM and P'OM are geometrically equal. .-. sin(-A) = .sinAOP' = PMVt OP'^ PM OP = — sin A, cos ( — A) = cos AOP' = —~- = — — = cosA, tan ( - A) = tan AOP' = ^^^ = ~-^^ = - tan A. ^ ^ OM OM 34. To prove 8in(270° + A) = sin(270°- A) = -co8A, and cos (270° + A) = - cos (270°- A) g = 8inA. Let the angle AOP = A ; then the / angles AOQ and AOR, measured in the positive direction, =(270°— A) ^\ and (2^0°+ A) respectively. The triangles POM, QON, and EOL are geometrically equal. ^ EL = QX = OM. ^- = QN^-J)M OE OQ OP Also, .-. sin (270°+ A) = sin (270°- A) = - cos A. ; OL -ON PM OE .-. cos (270° + A): OQ OP - cos (270°- A) = sin A. h VALUES OF THE FUNCTIONS. 37 35. Table giving the Values of the Functions of Any Angle in Terms of the Functions of an Angle less than 90". — The foregoing results, and other similar ones, which may be proved in the same manner, are here collected for reference. Quadrant II. sm(180°-A)= sinA. sin(90°+A)= cosA.' cos(180°- A) = — cosA. tan (180°- A) = - tan A. cot (180°- A) = - cot A. sec (180°- A) = - sec A. cosec (180°— A) = cosec A. cos(90°+ A) = -sinA. tan(00°+ A) = -cotA. cot(9()°-f A)=.-tanA. sec (1)0° + A ) = — cosec A. cosec (90° -f A ) = sec A. Quadrant III. sin ( 180°+ A) = cos (180°+ A) = tan (180°+ A) = cot (180°+ A) = sin A. cos A. tan A. cot A. sec A. sin (270°- A) = cos (270°- A) = tan (270°- A) = cot (270°- A) = sec (270°- A) = sec (180°+ A) = cosec (180°+ A) = - cosec A. cosec (270°- A) = — cos A. — sin A. cot A. tan A. — cosec A. — sec A. Quadrant IV. A) = -sinA. sin(270°+ A) = -cosA. cos (270°+ A) = sin A. tan(270°+ A) = -cotA. cot (270°+ A) = - tan A. sec (270°+ A) = cosec A. sin (SCO' cos (360°- A) = cos A. tan(360°- A) = -tanA. cot(360°- A)=:— cotA. sec (360°- A) = sec A. cosec (360°— A) = — cosec A. cosec (270°+ A) = — sec A. Note. — These relations* may be remembered by noting the followinp: rules : When A \» associated with an ei'en multiple of 90°, any function of the angle is numerically equal to the same, function of A. When A is associated with an odd multiple of 90°, any function of the angle is uumericaliy equal to the corresponding cofunction of the angle A. The sign to be prefixed will depend upon the quadrant to which the angle belongs (Art. 5), regarding A as an acute angle. * Although these relations have been pro%'ed only in case of A, an acute angle, they are true whatever A may be. I) i 88 PLANE TRIGONOMETRY. ;i ■ ij il ■ hi ' !l! i ■ li I 'i ThuB, cos (270° -A) = — sinA; the angle 270° — A being in the 3d quadrant, and its cosine uegiUivc in consequence. For an angle in the i^^irst qiittdrant all the functions are positive. Second quadrant all are negative except the sine and cosecant. Third quadrant all are negative except the tangent and cotangent. J Fourth quadrant all are negative except the cosine and secant. 36. Periodicity of the Trigonometric Functions. — Let AOP be an angle of any magnitude, as in the figure of Art. 18 ; then if OP revolve in the positive or the negative direction through an angle of 360°, it will return to the position from which it started. Hence it is clear from the definitions that the trigonometric functions remain un- changed when the angle is increased or diminished by 360°, or an}' multiple of 360°. Thus the functions of the angle 400° are the same both in numerical value and in algebraic sign as the functions of the angle of 400°— 360°, i.e., of the angle of 40°. Also the functions of 360°+ A are the same in numerical value and in sign as those of A. In general, if w denote any integer, either positive or negative, the functions of n x 360°+ A are the same as those of A. Thus the functions of 1470°= the functions of 30°. If denotes any angle in circular measure, the functions of (27i7r + 0) are the same as those of 0. Thus sin (2w7r + d) = s'mO, cos (2 wtt -\-6) = cos 0, etc. By this proposition we can reduce an angle of any magni- tude to an angle less than 360° without changing the values of the functions. It is therefore unnecessary to consider the functions of angles greater than 360° ; the formulae already established are true for anyles of any magnitude whatever. EXAMPLES. Express sin 700° in terms of the functions of an acute angle. sin 700°= sin (360°+ 340°) = sin 340°= sin (180°+ 160°) = sinl60°=~sin2()°. l«Mi •«»« EXAMPLES, 39 Express the following functions in terms of the functions of acute angles : 1. sin 204°, sin 510°. Aiis. -sin 24°, sin 30°. 2. cos (-800°), cos 359°. cos 80°, cosl°. G. tan 500°, tan 300°. -tan 40°, -cot 30°. Find the value of the sine, cosine, and tangent of the following angles : 4. 150°. 5. - 240°. C. 330°. • 7. 225°. Ans. iV3, iV3, 1 "2' 1 V3' -V3. 2 V3 ._JL _1_ 1 V2 V2 Find the values of the following functions : 8. sin 810°, sin (-240°), cos 210°. ^><6-. 1, |V3, -^VS. . 9. tan (-120°), cot 420°, cot 510° 10. sin 930°, tanG420°. 11. cot 1035°, cosec570°. V3, -^, 1. V3 __1 J_ 2' V3* -1, -2. 37. Angles corresponding to Given Functions. — When an angle is given, we can hud its trigonometric functions, as in Arts. 26 and 27 ; and to each value of the angle there is but one value of each of the functions. But in the converse proposition — being given the value of the trigonometric functions, to find the corresponding angles — we have seen (Art. 36) that there are many angles of different magnitude which have the same functions. If two such angles are in the .same quadrant, they are represented geometrically by the same position of OP, so that they differ by some multiple of four right angles. i 40 PLANE TRIGONOMEJRY. ^il ri^'i If we are given the value of the sine of an angle, it is important to be able to find all the angles which have that ' value for their sine. , , . 38. General Expression for All Angles which have a Given Sine a. — Let O be the centre of the unit circle. Draw the diameters AA', BB', at right angles. From draw on OB a line ON, so that its measure is a. Through N draw PP' parallel to AA'. Join OP, OP', and draw PM, P'M', perpendicular to AA'. Then since MP = M'P' = ON = a, the sine of AOP is equal to the sine of AOP'. Hence the angles AOP and AOP' are supplemental (Art. 30), and if AOP be denoted by a, AOP' will = tt - «. Now it is clear from the ligure that the only 2>ositive angles which have the sine equal to a are « and tt — a, and the angles formed by adding any multiple of four right angles to a and tt — «. Hence, if 6 be the general value of the required angle, we have . ;- 6 = 2mr-^a, or 6 = 2mr -\-7r — a, .... (1) where n is zero or any positive integer. Also the only negative angles which have the sine equal to a are — (7r + «)j and — {2v — n), and the angles formed by adding to these any multiple of four right angles taken negatively ; that is, we have e=27i7r-(Tr + a), 6 = 2mr - {2ir - a), . . (2) where n is zero or any negative integer. Now the angles in (1) and (2) may be arranged thus: 2rnr + a, (2n + l)7r — «, (2n — l)7r — «, {2n—2)ir + a, all of which, and no others, are included in the formula ^ = «7r + (-!)"«, (3) GENERAL EXPRESSION FOR ANGLES. 41 where n is zero, or any positive or negative integer. There- fore (3) is the general expression for all angles which have a given sine. Note. — The same formula dctertnineB all the angles wbicb have the same cosacant as a. 39. An Expression for All Angles with a Given Cosine a. — Let be the centre of the unit circle. g Draw AA', BB', at right angles. From O draw OM, so that its meas- ure is a. Through M draw PP' parallel to BB'. Join OP, OP'. Then since OM — a, the cosine of AOP is equal to the cosine of AOP'. Hence, if AOP = «, A()P'=-«. Now it is clear that the only angles which have the cosine equal to a are a and — a, and the angles which differ from either by a multiple of four right angles. Hence if 9 be the general value of all angles whose cosine is a, we have 6 = 2mr±a, "- ' ' ; where n is zero, or any positive or negative integer. Note. — The same formula determines all the angles which have the same secant or the same versed sine as a. 40. An Expression for All Angles with a Oiven Tangent a. — Let O be the centre of the unit .-p circle. Draw AT, touching the circle at A, and take AT so that its measure is a. Join OT, cutting the circle at PA'j- and P'. Then it is clear from the figure that the only angles which have the tan- ^ gent equal to a are a and tt -f a, and the angles which differ 42 PLANE TRIGONOMETRY. w ; I from either by a multiple of four right angles. Hence if 6 be the general value of the required angle, we have = 2 iiTT + «, and 2 mr + tt + «. (1) Also, the only negative angles which have the tangent equal to a are — (tt — «), and — (27r — «), and the angles which differ from either by a multiple of four right angles taken negatively ; that is, we have ^ = 27177 — (tt —«), and 2mr — {2tt — a), . . (2) where n is zero or any negative integer. Now the angles in (1) and (2) may be arranged thus : 2n7r + «, (2?l + l)7r4-«, (2?J — l)7r4-«, (2?J-2)7r + «, all of which, and no others, are included in the formula ^ = W7r + «, (3) where n is zero, or any positive or negative integer. There- fore (3) is the general expression for all angles which have a given tangent. Note. — The same formula determines all the angles which have the same cotan- gent as a. EXAMPLES. 1, Find six angles between — 4 right angles and -f 8 right angles which satisfy the equation sin A = sin 18°. We have from (3) of Art. 38, e = n7r + {-!)"—, or A = «xl80°+(-l)»18°. Put for n the vahu's — 2, — 1, 0, 1, 2, 3, successively, and we get A = - 360° + 18°, - 180° - 18°, 18°, 180° - 18°, 360°+ 18°, 640°- 18°; that is, 342°, -198°, 18°, 162°, 378°, 522° NoTK. — The student should draw a figure in the above example, and in each example of this Liud which he workt). TRIGONOMETRIC IDENTITIES. 43 2. Find the four smallest angles which satisfy the equations (1) sinA = -, (2) sinA = — , (3) sinA= ---' ^ 2 -y/2 ■" (4) siuA = -- Ans. (1) 30°, 150°, - 210°, - 330° (2) 45°, 135°, -225°, -315° (3) 00°, 120°, - 240°, - 300° (4) -30°, -150°, 210°, 330°. 41. Trigonometric Identities. — A trigonometric identitj'^ is an expression which states in the form of an equation a relation which is true for all values of the angle involved. Thus, the relations of Arts. 13 and 23, and all others that may be deduced from them by the aid of the ordinary formulae of Algebra, are universally true, and are therefore called identities; but such relations as sin ^ = 4, cosO=i, are not identities. ' • EXAMPLES, 1. Prove that sec 6 — tan 6 • sin = cos 6. Here see $ — tan^ sin 6 = ' sin d cos 6 cos ^ l-sin' 'g cos 6 cos^d (Art. 24) (Art. 24) cosd = cos 6. 2. Prove that cot ^ — sec ^ cosec ^ (1 — 2 sin- 6/) = tan B. cot^ — seed cosecd(l — 2 sin^^) cos^ 1 1 sin^ cos^ sin ^ cos'^g-l + 2sin'-tf sin ^ cos 61 (l-2sin-^) (Art. 24) 44 PLANE TRIGONOMETHY, l! Ill l':i! ''\ ^! cos^^j-(siii- ^ + cos- ^1+2^^ sill cos ^ (Art. 24) sin^ sin 9 cos ^ siii^ cos = tan ^. Note. — It will be observed that in Bolving Uieee examples we first express the other fiinctioiiH in tcrnia of the sine and coHine, und in most cases the beginner will find this the siinplest course. It is generally advisable to begin with the most com- pliculed side and work towards the otlier. Prove the following identities : o. cos 6 tan ^ = si n ^. 4. cos^ = sin^ cot0. 5. (tan^ + cot^),sin^cos^ = l. 6. (taii^ — cot6) sin^cos^ = sin'*^ — cos^^. 7. siii^ B -=- cosec- B = sin* B. 8. sec" B - tan-* B = sec- B + tan^ B. 9. (sin B - cos f))-= 1-2 sin B cos B. 10. l-tan*^ = 2sec-^-sec*^. 11. l+S'^ = (eosec B-\-cotB)\ (> 12. (sin^ + o.ohB)--\- (sin^ - cos 5)-= 2 13. sin"^-cos"^=siir6'- cos-^. 14. sin- ^ + vers- ^::= 2(1- cos e). 15. cot'^e-cos2^ = cot2dcos2d. EXAMPLES. In a right triangle Al^C (see tignre of Art. 15) given 1. (( = P' + ptji c = 3 3 5' r V7 _3_ I Z*^ 1 sin A = — ; find cos A and tan A. \-, V3 >'3' V3 cos A = b; find tan A and cosec A. vn^'^ 1 sin A =.6; find cos A and cot A. tan A = - ; find sin A. 5 8 » cot A = — ; find sec A and sin A. 15 12 sin A = — ; find cos A. 13' cos A = .28; find sin A. tan A =-; find sin A. 3' sin A =-; find cos A. -» ' VI - b' 4 4 5' 3 4 V41 8 15 17' 17 6 13 .96 . 4 fV2. ^ I' ! I - 46 PLANE TRIGONOMETRY. 21. Given tanA = ,^; lind sinA and secA. Atis. -, -• 22. 23. " taiid =-; tind sin^ and cosd. b a (( COS0 =-; find sin^ and cot 6. a 6. ■V¥^l 1 24. If sin ^ = a, and tan ^ = 6, prove that (l-a^)(l + &^) = l. Express the following functions in terms of the functions of acute angles less than 45° : 25. sin 168°, sin 210°. 26. tan 125°, tan 310°. 27. sec 244°, cosec281°. 28. sec 930°, cosec (-600°). 29. cot 460°, sec 299°. 30. tan 1400°, cot (-1400°). Ans. sin 12°, -sin 30° -cot 35°, -cot 40° — cosec 26°, — secll° -sec 30°, sec 30° -tan 10°, cosec 29° -tan 40°, cot 40° Find the values of the following functions : V3 1 31. sin 120°, sin 135°, sin 240°. Ans. 32. cos 135°, tan 300°, cosec 300° V3 V2' V2 -V3, V3 33. sec315°, cot330°, tan 780° 34. sin 480°, sin 495°, sin 870° 36. tan 1020°. sec 1395°, sin 1486° V2, -V3, V3. V3 1 1 2 ' V2 2 -V3, V2, -K- V2 EXAMPLES. 47 36. sin (-240°), cot (-675°), cosec (- 690°). Ans. ^, 1, 2. 37. cos(-300°), cot(-316°), cosec (- 1740°). 1 *> 1 ■" 2\ va 38. tair^660°, cos" 1020°. -3V3, -- Find the value of the sine, cosine, and tangent of the following angles : 39. -300°. 40. - 135°. 41. 750°. 42. -840°. 43. 1020°. 44. (2n + l)^-|. 45. (2?i-l),r + v A V3 1 Alls. - ■x/.S ' , 1 TT^l V2 V2 1 V3 1 2' 2 ' V3 _V3 _1 /- "2 ' 2' "^"^ V3 1 /- ~T'' 2' "^^ V3 _1 _ /. 2 ' 2' ^"^ 1 V3 1 2 ' V3 Trove, drawing a separate figure in each case, that 46. sin340°=sin(-160°). , 47. sin (-40°) = sin 220°. 48. cos320°=-cos(140°). A9. cos (-380°)= -cos 560°. 50. cos 195°= - cos ( - 15°). 61. cos380°=-cos660°. rr 48 PLANE TRIGONOMETRY. 1 • ;' ! ill jiiiKj 1^ 41 m > I 52. • cos ( - 225°) = - cos ( - 45"). 53. cos 1()0«° =. - cos 1 185°. 54. Draw an angle whose sine is -• 55. " " « « cosecant is 2. 56. " « " * tangent is 2. 57. Can an angle be drawn whose tangent is 427 ? 58. 59. (( u U <( (( (( (( 9 — Wii TT 7 ' 27 TT 7 (1) 275°; (2) -91°; (3) -193°; (4) (5) -1000°; (6) 2mr-\- 350°; Sir + ; («) +, -, - Ans. (1) -,+,-; (2) -,-,+; (3) + (4) +,+,+; (5) +, +, Prove the following identities : 62. (mi'd-\-Gos'e)-=l. 63. (sin^^ - (iOH'eyr= 1 - 4cos=^^ -j- 4cos*d. 64. (sin e + cos 0)-= 1 + 2 sin cos 0. 65. (sec^-tan6')(sec0 + tau^) = l. 411 EXAMPLES. 49 66. (cosec $ — cot B) (cosec ^ + cot ^) = 1. * 67. aio!^d-\-Gos'^d = {suid-\-cosd){i-sm6cos$). 68. sin''^ + cos«^ = sm^0 + cos*^-sin-^cos''^. 69. 8iii=* $ tan'' d + cos=^ cot^ 6 = tan- 6 + cot'^ 6-1. 70. sin ^ tan''^+ cosec^ sgc-6= 2 tan ^ sec^— cosec ^4- sin 0. 71. cos''^ — sin''d:=(cos^-sine)(l + sin^cosd). , 72. sin«^ + cos"^ = l-asin2^cos2d. 73. tan« + tan/? = tanrttan/3(cot«4- tJot/8). 74. cot « + tan jS = cot « tan /3 (tan « + cot /8) . 75. 1 — sin« = (l +sin«)(sec«— tan«). 76. 1 4-cos« = (1 — cos«)(cosec« — cot«). » 77. (1 4-sin« + cos«)^=2(l + sin«)(l -f cos«). 78. (1 — sin« — co8a)^(l + sin«+cos«)*= 4sin''«cos*«. 79. 2 vers a — vers^« = sin^t. . , • 80. versa (1 + cos a) = sinVx. 50 PLANE TRIGONOMETRY. mt 1 ' |nj ' i ' 1 )• m CHAPTER III. TRIGONOMETEIO FUNOTIONS OF TWO ANGLES. 42. Fundamental Formulee. — We now proceed to express the trigonometric functions of the sum and difference of two angles in terms of the trigonometric functions of the angles themselves. The fundamental formulix, first to be established are the following : «- sin (a; + y) = sin a; cos y + cos x sin y . ... {\\ cos(a + 2/) = cosa;cos2/ — sinxsiny . . . . (2) ^sin (a — 2/) = sinajcost/ — cosiKsiny . . . . (3) cos (a; — y) = cosxcosy -f sinajs'ny . . . . (4) NoTK. — Here x and y are iinglcs; bo that {x + y) and {x — y) are also angles. Hence, Bin (x + y) is the Bine pf an angle, and is not the same as Bin x + Bin y. Sin (X + 2/) is a single fraction. Bin X + Bint/ is the sura of two fractions. 43. To prove that sin (a; 4- y) = sin x cos y-\-Q,osx sin y, and cos {x-\-y) = cos x cos y ~ sin x sin y. Let the angle AOB = x, and the angle BOC = y j then the angle AOC = a; + y. In OC, the bounding line of the angle (x-i-y), take any point P, and draw PD, PE, perpendicular to A and OB, respectively ; draw EH, EK, perpendicular to PD and OA. !: FUNDAMENTAL FORMULA. 61 Then angle Eril = mf- REV = HEO = AOE = x, . , . , DP EK + l'H EK^PH 8m(xH-2/)=— = -——— = -- + -^ J 4^ OP OP _ek oe*. ph pe "oe'op pe'op '" ••''" . " s= sin a; cos 2/ + cos a; sin y. , . , OD OK -HE OK HE ^OK OE_HE PE ..<•>'•. ■■ ""OE 'op PE'OP '^^^ ' ; . ' = cos X cos y — sin x sin y. Note.— These two formr.lw have been obtained by a construction In which x + y is an acute angle; but the proof is perfectly general, and applies to iingieH of any magnitude whatever, by paying due regard to the algebraic signs. For example, Let AOB = a;, as before, and BOC=y; v g then AOC, measured in the positive direc- tion. Is the angle x + y. In OC, take any point P, »nd draw PD, PE, perpendicular to O A and OB produced ; draw Ed and EK perpendicular to PD and OA. Then, angle EOK = IW-a;; KPH==KOK = 180°-x; and COE=y-180°. DP EK - PH sin (x + y) = — OP OP EK^PH OP OP EK OE ^ PH PE ' OE * OP PE ' OP = - sin (180'^ - X) cos (y - 180°) + cos (180° - x) sin (y - 180°) = sin X cos y + cos x sin y. (Art. 36) '^ "' OP OP OP OP ^OK OE EH PE OE ' OP >E OP * The introduced line OE is the only line in the figure which Is at once a side of two right triangles (OEK and OEP) into which EK and OP enter. A similar remark applies to PB. r^' 1 6J2 PLANE TRIGONOMETRY. = COB (180° X) COB (y - 180") + sin (180=' - x) sin {y - 180'*) = C0BXC0B^ — Binxsiny. (Art. 35) The student should notice that the words of the two proofs are very nearly the Mme. 44. To prove that sin (x — y) = sin x cos y — cos x sin ?/, and cos (x — y) = cos x cos y + s\nx sin y. Let the angle AOB be denoted by x, and COB by y ; then the angle AOC = x-y. In OC take any point P*, and draw PD, PE, perpendicular to OA and OB respectively; draw EH, EK, perpendicular to PD and O OA respectively. Then the angle EPH = 90° - HEP = BEH = AOB = x. -. . DP EK - HP EK HP sin { X — v ) = — = = — ^ ^^ OP OP OP OP ^EK (2E_HP PE oe'op PEOP = sin .13 cos 2/ — cos 05 sin y. noc^^ ,A ^^I^ OK + EH OK^EH cos (a; — y) = — = ' = ^ -^^ OP OP OP OP ^OK OE EH PE oe'op PE OP = cos X cos y -f sin x sin y. Note 1. — The sign in the ezprcBHion of the sine is the same as it is in the angle expanded ; in the cosine it is the opposite. * P is taken In the lin« boundtns the angle undsr consideration; i.«., AOO. 4VJ SINE AND COSINE. 53 NoTX 2. — In this proof the angle x -y \is acute ; but the proof, like the one given in Art. 43, upplies to angles of any magnitude whatever. For example, Let AOB, measured in tlie positive x direction, =x, and BOC = y. Tlien \ AOC = x-y Hi i.E In OC talce any point P, and draw O PD, PE, perpendicular to OA and OB produced: draw EH, EK, perpendicu- lar to DP and AO produced. Then, angle EPH = EOK = AOB = 360^- a?, and POE = 180° - y. OP OP co»(x-y) = - ^EK OE_HP_PE OK OP PE " OP = Bin (360° - X) cos (180° -y)- coi (SflO" - x) sin (180° - ^0 = (— sin 3!) ( - cos y) — cos xainy = sin X cos y — cos x nin y, OD OK + HK OP OP OK OE_HK_PE OE ' OP PEOP = - cos (360° - X) cos (180° - y) - sin (360° - x) lin (180° - y) = (— cos X) (— cos y) - (— sin x) sin y = cos X cos y + sin X sin y. NoTK 3. — The four fundnnientai formulae Just proved are very Important, and must be committed to memory. It will be convenient to refer to them as the ' x,y' formulie. From any one of them, all the others can be deduced in the following manner : Thus, from cos (x — y) to deduce sin (x + y) . We have COS (aj — y) = cosa;cosi/ + sinajsiny . . . . (1) Substitute 90"— x for x in (1), and it becomes cos 1 90°- (x + y) I = cos (90°- x) cosy + sin (90°- a;) siny. .'. sin (a; 4-y ) = sina;cos2/ + cosajsiny. (Art. 29) The student should make the substitutions indicated below, and satisfy himself that the corresponding i -ults follow : 64 PLANE TRIGONOMETRY. m i I li u t( it (t From sin (ic + y) to deduce cos {x + 2/) substitute (90°+ a;) for x. ti tt « cos{x — y) " (90° -a;) for a;. cos(a; + 2/) " " sin(x + y) " (90° + x) ^ or ». « sin (a; -2/) " (90°- ») for a;. « cos(a;-y) " —yioxy. etc. etc. etc. EXAMPLES. 1. To find the value of sin 15°. sinl5°=sin(45°-30°) = sin 45° cos 30° - cos 45° sin 30° . V2' 2 V2'2 ^ V3-l 2V2 V3 + 1 . 2V2 ' V3-1 2V2 * V3 + 1 2V2 2. Sh^w that sin 75°= 3. Show that cos 75°= 4. Show that cos 15°= 3 5 5. If sina; = \ and cosy = cos {x — y). 6. If sin X = -, and cos y = .^ cos (x—y). find sin (a; -f .v) and . 63 , 50 ^"*- 65' ""^ 65' find sin (a; -f y) and Ans. 1, and -~ • IVH^ ^ FORMULAE FOR TRANSFORMATION. 65 45. Formulae for the Transformation of Sums into Prod- ucts. — From the four fundamental formuUe of Arts. 43 and 44 we have, by addition and subtraction, the following: sin (a; 4- 2/) 4- sin (.1; — 2/) = 2 sin ic cosy . . . (1) sin (05 4-?/) — sin (a; — y) = 2 cos a; sin 2/ . . '. (2) cos {x 4- y) 4- cos (a; — y) = 2 cos x cos ?/ . . . (o) cos (x — y) — cos {x 4- 2/) = 2 sin x sin y . . . (4) These formulae are useful in proving identities by trans- forming products into terms of first degree. They enable us, when read from right to left, to repla(^e the product of a sine or a cofdne into a sine or a cosine by half the sum or half the difference of two such ratios. Let «4-2/ = A, and x — y = li. .-. a; = i(A + B), and y = ^(A-B). Substituting these values in the above formulae, and putting, for the sake of uniformity of notation, x, y instead of A, B, we get sina;4- sin2/ = 2sin^ (a;4-2/)cos^(a; — 2/) . . (5) sin X — sin y = 2 cos ^ (a; 4- y) sin ^{x — y) . . (6) cosa;4-cos2/ = 2cos } (a;4-y) cos^(.r — 2/) . • (7) C0S2/ — cosa; = 2sin ^ (a; + 2/) sin^(a; — y) . . (8) The formulae are of great importance in mathematical investigations (especially in computations by logarithms) ; they enable us to express the sum or the difference of two sines or two cosines in the form of a product. The student is recommended to become familiar with them, and to com mit the following enunciations to memory : Of any two angles, the Sum of the sines = 2 sin ^ sum -cos^difP. Diff. " " " =2cosisum.sinidiff. ~ T liii I U » 66 PLANE TRIGONOMETRY. Sum of the cosines = 2 cos ^ sum • cos J diff. Diff. " " " = 2 sin I sum. sin ^ diff. EXAMPLES. 1. sin 5 a; COS 3 a; = ^- (sin 8a; + sin 2 a;). For, sin5a;cos3a; = ^ |sin (5a; + 3a;) + sin (5a; — 3a;)| = ^ (sin 8 a; + sin 2 x) . 2. Prove sin ^ sin 3^ =^ (eos2e- cos4tf). 3. " 2sin^cos<^ = sin (^ + <^) + sin (^ — <^). 2sin2^cos3<^ = sin(2^-f 3<^)-f sin(2^-3<^). sin 60° + sin 30° = 2 sin 45° cos 15°. sin 40° - sin 10° = 2 cos 25° sin 15°. sinl0^+sin6^= 2sin8dcos2tf. sin 8 « — sin 4 a = 2 cos 6 a sin 2 a. sin3a; + sina; = 2 sin 2 a; cos a;. sin3a; — sina; = 2 cos 2 a; sin a;, sin 4 a; 4- sin 2 a; = 2 sin 3 x cos a;. 46. Useful Formnlee. — The following formulse, which are of frequent use, may be deduced by taking the quotient of each pair of the formulae (5) to (8) of Art. 45 as follows : 1 sina;-}- siny _ 2 sin^(a; -}-y) cos^(a; — y) sina;— sin y 2cos^(a;-|- y) sin^(x' — y) = tan ^(a; -|- ?/) cot ^(a; — y) = ^-^te±ll. (Art. 24) tan^(x — y) The following may be proved by the student in a similar manner : „ sin X -f sin y . i / , v 2. — -T '^ — tan ^ ( a; -f .?/ ) , cos X + cos y 5. 6. 7. 8. 9. 10. 11. THE TANGENT OF TWO ANGLES. 67 o sin a; + sin V 1.1/ \ cos y — cos X cosaj + cosy ' ^ sin X — sin w ^1. 1 / , \ cos 2/ — cos a; ^ cos X 4- cos V 1. 1 / , \ i. 1 / \ cos 2/ — cos x 47. The Tangent of the Sum and Difference of Two Angles. — Expressions for the value of tan (a* + y), tan {x — y), etc., may be established geometrically. It is simpler, however, to deduce them from the formulae already established, as follows : Dividing the first of the 'x, y^ formulae by the second, we have, by Art. 23, tan (X + y)= ^^^-^tl) = sin a; cos y 4- cos a; si n. y cos (a; + 2/) cosajcosy — sinajsiny Dividing both terms of the fraction by cos a; cosy, sin a; co sy cos a; sin y . . . cos a; cosy cos a; cosy tan (a; -f- y) = '- : :— i cos a; cosy sinajsiny cos a; cosy cos a; cos a; ^ taji^JJany ,^j.^ 23) . . . (1) 1 — tan X tan y In the same manner may be derived ta„(x-!,) = i?Sfn^ (2) 1 -f tan X tan y Also, cot(x + j,) = 55tf-«2^ (3) cotaj + coty and cot (a: - y) = ^-5t?-£5*y±l (4) coty — cotx 4 1 !■* Hi' \\ 58 PLANE rUlGONOMETRY. EXERCISES. Prove the following : 1. tan (a; + 45°) = ^^"^ "^^- 1 — tanaj tan x — 1 2. 3. 4. tan (x - 45°) = 1 4- tan X si n (x 4- y ) _ tan x -f- tan y sin {x — 2/) tan a; — tan y cos (a; — y ) __ ta n a; tan y-\-l cos (» -f y) 1 — tan X tan y 6. sin (a; + y) sin (x — y) = sin'' a; — sin''3/ = co&'y — cos' a;. 6. cos (a; -f y) cos (a; — y) = cos* a; — sin'''y = cos'y — sin'a: 7. tan«±tan2/=«i'li^2/). cos a; cosy 8. cota.±coty =«^"(2/±^). smajsiny 9. sin 2 a; cos 2a; — : = sec a;. sma; cos a; 10. If tan X = ^ and tany = \, prove that tan (a; + y) = f, and tan (x — y) = |. 11. Prove that tan 15° = 2 - V3. 12. If tan x= |, and tan y = ^, prove that tan (x -f y) = 1. What is (x -f y) in this case ? 48. FormulsB for the Sum of Three or More Angles. — Let X, y, z be any three angles ; we have by Art. 43, EXAMPLES. 69 sin (x + y + z) = sin (x -\- y) cos z -f cos (a; + y) sin z = sin a; cos y cos z + cos x sin y cos « + cos a; cos 2/ sin 2 — sin a; sin y sin 2 . . (1) In like manner, cos (a; + y + «) = cosa;co8y C0S2 — sin a; sin y cos z — sin a; cos 2/ sin 2 — cos a; sin y sin 2 . . (2) Dividing (1) by (2), and reducing by dividing both ms of the fraction by cos x cos y cos 2, we get tan(x+y+2)= tana; + tany + tan^-_^ "na- tany tan2 .3. 1 — tan X tan y — tan v tan 2 — tan 2 tan x 1 EXAMPLES. 1. Prove that sin a; + sin y + sin 2 — sin (a: + y + 2) = 4sin|(a; + y) sm^(y + 2) sin ^(2 + x). By (6) of Art. 44 we have sin a; — sin {x-\-y + 2) = — 2cos|(2a;-f y -I-2) sin^(2/ + 2), and siny 4-sin2 = 2sin^(y 4- 2)cos^(?/ — 2). .-. sin a; + sin 2/ + sin 2 — sin (a; 4- y + 2) =2sin|(y+2)cos|(y— 2)-2cos|(2x- :)s'm^{y+z) =2sin|(i/ + 2)|cos^(.V — 2)-cosi ''t ' y + z)l =28in^(y 4- 2) 2sin^(aj + y) sin |(a: -f 2) =4 sin i (a; 4- y) sin ^(y 4- 2) sin 1(2 -I- a;). Prove the following : 2. cos X 4- cos y 4- cos 2 + cos (a; 4- y 4- 2) = 4 cos ^ (y 4- 2) «os ^ (2 -I- a;) cos ^ (a; 4- y) • m i "I iiii ; ! '* til 60 PLANE TRIGONOMETRY. 3. 8iii(«-f y — 2) = sin a; cos y cos 2 + cos a; sin y cos 2 — cos X cos y si» 2 + sin a; sin y sin 2. 4. sina; + siny — sin2 — sin (aj-f- y — 2) = 4 sin ^ (a? — 2) sin ^ (y — 2) sin ^{x-}- y) . 6. sin (3/ — 2) 4- sin {z — x)-\- sin (x — y) + 48in^(.?/— 2) sin ^(2 — x) sin J (a; — y) = 0. 49. Functions of Double /Ingles. — To express the trigo- nometric functions of the angle 2x in terms of those of the angle x. Put y = a; in (1) of Art. 42, and it becomes sin 2 a; = sin a; cos x + cos x sin x, or sin 2 a; = 2 sin a; cos a; (1) Put y = a; in (2) of Art. 42, and it becomes cos2a; = cos*a; — sin'a; (2) = 1— 2sin-a; (3) or =2cos''a;— 1. (4) Put y = a; in (1) and (3) of Art. 47, and they become taii2a;= ^^^"^, (5) l-tan*a; ^ ^ cot2a; = ^^*'^~^ (6) 2cot« ^ ^ Transposing 1 in (4), and dividing it into (1), we have . " ".r ^^ tan a?. • • • • • • • • * lil l + cos2a; ^ ' EXAMPLES. 61 Note. — These ieven formulsp are very important. The •tudent muit notice that X it nny nnRJe, and therefore these formulee will be true whatever we put for x. Thus, If we write - for x, we got •ln«->2sin?eos£ (8) 3 2 ^ ' cos x== COB* --sin*- (9) 2 a or «=l-2sln»- = 2cos«?-l (10) and so on. EXAMPLES. Prove the following : 1. 2cosec2x =seca;coseca;. . . .; . 2. _.5^??^ =sec2x. cosec X — 2 o 2 tan a; . „ 3. — =sin2x. 1 + tan* a; . 1 — tan'a; „ 4. — = cos 2x. — 1 4- tan'a; 6. tan a; + cot a; = 2 cosec 2 a;. 6. cot X — tan x = 2 cot 2 a;. 7 sin a; . x 7. =tan-' 1 4- cos a; 2 Q sina; .x 8. ;; = cot-« 1 —cos a; 2 9. Given sin 45**=-^; find tan 22^**. Ans. V2-1. V2 10. Given tan a; = - : find tan 2 a;, and sin 2 a;. — , — 4' 7' 25 50. To Express the Functions of 3 a; in Terms of the Func- tions of X. Put y = 2a;in (1) of Art. 42, and it becomes V ■i '. 62 PLANE TRIGONOMETRY. 'ml 8in3a5 = siii (2a; + x) = sin2a;co8a/ -f- Q,os2xs\nx = 28ma;co82a;-f-(l — 2sin*a;)sina; (Art. 49) = 2sina;(l — 8in''a;) + 8ina; — 2sin'a; = 38inx — 48in''aj. cos 3a; = cos (2x-f-x) = cos 2 a; cos a; — sin 2 a; sin a; = (2 cos' a;— 1) cos a; — 2 sin* a; cos a; (Art. 49) = 4 cos' a; — 3 cos a;. tan3a; = tan 2a; + a; _ tan 2a; + tana; 1 — tan2a; tanx 2tana; 1 — tan' a; - -f- tan X 1 _ ^t3,n'a; 1 — tan' a; 3 tan a; — tan' a; 1 —3 tan' a; or \ i- ::! hi'li EXAMPLES. Prove the following : ^ sin3a; „ o ■ ^ 1. — = 2cos2aj-fl. sin a; o sin 3a; — sin a; iS. — = tan X. cos 3 a; -f- cos a; Q sin3a;4-cos3a; o • o -t 3. 2__ = 28in2a; — 1. cos » — sin a; t •''■■ FUNCTIONS OF TUB UALF ANGLE. 68 4. 1 4- ! = cot2x. tan 3x— tana; cota; — cot3a; 6. L=L^?^ = (i4.2cosa;)«. 1 — cos a; ' 51. Functions of Half an Angle. — To express the f unc- tions of - in terms of the functions of x. 2 Since cosa;= 1 — 2sin*-, 2 • or =2co82~-l ... [Art. 49, (10)] . « a; 1 — ros.T ,^v ••• 8»" 2 = ^ (1) and cos2- = -^!l^ (2) ^ . a; , /I —cosa; ,ov Or sin- = ±^ ^ (3) , a; , /I + cos X ... and cos- = ±-i/ — ^— (4) T> J- • • i. a; , /I— cosa; , 1— cosa; ,t!\ By division, tan - = ± \\- = ± . . (5) ^ ' 2 \l + cosa; sina; ^ ' By formulsB (8), (4), and (5) the functions of half an ngle may be found when the cosine of the whole angle is given. 52. If the Cosine of an Angle be given, the Sine and the Cosine of its Half are each Two-Valued. By Art. 51, each value of cosa; (nothing else being known about the angle x) gives \wo values each for sin- and cos-, one positive and one negative. But if the value of x be 64 PLANE TRIGONOMETRY. t ViTJii given, we know the quadrant in which - lies, and hence we know which sign is to be taken. Thus, if X lies between 0° and 360°, ^ lies between 0° and 180°, and therefore sin- is positive; but if x lies between 360° and 720°, - lies between 180° and 360°, and hen(!e sin- is negative. Also, if x lie between 0° and 180°, cos- is positive; but if x lie between 180° and 360°, cos- is negative. The case may be investigated geometncally thus : Let OM = the given cosine (radius being unity, Art. 16), = cos X. Through M draw PQ per- pendicular to O A ; and draw OP, OQ. Then all angles whose cosines are equal to cos a; are terminated either by OP or OQ, and the halves of these angles are terminated by the dotted lines Op, Oq, Or, or Os. The sines of angles ending at Op and Oq are the same, and equal numerically to those of angles ending at Or and Os ; but in the former ease they are positive, and in the latter, negative ; hence we obtain two, and only two, values of sin - from a given value of cos X. Also, the cosines of angles ending at Op and Os are the same, and have the positive sign. They are equal numeri- cally to the cosines of the angles ending at Oq and Or, but the latter are negative ; hence we obtain two, and only two, values of cos- from a given value of cos a;. Also, the tangent of half the angle whose cosine is given is two-valued. This follows immediately from (5) of Art. fil. Th FUNCTIONS OF THE HALF ANGLE. 66 53. If the Sine of an Angle be given, the Sine and the Cosine of its Half are each Four-Valued. We have and By addition, 2sin-co8- = 8iiia5 . 2 2 snr - 4- cos^ - = 1 . . sin - + cos - ] = 1 4- sin x. , 2 '-^ ' (Art. 49) (Art. 23) By subtraction, [ sin^ — cos*- | = 1 — sin X. sin- 4- cos- = ± Vl -f sin a; 2 2 and • X X sm — cos-: 2 2 ± Vl sin a; .-. 2sin^= ± Vl + 8inx± Vl — sina; 2 and 2 cos X ± Vl + sina; T Vl — sina; (1) (2) (3) (4) Thus, if we are given the value of sina; (nothing else being known about the angle x), it follows from (3) and (4) that sin- and cosp have each four values equal, two by two, in absolute value, but of contrary signs. T/te case may be inventigated geometrically thus : Let ON = the given sine (radius being unity) = sina;. Through N draw PQ parallel to OA ; b and draw OP, OQ. Then all angles whose sines are equal to sina; are terminated either by OP or OQ, and the halves of these angles are termi- nated by the dotted lines Op, Oq, Or, '* or 0.9. The sines of angles ending at Op, Oq, Or, and Os are all different « B' rr 66 PLANE TRIGONOMETRY. m ' ij I ^'? in value; and so are their cosines. Hsnce we obtain four values for sin-, and four also for cos-, in terms of x. When the angle x is given, there is no ambiguity in the calculations ; for - is then known, and therefore the signs and relative magnitudes of sin- and cos- are Known. Then equations (1) and (2), which should always be used, im- mediately determine the signs to be taken in equations (3) and (4). Thus, when - lies between — 45** and 4- 45**, cos- > sin -, and is ponitive. Therefore (1) is positive, and (2) is negative' and hence (3) and (4) become 2sin--= Vl + sina;— Vl — sin a?, 2cos- = Vl 4- sin x -f Vl — sin x. X X When - lies between 45** and 135°, sir. >cos-, and is 2 ' ^ ,2 positive. Therefore (1) md (2) are both positive; and hence (3) and (4) become 2 sin -' = Vl + sino; -f Vl — sina;, 2cos-= VI -f-sina;— VI — sina. And so on. 54. If the Tangent of 9:1 Angle be given, the TangeDt of it' Half is Tvo- Valued. We have tantf = 2 tan ^ 1-tan'^ 2 (Alt. 49) Bl'l I! m FUNCTIONS OF THE HALF ANGLE, 67 Put tan - = x\ thus 2 (1 - ar') tan d= 2a;, a^H -xss 1. tautf ., tan? = :c=Ill^^l-±^!-^. 2 . tantf Thus, given tan^, we find tico unequal vahies for tan-, one positive and one negative. This result may be proved geometrically, an exercise which we leave for the student. 55. If the Sine of an Angle be given, the Sine of One- Third of the Angle is Three- Valued. We have Put sin3x* = .'isinx — 4sin'''a; . . (Art. 50) jc = -, and we get o • 8intf = 38ln 4sin''-, :i 3 • a cubic equation, which therefore has three : ^obs. s i '4 EXAMPLES. ' 1. Determine the limits between which A must lie to satisfy tue equation 28inA = — Vl 4-sin2A — VI — sin2A. By (1) and (2) of Art. 63, 2 sin A can have this value only when sin A 4- cos A = — Vl + sin 2 A, aRil sinA — cosA = — Vl — sin2A; i.e., when sin A > cos A and relative. ; I . 68 PLANE TRiaONOMETR Y. • m'i Therefore A lies between 225"* and .'U5°, or between the angles formed by adding or subtracting any multiple of four right angles to each of these ; i.e., A lies between 2nir + — and 2n7r + ^, 4 4 where n is zero or any positive or negative integer. 2. Determine the limits between which A must lie to satisfy the equation 2cosA = VI + sin2A — vl — sin 2 A. By (1) and (2) of Art. 53, 2cosA can have this value only v;hen cos A + sin A = VI + sin 2 A, and cos A — sin A = — VI — sin 2 A i.e., when sin A > cos A and positive. Therefore A lies between 2n7r-f 7 and 2n7r + ^, 4 4 where n is any positive or negative integer. 3. State the signs of (sin d -}- cos 0) and (sin ^ — cos 6) when $ has the following values: (I) 22°; (2) 191°; (3) 290°; (4) 345°; (5) -22°; (0) -275°; (7) -470°; (8) 1000°. Ana. (1) +, -; (2) -, +; (II) -, _; (4) +, -; (5) +, -; («) +, +; (7) -, -; (8).-,-^. 4. Prove thjit the formul.T? which give the values of sin - and of cos- in terms of sinx are unaltered when x •> «> has the values ( 1 ) 92°, 268°, 900°, 4 htt -f | tt, or (4 n + 2) tt - | tt ; (2) 88°, - 88°, 770°, -770°, or 4nir ± 5. o VALUES OF 5. Find the limits betw» 2 sin A = Vl~-^ 56. Find the Values o: (4), aud (5) of Art. 61, siu22i°=^' cos22i 001° ♦ -- OMETRY. of 36°. , . . [(3) of Art. 49] , and cosine, respec- '7T, and 3x-=l08°.^ /^ ANGLES OF A TRIANGLE. 71 We have A + B + C = 180". .-. 8in(A-f-B) =sinC, and sin— ^- = cos^- (Arts. 15 and 30) ,, . A + li „ A-B '> QUI ' COS 2 2 Now 8inA + sinB=2sm— ■'^— -cos^^^^-jY"^ . (Art. 45) = 2 cos 5 cos ^^^^^ . . (Art. 15) 2 ^ and 8inC = 2sin5^eos^ . . . . (Art. 49) 2 2 = 2 COS ^-±^008^ . . (Art. 15) 2 2 ,.•. sin A 4- sinB + sinC = 2cos|cos^~- +2co8|cos^^ „ Of A - B . ^ A -f- B\ = 2cos-f cos — hcos — ^ J = 2cos^/'2cos|cos|y (Art. 45) .ABC ,1^ = 4cos-cos-cos- . . . . (1) Z M Z Again, cos A -f cos B = 2 cos -^ cos ^— . (Art. 45) 2 ^ and o . C A-B c=2sm^-cos— ^; cosC = l-2sin'^^ .... (Art. 49) .'. cosA+ cosB + cosC=l + 2sin-Yco3— j^ sin^ = l + 2sinf(oosAr-B_eosA±B) = l+4sm^smj;jsm5 . . (2) 2 <^ M 1 72 PLANE TRIGONOMETRY Again, tan(A + B) = -tanC (Art. 30) _ U nA -f - tan B . . . (Art. 47) 1 — tan A tan 13 ••• tan A + tanB = — tan C (1 — tan A tan B). . •. tan A -f tan B -f tan C = tan A tan B tan C . . . . (,*3) Note. — The ntuilcnt will observe that (1), (2), and (3) follow dlreclly from Kxttiiiplea 1 and 2, and formula (3), renpectivel^, of Art. 48, by puttiug A + B + C = 180°. IH ill :t- EXAMPLES. Prove the following statements if A + B -f C = 180°: 1. co8(A + B-C) = -cos2C. ABC 2. sin A + sin B — sin C = 4 sin - sin — cos - • 2 2 2 3. sin2A + sin2B + sin2C = 4sin A sinB sinC. 4. sin 2 A -I- sin 2 B — sin 2 C = 4 sin C oos A cos B. 6. tanZA — tan4A — tan3A = tan7 A tan4 A tan3A. 6. sinA — sinB + sinC = 4sin — cos sin — 2 2 2 „ , A , .B , .C .A .B ,C 7. cot — h cot — f- cot - = cot cot cot -- 222 222 8. tan A — cot B = sec A cosec B cos C. 60. Inverse Trigonometric Functions. — The equation 8in^ = .T means thiit 6 is the angle whose sine is ic; this may be written d = sin^ir, where sin^x* is an abbreviation for the angle (or arc) tvhose sine is x. So the symbols cos 'x, tan~'a;, and sec" \?/, are read "the angle (or arc) whose cosine is a:," " the angle (or arc) whose tangent is a;," and " the angle (or arc) whose secant is y." These angles are spoken of as being the inverse sine ofx, the INVERSE TRIGONOMETUIC FUNCTIONS. 73 inverse cosine of x, the inverse tangent of x, and the inverse secant ofy, resjH'otively. Such expressions are called inverse lri(jonumetric functions. Note. —The Htudent muat bu careful to notice that — 1 in not au exponent, 8in''x Is not {•In X)"', which = -i—. ■Inx 1 ^3 1 1 Notice bUo that Bin'* — = coa'' - la not uu identity, but ia true ouiy for Ui« jiur- tioulur an({lu )H)<^. This Udtntion ta oniy aniiloyou» to the uiie of rxpuncnts In mnltlplioatlon, where we have tt''t It annuls It; and so for other functions. The French method of writing inverse functions is arc sinx, arc cosx, arc tanx, and so on. EXAMPLES. 1. Show that 30" is one value of sin'^. We know that sin 30°= ^. .-. 30° is an angle whose sine is h', or 30°= sin '^. L*. Prove that tan"' i -|- tan"' \ = 45°. tan~'.} is one of the angles whose tangent is ^, and tan~'^ is one of the angles whose tangent is ^. Let ct = tan"' ^, and /3 = tan~' ^ ; then tana = ^ and tan ft = ^. xNow ' tan (« + )«) = i^» « +-t^»A^ 1 — tan«taniC • f * :tan/3 = _i±i-. = i But tan 45° = 1, . •. « + /8 = 45° ; that is, . taH-' ^ + tan"' | = 45°. Therefore 45° is one value of tan~U 4- tau-*|. (Art. 47) 74 PL A Nt: TltlGONOMETH Y. 3. Prove that tau-'aj + tan-'w=: tan-'~i^. Let Now tau~' X = A. . •. tan A = ar. tan~'y = B. .♦. tanB = y. tan (A -I- B) tan A 4- tanB 1 — tan A tan B l-xy .-. A + B = tan-' ,^ti^. i-xy .-. tan-'a? + tan-'« = tan"' ^'^^ . 1-xy Any relations which have been established among the trigonometric functions may be expressed by means of the inverse notation. Thus, we know that 4. cos X = Vl — sink's;. This may be written a; = cos"' Vl — siu^a; . . (1) Put 8inar = ^; then « = sin-'^. Thus (1) becomes sin"' $ = cos*' Vl — ^. 6. By Art. 49, cos2d = 2cos'tf - 1, which may be written 26 = coS-'(2co8-tf — 1). Put cos^ = a?. .-. 2cos-'iC = cos-*(2x' — 1). 6. By Art. 49, sin 2 ^ = 2 sin tf cos 9, which may be written 2d = sin"' (2 sin tf cos d). o u Put sin^ = x. .-. 2siu"'a; = sin-'(2a;Vl — «')• TABLE OF USEFUL FORMULAE. 76 7. Prove 8. sin-'a-scos"' Vl — a:* = tan-' * z' (( tan-'a: = sin-' X = COS"' 9. « 10. " 11. « 12. « 13. " 14. '* Vi-f«* 2tan-'x = tan-'-^^. v^^ 1 — «* Vi + j;' sin (2 sin-' x) = 2x Vl — x*. " tan-'- + tan-'- = -. 7 G 4 • co8-'- + 28in-'?=120*. ; 2 2 cot-' 3 + cosec-' VC = -• 4 Ssin-'a; = sin-'(3a; ~ 4.r') 61. Table of Usefttl FormulsB. — The following is a list uf important formulae proved in this chapter, and summed up lor the convenience of the student : 1. sin (a? -\-y) = sin x cosy + cos x sin y . . (Art. 43) 2. co8(x + y) =cosxcosy — sinxsiny. 3. sin(x — y) = sin x cosy — cos xsiny . . (Art. 44) 4. cos(x — y) = cos T cos y-f sin xsiny. 5. 2 sin X cosy = sin (x -f y) -f sin (x — y) . (Ar«. 45) 6. 2cosxsiny = sin (x + y) — sin (x — y). 7. 2cosxcosy =cos (x + y)4- cos (x — y). 8. 2 sin X sin • =cos(x — y) — cos(x-fy). 9. sinx-H siny = 28in^(.r + y)cos4^(x — y). 10. sinx — iiny = 2co.s^(x + y) si;,^(x — y). m \ ■i i7 ' 76 PLANE TRiaONOMETRY. 11. cosx + cosy = 2co8 J(x-|- 1/) co8^(aj — y). 12. cosy — cos a; = 28iiii(ic + y) sini(a; — y). 13. 8nix-t8iny^tanHar + .v). (^,t. 4c) 8111 JB — amy tan^vx — y) 14. tan(x + y) = tana^ + tan ;^ ^^rt. 47) 1 — taiia;tany ^ r i. / V tan X — tan y 15. tau (a; — y) = ■ '^ • 1 -f tan X tau y -« i. / . ' cotxcoty — 1 16. cot (x-fy> = — ^ •• cotaj-f coty 17. cot(x-y) :^g2 t^coty-Hl . cot» — coty 18. tan(a;±4o'') = *-^-5-^^ (Art. 47) ^ -^ tan a; ± 1 ^ ' 19. sin {x-\- y) sin (a; — y) = sin'' a; — sin'y = cos^y — cos'*. 20. cos (x4- y) cos (x — y) = cos*x — sin'y = cos^y — sin*x. 21. tanx±tany = ?ilL(^±J^. coax cosy 22. cotx±coty=:»i- "Cy^^> . sin X sin y 23. sin2x = 2sinxcosx = 2tanx . . . (Art. 49) 1 -f tan'x 24. co82x = cos'x — sin^'x = 1 — 2sin''x = 2cos'x — 1 1 — tan'x 1 -f taii'x Of 1 — oor2x 28in'x . « 26. ^ ^ — =! tan*x. 1 +eo8 2x 2co8'x SXAMPLES. 77 26. tan 2 a; = 27. cot2a5 = 2tan« I I II !■!■■■ » 1 — ttin'x voM^x — 1 2cotic 28. sin.Sjj ssiiHina; — •Isin'j! 29. cos ti a; = i cob"' x — 3 cos a;. 3 tana; — tan" a? 30. tan 3 a; = l-3tan«x 32. co8»? =l±i^. 2 2 33. tan-' x + tan"' y = tan"' :^^ . . (Art. nO) 31. 8in»- =t ^"*^"^^ \ . (Art. r.l) 2 2 ^ • • • (Art. 6(») and sin (a — /3). EXAMPLES. 12 1. If sin a = ; , and sin /3 = -, find a value for sin ( « -f ;8)» Ans. ^^1^2 VO-4V2. 9 ' 9 2. If cos rt = r> ainl cos B — —-, find a value for -sin ( « -f fl ) , o 41 and cos (« + /?). : , . IfHi l.'!;5 205 205 3 2 3. If cos « = *-, and c^s B = -, find a value for sin (a 4- j8), 4 5 and 8in(« — /3). vln."». - 2 V7 4-3V21 2V7-3V21 20 20 4. If sin « = -, and sin B = -, find a value for sin {u 4- fi), 5 5 and co8.(rt4-/3). .^ns. 1, !^4 25 m IMAGE EVALUATION TEST TARGET (MT-3) 1.0 I.I 1.25 12.5 M 112.2 .^~ m 2.0 ^ 1.4 III 1.6 Photographic Sciences Corporation 23 WEST MAIN STREET WEBSTER, N.Y. 14580 (716) 872-4503 r ^ ^i$ 78 PLANE TRIGONOMETRY. |i iHiii i Ml li:.'*!- )4 I > 'flr \ !§![ 5. If sin« =.6, and sinyS = — , find a value for sin(«— /8), and cos (a + ^)- ,16 33 -^^"- 65' 65' 6. If sin«=— -, and sinfl = , show that one value V5 , ., VlO .-.■-^•■■•;: of « + j8is 45°. ' 7. Prove cos ^ + cos 3 ^ =2eos2dcos0. 8. " 2cosacos/3 =cos («-/?) + cos (« 4-/3). 2sin3ecos5^ =sm8d-sin2^. 9. 10. 11. 12. a " 2cos|^cos- =cos^ + cos2^. " sin4dsin^ = |(cos3d — cos5^). " 2cosl0°sin50°=sin60^-f sin 40". 13. Simplify 2cos2^cos^-2sin4esin^. Ans. 2 cos 3^ cos 2^. 14. Simplify sm — cos - — sin -r- cos — • z z z z Prove the following statements : ,t : ! 15. cosSrt — cos7a = 2sin5«sin2«. 16. sin 60°+ sin 20°= 2 sin 40° cos 20°. 17. sin3^4-sin5^ =2sin4^cos0. 18. sin 7 — sin 5 ^ = 2 cos 6 ^ sin d. 19. cos50 + cos9^ = 2cos7^cos2d. 20. «^"^^ + ^"^^ =tan^. costf + cos2e 2 21. cos (60° + A) + cos (60° - A) = cos A. cos 4 tf sin 2^. w EXAMPLES. 79 22. cos(45°-f- A) + cos(4.r- A)= V2cosA. 23. sin(45°+ A)-siu(45°- A) = V2sinA. 24. cos2^ + cos4^ = 2cos3^cos0. 25. cos4 6 — cos6^ = 2sin5^sin(?. 26. cos^ + cos3^ + cos5^ + cos76 = 4cos0cos2dcos4^. 27. cot« + tany8 =^ («-^) . sin« cos^S . 1. /D cos (« 4- )8) cot a — tan /8 = -^— ^^ — '-J^ < sin a cos /8 28. 29. sin (A -45°) sin A — cos A 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. V2 sin (A + 45°) = sin A -f cos A. cos (A + 45°) + sin (A - 45°) = 0. tan (g — <^)4- tan<^ _ ^^^ ^ 1 — tan {6 — ^) tan <^ tan ( ^ + <^) 4- tan (^ _ ^^^ , l + tan(^+<^) tan^ cos (^ + <^) - sin (^ - <^) = 2 sin /"- - ^"j cos sin 7i6 cos d + cos n6 sin ^ = sin (n + 1 ) d. \^ 4/ 1-cot^ 4- cot /'^ + ?^ = 0. cot r^ - ^) + tan (^ + ^) = 0. tan(n + l)<^-tan??0 ^ ^^^^ , 1 + tan (n-{-l) (f) tan n *'"(*~i) ■^ ■■,,*■ 80 PLANE TRIGONOMETRY. If mmi 40. If tanx = 1, and tany = — -, prove that V3 tan {x -\- ij) = 2 -\- V3. 41. If tana m 1 , and tan/S = - — -7-^, prove that m -f- 1 2 m + 1 tan(a + /3)=l. 42. If tan « = m, and tan /? = ?t, prove that 1 ~ mn cos (« + 18) = V(l+m2)(l+rO ' 43. If tan^ = (a -f- 1), and tan ^ = (a — 1), prove that T».' Prove the following statements : 44. cos {x — y-\-z) = cos a; cos y cos 2 + cos ic siny sing — sin 03 cos y sin 2; + sin a; sin ?/ cos «. 45. sin (« — 1/ — 2) = sin.r + siny 4- sing -f 4sin^(.r— ?/) sini(.r— «) sin^(//-f 2). 46. sin (a; + y — 2) + sin (ic -|- 2: — y) + sin (?/ + 2 — .t) = sin (a; + y + 2!) + 4 sin a; sin 2/ sin 2. 47. sin2a;4- sin2?/ + sin22— sin2 (a; -f y +2) = 4 sin {x + ?/) sin (?y + 2) sin (2 -f- a?)- 48. cos 2 a; + cos 2?/ 4- cos 22 + cos 2 (a;4- y + 2) ' = 4 cos {x -f y) cos (2/ + 2) cos (« + aj) . 49. cos {x + y — 2) 4- cos (y + 2 — x) + cos (2 + a; — y) + cos (a; + y + 2) = 4 cos a; cos y C0S2. EXAMPLES. 81 50. sin^a; sin' 2/ + siirz + sin^ (x^y + z) = 2|1 — cos (x + y) cos (y+z) cos (a+x) |. 51. cos^ + cos^y + cos^a; + cos^ {x-\-y — z) = 2| 1 + cos (a; + y) cos (a;— 2) cos (2/— 2;) |. 52. cr'6x sm(y — %) -\- cos y sm{z—x)-{- cos zsin(x—y) = 0. sina;sin(?/ — z) + sin y sin (2— re) + sin2sin(x — y) = 0. ^64. cos (a; + 2/) cos (x — ?/) + sin (y + 2) sin (?/ — «) — cc s < ^ -I- 2) cos (a; — 2) = 0. 55. 2-se.c^g sec^^ = cos 2 1;. 56. cos2d(l-tan2e) = cos2^. 57. cot2^ = 68. sec2^ = 2cotd cot^^ + l cot^^-l* 59. (-1 + 6V cos - 1 = 1 + sin 0. 2 6 60. f sin- — cos- 1 = 1 — sin^. 2 61. 62. 63. 2 14-secg sec^ cos 20 00 = 2cos^ - 2 1 - tan l + sin20 l + tan0 COS0 l-sin0 1 + tan - l-tan| y^ :-; 'r ■ i < 82 PLANE TRIGONOMETRY, !■•'■? ■.-.It % I :t ;. 1,1 ih f 64. 65. m. 67. 68. 69, 70. 71. 72. 73. 74. 1 + sm a; + cos x ,x T— ' — ^ ■ = cot-- 1 -f- sin X — cos X 2 cos^a;4- sin^a; cosa; + sinic cos'^a; — sin^x cos a? — sin a; cos*^ — sin*^ sin3g cos3g sin 6 cos 6 cos3^ , sin30 2 — sin 2 a; 2 2 + sin 2 35 = cos 2 6. = 2. sin 6 sin4^ _ sin2d" 12 + sm cosd 2 cos 2d. 57r = 2cot2d. cos 12 sm- TT 12 TT cos — 12 = 2V3. tan (45° + a;) - tan (45° - a;) = 2 tan 2 x. tan (45° - «) + cot ( 45° -x) = 2 sec 2 a;. tan2(45°4-«)-l • o — -^^ — ^ =sm2x. , ; , tan2(45°+ic) + l , , . , , cos (x -f 45°) o i- o \—^ — —-i- = sec 2 a; — tan 2 a;. cos {x — 45 ) 76. 76. tan x = 77. tan x — 78. sinx-f sin2a5 I— ■■■■ '■ ' ■ .1 — ■ • 1 4- cos a;-}- cos 2 a; sin 2 a; — sin a; 1 — cosaj + cos 2 a; cos 3 a; = 2 cos 2 a; — 1. cos a; X. i iMi^:if r EXAMPLES. 83 \ r,n 3 sin a; — sin 3 a: . , 79. .^ = tau^ X, cos3x + 3 cos a; cot'^'ic — 3 cot a; 80. cot 3 a; = 3cot2.'«-l 81. J- ~^Q^^^ =(l + 2cosa;)l 1 — cosa; Qo sin x + cos a; 1.0.0 82. — — I— ^ — = tan 2 a; -f sec 2 a;, cos X — sin X go cos 2 a; + cos 12 a; cos 7 a; — cos 3 a ; 2sin4a; ^ cos 6 a; 4- cos 8 a; cosa; — cos3a; sin2x 84. sin2a; sin22/ = sin^(a; + y) — sin^(a; — y). 85. tan50°+cot50°=2secl0°. 86. sin 3 a; = 4 sin a; sin (60" + a;) sin (60° —x). 87. cot|-tan- = 2. 8 8 88. tan4g=:^^^"^(^-^^^'^). l-6tan2^ + tan*d 89. 2cos^=V2 + V2. 90. (3sin^ - 4siiv''^)2+ {4.cos'6 - ScosOy = 1. 91. sin 2 g cos ^ 9 (1 + cos 2 ^) (1 + cos 0) ~ ^^2 12 1 92. If tan = -, and tan <^ = — , prove tan (2 ^ + <^) = -• 93. Prove that tan- and cot- are the roots of the equation Q? — 2x cosec ^4-1 = 0. : ■■; ' '; t r it 84 PLANE TRIGONOMETRY. i'1 94. If tan0= -, prove that a cosO + b Vcos2e 95. Find the values of (1) sin 9°, (2) 008 9°, (3) sin 81°, (4) cos 189°, (5) tan202i°, (6) tan97i°. Ans. (1) UV3+V§-^5-V5), (2) I (V3+ V5 + V5 - V5), (3) sin 81°= cos 9°, (4) cosl89°=-cos9°, (5) V2-I, 'r ^ :^--^-- "^^■.;; (6) _(V3+V2)(V2 + 1). 96. If A = 200°, prove that •,.:.« (1) 2sin^ =+ Vl4-sinA + VI -sin A. (2) ta„| = ^±^'-±^^*-^- ^ ^ 2 tan A 97. If A lies between 270° and 360°, prove that A (1) 2sin— = + Vl -sin A— Vl + sinA. (2) tan - = — cot A + cosec A. 98. If A lies between 450° and 630°, prove that (1) 2sin— = - Vl + sinA — Vl -sin A. ■>;.:^ J- (2) 2 cos Vl + sin A 4- Vl — sin A. Ji li \ EXAMPLES. 85 Prove the following statements, A, B, C being the angles of a triangle. 99. ^'°f -"'» -! = tang tan ^—g. Sin A +sinB 2 *2 sin3B-s in3C _ tan 3 A cos3C-cos3B~ 2 A 2 "2 ■ ^'"2 100. 101. sin ^ cos ^ + sin I cos ^+ sin ^ cos? 9. '}. 9. 9 9 O ''''""''''''' A B C ' = 2cos — cos — cos - • 2 2 2 102. cos'' I + cos''?- - cos^^ = 2 cos A cos? sin?. 2 ,2,. 2 2 2 2 103. sin A cos A — sin B cos B + sin C cos C = 2cos A sinB cosC. ;^ 104. cos2A + cos2B + cos2C = -l -4cosAcosBcosC. 105. sin2A-sin2B + sin2C = 2sinAcosBsinC. V '^ 106. tan I tan ? + tan ? tan ^ + tan - tan ? = 1. ' - ' 2 2 2 2 2 2 Prove the following statements when we take for sin"^, COS"', etc., their least positive value. 107. sin-i^ = cos-^^ = cot-V3. 108. 2tan-Hcos2^) = tan-'/'^5J^!^7-^^V : ' 109. 4tan- -ll_tan-iA__![. 5 239 4 110. sin-i^H-sin-'l4-sin-'^ = ^. 5 17 85 2 rr 8(5 PLANE TlilGONOMETRy. 111. tan-^ V5 (2 - V3) - cot"' V5 (2 + Va) = cot-' V5. 112. sec-V3 = 2cot-W2. ,v 113. 2cot-^a; = (!Osec 2x \/3-fV2 . -1 3 Sit V3-V2 ^^ * 115. sin-^-i= + cot->3 = '- 116. cos-^|i^ + 2tan-'p = sm-^-. :.; 117 Ue = sin-'-, and <^ = cos-^| then ^ + = 90°. 5 o 1 — 7? 118. Prove that cos (2 tan"' a;) = ^"+^2* <,'.^ ■, ■;* 119. 120. 121. 122. « « tan-^- + cosec-^\/lO = |- ^ > ,;^ <( (( (( 2 i5 • _i33 " 2tan-'|-cosec-'- = sm — • -a^i " 2tan-i^ + cos-*- = ^- " sin-' (cos a;) + cos"' (sin y) -\- x -[- y = tt. 123. « " tan-'r-^+tan-'^-^ + tan-'5^=/'7r. 1 + ^ 124. 126. « « « tan-'— "-4-tan 1-a- TT a; « sin-^a — sin-^2/ = ?i7r + 2a;-l 4 = cos-' {xy ± Vl-a;'-2/' + ^"V)- K:.„. i I NA.TUHE AND USE OF LOaASITIIMS. 87 ■y / CHAPTER IV. / LOGARITHMS AND LOGAEITHMIO TABLES. — TEIQO- / NOMETEIO TABLES. 62. Nature and Use of Logarithms. — The numerical cal- culations which occur in Trigonometry are very much abbreviated by the aid of logarithms ; and thus it is neces- sary to explain the nature and use of logarithms, and the manner of calculating them. The logarithm of a number to a giver base is the exponent of the power to which the base must \)e ra'sed to give the number. Thus, if a' = 711, x is called the " logarithm of m to the base a," and is usually written « = log„?>i, the base being put as a suffix.* The relation between the base, logarithm, and number is expressed by the equation, ' • ,v ' ■. ' (base)'''^ = number. Thus, if the base of a system of logarithms is 2, then 3 is tb logarithm of the number 8, because 2^^ = 8. If the base be 5, then 3 is the logarithm of 125, because 53 = 125. ■•■•'' ' ' ^ 63. Properties of Logarithms. — The use of logarithms depends on the following j^^'operties which are true for all logarithms, whatever may be the base. ♦From the definition it follows that (1) logaa*=x, and conversely (2) a'°Ko •» = »». Taking the logarithms of both sides of the equation «* = m, we have log^ a' = ar = log to. Conversely, taking the exponentials of both sides of x = log„ m to base a, we have a* = o'°!?o "» = TO. a' = m and a; = log„TO are thus seen to be equivalent, and to express the same relation between a number, m, and its logarithm, :c, to base a. ■' f 88 PLANE TRIGONOMETRY. ■ " ''r I A (1) The logarithm of 1 is zero. For a" = 1, whatever a may be ; therefore log 1=0. (2) The logarithm of the base of any system is unity. , For a'= a, whatever a may be ; tlierefore log„a = 1. ■, (.3) TJie logarithm of zero in any system ivhose base is greater than 1 is minus infinity, 11 For a~* = — = = ; therefore log = — oo . a* 00 (4) The logarithm of a product is equal to the sum of the loganthms of its factors. For let a; = log„m, and .v=log„w. and n = a'. r -I .'. m = a, ., ,, .*. m?i = a'+''. , .-. Iog„wn = a; + y = log„m4-log„w. ' Similarly, log„mrjp = log„m + log„n + log„j9, and so on for any number of factors. ' i . , Thus, Iog60 = log(3x4x5), .! 7l;.^ ■ • =log3 + log4-j-log5. ' (5) The logarithm of a quotient is equal to the logarithm of the dividend minus the logarithm of the divisor. ' > For let x = log„m, and y = \ogan. m = a' and n = a*. m n m = a x-y log« - = x — y=i log„ m — log„ w. 17 log— = log 17— log 5. o PROPERTIES OF LOGARITHMS, 89 (6) The logarithm of any poiver of a number is equal to the logarithm of the number iriultiplied by the exponent of the poiver. For let fl; = log.,m. .-. m — a*. .'. m>'=a''*. .'. log^m'' = px = plog^m. (7) The logarithm of any root of a mimber is equal to the logarithm of the number divided by the index of the root. For let a;=log„?)i. .-. m = a'. 1 X ^ X 1 .'. log(m'-) = - = - logaWi. r r It follows from these propositions that by means of logarithms, the operations of midtipUcation and division are changed into those of addition and siibt r action ; and the operations of involution and evolution are changed into those oi niultiplicationimCi division. 1. Suppose, for instance, it is required to thid the product of 246 and 357; we add the logarithms of the factors, and the sum is the logarithm of the product : thus, ■ log,o246 = 2.39093 logio357 = 2.55267 4.94360 which is the logarithm of 87822, the product required. 2. If we are required to divide 371.49 by 52.376, we pro- ceed thus: ' log,o371.49=: 2.56995 : logio52.376 = 1.71913 [.■,_:::■:■[ 0.85082 ■■;■;.; . ,„,: ■ r'. which is the logarithm of 7.092752, the quotient required. " $ :« \1 -! .,1 90 PLANE TRIGONOMETRY. 3. If we have to find the fourth power of 18, we proceed thus: logiol3= 1.11394 ■ ■ 4.45576 which is the logarithm of 285C1, the number required. 4. If we are to find the fifth root of 1G807, we proceed thus : 5)4.22549 = logi„ 16807, 0.845098 which is the logarithm of 7, the root required. 5. Given logio2 = 0.30103 ; find logi„128, log,,, 512. Ans. 2.10721, 2.70927. 6. Given logio3 = 0.47712 ; find logi„81, logi„2187. Ans. 1.90849, 3.33985. _ 7. Given log,o3; find log,,, W'. ... , , 0.28627. 8. Find the logarithms to the base a of 10 v/a. Vet-. 9. Find the logarithms to the base 2 of 8, 04, ^, .125, .015625, V64. Ans. 3, 6, — 1, 3, 6, 2. 10. Find the logarithms to base 4 of 8, V 16, V.5> ^.015625. Ans. |, |, - \, - 1. Express the following logarithms in terms of log a, log?>, and logo: , , ,. Ans. 6 log a -f- log 6 4- 3 log c. 11. log -/ {on'' cy. 12. log^a^^'FZ. 13. log t' f'St • (a-n-'c-*)^ floga+flog6 + |log, iloga. SYSTEMS OF LOGARITHMS. 91 64. Common System of Logarithms. — There are two systems of logarithms in use, viz., the Naperian* system and the common system. ' . . The Naperian system is used for purely theoretic investi- gations; its base is e = 2.7182818. • The common system f of logarithms is the system that is used in all practical calculations ; its base is 10. By a system of logarithms to the base 10, is meant a suc- cession of values of x which satisfy the equation ■ -. .." ; •. -7)1 = 10', for all positive values of m, integral or fractional. Thus, if Ave suppose m to assume in STiccession every value from to 00, the corresponding values of x will form a system of logarithms, to the base 10. . Such a system is fo -med by means of the series of loga- rithms of the natural numbers from 1 to 100000, which con- stitute the logarithms registered in our ordinary tables. J/'* Now and so on. Also, 10" =1, 10^ = 10, 10' = 100, lO'' = 1000, logl =0; log 10 =1; log 100 =2; log 1000 = 3. 10-'=, V =•!. ••• log.l =-1; _ , . 10-^= ik =-01. ••• log-01 =-2; ■' " ; ■ io-»= i-^o = -^^1? •*• log.OOl =-3. and so on. Hence, in the common system, the logarithm of any number between 1 and 10 is some number between and 1 ; i.e., -}- a decimal ; * So cnllerl from its inventor, 7?«»'on iV«/>i>/*, a Scotch motliematlcian. t First introduced in lOl;') by liriygs, a contemporary of Napier. t 1 t! I 92 PLANE TRIGONOMETBY. IV. ' 10 and 100 is some number between 1 and 2 ; i.e., 1 + a decimal; . -^ ;i 100 and 1000 is some number between 2 and 3 ; i.e., 2 -f a decimal; " • ^ 1 and .1 is some number between and — 1 ; i.e., — 1 -f- a decimal ; .1 and .01 is some number between — 1 and — 2 ; i.e., — 2 + a decimal ; .01 and .001 is some number between — 2 and — 3 ; i.e., — 3 + a decimal ; and so on. It thus appears that (1) The (common) logarithm of any number greater than 1 is positive. (2) The logarithm of any positive number less than 1 is negative. (3) In general, the common logarithm of a number con- sists of two parts, an integral part and a decimal part. The integral part of a logarithm is called the characteristic of the logarithm, and may be either positive or negative. The decimal part of a logarithm is called the mantissa of the logarithm, and is ahcays kept 2wsitive. Note. — It is convenient to keep the decimal part of the lognrithma alwaya posi- tive, in order tliat numbers couBisting of the same digits in the same order may correspond to the same mantissa. It is evident from the above examples that the character- istic of a logarithm can always be obtained by the following rule : Rule. — The characteristic of the logarithm of a number greater than unity is one less than the number of digits in the whole number. The characteristic of the logarithm of a number less than unity is negative, and is one more than the number of ciphers immediately after the decimal point. RULES FOR THE CHARACTERISTIC. 93 Thus, the characteristics of the logarithms of 1234, 123.4, 1.234, .1234, .00001234, 12340, are respectively, 3, 2, 0, - 1, -6, 4. ^^ ■ •'" ' ■ ' Note. — When the characteristic Is uegative, the minus sign is written over it to indicate that the characteristic alone is negative, the mantissa being always positive. Write down the characteristics of the common logarithms of the following numbers : 1. 17601, 361.1, 4.01, 723000, 29. Ans. 4, 2, 0, 5, 1. 2. .04, .0000612, .7963, .001201, .1. Ans. -2, -5, -1, -3, -1. 3. How many digits are there in the integral part of the numbers whose common logarithms are respectively 3.461, 0.30203, 5.47123, 2.67101 ? 4. Given log 2 = 0.30103 ; find the number of digits in the integral part of 8^", 2'^, 16"^, 2'''. Ans. 10, 4, 25, 31. 65. Comparison of Two Systems of Logarithms. — Given the logarithm of a number to base a ; to find the logarithm of the same number to base b. Let m be any number whose logarithm to base b is required. Lfll x = \ogf,m; then b' = m. V^ loga(60 = log„m; or a;log„6 = log„m. 1 .*. 05 = log„6 X log<.m. or logsm = log„6 (1) •Hence, to transform the logarithm of a number from 1 base a to base b, we multiply it by loga^ 94 PLANE TRIGONOMETRY. This constant multiplier log„6 is called the modulus of tl i I? f the system of wJiich the base is b with reference to the system of which the base is a. If, then, a list of logarithms to some base e can be made, we can deduce from it a list of common logarithms by mul- tiplying each logarithm in the given list by the modulus of the common system Putting a for m in (1), we have c. r . : log»a = J^ = r^, by (2) of Art. 63. log„6 log„6 r ; ,-. logja X log„&=l. ; ,,„^ ^,, ;,„^,^, ,,,.;, ,,^,,^,, EXAMPLES, ■' ,»..■, - 1. Show how to transform logarithms with base 5 to logarithms with base 125. Let m be any number, and let x be its logarithm to base 125. Then m = 125' = (5»)' = 5^^ .-. 3a; = \og,m. .-. a; = logi25m = |log5m. ;^ • Thus, the logarithm of any number to base 5, divided by 3 (i.e., by logs 125), is the logarithm of the same number to the base 125. Otherwise by the rule given in (1) . Thus, log5l25 o Show how to transform 2. Logarithms with base 2 to logarithms with base 8. Ans. Divide each logarithm by 3. iti TABLES OF LOGARITHMS. 95 iW 3. Logarithms with base 9 to logarithms with base 3. Ans. Multiply each logarithm by 2. 4. Find logs 8, logjl, log8 2, logyl, log32l28. Ans. 3, 0, i, 0, i. 66. Tables of Logarithms. — The common logarithms of all integers from 1 to 100000 have been found and registered in tables, which are therefore called tabular logarithms. In most tables they are given to six places of decimals, though they may be calculated to various degrees of approximation, such as five, six, seven, or a higher number of decimal places. Tables of logarithms to seven places of decimals are in common use for astronomical and mathematical calculations. The common system to base 10 is the one in practical use, and it has two great advantages : (1) From the rule (Art. 64) the characteristics can be written down at once, so that only the mantissse have to be given in the tables. (2) The mantissse are the same for the logarithms of all numbers which have the same significant digits, in the same order, so that it is sufiicient to tabulate the mantissse of the logarithms of integers. For, since altering the position of the decimal point with- out changing the sequence of figures merely multiplies or divides the number by an integral power of 10, it follows that its logarithm will be increased or diminished by an integer; i.e., that the mantissa of the logarithm remains unaltered. In General. — If N be any number, and p and q any integers, it follows that N x 10^ and N -f- 10' are numbers whose significant digits are the same as those of N. Then log (N x 10^) = log N 4- i> log 10 = logN -{-p. (1) Also, log (N -5- 10') = logN - gloglO = logN - q. (2) 96 PLANE TRIGONOMETRY. !■■ . ( ar* In (1) the logarithm of N is increased by an integer, and in (2) it is diminished by an integer. That is, the same mantissa serves for the logarithms of all numbers, whether greater or less than unity, which have the same significant digits, and differ only in the position of the decimal point. This will perhaps be better understood if we take a particular case. From a table of logarithms we find the mantissa of the logarithm of 787 to be 895975 ; therefore, prefixing the char- acteristic with its appropriate sign according to the rule, we have , , , . . log787 =2.895975. . , , : / . Now log 7.87 = log ^ = log 787 -2 = 0.895975. , Also, log.0787 =log(j55^)^ ^og787-4 , ' ' . - J': . . . =2.895975. '.';. , '.r : . ,.: ■. Also, log 78700 = log (787 x 100) = log 787 + 2 ■ r.'.-' <-:.,: ' =4.895975. '^■^-'' : . '■ ' Note 1. — We do not write log,o 787 ; for so long as we are treating of logarithms to the particular base 10, we may omit the suffix. Note 2. — Sometimes in woriiing with negative logarithms, an arithmetic artifice will be necessary to make the mantissa positive. For example, a result such as — 2.69897, in which the whole expression is negative, may be transformed by sub- tracting 1 from the characteristic, and adding 1 to the mantissa, Thus, . , - 2.69897 = - 3 + (1 - .69897) = 3.30103. NoTB 3. — When the characteristic of a logarithm is negative, it is often, espe- cially in Astronomy and Geodesy, for convenience, made positive by the addition of 10, which can lead to no error, if we are careful to subtract 10. Thus, instead of the logarithm 3.608582, we may write 7.603582 — 10. In calculations with negative characteristics we follow the rules of Algebra. EXAMPLES. 97 1. Add together EXAMPLES. 2.2143 . 1.3142 6.9068 2. From take 7.4353 Ans. 3.24569 5.62493 1.62076 the 1 carried from the last subtraction in decimal places changes — 5 into — 4, and then — 4 subtracted from — 3 gives 1 as a result, 3. Multiply 2.1528 by 7. 2.1528 • 7 ., 13.0696 . . : ■ the 1 carried from the last multiplication of the decimal places being added to — 14, and thus giving — 13 as a result. Note 4. — When a logarithm with negative characterietic has to be divided by a number which ie not an exact divisor of the characteristic, we proceed as follows in order to keep the characteristic integral. Increase the characteristic numerically by a number which will make it exactly divisible, and prefix an equal positive number to the mantissa. 4. Divide 3.7268 by 5. Increase the negative characteristic so that it may be exactly divisible by 5 ; thus - 3.7268 ^ 5 -f 2.7268 ^ j- 5 5 5453. Given that log2=.30103, log3=:.47712, and log7=.84510; find the values of 5. log6, log42, logl6. ^ri5. .77815, 1.62325, 1.20412. « ll' 98 PLANE TRIGONOMETRY. i n fi Ipifj 6. log 49, log 36, log 63. Ans. 1.69020, 1.55630, 1.79934. 7. log 200, log 600, log 70. 2.30103,2.77815,1.84510. 8. log 60, log. 03, log 1.05, log. 0000432. Note. — The logarithm of 5 and its powers can always be obtained from log 2. Ans. 1.77815, 2.47712, .02119, 5.63548. 9. Given log 2 = .30103 ; find log 128, log 125, and log 2500. Ans. 2.10721, 2.09691, 3.39794. Given the logarithms of 2, 3, and 7, as above ; find the logarithms of the following : 10. 20736, 432, 98, 686, 1.728, .336. Ans. 4.31672, 2.63548, 1.99122, 2.83632, .23754,1.52634. 11. V.2, (.03) ^ (.0021)^, (.098)^ (.00042)^ (.0336)*. Ans. 1.65052, 1.61928, 1.46444, 4.97368, 17.11625, 1.26317. 67. Use of Tables of Logarithms * of Numbers. — In our explanations of the use of tables of common logarithms Ave shall use tables of seven places of decimals.f These tables are arranged so as to give the mantissie of the logarithms of the natural members from 1 to 100000 ; i.e., of numbers containing from one to five digits. A table of logarithms of numbers correct to seven deci- mal places is exact for all the practical purposes of Astron- omy and Geodesy. For an actual measurement of any kind must be made with the greatest care, with the most accurate instruments, by the most skilful observers, if it is to attain to anything like the accuracy represented by seven signifi- cant figures. * The methods by which these tables are formed will be given in Chap. VIII. t The student should here provide himself with logarithmic and trigonometric tables of seven decimal places. The most convenient seven-flgure tables used in this country are Stanley's, Vega's, Bruhns', etc. In the appendix to the Elementary Trigonometry are given five-flgurod tables, which are sufficiently near for most prac- tical application!. ,Jt USE OF LOGARITHMIC TABLES. 99 If the measure of any length is known accurately to seven figures, it is practically exact ; i.e., it is known to within the limittt of obser- vation. If the measure of any angle is known to within the tenth part of a second, the greatest accuracy possible, at present, in the measurement of angles is reached. The tenth part of a second is about the two- millionth part of a radian. This degree of accuracy is attainable only with the lai'gest and best instruments, and under the most favorable conditions. On page 101 is a specimen page of Logarithmic Tables. It consists of the mantissse of the logarithms, correct to seven places of decimals, of all numbers between 62500 and 63009. The figures of the ninnher are those in the left column headed N, followed l)y one in larger type at the top of the page. The first three figures of the mantisste 795, 796, 797 •••, and the remaining four are in the same hori- zontal line with the first four figures of the number, and in the vertical column under the last. Logarithms are in general incommensurable numbers. Their values can therefore only be given approximately. Throughout all approximate calculations it is usual to take for the last figure whicli we retain, that figure which gives the nearest aj^proach to the true value. When only a cer- tain number of decimal places is required, the general rule is this: Strike out the rest of the Jirjures, and increase the last figure retained by 1 if the first figure struck off is 5 or greater than 5. 68. To find the Logarithm of a Given Number. — When the given number has not more than five digits, we have merely to take the mantissa immediately from the table, and prefix the characteristic by the rule (Art. 64). Thus, suppose Ave require the logarithm of 62541. The table gives .7961648 as the mantissa, and the characteristic is 4, by the rule ; therefore log 62541 = 4.7961648. Similarly, log .006281 = 3.7980288 . . (Art. 64) 100 PLANE TRIGONOMETRY. WM My t 1 Suppose, however, that the given number has more than five digits. For example : Suppose we require to find log 627G1.6. We find froxn the table log62761 = 4.7976899 log 62762 = 4.7976968 and diff. for 1 = 0.0000069 Thus for an intrease of 1 in the number there is an in- crease of .0000069 in the logarithm. Hence, assuming that the increase of the logarithm is proportional to the increase of the number, then an increase in the number of .6 will correspond to a.i increase in the logarithm of ,6 x .0000069 = .0000041, to the nearest sev- enth decimal place. Hence, log 62761 = 4.7976899 diff. for .6 = 41 log 62761.6 = 4.7976940 This explains the use of the column of proportional parts on the extreme right of the page. It will be seen that the difference between the logarithms of two consecutive num- bers is not always the same ; for instance, those in the upper part of the page before us differ by .0000070, while those in the middle and the lower parts differ by .0000069 and .0000068. Under the column with the heading 69 we see the difference 41 corresponding to the figure 6, which implies that when the difference between the logarithms of two consecutive members is .0000069, the increase in the logarithm corresponding to an increase of .6 in the number is .0000041 ; for .06 it is evidently .0000004, and so on. Note. — We assume in Uiis method that the increase in a logarithm is propor- tional to the increase in the number. Although this is not strictly true, yet it is in most cages sufllciently exact for practical purposes. Had we talven a whole number or a decimal, the process would have been the ■ame. 1* TABLE OF LOOARITHMS. 101 N. 6250 51 52 58 54 55 56 57 58 59 6260 61 62 63 64 65 66 67 68 69 6270 71 72 73 74 75 76 77 78 79 6280 81 82 83 84 85 86 87 88 89 6290 91 92 93 94 95 96 97 98 99 6800 N. 795 8800 9495 796 0190 0884 1579 2273 2%7 3662 4356 5050 8870 9564 0259 0954 1648 2343 3037 3731 4425 4494 5119 5188 796 5743 6437 7131 7824 8517 9211 9904 797 0597 1290 1983 797 2675 6506 7200 7893 8587 9280 9973 0666 1359 2052 2745 3368 4060 4753 5445 6137 6829 7521 8213 8905 797 9596 798 0288 0979 1671 2362 3053 3744 4435 5125 5816 798 6506 7197 7887 8577 9267 9957 799 0647 1337 2027 2716 799 3405 O 8939 9634 0329 1023 1718 2412 3106 3800 5813 3437 4130 4822 5514 6207 6899 7590 8282 8974 9666 0357 1048 1740 2431 3122 3813 4504 5194 5885 6575 7266 7956 8646 9336 0026 0716 1406 2096 2785 3474 1 5882 6576 7269 7963 8656 9349 0043 0736 1428 2121 2814 3507 4199 4891 5584 6276 6968 7660 8351 9043 5951 6645 7339 8032 8725 9419 0112 0805 1498 2191 97^ 0426 1118 1809 Z500 3191 3882 4573 5263 5954 6645 7335 8025 8715 9405 0095 0785 1475 2164 2854 9009 9703 0398 1093 1787 2481 3176 3870 4564 5258 9078 9773 0468 1162 1857 2551 3245 3939 4633 532^ 6021 2883 3576 4268 4961 5653 6345 7037 7729 8421 6714 7408 8101 8795 9488 0181 0874 1567 2^60 2952 3645 4337 5030 5722 6414 7106 7798 8490 911219181 9804 9873 0495 1187 1878 2569 3260 3951 4642 5333 6023 6714 3543 7404 8094 8784 9474 0164 0854 1544 2233 0565 1256 1947 2638 3329 4020 4711 5402 6092 6783 7473 8163 8853 9543 0233 0923 1613 2302 2923 2992 3612 3681 9148 9842 0537 1232 1926 2620 3314 4009 4703 5396 6090 6784 7477 8171 8864 6 9217 9912 0606 1301 1995 2690 3384 4078 4772 5466 6160 6853 7547 8240 8933 9557 9627 0250 0320 0943 1636 2329 3022 3714 4407 5099 5 791 6483 7175 7867 8559 9251 9942 0634 1325 2016 2707 3398 4089 4780 5471 6161 6852 7542 8232 8922 9612 0302 0992 1682 2371 3061 3750 1013 1706 2398 3091 3784 4476 5168 5860 6553 7245 7936 8628 9320 0011 0703 1394 2085 2776 3467 4158 4849 5540 6230 6921 7611 8301 8991 9681 0371 1061 1751 2440 3130 3819 6 9m 9981 0676 1370 2065 2759 3453 4147 4841 5535 6229 6923 7616 8309 9003 9696 0389 1082 1775 2468 3160 3853 4545 5237 5930 6622 7314 8006 8697 9389 0080 0772 1463 2154 2846 3536 4227 4918 5609 6299 6990 7680 8370 9060 9750 0440 1130 1820 2509 3199 3888 8 9 9356 9426 0051,0120 0745 1440 2134 0815 1509 2204 28292898 35233592 •12174286 49114980 5605:5674 629816368 6992 7061 7685 7755 8379 8448 9072 9141 9765 9835 0458 0528 1151 1844 2537 3229 3922 4614 5307 5999 6691 7383 8075 8766 9458 0150 0841 1532 2224 2915 3606 4296 4987 5678 6368 7059 7749 8439 9129 9819 0509 1199 1889 2578 3268 1221 1913 2606 3299 3991 4684 5376 6068 6760 7452 8144 8836 9527 0219 0910 1601 2293 2984 3675 4366 5056 5747 6437 7128 3957 7818 8508 9198 9888 0578 1268 1958 2647 3337 4026 9 P.P. 70 7.0 14.0 21.0 28.0 35.0 42.0 49.0 56.0 63.0 69 6.9 13.8 20.7 27.6 34.5 41.4 48.3 55.2 9 62.1 68 6.8 13.6 20.4 27.2 34.0 40.8 47.6 54.4 61.2 i *. Sf:# p.p. \m 102 PLANE TRIGONOMETRY. > Tbua, Huppotc we require to And log 0-27U10 uiul lug .627010. TIil- inanllijiiu In exactly the auine us before (Art. 00), and the only dltt'urenue tu be iiiudc io the tlual reault it to cbauge tbe ciiaracteriatiu uccurding to rule (Art. 04). Tbu» log 027010 = u.7U7e942, and log .627016 = 1.7970942. 69. To find the Number corresponding to a Oiven Loga- rithm. — It' the decimal part of tlie logarithm is found ex- actly ill the table, we can take out the corresponding number, and put the decimal point in the number, in the place indicated by the characteristic. Thus if we have to tind the number whose logarithm is 2.7982915, we look in the table for the mantissa .7982915, and we find it set down opposite the number 62848 : and as the characteristic is 2, there must be one cipher before the first significant figure (Art. G4). Hence 2.7982915 is the logarithm of .062848. Next, suppose that the decimal part of the logarithm is not found exactly in the table. For example, suppose we have to find the number whose logarithm is 2.7974453. We find from the table log62726 = 4.7974476 log62725 = 4.7974407 diff . for 1 = ".0000069 Thus for a difference of 1 in the numbers there is a ditt'erence of .0000069 in the logarithms. The excess of the given manl.ssa above .7974407 is (.7974453 - .7974407) or .0000046. Hence, assuming that the increase of the number is proportional to the increase of the logarithm, we have .0000069 : .0000046 : : 1 : number to be added to 627.25. ... number to be added = -^^-^^^' = 1^'=. 667 fi9)46.0(.666 .0000069 69 41 4 .-. log62725.667 = 4.7974453, and .-. log627.25667 = 2.7974453; therefore number required is 627.25667. 4 60 414 ~460 ARITHMETIC COMPLEMENT. 108 We might have saved the labor of dividing 40 by 09, by using the table of proportional parts as follows : given mantissa = .7074453 mantissa of 02725 = .7974407 diff. of mantissie = 40 proportional part for .0 = 41.4 r- .' I' '. (( it ''•' 4.6 n (C .00 = 4.14 .46 « (( .000 = .414 . . and so on number = 027.250000.. •. 1 69a. Arithmetic Complement. — By the arithmetic com- plement of the logarithm of a number, or, briefly, the cologarithm of the number, is meant the remainder found by subtracting the logarithm from 10. To subtract one logarithm from another is the same as to add the co- logarithm and then subtract 10 from the result. , . .. Thus, a 6 = a -4- (10 - 6) - 10, where a and b are logarithms, and 10 — 6 is the arithmetic complement of 6. When one logarithm is to be subtracted from the sum of several others, it is more convenient to add its cologarithm to the sum, and reject 10. The advantage of using the cologarithm is that it enables us to exhibit the work in a more compact form. The cologarithm is easily taken from thr table mentally by subtracting the last significant figure on the right from 10, and all the others from 9. 104 PLANE TRIGONOMETRY. 1. Given find 1 2. Given wi 1 find 1 3. Given m find f 4. Given • find - 6. Given find the number 6. Given .1 ':•' find the number 7. Given find the number 8. Given . I find the number Ans. 4.7201799. EXAMPLES. log 52502 =4.7201758, log 52503 =4.7201841; log 52502.5. log 3.0042 = 0.4777288, log 3.0043 = 0.4777433; J log 300.425. ■ ' 2.4777360. ■ log 7.6543 = 0.8839055, ■ log 7.6544 = 0.8839112; log 7.65432. .8839066. log 6.4371 = 0.8086903, log 6.4372 = 0.8086970; log 6437125. 6.8086920. log 12954 =4.1124039, log 12955 =4.1124374; whose logarithm is 4.1124307. 12954.8. log 60195 = 4.7795532, ' - log 60196 =4.7795604; whose logarithm is 2.7795561. 601.95403. ' log3.7040 = .5686710, ^ ' ' ' log 3.7041 = .5686827 ; , ;- , , whose logarithm is .5686760. 3.70404. log 2.4928 = .3966874,. log 2.4929 = .3967049 ; whose logarithm is 6.3966938. 2492837. NATURAL TRIGONOMETRIC FUNCTIONS. 105 find 9. Given log 32642 == 4.5137768, log 32643 =-.4.5137901; log 32642.6. '' 10. Find the logarithm of 62654326. Use specimen page. Ans. 4.5137835. 7.7969510. 11. Find the number whose logarithm is 4.7989672. Ans. 62945.876. 70. Use of Trigonometric Tables. — Trigonometric Tables are of two kinds, — Tables of Natural Trigonometric Functions and Tables of Logarithmic Trigonometric Functions. As the greater part of the computations of Trigonometry is carried on by logarithms, the latter tables are by far the most use- ful. We have explained in Art. 27 how to find the actual numerical values of certain trigonometric functions, exactly or approximately. ' ' Thus, sin 30° = -; that is, .5 exactly. Also, tan 60° = V3 ; that is, 1.73205 approximately. A table of natural trigonometric functions gives their approximate numerical values for angles at regular intervals in the first quadrant. In some tables the angles succeed each other at intervals of 1", in others, at intervals of 10", but in ordinary tables at intervals of 1': and the values of the functions are given correct to five, six, and seven places. The functions of intermediate angles can be found by the principle of proportional parts as applied in the table of logarithms of numbers (Arts. 68 and 69). It is sufficient to have tables which give the functions of angles only in the first quadrant, since the functions of all angles of whatever size can be reduced to functions of angles less than 90° (Art. 35). ■ '■•■«. 106 PLANE TRIGONOMETRY. '141 71. Use of Tables of Natural Trigonometric Functions. — These tables, which consist of the actual numerical values of the trigonometric functions, are commonly called tables of natural sines, cosines, etc., so as to distinguish them from the tables of the logarithms of the sines, cosines, etc. We shall now explain, first, how to determine the value of a function that lies between the functions of two con- secutive angles given in the tables ; and secondly, how to determine the angle to which a given ratio corresponds. S' 72. To find the Sine of a Given Angle. Find the sine of 25° 14' 20", having given from the table sin2o°lo' = .42GoG87 , ^ - sin 25° 14' = .4203056 !.; : d = .0000877. diff. for 1' = .0002631 ' Let (? = diff. for 20" ; and assuming that an increase in the angle is proportional to an increase in the sine, we have 60 : 20 :: .0002031 : (^ 20 X .0002031 60 .-. sin 25° 14' 20'- =.4203056 + .0000877 • =.4263933. ' Note. — We asnumed here tliaf. iin increase in tlie angle ia proportional to the increase in the corresponding sine, wliich is suiiiciently exact for practical purposes, with certain exceptions. 73. To find the Cosine of a Given Angle. Find the cosine of 44° 35' 25", having given from the table cos 44° 35' = .7122303 cos 44° 30' = .7120260 dife. for 1' = .0002043 observing that the cosine decreases as the angle increases from 0° to 90°. m th CO hi f EXAMPLES. 107 Let d — decrease > 'f cosine for 25" ; then GO : 25 : : .0002043 : d. 25 .-. rf = ~ X .0002043 = .0000851. m .'. cos 44° 35' 25" = .7122303 - .0000851 '■ ' =7121452. Similarly, we may find the values of the other trigono- metric functions, remembering that, in the lirst quadrant, the tangent and secant increase and the cotangent and cosecant decrease, as the angle increases. 1. Given find 2. Given find 3. Given find 4. Given find •y, , 5. Given find EXAMPLES. sin 44° 35'= .7019459, sin 44° 36'= .7021531; sin 44° 35' 25". sin42°15'=.G723GG8, sin 42° 1G'= .0725821; sin 42° 15' IG". sin 43° 23'= .G8G87G1, sin43°22'=.G8GG647; sin 43° 22' 50". sin 31° G' =.5105333, sin 31° 7' =.51G7824; sin 31° G' 25". cos74°45'=.42G5G87, cos 74° 40' =.4263056; cos 74° 45' 40". Ahs. .7020322. .6724242. .6868408. .5166371. .4263933. ■* M 1 '^^^ H m 6. Given 1 1 ^"^ I m 7. Given -■1* ).■ (;; find PLANE TRIGONOMETRY. COS 41° 13'= .7522233, cos 41° 14'= .7520316; cos 41° 13' 26". cos 47° 38'= .6738727, cos 47° 39'= .6736577; cos 47° 38' 30". Ans. .7521403. .6737652. 74. To find the Angle whose Sine is Given. Find the angle whose sine is .5082784, having given from sin 30° 33' = .5082901 sin 30° 32' = .5080396 ' ' . the table diff. for 1' = .0002505 given sine = .5082784 sin 30° 32' = .5080396 diff. = .0002388 Let d = diff. between 30° 32' and required angle ; then .0002505 : .0002388 : : 60 : d. d = 2388 X 60 _ 6552 2505 167 = 57.2 nearly. .♦. required angle = 30° 32' 57".2. 75. To find the Angle whose Cosine is Given. Kind the angle whose cosine is .40432§1, having given lioni the ^^,ble cos 66° 9' = .4043436 cos 66° 10' = .4040775 ',.'■-' diff. for 1' = .0002661 cos 66° 9' = .4043436 given cosine = .4043281 diff. = .0000155 > k EXAMPLES. 109 Let d = diff. between 66° 9' and required angle ; then .0002661 : .0000155 : : 60 : d. 2661 Required angle is greater than 66° 9' because its cosine is /ess than cos 66° 9'. ,'. .-. required angle = 66° 9' 3".5. . - EXAMPLES. sin 44° 12' = .6071651, ' . sin 44° 11'= .6969565; whose sine is .6970886. Ans. 44' 11' 38". V sin 48° 47' = .7522233, ' sin 48° 46' = .7520316; whose sine is .752140. 1. Given find the angle 2. Given find the angle 3. Given 48° 46' 34" find the angle 4. Given find the angle 5. Given find the angle 6. Given sin 24° 11' = .4096577, sin 24° 12' = .4099230 ; whose sine is .4097559. ; '"' «os 32° 31' =.8432351, cos 32° 32' = .8430787; whose cosine is .8432. cos 44° 11' = .7171134, cos 44° 12' = .7169106; whose cosine is .7169848. 24° 11' 22".2. 32°31'13".5. 44° 11' 38 cos 70° 32' = .3332584, cos 70° 31' = .3335326 ; find the angle whose cosine is .3333333. r- 70° 31' 43".6 110 PLANE TRIGONOMETRY. 76. Use of Tables of Logarithmic Trigonometric Func- tions. — Since the sines, cosines, tangents, etc., of angles are numbers, we may use the logarithms of these numbers in numerical calculations in \vhit;li trigonometric functions are involved ; and these logarithms are in practice much more useful than the numbers themselves, as with their assistance we are able to abbreviate greatly our calcula- tions ; this is especially the case, as we shall see hereafter, in the solution of triangles. In order to avoid the trouble of referring twice to tables — first to the table of natural functions for the value of the function, and then to a table of logarithms for the logarithm of that function — the log- arithms of the trigonometric functions have been calculated and arranged in tables, forming tables of the logarithms of the sines, logarithms of the cosines, etc. ; these tables are called tables of logarithmic sines, logarithmic cosines, etc. Since the sines and cosines of all angles and the tangents of angles less than 45° are less than unity, the logarithms of these functions are negative. To avoid the inconvenience of using negative characteristics, 10 is added to the logarithms of all the functions before they are entered in the table. The logarithms so increased are called the tabular logarithms of the sine, cosine, etc. Thus, the tabular logarithmic sine of 30° is ^ ^ , , 10 + log sin 30° = 10 -}- log^ = 10 - log 2 = 9.6989700. In calculations we have to remember and allow for this in(;rease of the true logarithms. When the value of any one of the tabular logarithms is given, we must take away 10 from it to obtain the true value of the logarithm. Thus in the tables we find , ■ . ' log sin 31° 15' = 9.7149776. Therefore the true value of tlie logarithm of the sine of 31° 15' is 9.7149776 - 10 = i.7l4977(;^ Similarly with the logarithms of other functions. ^^' * TABLES OF LOGARITHMIC FUNCTIONS. Ill Note. — Eiigligh authors uauiilly denote tlieso tabular logarithms by the letter L. Thus, IjsIii a denotes the tubular loguriiliin of the Bine of A. French authors use the logurithuis of the tables diminished by 10. Thus, log sin A = 1.8598213, instead of 9.8598213. The Tables contain the tabular logs of the functions of all angles in the first quadrant at intervals of 1' ; and from these the logarithmic functions of all other angles can be found.* Since every angle between 45° and 90° is the complement of another angle between 45° and 0°, every sine, tangent, etc., of an cngle less than 45° is the cosine, cotangent, etc., of another ingle greater than 45° (Art. 16). Hence the degrees at th^^/ top of the tables are generally marked from 0° to 45°, and those at the bottom from 45° to 90°, while the minutes are marked both in the first column at the left, and in the last column at the right. Every number tliere- fore in each column, except those marked diff., stands for two functions — the one named at the top of the column, and the complemental function named at the bottom of the column. In looking for a function of an angle, if it be less than 45°, the degrees are found at the top, and the minutes at the left-hand side. If greater than 45°, the degrees are found at the foot, and the minutes at the right-hand side. On page 113 is a specimen page of Mathematical Tables. It gives the tabular logarithmic functions of all angles between 38° and 39°, and also of those between 51° and 52°, both inclusive, at intervals of 1'. The names of the functions for 38° are printed at the top of the page, and those for 51° at the foot. The column of minutes for 38° is on the left, that for 51° is on the right. Thus we find log sin 38° 29' =9.7939907. log cos 38° 45' = 9.8920303. log tan 51° 18' = 10.0962856. ', . jktSi^' ' i: Wl^f wm * Many tables are calculated for anglet at iotervaU of 10". 112 PLANE TRIGONOMETRY. ^m . :.; ■ '■ *t ^ iNl' • ■ i 1 ' .; iL. ' r iir t-'- ■■■' !l M : ■ ■, I ■ 1 :i .1.: I m i: 77. To find the Logarithmic Sine of a Given Angle. Find log sin 38° 52' 40". We have from page 113 log sin 38" 53' = 9.7977775 log sin 38° 52' = 9.797()2()8 diff. for 1'= .0001567 ' -.^' Let d = diff. for 46", and assuming that the change in the log sine is proportional to the change in the angle, we have 60 : 46 :: .0001567 : d. ... d = 4^_x_M>lf!^=. 0001201. r • . . 60 .-. log sin 38° 52' 4.6" =0.7976208 + .0001201 = 9.7977409. 78. To find the Logarithmic Cosine of a Given Angle. Find log cos 83° 27' 23", having given from the table ^ log cos 83° 27'= 9.0571723 ; • . "-■: : ■■ ' log cos 83° 28' = 9.0560706 diff. fori' = .0011017 ' ■ Let d= decrease of log cosine for 23"; then 60 : 23 :: .0011017 : d. . .: -: ^ . <- 23 X .0011017 •. d = 60 = .0004223, nearly. .-. log cos 83° 27' 23"= 9.0571723 - .0004223 = 9.0567500. EXAMPLES. 1. Given log sin 6° 33'= 9.0571723, { find log sin 6° 32'= 9.0560706: log e;« fi° ^'}' 'A7II sin 6° 32' 37 Ans, 9.05675. \. 38 Peg. TABLE OF LOGARITHMS. 113 Sine. 9.7893420 9.7895036 9.7896652 9.7898266 9.7899880 .?790I493_ 9.7903104 7 i 97904715 8 9 10 II 12 13 IS 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 f 9.7906325 9.7907933 9.7909541 9.791 1 148 9.7912754 9-7914359 97915963 9.7917566 16 9.7919168 17 I 9.7920769 18 j 9.7922369 19 I 97923968 20 97925 566 21 ' 9.7927163 22 I 9.7928760 97930355 97931949 97933543 97935135 97936727 97938317 97939907 ^.7941496 9 9 9 9 _9 9 9 9 9 9 7943083 ,7944670 ,7946256 ,7947841 79494?S ,7951008 7952590 795417 I 7955751 7957330 7958909 ,7960486 ,7962062 ,7963638 7965212 ,7966786 7968359 ,7969930 ,7971501 _797397i 9.7974640 9.7976208 9.7977775 97979341 9.7980906 9.7982470 9.7984034 97985596 9.7987158 9.7988718 Diff. 616 6x6 614 614 613 611 611 610 608 608 607 606 60s 604 603 602 601 600 599 598 597 597 595 594 594 592 i 592 590 590 589 587 587 586 585 584 583 582 581 580 579 579 577 576 576 574 574 573 571 571 570 569 568 567 566 565 564 564 562 562 560 Tang. I Diff. { Ootang. Diff. 9.8928098 9.8930702 9-8933306 9.8935909 9.893851 1 9.8941 114 9-8943715 9.8946317 9.8948918 9.8951519 9.89541 19 9.8956719 9-8959319 9.8961918 9.8964517 9.89671 16 9.8969714 9.8972312 9.8974910 9.8977507 9.8980104 9.8982700 9.8985296 9.8987892 9.8990487 9.8993082 9-8995677 9.8998271 9.9000865 9.9003459 9.9006052 9.900864s i 9.901 1237 I 9.9013830 ; 9.9016422 9-90}9pj3 ' 9.9021604 I 9.902419s 9.9026786 ; 9.9029376 /I 9.9031966 9-9034555 ' 9.9037144 9-9039733 9.9042321 9.9044910 9.9047497 9.9050085 9.9052672 9-9055259 9.9057845 9.9060431 9.9063017 9.9065603 9.9068183 9^9070773 9-9073357 9.9075941 9.9078525 9.908 1 109 9.9083692 Oosine. | Diff. ! Ootang. 2604 2604 2603 2602 2603 2601 2602 2601 2601 2600 2600 2600 2599 2599 2599 2598 2598 2598 2597 2597 2596 2596 2596 2595 2595 2595 2594 2594 2594 2593 2593 2592 2593 2592 2591 2591 2591 2591 2590 2590 2589 2589 2589 2588 2589 2587 2588 2587 2587 2586 2586 2586 2586 2585 2585 2584 2584 2584 2584 2583 0,1071902 0.1069298 0.1066694 0.1 06409 1 0.1061489 0.1058886 0.1056285 0.1053683 0.1051082 0.104848 1 0.1045881 0.1043281 0.104068 1 0.1038082 0.1035483 0.1032884 0.1030286 0.1027688 0.1025090 0.1022493 0.1019896 0.1017300 0.1014704 0.1012108 0.1009513 0.1006918 0.1004323 0.1001729 0.099913s 0.0996541 0-0993948 0-0991355 0.0988763 0.0986170 0.0983578 0.0980987 0.0978396 0.0975805 0.0973214 0.0970624 0.0968034 0.0965445 0.0962856 0.0960267 0.0957679 0.0955090 0.0952503 0.0949915 0.0947328 0.0944741 0.0942155 0.0939569 0.0936983 0.0934397 0.0931812 0.0929227 0.0926643 0.0924059 0.0921475 0.0918891 0.0916308 Diff. i Tang. 987 988 988 989 990 990 991 992 992 992 993 994 995 995 995 997 996 998 998 998 1000 999 001 001 001 003 002 004 004 004 005 006 007 007 007 008 009 010 010 010 on 012 013 013 013 014 015 016 016 oi6 018 017 019 019 020 020 021 021 022 023 Oosine. 9.8965321 9.8964334 9.8963346 9.8962358 9.8961369 9.8960379 9.8959389 9.8958398 9.8957406 9.8956414 _9:8955422 9.8954429 9-8953435 9.8952440 9.8951445 9.8950450 9.8949453 9-8948457 9.8947459 9.8946461 9.8945463 9.8944463 9.8943464 9.8942463 9.8941462 9.8940461 9-8939458 9.8938456 9.8937452 9.8936448 9.8935444 9.8934439 9-8933433 9.8932426 9.8931419 9.8930412 9.8929404 9-8928395 9-8927385 9.892637s 9.8925365 9-8924354 9.8923342 9.8922329 9.8921316 9.8920303 9.8919289 9.8918274 9.8917258 9.8916242 9.8915226 Diff. 9.8914208 9.8913191 9.8912172 9.8911153 9.8910133 9.8909113 9.8908092 9.8907071 9,8906049 9.8905026 60 59 58 57 56 55 54 53 52 51 50 49 48 47 46 45 44 43 42 I 41 J40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 IS 14 I 13 ! 12 i 11 10 9 8 7 6 5 4 3 2 I o Sine. f ■ tf'-: : 51 Degr. 114 If 2. Given find 3. Given find 4. Given find PLANE TRIGONOMETRY. log sin 55" 33'= i).91()2530, log sin 55° 34'= U.U1G34()G ; log sin 55° 33' 54". log cos 37° 28'= 9.8()0(>604, log cos 37° 29'= 9.8995636 ; log cos 37° 28' 36". log cos 44° 35' 20'= 9.8525789, log cos 44° 35' 30"= 9.8525582 ; logcos44°35'25".7. Ans. 9.9163319. 9.8996023. 9.8525671. See foot-note of Art. 76. 5. Given log cos 55° 1 1 ' = 9. 7565999, log cos 55° 12'= 9.7561182; logcos 55° 11' 12". logtan27° 13'= 9.7112148, log tan 27° 14'= 9.7115254 ; log tan 27° 13' 45". find 6. Given find 9.7565636. 9.7114477. 79. To find the Angle whose Logarithmic Sine is Oiven. Find the angle whose log sine is 8.8785940, having given from the table logsin4° 21'= 8.8799493 log sin 4° 20'= 8.8782854 diff. forl'= .0016639 given log sine = 8.8785940 log sin 4° 20'= 8.8782854 diff. = ~T0003086 Let d = diff. between 4° 20' and required angle ; then .0016639 : .0003086 : : 60 : d. , 3086 X 60 .- , , . •, d= ~ — = 24, nearly. 16639 ' ^ .'. required angle = 4° 20' 24". EXAifPLES. 116 80. To find the Angle whose Logarithmic Cosine is Given. Find the angle whoso log cosine is 9.8934342. We have from page 113 log cos 38° 31'= 9. 8934430 log cos 38° 32' = 9.8933433 cliff, for 1'= .0001006 log cos 38° 31'= 9.8934439 given log cosine = 9.8934.342 cliff. = .0000097 Let d = (liff. between 38° 31' and required angle ; then .OOOIOOG : .0000097 :: GO : d. " " '"' .^^^)00O()97 .,^97x«0^ .0001006 1006 .-. required angle = 38° 31' 5".8. Note. — In using both the tables of the nntural sines, cosines, etc., and the tables of tlie loffarithmk; sines, cosines, etc., the stiuient will remember that, in the first quadrant, as the uiikIu increases, the sine, taugvMit, and secant increase, but the cosine, cotangent, and cosecant decrease. EXAMPLES. , V ':::,■..■•.:■'■' 1. Given log sin 14° 24'= 9.3956581, log sin 14° 25'= 9.3961499 ; find the angle whose log sine is 9.3959449. Ans. 14° 24' 35". 2. Given log sin 71° 40'= 9.9773772, log .sin 71° 41'= 9.9774191; find the angle whose log sine is 9.9773897. 71° 40' 18". 3. Given log cos 28° 17'= 9.9447862, log cos 28° 16'= 9.9448541 ; find the angle whose log cosine is 9.9448230. 28° 16' 27".5. It ~»t V 116 PLANE TRIGONOMETRY. 4. Given log cos 80° Tni' = '.). 1 ... BXAMPLBS, 1. Given log,,, 7 = .8450980, find log,,, 343, log,„ 2401, and logio 16.807. Ans. 2.5352940, 3.3803920, 1.2254900. 2. Find the logarithms to the base 3 of 9, 81, \, ^j, .1, -^j. Ans. 2, 4, - 1, - 3, - 2, - 4. 3. Find the value of log2 8, loga.5, logy 243, log5(.04), logiolOOO, log,o .001. Ans. 3, - 1, 5, - 2, 3, - 3. 4. Find the value of log„ a*, \og(,Vb% logs 2, log^; 3, logmolO. ^ns. I, I, I, I, |. Given log,o 2 = .3010300, log,o 3 = .4771213, and logio7== .8450980, find the values of the following : . . ' .,..,-...,*•= *•••.■,»• . 5. Iogio35, logiul50, logio.2. Ans. 1.544068, 2.1760913, 1.30103. 6. logio 3.5, log,„ 7.29, log,o. 081. _ Ans. .5440680, .8627278, 2.9084852. 7. log,o|, log,„3^1og,o\/-i?. Ans. .3679767, 2.3856065, .0780278. J. 118 PLANE TiilGONOMETRY. 8. Write down the integral })ai't of the common loga- rithms of 796o, .1, 2.61, 79.6341, 1.0006, .00000079. Ans. 3,-1, 0, 1, 0, - 7. 9. Give the position of the first significant figure in the numbers whose logarithms are 2.4612310, 1.2793400, 6.1763241. 10. Give the position of the first significant figure in the numbers whose logarithms are 4.2990713, .3040595, 2.5860244, 3.1760913, 1.3180633, .4980347. Ans. ten thousands, units, hundreds, 3rd dec. pi., 1st dec. pi., units. , . . 11. Given log 7 = .8450980, find the number of digits in the integral part of 7'", 49«, 343'"", (Y)^", (4.9)'^ (3.43)'". Ans. 9, 11, 85, 4, 9, 6. 12. Find the position of the first significant figure in the numerical value of 20^, (.02)^, (.007)^ (3.43) ''o, (.0343)«, (.0343)t'5. . , Ans. tenth integral pi., 12th dec. pi., 5th dec. pi., units, 12th dec. pi., 1st dec. pi. Show how to transform 13. Common logarithms to logarithms with base 2. Ans. Divide each logarithm by .3010."). 14. Logarithms with base 3 to common logarithms. Ans. Multiply each log by .4771213. 15. Given logio2 = ,3010300, find logglO. 3.32190. 16. Givenlog,o7 = .S450980, findlog^lO. \ 1.183. 17. Given log,o 2=- .3010300, find logs 10. 1.10730. 18. The mantissa of the log of 85762 is 9332949; find (1) the log of a/.0085762, and (2) the number of figures in (85762)", when it vi multiplied out. Ans. (1) 1.8121177, (2) 55. EXAMPLES. 119 19. What are the characteristics of the logarithms of 3742 to the bases 3, G, 10, and 12 respectively ? . ^;l ■ / : Ans. 7, 4, 3, 3. 20. Prove that 7 log f| + G log | + 5 log f + log || = log 3. 1 Aus. 2 + 21. Given log.o 7, find log, 490. 22. From 5.3429 take 3.G284. 23. Divide 13.2G15 by 8. 24. Prove that G log | + 4 log ^\ + 2 log -\^- = 25. Find log |297VIl| » to the base 3Vn. Given log 2 = .3010300, log 3 = .4771213. logio 7 3.7145. 2.407G. 1.8. 2G. Fijid log21G, G480, 5400, A Ans. 2.3344539, 3.8115752, 3.7323939, r.G478174. 27. Find log .03, G"', (5^)"^. Ans. 2.4771213, I.740G1G2, 1.G36500G. 28. Find log. 18, log 2.4, log^^g-. , A71S. i.255272G, .3802113, 1.2730013. 29. Find log (G.25)^, log4VJ005. .113G971, 1.45154. log 5G321 = 4.750G704, log 56322 = 4.750G781; log 5632147. 6.7506740. log 53403 = 4.7275657, log 53402 = 4.7275575; log 5340234. 6.7275603. log564l2 = 4.7513715, log 56413 = 4.7513792 ; log 564.123. 2.7513738. 30. Given find' 31. Given find 32. Given find 120 PLANE TRIGONOMETRY. 33. Given find 34. Given find 35. Given find 36. Given find 37. Given find :■ • ■■ ., 38. Given find 39. Given iind the number 40. Given find the number 41. Given find the number Ans. 4.9413333. log 87364 = 4.9413325, log 87365 = 4.9413375; log .0008736416. log 37245 = 4.5710680, log 37246 = 4.5710796; log 3.72456. .5710750. log 32025 = 4.5054891, log 32026 = 4.5055027; log 32.025613. 1.5054974. log 65931 = 4.8190897, log 65932 = 4.8190962; ■ ; log .000006593171. _^., 6.8190943. log 25819 = 4.4119394, ; ' log 25820 = 4.4119562 ; : ; x %> log 2.581926. .4119438. log 23454 = 4.3702169, , ft ; *:: log 23453 = 4.3701984; log 23453487. ■ , v , ^ T.3702074. log 45740 = 4.6602962, ^ . : : ' .v> log 45741 = 4.6603057; whose logarithm is 4.6602987. 45740.26. log 43965 = 4.6431071, - r log 43966 = 4.6431170; ^ ' whose logarithm is 4.6431150. .000439658. log ^6891 = 4.7550436, log 56892 = 4.7550512; whose logarithm is .7550480. 5.689158. m EXAMPLES. 121 42. Given log 34572 = 4.5387245, log 34573 = 4.5387371; find the number whose logarithm is 2.5387359. Alls. 345.7291. 43. Given log 10905 = 4.037G257, , log 10906 = 4.0376655 ; find the number whose logarithm is 3.0376371. 1090.5286. 44. Given log 25725 = 4.4103554, log 25726 = 4.4103723; find the number whose logarithm is 7.4103720. Alls. .00000025725982. In the following six examples the student must take his logarithms from the tables. 45. Required the prciuct of 3670.257 and 12.61158, by logarithms. ' Aiis. 46287.74. 46. Required the quotient of .1234567 by 54.87645, by logarithms. Ans. .002249721. 47. Required the cube of .3180230, by logarithms. Ans. .03216458. 48. Required the cube root of .3663265, by logarithms. A71S. .7155316. 49. Required the eleventh root of 63.742. 50. Required the fifth root of .07. 51. Given sin 42° 21'= .6736577, • sin 42° 22' =.6738727; ■ '■- sin 42° 21' 30". 1.45894. .68752. find -''■ • 52. Given find .6737652. sin 67° 22'= .9229865, sin 67° ^3' = .9230984 ; sin 67° 22' 48".6. .9230769. 122 PLANE TRIGONOMETRY. 53. Given find 54. Given find 55. Given find sin 7° 17' = .1267761, sin 7° 18' = .1270646; sin 7" 17' 25". cos 21° 27' = .9307370, cos 21° 28' = .9306306 ; cos 21° 27' 45". cos 34° 12' = .8270806, cos 34° 13' = .8269170; cos 34° 12' 19".6. Ana. .1268963. .9306572. .8270272. 66. Given sin 41° 48' = .0665325, ■-■■•' sin 41° 49' = .6667493; . find the angle whose sine is .6666666. 67. Given sin 73° 44' = .9599684, sin 73° 45' = .9600499; find the angle whose sine is .96. 58. Given cos 75° 32' = .2498167, ■' ^ 1, . cos 75° 31' =.2500984; find the angle whose cosine is .25. ! 69. Given cos 53° 7' = .0001876, cos 53° 8' = .5999549; find the angle whose cosine is .6. 60. Given log sin 45° 16' = 9.8514969, 1 log sin 45° 17' = 9.8516220 ; ' find log sin 45° 16' 30". 61. Given log sin 38° 24' = 9.7931949, log sin 38° 25' = 9.7933543 ; find log sin 38° 24' 27". 41° 48' 37 73° 44' 23".2. 76° 31' 21' 63° T 48".4. 9.8515594. 9.7932666. EXAMPLES. 123 62. Given find 03. Given find 64. Given find 65. Given find 66. Given find 67. Given find 68. Given find 69. Given find ■■ . 70. Given log sin 32° 28' = 9.7298197, log sin 32° 29' = 9.7300182 ; log sin 32° 28' 36". Ans. 9.7299388. log sin 17° 1' = 9.4603483. log sin 17° 0' = 9.4659353 J log sin 17° 0'12". ' find log sin 26° 24' = 9.6480038. log sin 26° 25' = 9.6482582; log sin 26° 24' 12". log cos 17° 31' = 9.9793796, log cos 17° 32' = 9.9793398 ; logcosl7°31'25".2. log tan 21° 17' = 9.5905617, log tan 21° 18' = 9.5909351 ; log tan 21° 17' 12". log tan 27° 26' = 9.7152419, log tan 27° 27' = 9.7155508; log tan 27° 26' 42". log cot 72° 15' = 9.5052891, log cot 72° 10' = 9.5048538 ; log cot 72° 15' 35". log cot 36° 18' = 1 0.1339650, log cot 36° 19' = 10.1337003 ; log cot 36° 18' 20". log cot 51° 17' = 9.9039733, log cot 51° 18' = 9.9037144 ; log cot 51° 17' 32". 9.4660179. 9.6480547. 9.9793629. 9.5906364. 9.7154581. 9.5050352. 10.1338768. 9.9038352. 124 PLANE TRIGONOMETRY. ■'. ' ■ .r.i:i il I tvyr,-,- illf i i 71. Given log sin 16° 19' = 9.4486227, log sin 16° 20' = 9.4490540 ; find the angle whose log sine is 9.4488105. Ans. 16° 19' 26". 72. Given log sin 6° 53' =9.0786310, log sin 6° 53' 10"= 9.0788054 ; find the angle whose log sine is 9,0787743. 73. Given logcos 22° 28' 20"= 9.9657025, log cos 22° 28' 10"= 9.9657112 ; find the angle whose log cosine is 9.9657056. 6° 53' 8 ' CM 22° 28' 16' In the following examples the tables are to be used : 74. Find log tp.n 55° 37' 53". Ans. 10.1650011. 7^. Findiogsin73°20'15".7. 9.9813707. 76. Find log cos 55° 11' 12". 9.7665636. 77. Find log tan 16° 0'2"r'. 9.4577109. 78. Find log sec 16° 0'27". 10.0171747. 79. Find the angle whose log cosine is 9.9713383. Ans. 20° 35' 16". 80. Find the angle whose log cosine is 9.9165646. Ans. 34° 23' 25". 81. Find log cos 34° 24' 36". 82. Find log cos 37° 19' 47". 83. Find log sin 37° 19' 47". 84. Findlogtan37°19'47". 85. Find log sin >° 18' 24".6. 86. Find log cos 32° 18' 24".6. 87. Find log tan32° 18' 24".6. 9.9164762. 9.9004540. 9.7827599. 9.8823059. 9.7279096. 9.9269585. 9.8009511. EXAMPLES. 125 Prove the following by the use of logarithms ; 88 (7014)'^- 1 ^ .9942207. (7.014)«+1 gg j^5:g_xj^00307j ^ .000232432. 90. W (2002)^'^'»x(i001g _9is^o.'^00000. \ 1001 X 2002 • Vfc 126 PLANE TRIGONOMETRY. CHAPTER V. SOLUTION or TRiaONOMETEIO EQUATIONS. 82. A Trigonometric Equation is an equation in which the unknown quantities involve trigonometric functions. The solution of a trigonometric equation is the process of finding the vahies of the unknown quantity which satisfy the equation. As in Algebra, we may have two or more simul- taneous equations, the number of angles involved being equal to the number of equations. EXAMPLES. 1. Solve sin^ = ^ This is a trigonometric equation. To solve it we must find some angle whose sine is We know that sin 30°= 1- 2 Therefore, if 30° be put for 6, the equation is satisfied. .-. $z= 30° is a solution of the equation. .-. ^ = n7r4-(-l)"- (Art. 38) 6 5 2. Solve cos ^ 4- sec ^ = -• The usual method of solution is to express all the func- ^■ions in terms of one of them. Thus, we put - for sec^, and get cos 9 1 cos^ + — — = - cos^ 2 TRIGONOMETRIC EQUATIONS. 127 This is an equation in which 9, and therefore cos^, is unknown. We proceed to solve the equation algebraically just as we should if x occu])ied the place of cos^, thus : CQS'O ^ cos 6 = — 1. '. cos^ = -±^ 4 4 2 or -. 1 2 The value 2 is inadmissible, for there is no angle whose cosine is numerically greater than 1 (Art. 21). cos ^ = • But cos 60° = 2' :; ; - .-. cos^ = cos60°. Therefore one value of 6 which satisfies the equation is 60°. 3. Solve cosec 6 - cot'^^ -f- 1 = 0, We have cosec - (cosec^ ^ - 1 ) + 1 = . . (Art. 23) cosec^^ — cosec ^ = 2. .'. cosec^ = - ± - Z Li = 2 or -1. . ; But cosec 30° = 2. .'. cosec ^ = cosec 30°. Therefore 30° is one value of which satisfies the equation . Find a value of B which will satisfy the following equa- tions: 4. cos^ = cos2) = ^, cos (6 4-c^) = 0. ^ = 60 ", = a (1) m cos are required. Dividing (1) by (2), we get * ' -^^ ' tan =-, ft which gives two values of , differing by 180°, and there- fore two values of m also from either of the equations b m = a sin cf> cos <^ The two values of m will be equal numerically with opposite signs. In practice, m is almost always ^jositive by the conditions of the problem. Accordingly, sin <^ has the sign of a, and cos the sign of 6, and hence <^ must be taken in the quad- rant denoted by these signs. These cases may be considered as follows : (1) Sin (f> and cos both positive. This requires that the angle <^ be taken in the first quadrant, because sin «^ and cos ^ are both positive in no other quadrant. or or TRIGONOMETRIC EQUATIONS. 129 (2) Sin ff) positive and cos <^ negative. This requires that j)(>sitive. This requires that be taken in the fourth quadrant, because only in this quadrant is sin<^ negative and cos t^ positive for the same angle. Ex. 1. Solve the equations msin. log m sin <^ = 2.52213 - r ■■ '^ log m cos <^ = 2.46252 . . " <^ = 48° 55'.2. m = 441.45. log tan <^ = 0.05061 log m sin <^ = 2.52213 log sin = 0.87725 log m = 2.64488 Ex. 2. Solve m sin <^ = - 72.631, and m cos ^ = 38.412. Ans. = 117° 52'.3, m = - 82.164. 84. Solve the equation , . a sin + 6 cos = c (1) a, b, and c being given, and ^ required. Eind in the tables the angle whose tangent is - ; let it be;8. ^. -: ^ ^ Then -=tan/8, and (1) becomes a , . or or "I a (sin sin (B\ _ , cos)3 = c; sin ( + ;8) = sin a, .-. <^ + /3 = mtt + (-!)'•« . . (Art. 38) .-. <^ = _^-|-rt7r + (-!)"«, where n is zero or any positive or negative integer. In order that the solution may be possible, it is necessary to have - cos /8 =, or < 1. a NoTS. — This example might bBve been solved by squaring both sides of the equation; but in solving trigonometric equations, it is important, if possible, to avoia squaring both sides of the equation. Thus, solve cos = ka\nO (3) If we square both sides wc get cos2 = A' slu2 fl = A»(l - cos* fl). .'. co8'9 = -^^; or cosO= ± — ^ — .......... (4) k Now if a be the least angle whose cosine = — , we get from (4) ^'l + k^ « = nir ± a (5) But (3) may be written cot B — k. .'. = nn + a (6) (6) is the complete solution of the given equation (3), while (5) is the solution of both cos fl = A; sin 6, and also of cos B = — k sin 9. Therefore by squaring both nu-m- bers of an equation we obtain solutions which do not belong to the given equation. EXAMPLES. 1. Solve 0.7466898 sin <^ - 1.0498 cos <^ = - 0.431689, when <^ < 180°. log 6 = 0.02112-* log a = 1.87314 log tan /? = 014798 - .-. /3 = 125° 25' 20". * The minus sign is written thus to denote that it belongs to the natural number and does not affect the logarithm. Sometimes the letter n is written instead of the minus sign, to denote the same thing. TBIGONOM-ETHIC EQ UA TIONS. 131 log sill 13 = 9.91111 ' log c= 1.63517 - colog 6 = 9.97888- logsin( + I3= 19° 34' 40" or 160° 25' 20". .-. «^ = -105°50'40" or 35°0'0". 2. Solve - 23.8 sin + 19-3 cos = 17.5 ((^ < 180"). Ans. <^ = 4° 12'.7 or - 106° 7'.9. 3. Solve 2 sin e + 2 cos ^ = V2. Ans. - --\-nn + (-1)"-- 4 6 4. " sin0 + V3cosd = l. 5. " sin^ — cos0 = 1. 6. " V3sin0-cos^ = V2. ^ir-{-7iir + (-iy\ir. 85. Solve the equation sin (« 4- ic) = m sino; in which a and m are given. From (1) we have sill (« 4- a?) 4- si^l x _ m-\-l (1) sin (« 4- a;) — sin a; m — 1 , ■' tanra;4-|j tan- ks',;' 2 -^ , / ,i\ wi4-li. « .'. tan (a; 4- 4a) = — ^2__^tan- m — 1 2 • ••..• (^^^ • • (Art. 46) . (3) which determines x -j- ^a, and therefore x. If we introduce an auxiliary angle, the calculation of equation (3) is facilitated. n'iff?. 132 PLANE TEIGONOMETRY. Thus, let m = tan<^ ; then we have by [(14) of Art. 61] m + 1 tan (6 + 1 i. / . t<-o\ — T-- = - — ^-3— = cot («^ - 4o ), m — 1 tan <^ — 1 which in (3) gives tan [a; + "]«= cot ( — 45°) tan^a This, with tan <^ = wi, gives the logarithmic solution. The logarithmic solution of the equation sin (a — a;) = m sin x is found in the same manner to be tan <^ = m, and tan [ a; — ^ j = cot (<^ -f 45°) tan -> which the student may show. Example. — Solve sin (106°+ x)=z- 1.263 sin a; (a; < 180°). log tan <^ = logm = log ( - 1.263) = 0.10140 -. .-. = m. Example. — Solve tan (23° 16'+ x) = .296 tana;, log tan <^ = logm = log(.296) = 1.47129. .-. ■■'■■». '■> .. cos(« -\-'2x) (Ex. 4 of Art. 47) .'. cos(rt + 2 a;) = m cos« where 1 -f- m = tan (45° - <^) cos « [ ( 16) of Art. 61] tan ^ = m. i.!i f, t i 134 PLANE TESONOMETBY. Example. — Solve tan (65° + a;) tana; = 1.519G(« < 180°). log tan tf> = login = 0.18173. .-. «^ = r.0='39'9"; 45° - <^ = - 11° 39' 9" ; log tan (45° -<^) = 9.31434- « = 65° 0' 0" ; log cos a = 9.62595 log cos (a + 2a;)= 8.94029- « + 2a; = 95° or 265°. . 2a; = 30° or 200°. .-. X = 15° or 100°. 88. Solve the equations m sin {B -\-x)=a (1) . m sin (<^ 4- a;) = 6 (2) for m and x, the other four quantities, 6, 4>, a, h, being known. Expanding (1) and (2) by (Art. 44), we get msin^cosa + mcos ^sin a; = a (3) wrin<^cosa;-f- mcos <^ sin a;= 6 (4) Multiplying (3) by sin «/!> and (4) by sin0, and subtracting the latter from the former, we liave m sin X (sin <^ cos 6 — cos <^ hin 6) = a sin <^ — 6 sin ^. a sin — 6 sin 6 ,^. sin((^-^) ^ To find the value of mcos.i;, multiply (3) and (4) by cos <^ and cos tf, respectively, and subtract the former from the latter. Thus m cos x(sin <^ cos — cos <^ sin 0)=h cos — a cos <^. ftcostf — acosd* ...^ .-. mcosa; = ^ ^ (6) sin(<^-e) ^ Having obtained the values of wsinx* and mcosa; from (5) and (6), in and x can be calculated by Art. 83. TRIGONOMETlilC EQUATIONS. 135 EXAMPLES. 1. Solve m COS (6 + x) = a, and w sin ( + .r ) = &, for m sin X and w cos x. a ■ l> t'■■-■' 186 PLANE TRIGONOMETRY. and quotients. An expression in the form of a jn'oduct or quotient is said to be adapted to logarithmic coftipntation. An angle, introduced into an expression in order to adapt it to logarithmic computation, is called a Subsidiary Angle. Such an angle was introduced into each of the Arts. 84, 85, 86, and 87. The following are further examples of the use of sub- sidiary angles : 1. Transform a cos 6 ±b sin 6 into a product, so as to adapt it to logarithmic computation. tan ; * thus acos^± &sin^ = af cos6 ± -sin0) . = a (cos 6 ± tan sin 0) Put ^ a a GOS cos(^T <^). a sin (^± <^). if Similarly, a sin 6 ±b cos 6 = QOStfi Transform a ±b into a product, a -f 6 = a ( 1 -f- J = « (1 + tan- <{>) ~ a sec^ , - = tan2, if - = surdi. a * The fundaraental foimulro coB(jr t y) and sin (a* + y) (Art. 42) afford o::amp1eg of one term equal to the Bum or difference of two terms ; hence we may transform an expression acosO t bsinO Into an equivalent product, hy conforming it to the for- muia* Just mentioned. Thus, comi)aring the identity, vi cos sin fl = m cos (0 ? 9) or m cos (6 ^^), witlincos0 t bH\n0, we will haven cos 9 h bfiine = mcoa(0 f if>) if we assume a-m cos (/> and 6 = See Art. 84. wising; i.e. (Art. 83), if tan 8in<^i - as above. ADAPTATION TO LOGARITHMIC COMPUTATION. 137 .-. log(a + ?>) = loga +21ogsec<^; and log (a — 6) = log a + 2 log cos cosO = a (1) rcos^sin^ = & (2) rsin, and $. Dividing (2) by (1), we have tan^ = -, a from which we obtain $. '' -..-'...■. ^ ' From (1) and (2) we have . rcos^ = -^ = -;^- (4) cos^ sin^ from which we obtain rcos<^. From (3) and (4) we obtain r and <^ (Art. 83). 1 1^- 1 i ' * 138 PLANE TRIGONOMETRY. 1. Solve for r, (f), 6. EXAMPLES. r COS COS 6 — — 53.953, rcos<^ sine = 197.207, rsin-{- 1) sin -{-c sin = a. Solving these equations for sin <^ and cos <^, we have . . be — a^ sm<^= , 0^ — ac , c- — ab coS(f> = -; ac — b^ which in cos'' <^ + sin^ <^ = 1, gives {be — a^y + (c^ — ab)''^ = {ac — Wy. 3. Eliminate <^ from the equations ycos -\-xcos = 2 a sin 2 <^. Solve for a; and y, then add and subtract, and we get X + y = «(sin<^ +cos<^)(l + sin 2^), x — y = a(sin<^ — cos y, (x-yy = a\l-8in2cl>y. ... (x + yy + {x-y)^ = 2J. 4. Eliminate a and /8 from the equations a = sin«cosj3sin 6 + cosacostf . . . (1) 6 = sinrt cos/8cos0 — cos« sin^ . . . (2) c = sinrt sin/3 sin^ (3) Squaring (1) and (2), and adding, we get a* + 6'' = sin^'rt cos'^/SH- cos'^rt (4) -^ = sm'us\n'l3 (5) I 140 PLANE TRIGONOMETRY, Adding (4) and (5), we have a- + t' +.''■' =1. sin=^^ 6. Eliminate ^ from the equations a sin ^ + h cos ^ = c, a COS ^ — 6 sin ^ = d. Ans. a? -\-W ^c^ -{■ d^. 6. Eliminate from the equations m = cosec 6 — sin tf, ?i = sec ^ — cos $. m^n^ (m) + n*) = 1. 7. Eliminate $ and <^ from the equations sin 6 + sin tf> = a, cos -\- cos = b, Gos{0-) = c. a2 + 62_2c = 2. 8. Eliminate a; and y from the equations tan X + tan ?/ = a, cot a; + cot y = b, x + y = c. cot c = 9. Eliminate from the equations X = cos 2 <^ + cos tji, y = sin2^ + sin <^. Ans. 2x=(x'^ + fy-3(x^-[-y^). EXAMPLES. Solve the following equations : 1. tan d + cot ^ = 2. 2. 2sin2^+V2cos^ = 2. 3. 3tan2^-4sin-^ = l. 4. 2sin2(9 4-V2sin^ = 2. a 1 — • b Ans. 45". 90% or 45°. 45^ 45". EXAMPLES. 141 5. cos2^-V3 cos ^4-^ = 0. 6. sin 5 ^ = 16 sin* ^. 7. sin9 6^ — sin^=sin4tf. 8. 2 sin ^ = tan 0. 9. 6cot20-4cos20 = l. 10. tan 6 + tan (^ - 45°) = 2. 11. cos^ + V3sin^ = V2. 12. tan (^ + 45°) = 1 + sin 2 ^.. 13. (cot0-tan0)''(2 + V3) = 4(2-V3) 14. cosec e cot ^ = 2 V3. 15. cosec $ + cot = V3. 16. sin- = cosec ^ — cot ^. 17. sin 5^ cos 3^ = sin 9^ COS 7^. 18. 8in2^ + cos2 26l = f. V3sin^-cos^ = V2. Ans. 30" nrr, or nir ± - ' 6 iwTT, or |j.7r ± -^ it) nTT, or 2 htt i: ; 2.w±| iiTT — -, or nir 4 2n7r±^ o 2n7r + ^7r, 2M7r, ^ig.n7r + (-l)- nTT ± :^, or 7i7r ± /^tt 19. 20. tan ^ + cot ^ = 4 21. 22. IT TT V3 V3 sin (^ + «^) = -^, cos (0 - <^) - 4' ^ 12 Solve m sin ^j!. «= 1.29743, and m cos <^ = 6.0024. Ans. (<^<180°) 142 PLANE TRIGONOMETRY. 'fi 23. Solve m sin «^= -0.3076258, and m cos 0=0.4278735. (m positive.) Ans. ^ = 324° 17' 6".6, m = 0.52098. . 24. Solve m sin <^ = 0.08219, and m cos = 0.1288. 25. Solve m sin = 194.683, and m cos = 8460.7. 26. If a sin ^4-6 cos d = c, and a cos ^ -j- & sin d = c sin 6* cos 6, show that sin 2 0{c^ — a- — ft'*) = 2 a6. Solve the following equations : 27. V2sin^4-V2cos^ = V3. Ans. _^-}-n7r + (- l)"!"- 28. 2 sin a; + 5 cos a; = 2. Sug. [2.5 = tan 68° 12']. .4715. a; = - 68° 12' + n 180° + ( - 1 )" (21° 48'). 29. 3cosa;-8sina; = 3. *Sug. [2.6 = tan 69° 26' 30"]. Ans. x = - 69° 26' 30" + 2 m 180° ± (69° 26' 30"). 30. 4 sin X - 15 cos x = 4. Sug. [3.75 = tan 75° 4']. Ans. X = 75° 4' + n 180° + (- 1)"(14° 56'). 31. cos (« 4- a;) = sin (« + x) -\- V2 cos p. Ans. a; = — « — ^4- Tnir ± /?. 32. cos ^ + cos 3 ^ + cos 5^ = 0. Ans. |(2ri4-l)7r, or ^(37i±l)7r. 33. sin 5 ^ = sin 3 ^ 4- sin ^ = 3 — 4 sin^ 6. Ans. Hit ± ^, or ^(2 71 4- 1)^. o nir 34. 2sin23'e4-sin26^ = 2. Ans. H2n + l)7r, or :^ + (- 1)"-^. 35. a(cos2^-l)4-2 6(cos(9 4-l) = 0. Ans. (2 7i4-l)7r, or cos~^-^^^^^ — a EXAMPLES. 143 36. Solve m sin(^ + a;) = a cos /8, and m cos(0 — x) = a sin /?, for msinic and mcosa;. (Art. 67.) Ans. m sm x = ^^-— — -, cos 2 m cos X = ^^ '-' cos U 37. Solve mcos(6' + <^) = 3.79, and m cos (^ -) = 2.00, for m and ^, when = 3r2TA. (Art. 67.) 38. Solve r cos <^ cos ^ = 1.271, r cos <^ sin 6 = — 0.981, r sin cos 6 = — 7.192, r cos <^ = 12.124, for r, ift, 6. 41. Solve cos (2 a; 4- 3 2/) = I, cos(3a; + 2//) = ^V3. Ans. X=^mr ±^jrir ±^'ir, y = ^ nir ± ^ir ± ^ir. 42. Solve cos3tf4-cos5e4-V2(cos^ + 3intf)cos^ = 0. Ans. A$±$ = 2mr±lTr, or |(27i + l)7r. 43. Solve cos3tf + sin3^ = cose + sin^. Ans. sin ^ = 0, or tan ^ = — 1 ± ■y/2. 44. Solve 3 sin ^ 4- cos ^ = 2 x, sin ^ + 2 cos ^ = a. Aris. e = 7rW, a;=|Vl0. Mi {'I ^^'. » !< i I'm <:! 144 PLANE TRIGONOMETRY, li 45. Solve l.L'dS sin <> = 0.1)48 + m sin (25° 27'.L'), 1.2G8 cos (^ = 0.281 + m cos (25° 27'.2). Ans. = 00° 53'.8, m = 0.872. 46. Transform x* -\- y* + z* - 'J fz- - 'J z'x- - '2 x'lf into a product. Ans. —{x-\-y-\-z){y-{-z — x){z+x—ij){x-\-y—z). 47. Eliminate 6 from the equations m sin 20= n sin ^, j:> cos 26 = q cos ^. ^n«. m^ +/)* = n^ -<- ^^ 48. Eliminate 6 and <^ from the equations x=:a cos"* ^ cos" tfi, y=b cos"* tf sin" , z = c sin" ^. A71S, f - l^H- ■,"»+ )"=l. \aj \bj \cj 49. Eliminate $ from the e; and show that " M* — 2 Mc cos ^ + c^ =: &* sec'' ^. EXAMPLES IN ELIMINATION. 145 54. Eliiniiiato 6 ami «/> from tlie e(|iuitiuns xcosO + y8m 6 = a, bain {6 + «^) = asiii , xoos{0-{-2) — y sin {$ + 2 <^) = a. nY 55. Eliminate from the e(j[uations a sec''^ + cos'*d' y = sec^^ + cos^^. 56. Eliminate 6 from the equations (a + 6) tan (^ — <^) = (a — 6) tan (^ 4- <^), 67. Eliminate from the equations • a A fzr~, — 2 cos^^ , sin'-'^ 1 or Ir x^ + 7f Ans. ^+-^ = 1. a^ b^ 58. Eliminate 6 and ^ from the equations a^ cos^ 6 — b"^ cop^ f{> = c^, a cos 6 -\-b cos 4> — r, a tan ^ = ?; tan . 59. Eliminate ^ from the equations »isin^ — ?rtcos^ = 2msin<^, , n sin 2$ — m cos 2 <^ = n. ^4ns. (wsin^ + mcos0)^ = 2?n,(m + 7i). 60. Eliminate a from the equations a; tan (« — /8) = 2/ tan (rt 4- /8), (a; — y) cos 2 « -f (a; + y) cos 2 /8 = 2. ^7}s. «^ + 4 a;y = 22 (ic 4- 2/) cos 2)3. II t it, W. '•"' !ii^ 146 PLANE TBIGONOMETRY, CHAPTER VI. BELATIONS BETWEEN THE SIDES OF A TKIANGLE AND THE FUNCTIONS OF ITS ANGLES. 93. FormulsB. — In this chapter we shall deduce forinulye which express certain relations between the sides of a tri angle and the functions of its angles. These relations will be applied in the next chapter to the solution of trianglen. One of the principal objects of Trigonometry, as its name implies (Art. 1). is to establish certain relations between the sides and angles of triangles, so that when some of these are known the rest may be determined. RIGHT TRIANGLES. 94. Let ABC be a triangle, right-angled at C. Denote the angles of the triangle by the let- ters A, B, C, and the lengths of the sides respectively opposite these an- gles, by the letters a, b, c* Then we have (Art. 14) the following relations : /^ o = csin A = ccosB = 6tan * = 6cotB . b —c sin B = c cos A = a tan B = sin C' * a b c a sin C sin A sin A sin B sin C a:b:c = sin A : sin B : sin C. 96. Law of Cosines. — In any triangle the square of any side is equal to the sum of the squares of the other two sides minus twice the product of these sides and the cosine of the included angle. In an acute-angled triangle (see first figure) we have (Geom., Book III., Prop. 26) •on' A ri2 BC = AC + AB - 2 AB X AD, or 0=* = 62-1- c'- 2c. AD. But AD = 6 cos A. .'. a' = 5' + (r* — 26c cos A. In an obtuse-angled triangle (see second figure) we have (Geom., Book III., Prop. 27) n2 or But BC = AC + AB' + 2 AB X AD, a^ = b^-{-c' + 2c-AI). AD s= 6 cos CAD = — b cos A. .'. a* = b^-\-(? — 2 be coa A. Similarly, 6* = c* -f- a* — 2 ca cos B, c' = o' + 6'-2a6cosC. Not*. — When one equnMon in the solution of triangleii has been obtained, the other two may generally be obtained by advancing the letters so tbat a becomes h, b becomes r, and c becomes n; the order is abc, bca, nib. It Is obvious that the forroule thus obtained are true, since the naming of the sides makes no difference, provided the right order is maintained. OBLIQUE TRIANGLES. 149 97. Law of Tangents. — In any triangle the sum of any two sides is to their difference as the tangent of half the sum of the opposite angles is to the tangent of half tfieir difference. By Art. 95, a : 6 = sin A : sinB. By composition and division, a + h _ sin A + sin B a — h sin A — sin B , ■ Similarly. ^+-« = *5tEi(liJlC) ^ b-c tan^(B-C) .: ^ ^ c + a _tan ^(C + A ) .^. •' c-a tan^(C-A) ^ ^ Since tan^(A + B) = tan(9()° ~iC)= cot^C, the result in (1) may be written a + & _ cot|C .^x a-b tan^(A-B) ' ^ - and similar expressions for (2) and (3). 98. To show that in any triangle c = a cos B -f & cos A. In an acute-angled triangle (first figure of Art. 90) we l^a.ve c = DB + DA = a cos B + 6 cos A. In an obtuse-angled triangle (second figure of Art. 96) we have c = DB-DA =.: a cos B — 6 cos CAD. .*. c = acosB + 6cos A. Similarly, 6 = c cos A -f- a cos C, a — b cos C -f- c cos B. *->^ ill .1 160 PLANE TRIGONOMETHY. EXAMPLES. 1. In the ti^iangle ABC prove (1) a-\-h:c= cos |(A — II) : sin \ C, and (2) a — 6 : c = sin ^ ( A — B) : cos | C. 2. If AD bisects the angle A of the triangle ABC, prove BD : DC = sin C : sin B. 3. If AD' bisects the external vertical angle A, prove BD':CD' = sinC:sinB. 4. Hence prove 1 ^ 2 cos j A cosHB^zCj. ' BC asinB and also 1 ^ 2sin^Asin^(C-B) D'C asinB 99. To express the Sine, the Cosine, and the Tangent of Half an Angle of a Triangle in Terms of the Sides. I. By Art. 96 we have cos A = 62 + c» a' Let then 26c = 1 — 2 sin * 9 •^^ 1 r» * _ 2 (Art. 49) 2sin*A = i_^±^_?: 2 26c ^ a8-(6-c)» ^ 26c — (ct + 6 — c ) {a - ■" 26c i^ 6tc), a + 6 4-c = 2«; a + 6 — c = 2(.9--c), and o — 6 + c = 2(s — 6). ., 28in'^ = 2i«-c)2(^-61. 2 26c sin ^v is-h)U-cX .... 6c (1) OBLIQUE TUIANGLEH. 151 Similarly, sin|^ = ^/5H3SEl0 . . . . . ." (O) II. cosA = 2cos2^-l (Art. 49) .'. 2cos^ = 1 H -^ — 2 26c 26c _ (g + 6 + c) (6 + c — g) 26c 2s.2(s-g) , " II- ■ — ■..- ^ z. » 26c A js(s — a) /,. Similarly, co3| = J'^'"** ) • • (5) „^„C /s(s — c) ,^. cos— = a/-^' ^ (6^ 2 \ g6 ^^ III. Dividing (1) by (4), we get -t=V^^^!^ (^) Similarly/tan? = x/5H3IZE«l (8) tau5=J5^;SZES (9) 2 \f s{s-c) ^ ^ Since any angle of a triangle is < 180°, the half angle is < 90° ; therefore the positive sign must be given to the radicals which occur in this article. J' *' ■? 'i 152 PLANE TRIGONOMETRY. 100. To express the Sine of an Angle in Terms of the Sides. sin A = 2 sin — cos — (Art. 49) \ be M be (Art. 99) /. sin A = - V« (s — a) (s — b){s — c). be Similarly, sin B = - Vs (s — a) (s — 6) (.s — c), ac 2 Cor. sin C = — Vs (s — a) (s — 6) (s — c). 1 , /o^2 sin A = -!- V"26V-f" 2cV+ 2a262_ a*- b*- c\ 2 be and similar expressions for sin B, sin C. EXAMPLES. In any triangle ABC prove the following statements : 1. a (6cosC — ccosB) = 6^^ — cl 2. (6 + c)cosA-f(c + a)cosB + (a + 6)cosC = a + 6 4-c. o sin A + 2sinB _ sinC a + 26 ~"7~' 4 sin' A — m sin'' B _ si n'C a^-mb^ ~~~^' 5. a cos A + 6 cos B — c cos C = 2 c cos A cos B. cos A sin B sin C + cosB 7. a8in(B-C) 8. tan ^ A tan ^ ; sin C sin A + b sin (C B-*-^. + -.- cosC sin A sin B ^2. -A) + csin(A-B) = 0. 9. tan^A-4-tan^B = (.s-6)H-(.'»~c). AREA OF A TRIANGLE. 153 101. Expressions for the Area of a Triangle. (1) Given two sides and their included angle. Let S denote the area of the tri- angle ABC. Then by Geometry, 2S = cxCD. But in either figure, by Art, 94, CD = Z>sinA. .-. S = 1 6c sin A. Similarly, S = J[^ ac sin B, S = ^ a6 sin C. (2) Given one side and the angles. Since which is a'.b = sin A : sin B (Art. 95) , a sin B sin A S = ^ n6 sin C, gives a- sin B sin C S = Similarly, S = 2 sin A h- sin A sin C c^ sin A sin B 2sinB 2 sin C (3) Given the three sides. sin A = ^ \/s{s - a) (s - b) (s - c) (Art. 100) be Substituting in we get S = ^ 6c sin A, S = \/s(s-a){s-b)(s-c). I 154 PLANE TinGONOMETRY. 102. Inscribed Circle. — To find the radius of the iusanbed circle of a triamjle. C Let ABC be a triangle, the centre of the inscribed circle, and r its radius. Draw radii to the points of contact D, E, F ; and join OA, OB, OC. Then a c D S = area of ABC =A AOB + A BOC + A CO A = ^ re 4- i »*« + i t'h a-\-h -\-c — r — = rs . . . . (Art. 99) ... r = ^ =J('-^)(''-miJ^l . (Art. 101) S \ 8 t 103. CircuniBcribed Circle. — To find the radius of the circumscribed circle of a triangle in ^q terms of the sides of the triangle. Let be the centre of the circle Ai described about the triangle ABC, and R its radius. Through O draw the diameter CD and join BD. D" Tken Z BDC = Z BAC = Z A. .-. BC = 2 R sin A, or « = 2 R sin A. a b c But .-. R = sin A = 2 sin A 2 sin B 2 sin C 6c Tj ahc . . (1) (Art. 101) . . (2) flli BADII OF THE ESCRIBED CIRCLES. 165 104. Escribed Circle. — To Jim! the radii of the escribed circles of a trianyle. A circle, which touches one side of a triangle and the other two sides produced, is called an escribed circle of the triangle. Let O be the centre of the escribed circle which touches the side BC and the other sides pro- duced,* at the points D, E, and F, respectively, and let the radius of this circle be rj. We then have from the figure A ABC = A AOB -f A AOC - A BOC. S n = 8 — a r,(s-a). (Art. 99) (1) Similarly it may be proved that if rg, r.j are the radii of the circles touching AC and AB respectively, S S ro = 8-6 r.= y '3 105. To find the Distance be- tween the Centres of the Inscribed and Circumscribed Circles^ of a Triangle. Let I and be the incentre and circumcentre, respectively, of the triangle ABC, I A and IC bisect the angles BAG and BCA ; * Often called the incentre and circumcen- tre of a triangle. 166 PLANE TRIGONOMETRY, {1^ ; therefore the arc BD is equal to the arc DC, and DOH bisects BC at right angles. Draw IM perpendicular to AC. Then ZDIC _A + C_ BCD + BCI = DCI. Also, that is, , DI = DC = 2Rsin^. A A AI = IM cosec — = r cosee — • 2 2 .-. DI.AI = 2Rr = EI.IF; (R + 0I)(R-0I) = 2Rr. .-. Ol' = R'^-2Rr. EXAMPLES. 1. The sides of a triangle are 18, 24, 30 ; find the radii of its inscribed, escribed, and circumscribed circles. Ans. 6, 12, 18, 36, 15. 2. Prove that the area of the triangle ABC is 2cotAH-cotB 3. Find the area of the triangle ABC when (1) a = 4, & = 10 ft., C = 30*. Ans. 10 sq. ft. (2) 6=5, c = 20 inches, A = 60°. 43.3 sq. in. (3) a = 13, 6 = 14, c = 15 chains. 84 sq. chains. >! T» 1 1 1 1 4. Prove - = — H \- — r Ti r-a rg - T. a sin 4 B sin 4 C 6. Prove r = -—, ^ ^ » cos^ A EXAMPLES. 157 6. ove that the area of the triangle ABC is represented by each of the three expressions : 2 II* sin A sin B sin C, ra, and Rr(sin A + sin B -f- sin C). 7. If A = 60°, a = V3, b = V2, prove that the area -=i(3 + V3). 8. Prove R(sin A + 8inB + sinC)«= s. 9. Prove that the bisectors of the angles A, B, C, of a triangle are, respectively, equal to A B C 2 be cos — 2ca cos — 2 ab cos — o 2 6 + c c + (t a + 6 106. To find the Area of a Oyclic* Quadrilateral. ^ Let ABCD be the quadrilateral, and a, b, c, and d its sides. Join BD. Then, area of figure = S = I ad sin A + ^ 6c sin C = ^(ad4-6c)sin A ... (1) Now in A ABD, BD' = a' + d''-2 ad cos A, and in A CBD, B& =b^ + c^ ~2bc cos C — ¥ -\-c^ — 2 be cos A. .: cosA = — — . 2 {ad -\- be) \ L 2{ad-{-bc) J y(2ad-f 2&c)'-(a'- b'- -c^ + d^y 2 {ad + be) * See Geometry, Art. 251. I I m m 168 PLANE THIQONOMETHY, ^ ^/[Ta + dy- (b- c)n [( 6 + 0)'- («"^~rf? ] 2(ad + 6c) y/(a + d + b — c) ( a+'d -fe+c) {b+c+a—d)(b + r^ ci 4- d ) ■^ 2{ad + bc) _ 2 V(< - g) (« - 6 ) (a -c)(a- d) ad + bc (where 2« = a -f- 6 -I- c 4-d). Substituting in (1), we have S ^ V(« - a) (a - b) (s - c) (s - d). The more important formula proved in this chapter are summed up Jis follows : 1. ^^ = T— 5 = -T-p; (Art. 95) sin A sinB smC 2. a»-6» + c»-26cco8A (Art. 96) 3 ^L±6^t^i(A + B) (Art. 97) a-b tani(A-B) "^ / 4. sin^A= J ^^"^^^^~^^ (Art. 99) 5. cos^A = \r-^^^ — ^• ^ be 6. taniA = \/^^^^SIEiI. ^ \ 8{8-a) 7. sin A = ^ V«(s -a) {8- b) (s - c) . . (Art. 100) be =:: _L V2 6 V + 2 c»a» + 2 a'6» - a^ - 6^ - c« . 26c 8. Area of A = V« (« - a) (a - 6) (s - c) . . (Art. 101) EXAMPLES. 169 9. Area oi A = Ua-\- b -\-c)= rs . . . . (Art. 102) 10. \ J.d -a){ »-h) { i-c) 11. K = ahc 4S (Art. 103) EXAMPLES. In a right triangle ABC, in which C is the right angle, prove the following : \ n-n sin^A — sin*H J.* sur A + sin^B . oB c — a 2. sin''-= -• 2 2c / A , . AV a + c 3. (cos- -f sin - ) =—1-. \, 2 2) c 4. 2 A b + c COS''— = — ■ — ♦ 3 2# 6. sin(A-B)-f-cos2A = 0. ^a — b.A — B 6. = tan • a + b 2 7. sin(A-B)4-8in(2A4-C) = 0. a 8. tan ^ A b-\-c 9. (sinA-sinB)'+(cosA4-cosB)' = 2. IsinA 10. Jl±^-f J?LzJ = -2^ + 6 Vcos2B In any triangle ABC, prove the following statements 11 / I I.N • C A-B 11. (a4-o)sin =cco8 — - — • i K liiM 160 PLANE TRIGONOMETRY. 12. 13. 14. (o — c) cos - = asm • ^ ^ 2 2 a(6*+o')co8A+6(c''+a»)co8B+c(a*+6=)cosC=3a6c. a — ft _ COS B — cos A c 1 + cos C 6 4- c cos li + cos C 16. i_i:.'^ = a 1 — cos A % 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. Woe sin B sin C = • ft + c ff -f-ft-f-c = (6 + c)cos A -)- (o -f a)cos B -|- (a -f ft)eo8 C. 6+c— a = (6 -f c)cos A — (c — a)co8B -|-(a — ft)co8C. a cos ( A+ B + C) - ft cos (B + A) - c cos ( A + C) = 0. cos A , cos B , cos (y a? -f ft' -|- c* H -, r a tan A ft c a si nC ft — a cos C 2aftc ft cos' — h c cos'— = «. 2 2 . B. C ft + c-o tan - tan - = ~ 2 2 ft+c+a A J^ tan - (ft + c — a) = tan - (c -f a — ft). 2 ^ c^ = (ffl -;- ft)'sin'^ 4- (a - ft)'cos'^. c(co8A4-cosB) = 2(rt + ft)v'n>^. ' c c(co8 A -~ cos ^) — l^(ft — a) cos'-* tan B + tan C = (a' 4- ft' - c') -. (a' - ft' + c^). EXAMl'LKS. 161 20. rt« -f. ftn 4- ,-2 = 2 (a& cos C + be cos A -f- ca cos B) . 30. C08»^ H- cos^J = (« - rt) -f- &(s - 6). C H ,'U. 6 sin'— + c sin* - = .s — a. 2 i 32. If /> is the length of the perpendicular from A on BC, sin A = -"• 6c 33. If A = 3B, then8inB = JJ^^--^. 34. It v6t' sin B sm C = j ^-^^^ " = C* 35. acos^30s ^^Josec-=:s. 36. If cos A = f, and cos B = |jj, then cos C = - \\. 37. If sin^B + sin='C = sin'A, then A = 1M)°. 38. If D is the middle point of BC', prove that 4Al)'=26»4--^'-'-al 39. If a = 26, and A = 3 B, prove that C = 60°. 40. If 1), P:, K, are the middh; points of the sides, BC, CA, AB, prove 4(Al)' 4- be' + CF') = 3(r.» + 6'^ + c»). 41. If a, 6, c, the sides of a triangle, are in arithmetic progression, prove , A, C 1 , . tan -tan 2 = 3. 42. If taq A - tan B ^ c - 6^ ^^^^ ^ ^ g^o tan A + tan B c 43. If cos L - ' . prove that B = C. 2sm{ . 162 PLANE TRIGONOMETRY. 41. If rt» = fc» - 6c 4- c", prove that A = 60°. 46. If the sides of a triangle are a, 6, and Va* + ab -\- 6-, prove that its grejitest angle is 120°. 4<). I'rove that the vertical angle of any triangle is divided by the median which bisects the base, into seg- ments whose sines are inversely proportional to the adja- cent sides. 47. If AD be the median that bisects BC, j^rove (1) (6« - c») tan ADB = 2 be sin A, and (2) cot BAD -f cot DAC = 4 cot A + cot 1^ -f cot C. 48. Find the area of the triangle ABC when a = (>25, 6 = 505, c = 904 yards. Ans. 151872 sq. yards. 49. Find the radii of the inscribed and each of the escribed circles of the triangle ABC when (1 = 13, 6 = 14, c=16. Ans. 4; 10.5; 12; 14. 60. Prove the aroa S = ^ a' sin B sin C coscc A. 61. 62. « <( i( li « " = Vrrirjr,. « « 2o6c / „A„„B_ C [( — cos cos — cos } a-f 6-fcV '^ '* '* 63. Prove that the lengths of the sides of the pedal tri- angle, that is, the triangle formed by joining the feet of the perpendiculars, are a cos A, b cos B, c cos C, respectively. 64. Prove that the angles of the pedal triangle are, respectively, tt — 2 A, tt — 2 1?, tt — 2 C. 66. Prove r,rjr, = r' cot* - cot" - cot' -• 2 2 2 t^a iJ ABC c6. Prove r, cos - = acos— cos — • 2 2 2 67. Prove that the area of the incircle ; area of the tri- angle : : w : cot — cot - cot -• , - > * 9 2 2 EXAMPLES. 163 Prove the following statements : 68. If a, b, c, are in A. 1*., then ac — 6 rR. 59. If the altitude of an isosceles triangle is equal to the base, R is five-eighths of the base. 60. 6c = 4ir(cosA + coslUosC). 61. If C is a right angle, 2r + 2R = a-f6. 62. r^r^ + rgr, + r,r2 = 5^. 63. l + l + i = _L. he ca ah 2rR C 64. r^-f- r2 = ccot -• 65. r cos - = a sin - sin - • 2 2 2 66. If jp„ Pa, Pn be the distances to the sides from the circumcentre, then a .h . c aha P\ P2 Pi ^PiPiPa 67. The radiuc R of the circuincircle — ^ a/ abc ~2\sinA8inB B sin C n* h^ 68. S = -- sin 2 B 4- - sin 2 A. 4 4 69. 1 +-^.+ ^ 1 4R s — a a — b 8 ~ c s S 7(». a6rr = 4R(«-a)(s-6)(.s-c). 71. The distances between the centres of the inscribed and escribed circles of the triangle ABC are 4 R sin — , 4R8in— , 4R8'n -• . a . 8 72. If A is a right angle, i\ + r^ = c. 164 PLANE TRIGOyoMKrii ? . M I I =.■ ! 73. In an equilateral triangle .'i H = (}r = 2r,. 74. If r, r„ r,, r,, denote the radii of the inscribed and escribed circles of a triangle, tan»^ = ^. 2 r^rs 75. The sides of a triangle; a.vv in arithmetic progression, and its area is to that of an ec^ni lateral triangle of the same perimeter as 3 is to />. Find the ratio of the sides and the value of the largest angle. Ann. As 7, 5, 3 ; 120**. 76. If an equil.iteral triangle bo described with its angular points on the sidtns of a given right isosceles triangle, and one side parallel to the hypotenuse, its area will be 2 a*' sin' 15° sill 60", where a is a sitlo of the given triangle. 77. Tf h be the dilTerenco between tlu' sides containing the right angle of a right triangle, a!id S its area, the diam- eter of the circumscribing circle = V^"' + 48. 78. Three circles touch one another externally : prove that the sqmire of the anui of the triangle formed by jo'ii- ing their centres is equal to the product of the sum and product of their radii. 79. On the sides of any triangle equilateral triangles are described externally, and their centres are joined : prove that the triangle thus formed is equilateral. SO. Tf ()„ Oj, O, are the centres of the escribed circles of a tiiangle, then the area of the triangle 0,().jO., = area ol a . b triangle A IK) 1 -j- {' + a -f ft — c J + — b ■\- c — n a -\- (' — b 81. If the centres of the three escril)ed circles of a tri- angle are joined, then the area of the triangle thus formed a. 'k' is - -, where r is the radius of the inscribed circle of the 2r original triangle. at th( an so LI T ION OF TK I ANGLES. 165 CHAPTER VII. SOLUTION OF TRIANGLES. 107. Triangles. — Ju every triangle there are six elements, the three sides and the three angles. When any three ele- ments are given, one at least of the three being a side, thp other three can be calculated. The process of determining the unknown elements from the known is called the solution of triangles. NoTl. — If the three angles only of a triungle are given, it ii itnpouible to deter- mine the Hide*, for there in an infinite number of trianglea that are equiangular to one another. Triangles are divided in Trigonometry into right and oblique. We shall commence with right triangles, and shall suppose C the right angle. RIGHT TRIANGLES. 106. There are Four Cases of Right Triangles. I. Given one side and the hypotenuse. II. Oiven an acute angle and the hypotenuse. III. Oiven one side and an acute angle. IV. Given the two sides. Let AHC be a triangle, rig'.it-angled at C, and let a, h, and c, as before, he the sides opposite the angles A, B, and C, resjx'ctively. The fornmlie for the solution of right triangles are (1), (2), (.•{) of Art. 94. Ii ■I. I' 160 PLANK THIQONOMKTH V. 109. Case I. — Given a side and the hypotenuse, as a and c; to Jind A, B, b. We have sin A = a .'. log sin A = log a — log c, from which A is deterniiued ; then B = \H)° — A. Lastly, b = c cos A. .*. log b = log c -f log cos A. Thus A, B, and b are determined. Ex. 1. Given a = o3G, c = Ml ; find A, K, b. Sohdion by Natural Functions. We have sin A = - = .rmm7. I c 941 From a table of natural sines we find that A = 34" 43' 22". .-. B = r,r>M 6' 38". Lastly, 6 = c cos A = 941 x .821918 - = 773.425. Logarithmic Sohdion. log sin A = log a — log c. loga = 2.7291 648 log c = 2.9735890 log sin A* = 9.7555752 . .-. A = 34° 43' 22". ' .-. B = 55M6'38". log b = log c -f- log cos A. log c = 2.9735890 logcos A = 9.9148283 log ^ = 2.88841 79 .-. & = 773.424. Our two methods of calculation give results which do not quite agree. The discrepancies arise from the defects of the tables. • * 'fen In added so a» to get tli titbuhir lti|4.'i0.S0 iog6* = l.S()12180 .-. 6 = 72.047. Ex. 2. Given A = 37° 10', c = 8702 ; find a and 6. Ans. 5293.4; 0982.3. HI. Case in. — Oiveyi a aide and an acute angle, as A and a; tojind B, b, c. We have K = 90* - A. Also h- ^ and r- " y — ,■ - . , aiiu c — . tan A sin A .'. log b = log a — log tan A, and log c = log a — log sin A. Ex. 1. Given A = 32° 15' 24", a = 5472.5 ; find B, b, c. Solution. • B = 90° - A = 57° 44' 30". log b = log a — log tan A. log a = 3.7381858 log tan A = 9.8001090 log 6 = 3.9380768 .-. & = 8071.162. log c = log a — log sin A. log a ■-■= 3.7381858 log sin A = 9.7273070 log (J = 4.0108782 .-. c = 10253.04. Ex. 2. Given A = 34° 18', a « 337.0 j find B, b, c. " Ana. B = 55°42'; 6 = 348.31; c = 421.03. * Ten !• r^tcted becauite the Ubuiat .ugarUtiiuic (uucliuiu are tou large by tea (Art 76). HIGHT TllIANGLES. 1U9 112. Case IV. A, B, c. We have Also Given the two sides, as a and b; to find tan A a then B = 90° - A. c = a oosec A = a sin A .'. log tan A = lo{^ a — loj,' h, and logc = loga — lo}j;sin A. Ex. Given a = 22GG.35, h = r»4;«).24 ; find A, B, c. Solution. log tan A = log a — log b. log a = ;3.35rK{270 log b = ;i7355382 log tan A = 9.6197888 .-. A = 22" 37' 12". .-. B = 07° 22' 48". log c = log a — log sin A. log a = 3.3553270 logsinA = 9.r»8r)02(;() log c = 3.7703004 .-. = 5892.51. NoTK. — In thii exnmple we miKht have found r by nioana of the formula C-Va' + fc'; but wc would liuvo had to go through thi' jtrocess of aquariny the viihiea of a and b. If these vulueM are Dimple nuiiibiTH, it U often euMJcr to lliid c In this way; but this value of c is not adapted to loKarithnm. A fonnula which uonsists entirely oT fnctora is always preferred to one wkich consists uf termn, when any of those terms contain any power of the quantities involved. 113. When a Side and the Hypotenuse are nearly Equal. — When a side and the hypotenuse are given, as a and c in (!ase I., and are nearly equal in value, the angle A is very near 90°, and cannot be determined with much accuracy from the tables, because the sines of angles near 90° differ very little frr»m one another (Art. 81). It is therefore desirable, in this case, to find B first, by either of the following formultB . sm . B /l— cosB /. , wn\ ^^2=\ 2 — (Art. oO) (1) ■V c — a 2c 170 PLANE TIlWOyOMETUr. tan2=Jp^^ (Art. CO) =J"p (2) Then 6 = cco8A (3) or =V{,7+a){c-a) (4) Ex. 1. Given a = 4l>0".2105y, c = 4(>02.836; find B. c - a = ().Cli541, log (c - a) = i.7yr.lG48 2 e = 9206.072, log 2 c = 3. 1)040555 2 )5.83210 93 log sin 1^ = 7.9100547 B = 50'40".30. .'. ~ = 2S'20".18. NoTK. — Thu characterlatlc 5 In increaited numerically to 6 to muke ll dlvlHible by 'I (see Note 4 of Alt. 00). Teu in then added to the churact«ri«tiu 3, luuklntc it 7, «o ua to agree with tbe Table* (Art. 70). There is a slight error in the above value of B on account of the irregular differences of the log sines for angles near 0° (Art. 81). A more accurate value may he found by the principle that the sines of small angles are approximately proportional to the angles (Art. 130). EXAMPLES. The following right triangles must be solved by log- arithms. 1. Given « = (M), c = 100 ; find A, B, h. An». A = 30° ",2' ; B = 53° 8' j 6 = 80. 2. Given a=i;57.0<), (- = 240; fiiid A, B, &. An». A = :;.r ; B = 55° ; /* = 100.59. niOUT TRIANOLES. 171 3. Given a= 147, c= KS4; Hiul A, IJ, b. Ans. A=:5n'ao"; IJ = oG° 58' 1'5" ; 6 = llO.G7. 4. Given a = 10(), c = 2(M) ; tind A, li, 6. AiiB. A = 30°; H = ()()°; h=U)OV^. 6. Given A = 40'*, c= 100; find 1), a, b. Anit. li = 50''; (1 = 04.279; 6 = 70.004. 6. Given A = 30°, c = 160 ; find B, o, b. Ans. H = 60°; a = 7r»; 6 = 75V3. 7. Given A = 32°, (•= 1700; find IJ, a, 6. '• ■■ • ' ylHS. H = r>«°; a = l>32.00; />= 1492.57. 8. Given A = 35° 10' 25", e = 072.3412 ; find B, «, 6. ytn«. H = 54° 43' 35"; a = 388.20; 6 = 548.9. 9. Given A = 75°, a = 80; find B, 6, c. ylus. H = 15°; 6 = 80(2-V3); c = 80( VO -V2). 10. Given A = 30°, a = 520 ; find li, 6, c. yln«. B = 54°; 6 = 716.72; = 884.08. 11. Given A = 34° 15', a = 843.2; find B, 6, c. ^ns. B = 55° 45' ; c = 1498.2. 12. Given A = 07° 37' 15", b = 254.73 ; find B, a, e. Ans. B = 22° 22' 46" ; a = 018.00 ; c = 009.05. 13. Given a = 75, 6 = 75; find A, B, c. Ans. A = 45° = B J c = 75 V2. 14. Given a = 21, 6 = 20 ; find A, B, c. Ans. A = 40° 23' 60"; c = 29. 15. Given a = .100.43, 6 = 500 ; find A, B, c. J Ans. A = 31° ; B = 59° ; c = 683.31. 16. Given a = 4846, 6 = 4742; find A, B, c. Ans. A = 45° 30' 50". ■^ IMAGE EVALUATION TEST TARGET (MT-3) 1.0 I.I 1.25 iaiM 1125 fi^ IIIIIM 'f ■» II 22 %ti m 1.4 2.0 1.8 1.6 Photographic Sciences Corporation « ■1>^ \ # \ \ ^^ <^ '^^ <* ^.> 6"^ % O 1, which is impossible ; and therefore there is no triangle with the given parts. II. If a = 6 sin A, sin B = 1, and B = 90°; therefore there is one triangle — a right triangle — with the given parts. III. If a > 6 sin A, and < 6, sin B < 1 ; hence there are two values of B, one being the supplement of the other, i.e., one acute, the other obtuse, and both are admissible ; therefore there are two triangles with the given parts. IV^. If a>b, then A > 1^, and since A is given, B must be acute; thus there is only one triangle with the given parts. ■ •.. 174 PLANE TRIGONOMETRY. Tliese four cases may be illustrated geometrically. Draw A, the given angle. Make AC = b ; draw the per- pendicular CD, which = b sin A. With centre C and radius a, describe a circle. I. If ab sin A, and < 6, the circle cuts AX in two points B and B', on the \y^ same side of A ; thus there ^ \ A D /'g are two triangles ABC and AB'C, each having the given parts, the angles ABC, AB'C being supplementary. IV. If a>b, the circle cuts AX on opposite sides of A, and only the triangle ABC has the given parts, because the angle B'AC of the triangle AB'C is not the given angle A, but its supplement. These results may be stated as follows : , a<&sinA, no solution. ax= &sin A, one solution (right triangle). a > 6 sin A and < 6, two solutions. a^b, one solution. These results may be obtained algebraically thus : We have a'' = ?>- + c- - 2 6c cos A . . . (Art. Ofi) .-. c=b cos A ± Va^ — b^ sin^ A, OBLIQUE TBI ANGLES. 175 giving two roots, real and unequal, equal or imaginary, according as a >, =, or < 6 sin A. A discussion of these two values of c gives the same results as are found in the above four cases. We leave the discussion as an exercise for the student. Note. — When two sides and the angle opposite the greater are given, there can be no ambiguity, for the angle opposite the less must be acute. When the given angle is a right angle or obtuse, the other two angles arc both acute, and there can be no ambiguity. In the solution of trianglen there can be no ambiguity, except when an angle is determined by the sine or cosecant, and in no case whatever when the triaugle has a right angle. Ex. 1. Given a = 7, & = 8, A = 27° 47' 45"; find B, C, c. Solution. log 6 = 0.9030900 log sin A = 9.66868G0 colog a = 9.1549020 log a = 0.8450980 log sin C = 9.9375306 colog sin A = 0.3313140 log c= 1.1139426 log sin B = 9.7266780 .-. B = 32°12'15",orl47°47'45". .: c = 13. .-. C = 120°, or 4" 24' 30". Taking the second value of C as follows : ... ., log a = 0.8450980 log sin C = 8.8857232 ' ■ colog sin A = 0.3313140 ' > log c = 0.0621352 ' . ■ " i .-. c = 1.1538. Thus, there are two solutions. See Case III. Ex.2. Given a = 31.239, & = 49.5053, A = 32° 18'; find B, C, c. Ans. B = 56° 56' 56".3, or 123° 3' 3".7 ; C = 90°45' 3". 7, or 24°38'56".3; c= 58.456, or 24.382. n .«> 7' ^ " ! » 176 PLANE TRIG ONOMETR Y. 117. Case III. — Given two sides and the included angle, as a, 6, C ; find A, B, c. (1) tan A-B a a-\-b b .C - cot — 2 (Art. 114) Hence — - — is known, and — 4^— = 90'' .'. A and B are found (2) a sinC c = — - — — t or 2 b sinC C 2* sin A ' sin B ' and thus c is found and the triangle solved. In simple cases the third side c may be found directly by the formula c=:-\/aF+¥'-2ab'GosC . . . (Art. 96) or the formula may be adapted to logarithmic calculation by the use of c subsidiary angle (Art. 90). Ex. 1. Given a = 234.7, b = 185.4, C = 84° 36' ; find A, . , Solution. log{a - h) = 1.6928460 a = 234.7 b = 185.4 a-b= 49.3 a + b = 420.1 .-. ^ = 42° 18'. ^^4^ -=47'' 42'. 2 .-. A = 55° 2' 56", B = 40°21' 4", C = 84° 36', c = 286.0746. coiog(a +b)= 7.3766473 log cot ^ = 10.0409920 ^ 2 log tan B 2 A-B = 9.1104862 = 7° 20' 56". log b = 2.2681097 log sin = 9.9980683 colog sin 1? = 0.1887804 log c = 2.4549584 OBLIQUE TRIANGLES. 177 Ex. 2. Given a =■- .062387, b = .023475, C = 110° 32' ; find A, B, c. Ans. A = 52° 10' 33" ; B = 17° 17' 27" ; c = .0739635. 118. Case IV. Given the three sides, as a, b, c;Jind A, B, C. The solution in this case may be performed by the for- mulae of Art. 99. By means of these formulae we may compute two of the angles, and find the third by subtract- ing their sura from 180°. But in practice it is better to compute the three angles independently, 9,nd check the accuracy of the work by taking their sum. If only one angle is to be found, the formulae for the sines or cosines may be used. If all the angles are to be found, the tangent formulae are the most convenient, because then we require only the logarithms of the same four quantities, s, s — a, s — b, s — c,to find all the angles ; whereas the sine and cosine formulae require in addition the logs of a, b, c. The tangent formulae ( A.rt. 99) may be reduced as follows : tan- \ s(s -b)(s-c) {s — a) '■t i; ' , A r tan— = — o ^ 1 j{s-a)(s~b)(s-c) vK- 5 — Ct « S (Art. 102) s — a 14 B T Similarly, tan— = -? 2 s — b V ' ' tan — = 2 s-c Note. — The quantity r is the radius of the inscribed circle (Art. 102). V '■ V, •^w — T=- 178 PLANE TRIGONOMETRY. ■ W. -.1 I: Ex. 1. Given a = 13, 6 = 14, c = 15 ; find A, B, C. Solution. I -' I a =13 6 = 14 c = 15 2s = 42 5=21. s -a = 8, 5-6 = 7,* 5— c=6. log(s-a)= .9030900 log(s-6)= .8450980 log(s-c)= .7781513 colog 5 = 8.6777807 log 7-2 = 1.2041200 logr= .6020600. .-. log tan - = 9.6989700. ^ 2 .-. log tan - = 9.7569620; .-. log tan ^ = 9.8239087. ■ .-. A = 53°7'48".38; B = 59° 29' 23".18 ; /. C = 67° 22' 48".44. v -'uv,.,, Without the use of logarithms, the angles may be found by the cosine formulae (Art. 96). These -.nay sometimes be used with advantage, when the given lengths of a, b, c each contain less than three digits. Ex. 2. Find the greatest angle in the triangle whose sides are 13, 14, 15. Let a = 15, 6 = 14, c = 13. Then A is the greatest angle. Then oosX=='l±^ = ^,±§^f 26c 2x14x13 = ^ = .384615 = cos 67° 23', nearly (by the table of natural sines). .'. the greatest angle is 67° 23'. EXAMPLES. 179 EXAMPLES. 1. Given a = 254, B = 16°, C = 64°; find 6 = 71.0919. = 338.65, A = 53° 24', B = 66° 27'j find 2. Given c « = 313.46. 3. Given c b = 31.43. 4. Given a b and c. 5. Given a = 6. Given a = 7. Given a = 8. Given a = 9. Given 6 = 10. Given 6 = 11. Given a = 12. Given a = = 38, A = 48°, B = 64°; find a = 28.87, = 7012.5, B = 38° 12' 48", C = 60°; find Aits. 6 = 4382.82; c = 6135.94. 528, 6 = 252, A = 124° 34'; find B and C. Ans. B = 23°8'33"; C = 32° 17' 27". 170.6, 6 = 140.5, B = 40°; find A and C. Ans. A = 51° 18' 21", or 128° 41' 39" ; C = 88° 41' 39", or 11° 18' 21". 97, 6 .= 119, A = 50° ; find B and C. Ans. B = 70° 0' 56", or 109° 59' 4" ; C = 59°59' 4", or 20° 0'56". 7, 6 = 8, A = 27° 47' 45"; find B, C, c. A71S. B = 32° 12' 15", or 147° 47' 45" ; C = 120°, or 4° 24' 30"; c = 13, or 1.15385. , , 55, c = 45, A = 6° ; find B and C. Ans. B= 149° 20' 31"; C = 24° 39' 29". 131, c = 72, A = 40° ; find B and C. Ans. B = 108° 36' 30" ; C = 31° 23' 30". 35, 6 = 21, C = 50°; find A and B. Ans. A = 93° 11' 49"; B = 36° 48' 11". 601, 6=289, C=100° 19' 6"; find A and B. Ans. A = 56° 8' 42" 5 B = 23° 32' 12", ill Ui' mi iiiil' 180 , PLANE TRIGONOMETRY. 13. Given a = 222, h = 318, c = 406 ; find A = 32° 57' 8". 14. Given a = 275.36, &= 189.28, c=301.47 ; find A, B, C. Ans. A = 63° 30' 57"; B = 37° 58' 20"; C = 78° 30' 43". 16. Given a = 5238, 6 = 5662, c = 9384 ; find A and B. Ans. A = 29° 17' 16"; B = 31° 55' 31". 16. Given a = 317, h = 533, c = 510 ; find A, B, C. Ans. A = 35° 18' 0" ; B = 76° 18' 52" ; C = 68° 23' 8". 119. Area of a Triangle (Art. 101). rind the area : 1. Given a = 2. Given a = 3. Given & = 4. Given a = 6. Given 6 = 6. Given a = 7. Given a = 8. Given a = 9. Given a = 10. Given a = 11. Given a = 12. Given a = EXAMPLES. 116.082, 6 = 100, C = 118° 15' 41". Ans. 5112.25. 8, & = 5, C = 60°. 17.3205. 21.5, c = 30.456, A = 41° 22'. 216.372. 72.3, A = 52° 35', B = 63° 17'. 2644.94. 100, A = 76° 38' 13", C = 40° 5'. 3506.815. 31.325, B = 13°57'2", A = 53° 11' 18". Ans. 135.3545. .582, 6 = .601, c = .427. .117655. 408, 6 = 41, c = 401. ■ ' 8160. .9, 6 = 1.2, c = 1.5. .64. 21, 6 = 20, c = 29. 210. 24, 6 = 30, c = 18. -^^ - 216. 63.89, 6 = 138.24, c = 121.15. 3869.2. HEIGHT OF AN OBJECT. 181 'i ; MEASUREMENT OF HEIGHTS AND DISTANCES. 120. Definitions. — One of the most important applica- tions of Trigonometry is the determination of the heights and distances of objects which cannot be actually measured. The actual measurement, with scientific accuracy, of a line of any considerable length, is a very long and difficult operation. But the accurate measurement of an angle, with proper instruments, can be made with comparative ease and rapidity. By the aid of the Solution of Triangles we can determine : (1) The distance between points which are inaccessible. (2) The magnitude of angles which cannot be practically observed. . - ; (3) The relative heights of distant and inaccessible points. A vertical line is the line assumed by a plummet when freely suspended by a cord, and allowed to come to rest. A vertical plane is any plane containing a vertical line. A horizontal plane is a plane perpendicular to a vertical line. A vertical angle is one lying in a vertical plane. ; A horizontal angle is one lying in a horizontal plane. An angle of elevation is a vertical angle having one side horizontal and the other ascending. An angle of depression is a vertical angle having one side horizontal and the other descending. ' ' " . , By distance is meant the horizontal distance, unless other- wise named. By height is meant the vertical height above or below the horizontal plane of the observer. For a description of the requisite instruments, and the method of using them, the student is referred to books on practical surveying.* • * See Johnson's Surveying, Gillespie's Surveying, Clarke's Geodesy, Gore's Geodesy, etc. ill Hi mi 182 PLANE TBIOONOMETRY, ^y\ ^ 121. To find the Height of an Object standing on a Hcrizontal Plane, the Base of the Object being Accessible. Let BC be a vertical object, such as a church spire or a tower. From the base C measure a horizon- /^ tal line CA. At the point A measure the angle of elevation CAB, We can then determine the height of the object BC ; for BC = AC tan CAB. .^ EXAMPLES. 1. If AC = 100 feet and CAB = 60°, find BC. ' •' Ans. 173.2 feet. 2. If AC = 125 feet and CAB = 52° 34', find BC. Ans. 163.3 feet. 3. AC, the breadth of a river, is 100 feet. At tlie point A, on one bank, the angle of elevation of B, the top of a tree on the other bank directly opposite, is 25° 37'; find the height of the tree. Ans. 47.9 feet. 122. To find the Height and Distance of an Inaccessible Object on a Horizontal Plane. Let CD be the object, whose base D is inaccessible ; and let it be required /^ / to find the height CD, and its horizon- x^ /\ tal distance from A, the nearest acces- B ^ ' ^ — ^ sible point. (1) At A in the horizontal line BAD observe the Z DAC = a ; measure AB = a, and at B observe the Z DBC = ^. / Then CA = -^"^"^ sin (a — /8) (Art. 95) / HEIGHT OF AN OBJECT. 183 a ■i B and' rn^ r^ K • asinrtsinjS ,•. CD = CAsin u= ^, 8in(a — /?) Kr\ An«„„ a sin yS cos a AD = AC cos a = ^ • sin(rt — /3) rr.i.T ■):,Yi (2) When the line BA cannot be yneasured directly toward the object. At A 'Observe the vertical Z CAD = a, and the horizontal Z DAB = fi ; measure . AR = a, and at B observe y^ the /L DBiV = y. Then / f. \''/^ AD- "''^'^r . 0^ sin(y3 + y) CD = AD tan a _ a sin y tan a EXAMPLES. 1. A river 300 feet wide runs at the foot of a tower, which subtends an angle of 22° 30' at the edge of the remote ban V ; find the height of the tower. Ans. 124.26 feet. 2. At 3G0 feet from the foot of a steeple the elevation is half what it is at 135 feet ; find its height. Ans. 180 feet. 3. A person standing on the bank of a river observes the angle subtended by a tree on the opposite bank to be 60°, and when he retires 40 feet from the river's bank he finds the angle to be 30° ; find the height of the tree and the breadth of the river. , , ^ws. 20V3; 20. 4. What is the height of a hill whose angle of elevation, taken at the bottom, was 46°, and 100 yards farther off, on a level with the bottom, the angle was 31° ? ; Ans. 143.14 yards. 184 PLANE TRIGONOMETRY. 123. To find the Height of an Inaccessible Object sitiiated above a Horizontal Plane, and its tj Height above the Plane. Let CD be the object, and let A and B be two points in the horizontal plane, and in the same vertical plane with CD. At A, in the horizontal line B~ BAE, observe the A C AE = «, , and DAE = y ; measure AB = a, and at B observe the ZCBE = /8. ' ' <^ c , ■ . yy . . ■■ .. D • y^' <<'<' ,./-? . \ \ 3 a A E Then Also, ^^^^a^asuij (Art. 122) sin(a — /8) ^^^ocos^sin^ (Art. 122) sin {a— ft) r)F _ ^ ^os a sin ft tan y ~ sin (a -"ft) ' \ CD = ^- )sin a — cos a tan -y ( sm(a-ft)^ ^^ _ g sin/? sin { a — y) ' cos y sin {a — ft) ''•'^ EXAMPLES. 1. A man 6 feet high stands at a distance of 4 feet 9 inches from a lamp-post, and it is observed that his shadow is 19 feet long : find the height of the lamp. Ans. 7^ feet. 2. A flagstaff, 25 feet high, stands on the top of a cliff, and from a point on the seashore the angles of elevation of the highest and lowest points of the flagstaff are observed to be 47° 12' and 45° 13' respective! ■ find the height of the cliff. . . , ' . , Ans. 348 feet. 3. A castle standing on the top of a cliff is observed from two stations at sea, which are in line with it ; their HEIGHT OF AN OBJECT. 185 J. distance is a quarter of a mile : the elevation of the top of the castle, seen from the remote station, is 16° 28'; the elevations of the top and bottom, seen from the near station, are 52° 24' and 48° 38' respectively: (1) what is its height, and (2) what its elevation above the sea? ^ns. (1) 60.82 feet; (2) 445.23 feet. 124. To find the Distance of an Object on a Horizontal Plane, from Observations made at Two Points in the Same Vertical Line, above the Plane. Let the points of observation A and B be in the same vertical line, and at a given distance from each other ; let C be the point observed, whose horizontal distance CD and -s^l C D vertical distance AI) are required. Measure the angles of depression, 6BC, aAC, equal to a and /8 respectively, and denote AB by a. Or— B a A Then and BD = CDtan«, AD = CD tan /3. .'. a = CD (tan « — tan /3) . ^^^^ _ ^ cos a cos ji sin (« — /3) ' a cos a sin ^ sin (rt — (i) A jv u, i;us « sill p ill' I !in I i^^ii ^lili 'ill' lii! EXAMPLES. 1. From the top of a house, and from a window 30 feet Ijelow the top, the angles of depression of an object on the ground are 15° 40' and 10° : find (1) the horii,ontal distance of the object, and (2) the height of tlie house. Ans. (1) 288.1 feet ; (2) 80.8 feet. ■ 2. From the top and bottom of a castle, which is 68 feet high, the depressions of a ship at sea are observed to be 16° 28' and 14° : find its distance. \ns. 570.2 yards. 186 PLANE TRIGONOMETHY. 125. To find the Distance between Two Inaccessible Objects on a Horizontal Plane. Let C and D be the two inac- cessible objects. Measure a base line AB, from whose extremities C and D are visiblt!. At A observe the angles CAD, DAB ; and at B observe the angles CB A and CBD. Then, in the triangle ABC, we know two angles and th^ side AB. .-. AC may be found. In the triangle ABD we know two angles and the side AB. .*. AD may be found. Lastly, in the triangle ACD, AC and AD have been deter- mined, and the included angle CAD has been measured ; and thus CD can be found. EXAMPLES. 1. Let AB = 1000 yards, the angles BAC, BAD = 76° 30' and 44° 10', respectively ; and the angles ABD, ABC = 81° 12' and 40° 5', respectively : find tlie distance between C and D. . ^ns. 669.8 yards. 2. A and B are two trees on one side of a river ; at two stations P and Q on the other side observations are taken, and it is found that the angles APB, BPQ, AQP are each equal to 30°, and that the angle AQB is equal to 60°. If PQ = a, show that AB = -V2l. 6 126. The Dip of the Horizon. — Since the surface of the earth is spherical, it is obvious that an object on it will be visible only for a certain distance depending on its height ; and, conversely, that at a certain lieight above the ground the visible horizon will be limited. THE DIP OF THE HORIZON. 187 Let be the centre of the eartli, P a point above the sur- face, PD a tangent to the surface at « r— 71P D. Then D is a point on the terres- trial horizon ; and CPD, which is the angle of depression of the most dis- tant point on the horizon seen from P, is called the dip of the horizon at P. The angle DOP is equal to it. Denote the angle CPD by 0, the height AP by h, and the radius OD by r. Then A = OP - OA = r sec e — r _ r (1 - cos 6] ) cos^ h cos 1 — cos d ' PD = r tan 6 = 1 — cos^ = hcot^^ . . (Art. 48, Ex. 8) ' = PAx PB = /i(/i + 2r) . . . (Geom.) Also ] Since, in all cases which can occur in practice, h is very small compared with 2 r, we have approximately PD' = 2/ir. Let n •= tlie number * of miles in PD, Ji = the feet in PA, and r = 4000 miles nearly. Then A=^ = 2r 5280^2 8000 (5280 n)^ 8000 X 5280 ^ 5.28 ^2 ^ 2 8 W km i I i !i I'i'i I m ■;!• J b I! 'f i' I * It will be noticed thnt « is n number merely, and that the result will be in feel, Biace the luilea have been reduced to feet. 188 PLANE TRIGONOMETRY. That is, the height at which objects can be seen varies as the square of the distance. Thus, it n = l mile, we have /i = I feet = 8 inches ; if w = 2 miles, /i = -| . 2^ = | feet, etc., etc. Thus it appears that an object less than 8 inches above the surface of still water will be invisible to an eye on the surface at the distance of a mile. Example. From a balloon, at an elevation of 4 miles, the dip of the sea-horizon is observed to be 2° 33' 40": find (1) the diameter of the earth, and (2) the distance of the horizon from the balloon. Alls. (1) 8001.24 miles ; (2) 178.944 miles. 127. Problem of Pothenot or of Snellius. — To determine a point in the plane of a given triangle, at which the sides of the triangle subtend given angles. Let ABC be the given triangle, and P the required point. Join P with A, B, C. Let the given angles APC, BPC be denoted by a, (3, and the unknown angles PAC, PBC by x, y respectively ; then a and ^ are known ; and when x and y are found, the position of P can be determined, for the distances PA and PB can be found by solving the triangles PAC, PBC. We have a;-f?/ = 27r — a — /8 — C 6 sin X _ asin y __ pp sin /8 (1) Also sm a Assume an auxiliary angle <^ such that ' . , a. sin a tan <^ = —, ; b sin/S then the value of ^ can be found from the tables. EXAMPLES. 189 Thus, sin X sin y = tan <^» 1. Given a =51.303, c = 150; find A=20°, B = 70°, 2. Given a = 157.33, c=250; find A = 39°, ... B=51°, sin X — sin ?/ tan + 1 [(14) of Art. 61]. .'. tan ^(x — y) = tan ^{x-^-y) tan (c^ — 45°) [(13) of Art. 61]. = tan (45° - <^) tan L (a + ;8 + C) . (2) thus from (1) and (2) x and y are found. EXAMPLES. Solve the following right triangles : 5 ■'■''-■■ '■-■..'. 6 = 140.95. 6=194.28. 3. Given a=104, .,, c=185; , , . find A=34°12'19".6, B = 55° 47' 40 ".4, 6 = 153. 4. Given a=304, c=425; find A=45°40' 2".3, B=44°19'57".7, 6 = 297. 6. Given 6=3, c=5; find A=53° 7'48".4, B=36°52'll".6, a=4. 6. Given 6=15, c=l7; find A=28° 4'20".9, B=61°55'39".l, a=8. - 7. Given 6=21, - c=29; : . find A=43°36'10".l, B = 46°23'49".9, a=20. 8. Given 6=7, -; c=25; ^ * ? find A=73°44'23".3, B = 16°15'36".7, a =24. < «!i '^m ifvi- 190 PLANE TRIGONOMETRY. 9. Given 6=33, c=65; find A=oy°29'23".2, B = 30°30'36".8, a --56. 10. Given c=625, A=44°; find a =434.161, 6=449.587. 11. Given c=300, A=52''; find a =236.403, 6=184.698. 12. Given c=13, A= 67°22'48".5 > find B=22°37'll".5, a=12 6=5. 13. Given A=77°19'10".6, c=41; \ find B = 12°40'49".4, a=40, 6=9. 14. Given B=48°53'16".5, c=73; ' ,s ■ find A=41° 6'43".5, a =48, 6=55. 16. Given B = 64° 0'38".8, c=89; . ■ ^,Ui'; S find A=25°59'21".2, a=39. 6=80. 16. Given A = 77°19'10".6, a=40; find B = 12''40'49".4, 6=9, c=41. 17. Given A=87°12'20".3, a=840; find B= 2°47'39".7, 6=41, c=841. 18. Given A=32°31'13".5, a=336; find B=57°28'46".5, 6=527, c=625. 19. Given A=82°41'44", a=1100; find B= 7° 18' 16", 6=141, c=1109. 20. Given A=75°23'18".5, 6=195; '. --\ find B = 14°36'41".5, a=748, c=773. 21. Given B = 87° 49' 10", 6=42536.37; find A= 2° 10' 50", a =1619.626, c=42567.2. EXAMPLES. 191 22. Given A: find B: 23. Given A: find B: 24. Given ct: find A: 25. Given a: ' find A: 26. Given a: find A: 27. Given a: find A: 28. Given a= find A: 29. Given a- find A= 30. Given «= find A= 31. Given a= find A= 88° 59' 1° V, 35° 16' 25", 54°43'3J " 6: a-, b-- -. 7694.5, 6: = 42° 15', B. = 736, ^ b: -- 69°38'56".3, B: = 200, b: -. 18° 10' 50", B: =276, . 6: = 29°14'30".3, B: : 396, b: : 44° 29' 53", B: = 278.3, b: : 41° 30', B: =372, &= : 41° 16' 2". 7, B = : 526.2, b: : 51° 45' 18". 7, B = =2.234875; = 125.9365, 'c= 125.9563! =388.2647; =548.9018, c= 672.3412. = 8471; ..■■ .V, =47° 45', 0=11444. =273; =20° 21 '3". 7, c=785. :609; : 71° 49' 10", c=641. = 493; J ...... ... I . I :60°45'29."7, c=565. = 403; : 45° 30' 7", 0=565. = 314.6; . : 48° 30', c=420. :423.924; . . : 48°43'57".3, c=564. = 414.745; : 38°14'41".3, c=670. Solve the following oblique triangles : 32. Given B= 50° 30', C = 122° 9', find A= 7° 21', 33. Given A= 82°20', find C= 54° 20', 34. Given A = 79° 59', find C= 51° 20', 6=542.850, B= 43° 20', 6=331.657, B= 44° 41', 6=567.888, a =90; c= 595.638. a=479 ; c= 392.473. a=795; c= 663.986. % in liiiil ' 'ill; 192 PLANE I'RIGOIWMETRY. \ 3.^. Given B = 37° 58', C= 65° 2', a=999; find A = 77° 0', &= 630.771, c= 829.480. 36. Given A = 70° 55', C= 52° 9', a=6412 ; find B = 56° 56', 6=5686.00, c= 5357.50. 37. Given A = 48° 20', B= 81° 2' 16", b=o.7o ; find C = 50° 37' 44", a= 4.3485, c=4.5. 38. Given A = 72° 4', B= 41° 56' 18", c=24; '■ find C = 65° 59' 42", a = 24.995, 6=17.559. 39. Given A = 43°36'10".l, C = 124°58'33".6 ,6 = 29; • .; find B = 11°25'16".3, a = 101, c=120. 40. Given A = 69° 59' 2".5, C= 70° 42' 30", 6 = 149; find B = 39°18'27".5, tt = 221. c=222. 41. Given A = 21° 14' 25", «=345, 6=695; '■ find B = 46° 52' 10", C = 111° 53' 25", c= 883.65. or B'= 133° 7' 50", 0'= 25° 37' 45", c'= 411.92. 42. Given A = 41° 13' 0", a =77.04, 6 = 91.06; find B = 51° 9' 6", C= 87° 37' 54", c= 116.82, . or B'= 128° 50' 54", C'= 9° 56' 6", c' = 20.172. 43. Given A= 21° 14' 25", a =309, 6=360; find B = 24° 51' 54", C = 133° 47' 41", c=615.67. or B'= 155° L' 6", C'= 3°43'29'4 c'= 55.41. 44. Given B = 68° 10' 24", a =83. 856, 6=83.153; find A= 65° 5' 10", C= 45° 44' 26", c= 65.696. 45. Given B = 60° 0'32", a =27.548, 6=35.055; find A = 42° 53' 34", C= 77° 5' 54", c=. 39.453. 46. Given A = 60°, a=120, 6 = 80; find B = 35° 15' 52", C= 84° 44' 8", c= 137.9796 EXAMPLES. 198 •1' 47. Given find A= 50°, . B= 38° 38 '24", a =119, ,; I C= 91° 21' 36", 6 = 97; !; c= 155.3. ; 48. Given find or C= 65° 59', A= 72° 4' 48", A' =107° 55' 12", rt=25, B= 41° 56' 12", B'= 6° 5' 48". cce24; 1 6=17.56, 1 1. i 49. Given A= 18°55'28".7, find B= 67°22'48".l, a = 13, 6=37; or B' = 112°37'll".9. .; 50. Given find C= 15°11'21", A= 85° 11' 58", ci=232, ' B= 79° 36' 40", 6=229; c=61. 51. Given find C = 126° 12 '14", A= 32° 28' 19", rt=5132, ^ B= 21° 19' 27", 6=3476; . c= 7713.3. 52. Given find C= 55° 12' 3", A= 67°28'51".5, « = 20.71, B= 57°19'5".5, 6=18.87; c= 18.41. 53. Given find C= 12° 35' 8", A =136° 15' 48", a =8.54, B= 31° 9' 4", 6=6.39; c=2.69. - 54. Given find C= 34° 9' 16", A=133°51'34", a =3184, B= 11° 59' 10", 6=917; 0=2479.2. 55. Given find C= 32°10'53".8, A=136°23'49".9, a=101, B= 11°25'16".3, 6=29; c=78. ,":V^ 1 1, , 56. Given find C= 96°57'20".l, A,= 77°19'10".6, a=401, B= 5°43'29".2, 6=41; c=408. j 57. Given find C= 30° 40' 35", A=110° 0'57".5, a=221, B= 39°18'27".5, 6 = 149; j c=120. i V ■ 58. Given find C= 66°59'25".4, A= 79° 36' 40", . a =109, ' B= 33°23'54".6, 6=61; c=102. h - 1: 1, 69. Given find C=131°24'44", A= 33°23'54".6, * a =229, V- B= 15°11'21".4, 6=109; c=312. , v ii Mrikadi^itMkkMMM « w !•- 194 PLANE THWONOMETJiY. 60. Given 0=104" 3' 51", a=241, 6=169; find A =45° 46' 16". 5,B = 30° 9'52".5, c= 332.97. 61. Given a= 289, &=6()1, c=712; find A=23''32'12", B=56° 8' 42", C=100°19' 6". 62. Given a =17, 6=113, c=120; findA= 7°37'42", B =61° 55' 38", C = 110° 26' 40". 63. Given a= 15.47, 6=17.39, c= 22.88; find A=42°30'44", B = 49° 25' 49", C=88°3'27". ': 64. Given a =5134, 6=7268, c=9313; findA=33°15'39", B = 50°56' 0", C=95°48'21". 65. Given a =99, 6=101, c=158; find A =37° 22' 19", B=38°15'41", C = 104°22'0". 66. Given a=ll. 6=13, c=16; findA=43° 2'56", B = 53° 46' 44", C=83°10'20". 67. Given a =25, 6=26, c=27; findA=56°15' 4", B=59°51'10", = 63° 53' 46". 68. Given a =197, 6=53, c=240; find A =31° 53' 26". 8,B= 8°10'16".4, C = 139°56'16".8. | 69. Given a = 509, 6=221, c=480; 1 find A=84°32'50". 5, B=25°36'30".7, C=69°50'38".8. | 70. Given a =533, 6=317, c = 510; find A =76° 18' 52", B=35°18' 0".9 0=68°23'7".l. 71. Given a =565, 6=445, c=606; find A =62° 51' 32". 9,B=44°29'53", 0=72°38'34".l. 72. Given a =10, 6=12, c=14; find A=44°24'55". 2,B=57° 7' 18", = 78° 27' 47". 73. Given a = .8706, 6=.0916, c=.7902; find A =149° 49' 0". 4,B= 3° 1'56".2 ,0=27°9'3".4. f EXAMPLES. Find the area : 74. Given a=10, 6=12, = 60". 195 Ana. 30V3. 75. (( a =40, 6=60, C=30°. 600. 76. (( 6=7, c=5V2, A = 135°. 17^. 77. u a=32.5, 6=56.3, C =47" 5' 30". 670. 78. i( 6=149, A= 70° 42' 30", B = 39° 18' 28". 15540. 79. (I c= 8.025, B = 100° 5'23", C=3r 6'12". 46.177. 80. « a=5. 6=6, c=7. 12. 81. « a =625, 6=505, c=904. 151872. 82. (( a =409, 6=169, c=510. 30600. 83. u a=577, 6=73, c=520. : 12480. 84. il tt=52.53, 6=48.76, c=44.98. 1016.9487. 86. « a =13, 6=14, c=15. ' 84. 86. ({ a =242 yards, 6=1212 yards, c=l450 yards. Ans. 6 acres. 87. a a= 7.152, 6=8.263, c=9.375. 28.47717. 88. The sides of a triangle are as 2 : 3 : 4 : show that the radii of the escribed circles are as ^ : ^ : 1. 89. The area of ^ triangle is an acre ; two of its sides are 127 yards and 150 yards : find the angle between them. Ans. 30° 32' 23". 90. The adjacent sides of a parallelogram are 5 and 8, and they include an angle of 60°: find (1) the two diag- onals, and (2) the area. Ans. (1) 7, Vl29; (2) 20 V3. 91. Two angles of a triangular field are 22^° and 45°, and the length of the side opposite the latter is a furlong. Show that the field contains 2^ acres. !l; 196 '\ PLANE TRIGONOMETRY. HEIGHTS AND DISTANCES. 92. At a point 200 feet in a horizontal line from the foot of a tower, the angle of elevation of the top of the tower is observed to be 60° : find the height of the tower. Ans. 346 feet. 93. From the top of a vertical cliff, the angle of depres- sion of a point on the shore 150 feet from the base of the cliff, is observed to be 30° : find the height of the cliff. Ans. 86.6 feet. 94. From the top of a tower 117 feet high, the angle of depression of the top of a house 37 feet high is observed to be 30° : how far is the top of the house from the tower ? Alls. 138.6 feet. 95. The shadow of a tower in the sunlight is observed to be 100 feet long, and at the same time the shadow of a lamp-post 9 feet high is observed to be 3 V3 feet long : find the angle of elevation of the sun, and height of the tower. Ans. 60°; 173.2 feet. 96. A flagstaff 25 feet high stands on the top of a house; from a point on the plain on which the house stands, the angles of elevation of the top and bottom of the flagstaff are observed to be 60° and 45° respectively : find the height of the house above the point of observation. Ans. 34.15 feet. 97. From the top of a cliff 100 feet high, the angles of depression of two ships at sea are observed to be 45° and 30° respectively ; if the line joining the ships points directly to the foot of the cliff, find the distance between the ships. Ans. 73.2. 98. A tower 100 feet high stands on the top of a cliff ; from a point on the sand at the foot of the cliff the angles EXAMPLES. 197 jlitf; of elevation of the top and bottom of the tower are observed to be 75° and GO" respectively : find the height of the cliff. Ans. 8(5.6 feet. 99. A man walking a straight road observes at one mile- stone a house in a direction making an angle of 30° with the road, and at the next milestone the angle is 60° : how far is the house from the road ? Ans. 1524 yds. 100. A man stands at a point A on the bank AB of a straight river and observes that the line joining A to a post C on the opposite bank makes with AB an angle of 30°. He then goes 400 yards along the bank to B and finds that BC makes with BA an angle of 60°: find the breadth of the river. ' Ans. 173.2 yards. 101. From the top of a hill tlse angles of depression of the top and bottom of a flagstaff 25 feet high at the foot of the hill are observed to be 45° 13' and 47° 12' respectively : find the height of the hill. Ans. 373 feet. 102. From each of two stations, east and west of each other, the altitude of a balloon is observed to be 45°, and its bearings to be respectively N.W. and N.E. ; if the sta- tions be 1 mile apart, find the height of the balloon. Ans. 3733 feet. 103. The angle of elevation of a balloon from a station due south of it is 60°, and from another station due west of the former and distant a mile from it is 45° : find the height of the balloon. Ans. 6468 feet. 104. Find the height of a hill, the angle of elevation at its foot being 60°, and at a point 500 yards from the foot along a horizontal plane 30°. A^is. 250V3 yards. 105. A tower 51 feet high has a mark at a height of 25 feet from the ground : find at what distance from the foot the two parts subtend equal angles. : , :,• .^-. ^.'. ,::..« Jy V 11; f' 'ji'< If'; ■^: 198 PLANE , TRIGONOMETRY. 106. The angles of a triangle are as 1 : 2 : 3, and the per- pendicular from the greatest angle on the opposite side is 30 yards : find the sides. ; .4ns. 20 V3, 60, 40 V3. 107. At two points A, B, an object DE, situated in the same vertical line CE, subtends the same angle a ; if AC, BC be in the same right line, and equal to a and b, respec- tively, prove ; .. , ■ i .i DE = (a + ?j) tan «. 108. From a station B at the foot of an inclined plane BC the angle of elevation of the summit A of a mountain is 60°, the inclination of BC is 30°, the angle BOA 135°, £uid the length of BC is 1000 yards : find the height of A over B. 1 ;;,.j ' ^HS. 500(3 -f-V3) yards. 109. A right triangle rests on its hypotenuse, the length of which is 100 feet; one of tlie angles is 30°, and the inclination of the plane of the triangle to the horizon is 60°: find the height of the vertex above the ground. Ans. 2oV3cos 18°. 110. A station at A is due west of a railway train at B ; after traveling N.W. 6 miles, the bearing of A -from the train is S. 22|-° W. : find the distance AB. Ans. 6 miles. 111. The angles of depression of the top and bottom of a column observed from a tower 108 feet high are 30° and 60° respectively: find the height of the column. Ans. 72 feet. 112. At the foot of a mountain the elevation of its sum- mit is found to be 45°. After ascending for one mile, at a slope of 15°, towards the summit, its elevation is found to be 60° : find the height of the mountain. y3-fi V2 Ans. miles. 113. A and B are two stations on a hillside. The inclina- tion of the hill to the horizon is 30°. The distance between A and B is 500 yards. C is the summit of another hill in EXAMPLES. 199 the same vertical plane as A and B, on a level with A, but at B its elevation above the horizon is 15° : find the distance between A and C. Ans. 500(\/,S4-l). 114. From the top of a cliff the angles of depression of the top and bottom of a lighthouse 97.25 feet high are observed to be 2,3° 17' and 24° 19' respectively : how much higher is the cliff than the lighthouse ? Ans. 1942 feet. ■ 115. The angle of elevation of a balloon from a station due south of it is 47° 18' 30", and from another station due west of the former, and distant C71.38 feet from it, the elevation is 41° 14' : find the height of the balloon. Ans. 1000 feet. 116. A person standing on the bank of a river observes the elevation of the top of a tree on the opposite bank to be 51° ; and when he retires 80 feet from the river's bank he observes the elevation to be 46° : find the breadth of the river. . Ans. 155.823 feet. 117. From the top of a hill I observe two milestones on the level ground in a straight line before me, and I find their angles of depression to be respectively 5° and 15° : find the height of the hill. Ans. 228.6307 yards. 118. A tower is situated on the top of a hill whose angle of inclination to the horizon is 30°. The angle subtended by the tower at the foot of the hill is found by an observer to be 15°; and on ascending 485 feet up the hill the tower is found to subtend an angle of 30° : find (1) the height of the tower, and (2) the distance of its base from the foot of the hill. Ans. (1)280.015; (2) 705.015 feet. 119. The angle of elevation of a tower at a place A due south of it is 30° ; and at a place B, due west of A, and at a distance a from it, the elevation is 18° : show that the height of the tower is I •\/2 + 2 VO 200 PLANE TRIGONOMETRY. I'l: ,1 '( 120. On the bank of a river there is a cohimn 200 feet high supporting a statue 30 feet high. The statue to an observer on the opposite bank subtends an equal angle with a man 6 feet high standing at the base of the column: find the breadth of the river. Ans. 10V115 feet. 121. A man walking along a straight road at the rate of 3 miles an hour, sees in front of him, at an elevation of 60°, a balloon which is travelling horizontally in the same direc- tion at the rate of 6 miles an hour ; ten minutes after he observes that the elevation is 30° : prove that the height of the balloon above the road is 440 V3 yards. 122. An observer in a balloon observes the angle of depression of an object on the ground, due south, to be 35° 30'. The balloon drifts due east, at the same elevation, for 2^ miles, when the angle of depression of the same object is observed to be 23° 14' : find the height of the balloon. Ans. 1.34394 miles. 123. A column, on a pedestal 20 feet high, subtends an angle 45° to a person on the ground ; on approaching 20 feet, it again subtends an angle 45° : show that the height of the column is 100 feet. 124. A tower 51 feet high has a mark 25 feet from the ground : find at what distance the two parts subtend equal angles to an eye 5 feet from the ground. Ans. IGO feet. 125. From the extremities of a sea-wall, 300 feet long, the bearings of a boat at sea were observed to be N. 23° 30' E., and N. 35° 15' W. : find the distance of the boat from the sea-wall. Ana, 262.82 feet. 126. ABC is a triangle on a horizontal plane, on which stands a tower CD, whose elevation at A is 50° 3' 2" ; AB is 100.62 feet, and BC and AC make with AB angles 40° 35' 17" and 9° 59' 50" respectively : find CD. Ans. 101.166 feet. EXAMPLES. 201 127. The angle of elevation of a tower at a distance of 20 yards from its foot is three times as great as the angle of elevation 100 yards from the same point : show that the height of the tower is "^ — feet. 128. A man standing at a point A, due south of a tower built on a horizontal plain, observes the altitude of the tower to be 60°. He then walks to a point B due west from A and observes the altitude to be 45°, and then at the point C in AB produced he observes the altitude to be 30° : prove that AB = BC. 129. The angle of elevation of a balloon, which is ascend- ing uniformly and vertically, when it is one mile high is observed to be 35° 20' ; 20 minutes later the elevation is observed to be 55° 40' : how fast is the balloon moving ? Ans. 3 (sin 20° 20') (sec 55° 40') (cosec 35° 20') miles per hour. 130. The angle of elevation of the top of a steeple at a place due south of it is 45°, and at another place due west of the former station and distant 100 feet from it the elevation is 15° : show that the height of the steeple is 50(3^ -3-i) feet. 131. A tower stands at the foot of an inclined plane whose inclination to the horizon is 9°; a line is measured up the incline from the foot of the tower, of 100 feet in length. At the upper extremity of this line the tower sub- tends an angle of 54° : find the height of the tower. Ans. 114.4 feet. 132. The altitude of a certain rock is observed to be 47°, and after walking 1000 feet towards the rock, up a slope inclined at an angle of 32° to the horizon, the observer finds that the altitude is 77° : prove that the vertical height of the rock above the first point of observation is 1034 feet. 202 PLANE TRIGONOMETBY. 133. From a window it is obfserved that the angle of elevation of the top of a house on the opposite side of the street is 29°, and the angle of depression of the bottom of the house is 56°: find the height of the house, supposing the breadth of the street to be 80 feet. Aiis. 162.95 feet. 134. A and B are two positions on opposite sides of a mountain ; C is a point visible from A and B ; AC and BC are 10 miles and 8 miles respectively, and the angle BCA is 60°: prove that the distance between A and B is 9.165 miles. 135. P and Q are two inaccessible objects ; a straight line AB, in the same plane as P and Q, is measured, and found to be 280 yards ; the angle PAB is 95°, the angle QAB is 47^°, the angle QBA is 110°, and the angle PBA is 52° 20' : find the length of PQ. Ans. 509.77 yards. 136. Two hills each 264 feet high are Just visible from each other over the sea : how far are they apart ? (Take the radius of the earth = 4000 miles.) Ans. 40 miles. 137. A ship sailing out of harbor is watched by an observer from the shore ; and at the instant she disappears below the horizon he ascends to a height of 20 feet, and thus keeps her in sight 40 minutes longer : find the rate at which the ship is sailing, assuming the earth's radius to be 4000 miles, and neglecting the height of the observer. Alls. 40V330 feet per minute. 138. From the top of the mast of a ship 64 feet above the level of the sea the light of a distant lighthouse is just seen in the horizon ; and after the ship has sailed directly towards the light for 30 minutes it is seen from the deck of the ship, which is 16 feet above the sea : find the rate at which the ship is sailing. (Take radius = 4000 miles.) Ans. 8V|| miles per hour. EXAMPLES. 203 139. A, B, C, are three objects at known distances apart ; namely, AB = 1056 yards, AC = 924 yards, BC = 1716 yards. An observer places himself at a station P from which C appears directly in front of A, and observes the angle CPB to be 14° 24' : find the distance CB. Ans. 2109.824 yards. 140. A, B, C, are three objects such that AB=o20 yards, AC = 600 yards, and BC = 435 yards. Brom a station B it is observed that ABB = 15°, and BBC = 30° : find the distances of P from A, B, and C ; the point B being near- est to B, and the angle ABC being the sum of the angles ABB and BBC. Ans. BA = 777, BB = 502, BC = 790. 204 PLANE TRIGONOMETRY. CHAPTER VIII. OONSTRUOTION OF LOGAKITHMIO AND TEIGONOMETEIO TABLES. 128. Logarithmic and Trigonometric Tables. — In Chap- ters IV., v., and VII., it was shown how to use logarithmic and trigonometric tables; it will now be shown how to calculate such tables. Although the trigonometric func- tions are seldom capable of being expressed exactly, yet they can be found approximately for any angle ; and the calculations may be carried to any assigned degree of accu- racy. We shall first show how to calculate logarithmic tables, and shall repeat here substantially Arts. 208, 209, 210, from the College Algebra. 129. Exponential Series. — To expand e' in a series of ascendimj powers ofx. By the Binomial Theorem, /i 1 ly 11 1 , nx(nx-l) 1 \ nj n 12 n^ Similarly, 1\" [2 nx{nx — l){nx — 2) 1 . xltc ) x{x — \{x ) 13 (1) 1 + n . i_i A-i-Yi-'' 1 + 1 + n 12 + n, n 13 + (2) LOGARITHMIC SERIES. 206 and therefore series (1) is equal to series (2) however great n may be. Hence if n be indefinitely increased, we have from (1) and (2) 1 _(_ a; + - + ^ + ^ 4- ... = /^l 4- 1 4- i -f- i 4- 1 + ...Y. The series in the parenthesis is usually denoted by e; hence x' . x' x^ e'=l+x + ^ + '^ + ^^-^ (3) [2 ' [3 ■ |4 which is the expansion of e'^ in powers of x. This result is called the Exponential Theorem. If we put 03= 1, wo have from (3) , , _, •'■■:'. 6 = 1 + 1 ^ 1 + 1 + i 4- i + 1 + ... From this series we may readily compute the approxi- mate value of e to any required degree of accuracy. This constant value e is called the Napierian base (Art. 64). To ten places of decimals it is found to be 2.7182818284. Cor. Let a = e'' ; then c = log^ a, and a' = e". Substi- tuting in (3), we have c-af c^a^ e-=l + ca;4-^+^ + or a' = J + 0, log, a + ^Q|^% ?^5|iO! + (4) which is the expansion of a"" in powers of x. 130. Logarithmic Series. — To expand loge(l4-a;) in a series of ascending powers of X. ' ■ <_ • ,- By the Binomial Theorem, • ' a' = (l + a-l)' = l + a;(a-l) + "^^-^~^) (a-l)^ + ^^-^)(-^'--)(a-ir+..- 206 PLANE TRIGONOMETRY. + terms involving x-, x^, etc. Comparing this value of a' with that given in (4) of Art. 129, and equating the coefficients of x, we have Put a = I -f iB ; then •••••• (3) log.(l4-«) = a;-- + - a' + This is the Logarithmic Series; but unless x be very small, the terms diminish so slowly that a large number of them will have to be taken ; and hence the series is of little practical use for numerical calcul ition. If a;> 1, the series is altogether unsuitable. We shall therefore deduce some more convenient formulae. Changing x into — x, (1) becomes X- ar' cfi" log.(l — x) = — x Subtracting (2) from (1), we have log, ' — / ( 1! _L I I |_ Put l~x l-\-x n-\-l I{ XA h \ -'-5 5 X = 7 1 (2) (3) 2w + l' 1 —X n and (3) becomes W !i±l = 2r ^ 4- ^ + ^ + ^' n l2n + l 3{2n-\-iy o{2n-[-iy or log„ (n 4- 1) =log.n+2r-l |_2?H .+?:, -.+■ (4) i + l 3(2n + l)3 r)(2» + l)« This series is rapidly convergent, and gives the logarithm of either of two consecutive numbers to any extent when the logarithm of the other number is known. COMPUTATION OF LOGARITHMS. 207 131. Computation of Logarithms. — Logarithms to the base e are called Napierian Logarithms (Art. 64). They are also called natural logarithms, because they are the first logarithms which occur in the investigation of a method of calculating logarithms. Logarithms to the base 10 are called commoyi logarithms. When logarithms are used in theoretical investigations, the base e is always understood, just as in all practical calculations the base 10 is invariably employed. It is only necessary to com- pute the logarithms of prim,e numbers from the series, since the logarithm of a composite number may be obtained by adding together the logarithms of its component factors. The logarithm of 1 = 0. Putting w = 1, 2, 4, 6, etc., suc- cessively, in (4) of Art. 130, we obtain the following log,. 2 = 2 Napierian Logarithms : ■1.1.1 +J_H-J- + log. 3=log,2-f-2 log. 4 = 21og.2 log. 5= log. 4 +2 3 3 . S"^ 5 . 3« 7 . 3^ 9 . 3' 1.1.1,1 5 3.5' 5.5^ 7-r/ 9 3 . 93 5 . 9-^ 7 • 9^ log. 6 = log.2 + log.3 ,, - 13 3 . 13^^ 5 . 13* + + log. 7 = log/, + 2 log. 8 = 31og.2 log. 9 = 21og,3 log.l0 = log,r)4-log.2 And so on. = 0.69314718. = 1.09861228. = 1.38629436. = 1.60943790. = 1.79175946. = 1.94590996. = 2.07944154. = 2.19722456. = 2.30258509. The number of terms of the series which it is necessary to include diminishes as n increases. Thus, in computing 208 PLANE TRIGONOMETRY. the logarithm of 101, the first term ol the series gives the result true to seven decimal places. By changing b to 10 and a to e in (1) of Art. 65, we have or common log m = Napierian logm x .43429448. The n timber .43429448 is called the modulus of the common system. It is usually denoted by fi. Hence, the common logarithm of any number is equal to the Napierian logarithm of the same number multiplied by the modulus of the common system, .43429448. Multiplying (4) of Art. 130 by fi, we obtain a series by which common logarithms may be computed; thus, logio(ri + l) = Common Logarithms. Iog,o2 = fi log, 2 = .43429448 x .09314718 = .3010300. Iogio3 = /^log,3 = .43429448 x 1.09861228 = .4771213. log,o 4 = 2 logio 2 = .6020600. log,„5 = /ilogeO = .43429448 x 1.60943790 = .6989700. And so on. 132. If $ be the Circular Measure of an Acute Angle, sin 6, $, and tan 9 are in Ascending Order of Magnitude. With centre O, and any radius, de- scribe an arc BAB'. Bisect the angle BOB' by OA; join BB', and draw the tangents BT, B'T. Let AOB = AOB' = 6. Then BB' < arc BAB' < BT + B'T (Geom., Art. 246) .-. EC < arc BA < BT. LIMITING VALUES OF SIN 0. 209 BC BA BT • ■ OB OB OB' ' .-. sin ^ < 6/ < tcau d. 133. The Limit of ^^, when 6 is Indefinitely Diminished, is Unity. . - We have sin d<0 - 2 2 .00 $ sin - > - cos -• 2 2 2 (Art. 132) 210 PLANE TliWONOMETliY. .'. 2 sin - COS - > ^ COS"* -• 2 2 2 >/l_^ . . (Art. 132) .-. BmB^-?- 4 (2) cos^=l-2sin2-. ffi .'. cos^>l--. 2 ,. „ Also, sin|>^-^(|)^y(l). '^ , ...eos.l-?-and <1_.^ + ^. 2 2 lb * Note. —It may be proved that sin fl> S , as follows : 6 We have 3 sin - - sin = 4 sin^ - (Art. 50) (1) 3 3 .'. 3 sin - - sin ? = 1 sin' -- (by putting - for 0) (2) 3* 3 3' 3 3sini--sln-^ =4ein^ ?- (n) 3» 3"-^ S" Multiply (1), (2), ... (n) by 1, 3, ... S"-!, respectively, and add them, 3" sin - - sin 9 = 4^8in3 - + 3 sin^ - + - 3"-l sin^ -V 3" \ 3 3» 3V SINE AND COSINE OF lO" AND OF l\ 211 . (1) . (2) . (n) .to* ...*._ .,n»<4(3^ + - + ...^2,^) (Art «• 132) 33 V 3» ^in-2) tin — 3" If n B oe , then — ;: — = 1 4^,/, 1 1 \ 4*< 1 #a and 3,9a(l + ^ + ...^^_J = — ._ = -^-. '.' » (Art. 133) 3» .♦. «-sln9<-, and .-. •ln«>»-??. 6 6 Thii makeH tho limltB for sin closer than In (1) of this Art. 135. Te calculate the Sine and Cosine of 10" and of 1 '. (1) Let be the circular measure of 10". ' Then e = 10 1 180 X 60 X 60 3.141592653589793 64800 or ^ = .000048481368110 • • -, correct to 16 decimal places. = .000000000000032..., « ... e_i'=. 000048481368078..., « 4 (( (( u <( It (I Hence the two quantities 6 and 6 agree to 12 deci- mal places ; and since sin B ^ — -7 (Art. 134), .-. sin 10" = .000048481368, to 12 decimal places. We have cos 10" = Vl-sinno" = 1 - i sin2 10" = .9999999988248 .••, to 13 decimal places. Or we may use the results established in (2) of Art. 134, and obtain the same value. :^i 212 PLANE TRIGONOMETRY. (2) Let 6 be the circular measure of 1'. Then ' ^ • ' ' "'* 49 = ir - = .000290888208665, to 15 decimal places. (( 180 X 60 .-.- = .000000000006 to 12 « -■'■- 4 .-. tf-- = . 00029088820 toll « « 4 ■. / '■' Hence 6 and — — differ only in the twelfth decimal. ,-. sin 1' = .00029088820 to 11 decimal places. cos 1' = VT-sin21'=. 999999957692025 to 15 decimal places, Otherwise iiius : 1 - ^' =.999999957692025029 to 18 decimal places. 2 ^ and — = .00000000000000044 to 17 decimal places. 16 Butcosl'>l-^';-nd. . •> ■■ '^ 136. To construct a Table of Natural Sines and Cosines at Intervals of 1'. We have, by Art. 45, ' •■ . " ' sin {x + y) = 2 sin x cos y — sin (x — y), •'■■■'■ GOs{x -\- y) = 2Gosxcosy — c.os{x — y). 'm-.'- Suppose the angles to increase by 1'; putting 3/ = 1', we have, ;•,..•> I •' : ;,.,•■' :''',■■..,,.,..,';:,; /j jv sin(aj4- 1') = 2sina;cosl' — sin(aj — 1') . . . . (I) cos(.T + 1') = 2cos.'Bcos 1' ~ cos(a; — 1') . ... (2) Putting a;= 1', 2', rV, 4', etc., in (1) and (2), we get for the sines sin 2' = 2sin 1' cos 1' - sin 0' = .0005817764, ' ' sin 3' = 2 sin 2' cos 1' - sin 1' = .0008726646, sin 4' = 2 sin 3' cos 1' - sin 2' = .0011635526 ; and for the cosines cos 2' = 2 cos 1' cos 1' - cos 0' = .9999998308, cos 3' = 2 cos 2' cos 1' - cos 1 ' = .9999996193, cos 4' = 2 cos 3' cos 1' - cos 2' = .9999993223. ' .'^ We can proceed in this manner* until we find the values of the sines and cosines of all angles at intervals of 1' from 0°to30°. , , . .= ...( ,:,. 137. Another Method. Let a denote any angle. Then, in the identity, sin (n -f- 1)« = 2 sin na cos a — sin {n — l)a, put 2(1 — cos «) = k, and we get sin(w4-l)«— sin ?)«=sin 7J« — sin(?j — 1)« — A; sinn« * This method is due to Thomua Siiupsou, au Eugliitb geometriciaa. (1) '■ i-' b i\ !i 214 PLANE TRIGONOMETRY. II li? It This formula enables us to construct a table of sines of angles whose common difference is a. Thus, suppose « = 10", and let n = 1, 2, 3, 4, etc. Then sin 20" - sin 10" = sin 10" - A; sin 10", sin 30" - sin 20" = sin 20" - sin 10" - k sin 20", sin 40" - sin 30" = sin 30" - sin 20" - Tc sin 30", etc. These equations give in succession sin 20", sin 30", etc. It will be seen that the most laborious part of this work is the multiplication of k by the sines of 10", 20", etc., as they are successively found. But from the value of cos 10", we have A; = 2 (1 - cos 10") = .0000000023504, the smallness of which facilitates the process. • In the same manner a table of cosines can be constructed by means of the formula, cos (w + l)a — cos na = cos na — cos (ii — l)a — k cos na, which is obtained from the identity, cos {n-{-l)a = 2 cos na cos a — cos (?t — l)a, by putting 2(1 — cos u) = k, as before. 138. The Sines and Cosines from 30° to G0°. — It is not necessary to calculate in this way the sines and cosines of angles beyond 30°, as we can obtain their values for angles from 30° to 60° more easily by means of the formulae (Art. 45): sin (30° + «) = cos « - sin (30° - rt), cos (30° + «) = cos (30° - n) - sin «, by giving a all values up to 30°. Thus, sin 30° 1' = cos 1' - sin 29° 59', cos 30° 1' = cos 29° 59' - sin 1', and so on. TABLES OF TANGENTS AND SECANTS. 215 139. Sines of Angles greater than 45". — When the sines of angles up to 45° have been calculated, those of angles between 45° and 90° may be deduced by the formula sin (45° + a) - sin (45° - «) = V^ sin a . (Art. 45) Also, when the sines of angles up to 60° have been found, the remainder up to 90° can be found still more easily from the formula sin (60° + «) - sin (G0° - a) = sin «. Having com])leted a table of sines, the cosines are known, since • .^ao \ cos a = sin (90 — a) . Otherwise thus : When the sines and cosines of the angles up to 45° have been obtained, those of angles between 45° and 90° are obtained from the fact that the sine of an angle is equal to the cosine of its complement, so that it is not necessary to proceed in the calculation beyond 45°. Note. — A more modern method of calculating the sines and cosines of angles is to use series (3) and (4) of Art. 156. 140. Tables of Tangents and Secants. — To form a table of tangents, we find the tangents of angles up to 45°, from the tables of sines and cosines, by means of the formula tan a = sin a cos a Then the tangents of angles from 45° to 90° may be obtained by means of the identity * tan (45° 4- «) = tan (45° - a) + 2 tan 2 a. When the tangents have been found, the cotangents are known, since the cotangent of any angle is equal to the tangent of its complement. A table of cosecants may be obtained by calculating the reciprocals of the sines ; or they may be obtained more "^ Called Cagnoli's formula. HI f \-i 216 PLANE TRIGONOMETRY. easily from the tables of the tangents by means of the formula c: coser- a = tan - -}- cot a. The secants are then known, since the secant of any angle is equal to the cosecant of its complement. 141. FormulsB of Verification. — Formulce used to test the accuracy of the calculated sines or cosines of angles are called Formulce of Verification. It is necessary to have methods of verifying from time to time the correctness of the values of the sines and cosines of angles calculated by the preceding method, since any error made in obtaining the value of one of the func- tions would be repeated to the end of the work. For this purpose we may compare the value of the sine of any angle obtained by the preceding method with its value obtained independently. -"■ ■ ' < •\/5 — 1 Thus, for example, we know that sin 18° = — (Art. 4 57) ; hence the sine of 18° may be calculated to any degree of approximation, and by comparison with the value obtained in the tables, we can judge how far we can rely upon the tables. Similarly, we may compare our results for the angles 22^°, 30°, 36°, 45°, etc., calculated by the preceding method with the sines and cosines of the same angles as obtained in Arts. 26, 27, 56, 57, 58, etc. There are, however, certain well-known formulae of veri- fication which can be used to verify any part of the calcu- lated tables : these are Eider^s Formulce: ., . ,, sin (36° + A) - sin (36*' - A) + sin (72° - A) ., ; , . i,i,j. M . - sin (72° -f A) = sin A. / cos (36° + A) + cos (36° - A) - cos (7'>° -f A) , . , — cos (<2° — A) = cos A. LOGARITHMIC TEIGONOMETlilC FUNCTIONS. 217 Legendre's Formula: sin (54° + A) + sin (54° - A) - sin (18° + A) •"'-:;o :■ i.;; .. #s.kH 1^:^^« - sin (18° - A) = cos A. til The verification consists in giving to A any value, and taking from the tables the sines and cosines of the angles involved: these values must satisfy the above equations. To prove Euler's Formuloi : sin (36° + A) - sin (36° - A) = 2 cos 36° sin A VS + l • A = -^ — - — sin A 2 sin (72° + A) - sin (72° - A) = 2 cos 72° sin A 2 sin A (Art. 45) (Art. 58) (Art. 45) (Art. 57) Subtracting the latter from the former, we get sin A. Similarly, Euler's second formula may be proved. By substituting 90° — A for A in this formula we obtain Legendre's Formula. 142. Tables of Logarithmic Trigonometric Functions. — To save the trouble of referring twice to tables — first to the table of natural functions for the value of the function, and then to a table of logarithms for the logarithm of that function — it is convenient to calculate the logarithms of trigonometric functions, and arrange them in tables, called tables of logarithmic sines, cosines, etc. When tables of natural sines and cosines have been con- structed, tables of logarithmic sines and cosines may be made by means of tables of ordinary logarithms, which will give the logarithm of the calculated numerical value of the sine or cosine of any angle ; adding 10 to the logarithm so found we have the corresponding tabular logarithm. The logarithmic tangents may be found by the relation log tan A = 10 4- log sin A — log cos A ; • and thus a table of logarithmic tangents may be constructed. MV^ 218 PLANE TRIGONOMETRY. PROPORTIONAL PARTS. 143. The Principle of Proportional Parts. — It is often necessary to find from a table of logarithms, the logarithm of a number containing more digits than are given in the table. In order to do this, we assumed, in Chapter IV., the principle of proportional parts, which is as follows : The differences between three ntimhers are proportional to the corresponding differences between their logarithms, provided the differences between the numbers are small compared with the numbers. By means of this principle, we are enabled to use tables of a more moderate size than would otherwise be necessary. We shall now investigate how far, and with what excep- tions, the principle or rule of proportional increase is true. !l>l I 144. To prove the Rule for the Table of Common Loga- rithms. Wp have . ' }■■. log (n 4- rf) - logn = log^t_^ = log f 1 + '^ where /u, = .43429448 •••, a quantity < i- Now let n be 9,n integer not < 10000, and d not > 1 ; then , - is not greater than .0001. .• " n,, ^^^^-^ ^ .'. ;^ is not > 4(-0001)^ i.e., not > .0000000025: 2n^ ,. . ... . ,...,... ;,. . - ' and ^- is much less than this. '3n^ ; .-. log (n + rf) — log n = fx , correct at least as far as seven decimal places. ^ ■ '* RULE OF PROPORTIONAL PARTS. 219 Hence if the number be changed from n to n + d, the corresponding change in the logarithm is approximately fid n Therefore, the change of the logarithm is approximately proportional to the change of the number. 145. To prove the Rule for the Table of Natural Sines. sin {6 + h) — sin 6 = sin h cos 6 — sin 6 (1 — cos h) = sin /i cos 6(1 — tan ^ tan - I . (Art. 51) If h is the circular measure of a very small angle, sin h = h nearly, and tan - = - nearly. ,: .*. sin {0 + h) — sin d = h cos 6(1 — tan B tan - j = 7icos6 sin 6. 2 If h is the circular measure of an angle not > 1', then h is not > .0003 (Art. 135). .-. ^' is not > .00000005; and sin 6 is not > 1. .-. sin(6+/0 —sin d=h cos $, as far as seven decimal places, which proves the ]>roposition. Similarly, sin {6 — h) — sin d = — h cos 6, approximately. 146. To prove the Eule for a Table of Natural Cosines. cos {6 — h) — cos 6 = sin h sin 6 ~ cos 6(1 — cos h) h\ sin h sin 6(1 — cot 6 tan 2. If h is the circular measure of a very small angle, sin h — h nearly, and tan - = - nearly. ; .'. cos (6 — 7i) — cos 6— /i sin 6(1 — cot 6 tan ^ J = i'isin6— -cos 6. I , i « 220 PLANE TRIGONOMETRY. We may prove, as in Art. 145, that - cos is not > .00000005. .'. cos(0 — 7i) — GosO = f sind, as far as seven decimal places, which proves the proposition. Similarly, cos (^ + 7i)-- cos d = — /i sin^, approximately. 147. To prove the £11] e for a Table of Natural Tangents. tan {d + h) - tan = «i'L(^„+J) _ ^lil^ = sin h cos {6 + h) cos cos {0 -\- h) cos d , , __ tan A cos*^ ^ (1 — tan tan /t) / ' " ■■ If h is the circular measure of a very small angle, tan /i = A nearly. h sec'' tan {0 + h) — tan d = 1 — h tan 6 : i\!^t'i'f\^ = h sec- 6 + h^ sin $ sec^ $. .'. tan (^ -f- h) — tan ^ = 7i sec^ 6, approximately, unless sin $ sec^^ is large, which proves the proposition, n .•• Similarly, cot {6 — h) — cot 6 = h cosec" $, approximately. Sch. 1. If h is the circular measure of an angle not > 1', then h is not > .0003. Hence the greatest value of h^ sin sec"' is not > .00000009 sin 6 sec'^ $. Therefore, when d>'^, 4 we are liable to an error in the seventh place of decimals. Hence the rule is not true for tables of tangents calculated for every minute, when the angle is between 45° and 90°. , Sch. 2. Since the cotangent of an angle is equal to the tangent of its complement, it follows immediately that the rule must not be used for a table of cotangents, calculated for every minute, when the angle lies between 0° and 45°. BULE OF PROPORTIONAL PARTS. 221 rr 148. To prove the Rule for a Table of Logarithmic Sines. 7j2 sin {6 + h) — sin = h oos 9 - j- sin 6 sin ^ 2 .'. log sin (0 4- ^) — log sin ^ = ,*log^l + /tcot0-|'^ (Art. 145) = al ^ cot B — -- —-(hcot A ' D'-] (Art. 130) f.h\ = /x/i cot ^ - ^ (1 + cot^ ^) + . . . = fihcotd — ^cosec-0-\ If h is the circular measure of an angle not > 10", then h is not > .00005, and therefore, unless cot is small or cosec^ large, we have log sin (^ + ^) — log sin ^ = /i/i cot ^, as far as seven decimal places, which proves the rule to be generally true. Sch. 1. When 6 is small, cosec 6 is large. If the log sines are calculated to every 10", then h is not >. 00005, and fi is not > .5. .-. ^fxh^ cosec^O is not > 6 cosec^ 6 10 10 fmt In order that this error may not affect the seventh decimal place, ecosec'^^ must not be > 10^, that is, 6 must not be less than about 5°. ^ . >: , When $ is small, cot is large. Hence, when the angles fin 8 222 PLANE TRIGONOMETRY, are small, the differences of consecutive log sines are irregu- lar, and they are not insensible. Therefore the rule does not apply to the log sine when the angle is less than 5°. Sch. 2. When 6 is nearly a right angle, cot^ is small, and cosec 6 approaches unity. Hence, when the angles are nearly right angles, the dif- ferences of consecutive log sines are irregular and nearly insensible. 149. To prove the Rule for a Table of Logarithmic Co- sines. cos(^ — h) — cos 6 = h sin 6 — —cos . (Art. 146) . ... cos(^-/0 ^i^;,tan^-^. cos^ 2 .•. logcos(0 — /i) — logcosd /u. log ( 1 + /i tan ^ ] = J I ht^inO - — -Uhtan - ^Y = /x/itan ^ — ^-sec^^-f ••• • '< ',■ - In this case the differences will be irregular and large when is nearly a right angle, and irregular and insensible when B is nearly zero. This is also clear because the sine of an angle is the cosine of its complen^ent. 150. To prove the Rule for a Table of Logarithmic Tan- gents. tan {B + li) - tan B=h sec' Q -f W sin e sec"' B . (Art. 147) tan (B -\- h) t&nB 1 + -. h sin B cos $ + /i'sec2^. BULE OF PliOPOliTlONAL PARTS. 223 log tan {6 4- h) — log tan 6 —H- sin cos $ +h:^sec^e- 1/ Ji 2Vsin^cose = t!:^ + aJi'fseG^e sin ^ cos ^'^ V 2sin='^cos2tf .-. log tan {6 4- h) — log tan 6 ■{-hHec'e] +•• h ■] '»• sin 6 cos $ sm"2e 151. Cases where the Principle of Proportional Parts is Inapplicable. It appears from the last six Articles that if h is small enough, the differences are proportional to h, for values of 6 which are neither very small nor nearly equal to a right angle. The following exceptional eases arise : (1) The difference sin {$ + '0 — sin 6 is insensible when 6 is nearly 90°, for in that case h cos 6 is very small ; it is then also irregular ^ for ^W&mB may become comparable with h cos 0. (2) The difference cos {S -\- h) — cos $ is both insensible and irregular when 6 is small. (3) The difference tan (d -\- h) — tan 6 is irregular when 6 is nearly 90°, for hJ^ sin 6 sec^ ^ may then become compar- able with h sec*^ ^ ; it is never insensible, since sec 6 is not <1. .,.,,,,,,;-,..,,,^_ ,,.,,,._,.. .,.,„3.,.:v .;...:: (4) The difference log sin {$ + h) — log sin 6 is irregular when is small, and both irregular and insensible when ^ is nearly 90°. ^ (5) The difference log cos (0 + /a) — log cos ^ is insensible and irregular when ^ is small, and irregular when ^ is nearly 90°. , (6) The difference log tan (^ -f ^) — log tan ^ is irregular when is either small or nearly 90°. .. , „ ,, 224 PLANE TRIGONOMETliV. m A difference which is insensible is also irregular; but the converse does not hold. When the differences for a function are insensible to the number of decimal places of the tables, the tables will give the /mw ''flows when the angle is known, but we cannot use the tables to find any intermediate angle by means of this function; thus, we cannot determine 6 from the value log cos 0, for small angles, or from the value log sin 6, for angles nearly 90°. Wlien the differences for a function are irregular without being insensible, the approximate method of proportional parts is not sufficient for the determination of the angle by means of the function, nor the function by means of the angle ; thus, the approximation is inadmissible for log sin 0, when 6 is small, for log cos 6, when 6 is nearly 90°, and for log tan 6 in either case. (Compare Art. 81.) In these cases of irregularity without insensibility, the following three means may be used to effect the purpose of finding the angle corresponding to a given value of the function, or of the function corresponding to a given angle.* 152. Three N^thods to replace the Rule of Proportional Parts. ,v' j-^7 '--; '-i T '11 ,v, \■;^■. ,•■;,/. ;;,,, ■;•''; (1) The simplest plan is to have tables of log sines and log tangents, for each second, for the lirst few degrees of the quadrant, and of log cosines and log cotangents, for each second, for the few degrees near 90°. Such tables are generally given in trigonometric tables of seven places ; we can then use the principle of proportional parts for all angles which are not extremely near 0° or 90°. (2) Delambre's Method. In this method a table is con- structed which gives the value of log [- log sin 1" for every second for the first few degrees of the quadrant. * This article has been taken substantially from Hobson's Trigonometry. rii METHODS TO REPLACE THE RULE. 225 it; Let B 1)6 the circular measure of n seconds. Then, when 6 is small, we have 6 = nsin 1", approximately. 8in<> smw .•. log — — - = log — r— 7-, = log sin n" — log n — log sin 1". $ nsinl" .-. log sin n" = log n + Aog-~ + log sin 1"Y Hence, if the angle is known, the table gives the value of the expression in parenthesis, and log n can be found from the ordinary table of the logs of numbers ; thus log sin rt" can be found. If log sin n" is given, we can find approximately the value of n, and then from the ta^^le we have the value of the expression in parenthesis ; thus we can find log w, and then n from an ordinary table of logs of numbers. Rem. When $ is small (less than 5°), sin^ ^ ^ • 1. 1 = I — -, approximately $ 6 (Art. 134, Note) Hence a small error in will not produce a sensible error in the result, since log ^-— - will vary much less rapidly than e. ^ (3) Maskelyne^s Method. The principle of this method is the same as that of Delambre's. If ^ is a small angle, we have '. . sin tf = — — , approximately, 6 'i f and cos^=l— — , sin^ i u u (( (Art. 134) = (cos^)* log sin $= log + \ log cos 0, approximately. Ml 226 PLANE riilGONOMETRY. When ^ is a small angle, the differences of log cos 9 are insensible (Art. 149) ; hence, if 6 be given, we can find log^ accurately from the table of natural logarithms, and a,lso an approximate value of log cos 6 ; the formula then gives log sin at once. If log sin 6 be given, we must first find an approximate value of B from the table, and use that for finding log cos 6, approximately ; is then obtained from the formula. IMS' EXAMPLES. 1. Prove 1 + 2 +,- + ,-+ ••• = 2e. 2. Prove log^ = l-/^ — ? — + — ^ — + ... ^2 2 U •3.2''' 2.5.2* 3. Prove e.^\ 1+2 14-2 + 3 2 1^ [3 [4 4. Prove 1 = ,^ + i + ,^+ ... e [3 [5 [7 5. Prove tan + ^ tan'' ^ + ^ tan' ^ + = ilog| .os(.-l)^ (' ,cos ( B + IT 6. Prove logell = 2.39789527 •.•, by (4) of Art. 130. 7. Prove log. 13 = 2.56494935..-, « « 8. Prove log, 17 = 2.83321334..., " « 9. Prove log, 19 = 2.9444394 (( (( 10. Find, by means of the table of common logarithms and the modulus, the Napierian logarithms of 1325.07, 62.9381, and .086623. Ans. 7.18923, 3.96913, - 2.4578. EXAMPLES. 227 11. Prove that the limit of m sin — is 0, when m = oo. m 12. <( 13. « 14. ii 16. U 16. (( 17. (( 18, (( 19. 20. 21. t( a u u ii » (I <( n if a u " m tan — is 9, m u (( ii u — sin — IS 7rt 2 n ,.2 U Trr^ tan - is tt)^, n versa^ . a'' vers bO 6^' I cos - ) IS 1, V nj ( Sin .- ) IS 1, [ COS - ) IS 1, sin -\ n e n is 1, cos - I IS e 2j " (rff ii m = 30. n = oc. n = 30. « ^ = 0. « n = QO. n = QO. " Jl = 00. n = 00. " « — n = 00. is zero, when 7i = oo. 22. If is the circular measure of an acute angle, prove (l)cosd^ + |- 23. Given ^ = ^: prove that ^ = 4° 24', nearly. 24. Given d 1014 sing ^2165. $ 2166" prove that 6 = 3°, nearly. 228 PLANE TRIGONOMETRY. !! i- i ' 25. Given sin «/> = 71 sin 6, tan = cos ( — md) + V — 1 sin ( — m6) . .'. (cosd4- V— 1 sin^)" = cosw^+V— 1 sin n^, which proves the theorem when w is a negative integer. III. When n is a fraction, positive or negative. Let w = -, where p and q are integers. Then (cos^-HV^sin^)''=cosjj^+V^sinp^(by Land II.). But (cos-^ + V— isin-^j =cos|)^+V-lsinpe. .-. (cos e + V-1 sin ey = f cos -^ + V^^ sin -^^' .-. (cos e -\- V^^ sin ey = cos^^ + V^ sin^^ ; that is, one of the values of (cos ^ + V— 1 sin ^)» IS pd pd cos 1- V— 1 sin — 9 9 In like manner, (costf — V— 1 sin dy = cos nd — V— Isinn^. Thus, De Moivre's Theorem is completely established. It shows that to raise the binomial cos^-f V— 1 sin tf to I DE MOIVRE'S THEOREM. 231 any power, we have only to multiply the arc by the exponent of the power. This theorem is a fundamental one in Analytic Mathematics. 154. To find All the Values of (cos^ + V- Ising)^".— When n is an integer, the expression (c()s^ + V— Isin^)" can have only one value. But if n is a fraction = -, the expression becomes (cos ^ + V^^sin ey=yj (con ^ + V^^ sin d)", which has q different values, from the principle of Algebra (Art. 235). In III. of Art. 153, we found one of the values of (cos ^ + V — 1 sin $) ' ; we shall now find an expression which will give all the q values of (cos 6 + V— 1 sin 0)''. Now both cos 6 and sin 6 remain unchanged when 6 is increased by any multiple of 27r; that is, the expression cos^ + V— Isin^ is unaltered if for 6 we put (d-\-2rTr), where r is an integer (Art. 36). (cos^ + V^^sin^)' .1* ''■!> = [cos(e + 2r7r) + V-lsin(^4-2?>)]' ' ' '" p{e + 2r7r) , / — ^ . p{d-h2r7r) ., , V-qwVx = cos-^--^ ^+V~lsm^-^^ (Art.lo3)(l) The second member of (1) has q different values, and no more; these q values are found by putting r=0, 1, 2, ••• 7 — 1, successively, by which we obtain the following series of angles. ?)(^+2rjr) pd When r = 0, cos — - = cos 156. To develop sin 6 and cos 6 in Series of Powers of 6. Put nd = a in (2) and (3) of Art. 155; and let n be in- creased without limit while « remains unchanged. Then since 6 = -, must diminish without limit. Therefore the n above formulae may be written cos a = COS" e - '^^=^ cos"-^ 6 f'^^" \2_ \ d J 4. «(«-^)('^-2^)(«-3^) cos"-*g(g^y-... (1) and sin « = «cos"~*^( ] \ J _,(^_^)^(^_2g^^^^,._3 /sin^3_^ ... . (2) If w = QO, then ^ = 0, and the limit of cos and its powers is 1; also the limit of ( -— ) and its powers is 1. Hence (1) and (2) become ^ ^ a' . a* « cos« = l 1- , , — !-••• (3) 2 '14 [6 Sin a = « — , — • • • .'] ■ [5_ • ••••••#• \ / Sch. In the series for sin a and cos «, just found, a is t?ie circular measure of the angle considered. CONVERGENCE OF THE SERIES. 235 Cor. 1. If a be an angle so small that a^ and higher powers of a may be neglected when compared with unity, (3) becomes cos a = 1, and (4), sin « = «. If «^, u^ be retained, but higlier powers of a be neglected, (3) and (4) give sin a = a — — ; cos a = 1 — — (Compare Art. 134) Cor. 2. By dividing (3) by (4), we obtain ^3 2 a* 3 3 • o 3 • 3 • o • < 17 «^ , , —_ + etc. (5) 157. Convergence of the Series. — The series (3) and (4) of Art. 156 may be proved to be convergent, as follows : The numerical value of the ratio of the successive pairs of consecutive terms in the series for sin a are a* «' «* «' 2-3 4-5 0-7 8-9 etc. Hence the ratio of the (n -f l)th term to the nth term is -; and whatever be the value of a, we can take 2w(2« + l) n so large that for such value of n and all greater values, this action can be made less than any assignable quantity ; hence the series is convergent. Similarly, it may l)e shown that the series for cos a is always convergent. - - " 158. Expansion of cos"^ in Terms of Cosines of Multi- ples of 6, when n is a Positive Integer. Let .r = cos^ + t sin^; 1 1 then X cos $ -\-i sin 6 = cos 6 ~ i sin $. .: a; + - = 2 cos 6 : and x — =2i sin 6 X X (1) 236 PLANE TRIGONOMETRY. i^«ii liiiii if Also a;"=(cosd+isin^)"=cosn^4-tsmwd (Art. 153) (2) and — = (cos 6 — i sin 6) " = cos nd — i sin nB . . . (3) 1 1 .'. 2 cos n^ = a;"H — .and 2isin?i0 = a5" . (4) Hence (2cos^)" = (a; + .'K-i)", by (1), = a;" + nx"-- + '*^^^^ — ^) a?'-* -f- etc. + «a; <"-2) + aj-" n (n-1) == 2 cos wtf + n 2 cos ( ji - 2) e + "v, "'^ 2 cos (n - 4) ^ + etc. If .-. 2"-^ cos" d = cos 91^ + w cos (w — 2) ^ + '^-^^^^-^^ cos (?i - 4) ^ + etc. (5) Note. — In the expansion of (a; + «"■')" there are n + 1 terms; thus when n is even there is a middle term, the | - + l)th, which is independent of 9, and which is n( n-l)...(n-4n + l) ^^ ^2 / lln l-i" Hence when n is et'en the last term in the expansion of 2"^' cos** 6 is Hi When n is odd the last term in the expansion of 2"'^ cos" 9 is 159. Expansion of sin" '} in Terms of Cosines of Multiples of 6, when n is an Even Positive Integer. (2 i sin ^)» = /'a; - ^V'by (1) of Art. 158 = a;" - n^-'' + - ^^""^^ «"-♦ + .- 4. ^_(^ 1) a.-("-4) _ ,,x'-('-'»> + x-^ II' V. (5) EXPANSION OF SIN" 6. 237 =e-9-"e-^)-'^e*^^; (-irn(n-l)...(:-; + l + ••• + n [i± .. 2"-'(— l)'^sin"0 = cos w^ - n cos (h - 2) ^ + "^''~^^ cos (n - 4) J If (-irn(^-i)- :^+i + ri ^Li± 160. Expansion of sin" 6 in Terms of Sines of Multiples of 0, when n is an Odd Positive Integer. (2 i sin 0)" = (^a; - -Yby (1) of Art. 158 z=x" — nx' n-2 I n (n - 1) ^,._4 [2 X' n-4 n-1 li(n-i) V ^; .-. (2isin0)" = 2isinri0 — n2isin(n — 2)6 n(n-l)2,-sin(n-4)6 ^ [2 .1-1 (-1) '^ n(w-l)---K ^^±^9.-«infl [(4) of Art. 158] 1S¥ 238 PLANE TRIGONOMETRY. Whence dividing by 2t, we have w-l 2-'(_l) 2 sin"^ = sin 110 - 71 sin (n — 2)tf + ^^-f-^ sin (71 _ 4)^ + .- (-1)^.(n^)...^(», + .S)^.^ ^^ l''(H-J) ll » -^ EXAMPLES. Prove that 1. 128 co3''^=cos Se+Hcos(5e+'2S cos 4^+56 cos 20+35. 2. 64 cos^0 = cos 70 -f 7 cos 50 + 21 cos 30 + 35 cos 0. 161. Exponential Values of Sine and Cosine. Since e'=l + a; + ^ + ,^' + ^'+- . . . (Art. 129) Lf L? Li 0' • p« — 1 _ !L 4. ^ •• ^ "^ [2 + " = cos + i sin and (Art. 156) = cos 6 — i sin 0. .-. 2 cos = e*" + e- *«, and 2 i sin = e»« - e-»« . . (1 ) COS0 = e*'* + e -i0 -, and sin = otS ie 2t • • (2) which are called the exponential values* of the cosine and sine. Cor. From these exponential values we may deduce similar values for the other trigonometric functions. Thus, tan0 = -i— -^-- (3) * Called also Euler's equations, after Euler, tbeir discoverer. 6 + ^rt. 129) \rt. 156) . . (1) . . (2) Dsine and y deduce IS. Thus, . . (3) jr. f GREC, on Y ' 8 SERIES. 239 Sell. These results may i)e applied to prove any general fornnda in elementary Trigononu^try, and are of great im- portance in the Higher Mathematics. BXAAuPLBS. 1 . Prove We have sin2g l + cos2^ 2tsin2g 2 + 2cos2tf~2 + e2« + ^ = tan 6. o2i» ii» 2i» by (1) pi9 g- %9 Prove the following, by the exponential values of the sine and cosine. 2. cos 2 rt = cos^ « — sin'' a. 3. sin^ = — sin( — ^). 4. cos3^ = 4cos''e-3costf. Rem.— It we omit the i from the exponential values of the eine, coBine, and Un- gent 01 0, the reaulta are called respectively the hyperbolic sine, cosine, and tangent of 9, and are written sinh 9, cosh 9, and tanh 9, respectively. Thus we have ■inh = — f sin ifl, cosh 9 = cos i9, tanh = — i tan i9. Hyperbolic functions are so called, because they have geometric relations with the equilateral hyperbola analogous to those between the circular functions and the circle. A consideration of hyperbolic functions is clearly beyond the limits of this treatise. For an excellent discussion of such functions, the student is referred to Buch works as Casey's Trigonometry, Hobson's Trigonometry, Lock's Higher Trigonom- etry, etc. 162. Gregory's Series. — To expand 6 in jiowers of tan $ where 6 lies between — - and 4- -• 2 2 By (3) of Art. 161, we have i tan = -»■« l+ttan^^ 26'" 1-iidLue 2e-»« = e2»o. 240 PLANE TRIGONOMETRY. i .-. log e2»<'= log(l + i tan 6) - log(l - i tan 0). . : 2 W = 2 <■ (tan tf - ^ tan'^ 6 + ^ tan' - etc. ) (Art. 130 ) .-. ^=tan^- ^tan^0 + |tan*^-etc (1) which is Gregory^s Series. This series is convergent if tan ^ = or < 1, i.e., if 6 lies between v ^ and -, or between ^tt and Itt. 4 4 Sch. This series may also be obtained by reverting (5) in Cor. 2, Art. 15(1 Cor. 1. If tan 6 = x, we have from (1) x^ . x* tan~^a;= x 1 etc. 3 6 Cor. 2. If ^ = -, we have from (1) 4 (2) IT ^ = l-i + i-| + etc. (3) a series which is very slowly convergent, so that a large number of terms would have to be taken to calculate tt to a close approximation. We shall therefore show how series, which are more rapidly convergent, may be obtained from Gregory's series. 163. Enler's Series. . . (by Ex. 2, Art. 60) Puttf = tan-ii. .-. tan (9 = ,^, which in (1) of Art. 162 tan-' - + tan-' ^ = - 2 3 4 gives tan-' - = - — + — + etc. . 2 2 3.23^6.2* 7-2' • • • (1) Put $ — tan ' - .'. tan ^=-, and (1) becomes 3 3 ^ ^ tan * - = H ; + etc. 3 3 3.3^ 6.3' 7.3^ • • • • • (2) MACHINES SERIES. 211 (2) . (2) Adding (1) and (2) we have 4 V2 3-23 5-2* J \3 3.3^ 5.3* J ^ ^ a series which converges much more rapidly than (3) of Art. 162. 164. Machines Series. Since 2tan-^^=tan -^--i- (by Ex. 3, Art. 60)=tan-»-|, 5 1 — Alt i^ •^3 .-. 4 tan-' - = 2 tan"' 4 = tan"' — i— = tan"' ^ -.1 5 12 1 _ 2.5 Also, tan-' ^ - tan"' 1 = tan"' ^ } = tan 119 ... 4tan-'i-T=tan- 5 4 = 4 tan 239 119 239 tan" 239 .7r_./l__l ^1__ \ "I \5 3.53 5-5* "V -M _ + -i Y ;239)* y V239 3.(239)» 5(239)* In this way it is found that ir = 3.141592653589793 ••.. Cor. Since tan-' 1 + tan"' -1- = tan"' ^^-t^ NoTB. — The BcricB for tan-' ^ and tan"' '- are much more convenient for pur- 70 99 poses of numerical calculation than the series for tan~' — -• 239 Example. — Find the numerical value of tp to 6 figures by Machin's series. amimm^mgeom I - If. rt 242 PLANE TRIGONOMETRY. 165. Oiven sin 9 = x sin {6 + a); expand d in a Series of Ascending Powers of x. ] . . (Alt. 161) We have gi«_ e-»« = x[e^^+ ia e- -i9 -ia .-. e2»» -1 = X [e2»» . gia — e- *•] « • • Q2i9- 1 — xe- -ia 1 — ice*'" .-. 2'i$ = log(l — xe-*") — log(l — xe^) = a;(e»«» — e-»") + ^(e2»a_g-2ia) _j_ ^(^^ia_e-zia^ ... ^ ^ (Art. 130) .-. ^ = a;sina4-f sin2« + ^sin3a+--- (Art. 161) (1) Example. If a = tt — 2^, then a; =1. .*. (1) becomes ^ = sin 2^ - ^ sin40 + \ sin 6^ - \ sin 8^ + •••. 166. Oiven tan a; = n tan d ; expand x in Powers of n. g»9 g - i9 gia? _ g-»x gir _|_ g-w: />2ta; . = 71- plix + 1 = 71 gi« -I- g-W g2i« __ 1 (Art. 161) ,'iie + 1 . g2te = i^!!!±l)±ii(f'*!zill ^e2ifl_|.l)_„(g2»9_l) _ (1 + M)e2f9 -f 1 - n (l-7i)e2»» + l4-n ^g2«_|_i \^ 1 + nJ .-. 2ta; = 2 ttf + log(l + me- 2«) - log(l + me'^i«) = 2td - m(e2»» - e-2»«) -|- !^ (e4W _ g-4W) . m" a; = ^ — msin2d4- — sin4tf (Art. 161) \\\ RESOLUTION INTO FACTORS. 243 RESOLUTION OF EXPRESSIONS INTO FACTORS. 167. Resolve x" — 1 into Factors. Since cos 2 ?*7r ± V — 1 sin 2 rTr = 1, where r is any integer, and of = 1, .•. ic" = cos 2?*7r ± V— 1 sin 2r7r. 1^ .-. a; = (cos2r7r ±V-lsin2r7r)" 2r 2rTr' = cos ± V— 1 sin n n (Art. 163) (1) (1) When n is even. If r = 0, we obtain from (1) a real root 1 J if r = -, we obtain a real root — 1, and the two cor- responding factors are x—l and x + 1. If we put r = l,2,3...^-l, in succession in (1), we obtain ri — 2 additional roots, since each value of r gives two roots. The product of the two factors, which are and (X — cos — — V— 1 sin n (x — cos — + V^^ sin V n 2r7r\ n ) 2r^■^ n ( ■=.{ 05 — cos- n J 4- sin'' n = ar — 2a;co8 hi n (2) which is a real qiiadratic factor. .-. a;"-l= (o;2-l)raj*-2a; cos ^ -j-lYa;'- 2xcos ^ + iV- ../ar'_2a;cos*^7r+lY.x'-2a5C08^7r-|-lY.. (3) * Thia eipreaaloD give* the n nth root* of unity. Ff m ! i, 1* 244 PLANE TRIGONOMETRY. (2) TF/ien )i is odd. The only real root is 1, found by putting »' = in (1); the other 71 — 1 roots are found by 71—1 putting r = 1, 2, 3, — in (1) or (2) in succession. .'. af— l = (a5— l)f ar'— 2a;cos— + 1 jf ar^— 2a;cos — -f- 1 j-'- ^-far'— 2a;cos TT-fl jf ar*— 2a;cos tt+I ]••• (4) li Ml i M 11 lit) 168. Resolve a;" 4- 1 into Factors. Since cos (2r + l)7r ± V— 1 sin (2r + l)7r = — 1, where r is any integer, and a;" = — 1, .-. af = cos (2r + l)7r±V— 1 sin (2r + !)«•. .-. X = [cos (2r 4- l)7r ± V— 1 sin (2r -j- l)7r]" 2r+l .— ^ . 2r+l siCOS 7r±V— ISin TT . . . (1) n n which is a root of the equation a;" = — 1_; i.e.^ — 1 is a root. (1) When n is even. There is no real root; the n roots are all imaginary, and are found by putting r=:0,l,2,...^-l, successively, in (1), The product of the two factors. (: X — cos - 2r + l n — V— 1 sin '-■^') and I n ' ** / = a^-2a;co8?ili^7r4-l (2) n which is a real quadratic factor. RESOLUTION INTO FACTORS. 245 (4) .-. cc" + l = /V-2a;cos^ + lYar-2a;coSy^ + lY.. ../ar' - 2a; cos '-^^tt + iVar^ - 2a; cos ^^ tt + lY- (3) (2) When n is odd. The only real root is — 1 ; the ?t — 3 other n — 1 roots are found by putting r = 0, 1, 2, • • • in (1), in succession. .-. a;»+l= (a; + l/a^ - 2a; cos ^ + lYa-^ _ 2a; cos ^ + iV- • ••[ar^ — 2a; cos^?— li7r + lY^-^-2a;cos--^7r + l'] (4) 71 n EXAMPLES. 1. Find the roots of the equation ar* — 1 = 0. Aus. 1, cos \ (2 ?'7r) + i sin \ (2 rir), where r = 1, 2, 3, 4. 2. Find the quadratic factors of a^ — 1. Ans. {x" - 1) (ar^ - ^2 a; + 1) (^ + 1) (^f^ + v'2 a; + 1). 3. Find the roots of the equation a;^ + 1 = 0, and write down the quadratic factors of x* -\- 1. ■Ans. ± — ± V^^-^; (a^ - a;V2 + 1) (x- + a;V2 +1). V2 V2' 169. Resolve a;^ — 2 x" cos ^ + 1 into Factors. Let a;^-2a;"cos^ + l = 0. . •. x^" - 2 X" cos 6 + cos^ e = — sin^ 0. .'. a;" — cos 6= ± V— 1 sin ^ = ± « sin $. .: X = (cos B ±i sin ^)" = cos 2rw + ± I sm 2rw + e (1) 7t n since cos d is unaltered if for we put 6 -\- 2 rir. If we put r = 0, 1, 2, •••?? — 1, successively in (I), we find 2n differ- ent roots, since each value of r gives two roots. p lill'l ft !| III !f ift' ^1 lii :r .ill lifii 246 PLANE TBWONOMETRY. The product of the two factors in (1) cos — X I 05 — cos ' — \ n 2rn + d ..2r7r-^b — I sin + 1 sin 2rir + e ^) = ar2-2a;cos^':^^:+-^ + l n .*. 05** — 2 af cos ^ + 1 = /^a^-2a;cos^ + iyg'-2a;cos^^ + ^ + lY.. r ^ (2»i-4)7r + d \ ...far'-2a;cos^^ J +ll— (2) /»«- 2ajcos (2n-2)7r + d n + 1 !••• . . . . (3) a; Cor. Change x into - in (3) and clear of fractions, and a we get a;** — 2 a''af* cos tf + a**" = [ ar* — 2 aa; cos --\-aA'-' ../ar' - 2 oa; cos ^^!^±^ + a'Va;' - 2 aa; cos i^^^JtJ ^_ a^y • •• to w factors (4) EXAMPLES. Find the quadratic factors of the following : 1. a;*- 2a;* cos 60° 4- 1 = 0. Ans. {it' - 2x cos 15° + IXx** - 2 a; cos 105° + 1) X (a« - 2 a? cos 195° + 1) (ar* - 2 a; cos 286° + 1) = 0. 2. a;»°-2a'»cosl0° + l=0. Ans. («« - 2 a; cos 2° 4- 1) (x' -2x cos 74° + 1) X (ar* - 2 a; cos 146° + 1) («2 _ 2 a; cos 218° 4- 1) (ar' - 2 X' COS 2yo" -f 1) = 0. DE MOIVRE'S PROPERTY OF THE CIRCLE. 247 (2) 170. De Moivre's Property of the Circle centre of a circle, 1* any point in its plane. Divide the circumference into n equal parts BC, CI), DE, -.., begin- ning at any point B; and join O and P with the points of division B, C, D, •••. Let FOB = ; then will OB'" - 2 B" . OP" cos ne + OP :2n = PB' . PC . PD • ■ . to ?i terms. For, put OB = a, OP = x, and $ = -; then PB' = OP + OB'' - 20P • OB cos $ = ar* + a^ — 2 ax cos - n (1) ;2 n/-wT^ r\/~i a-[-2ir PC' = oF + OC - 20P ■ OC cos = x^ ^a'-2ax cos "-+15.; and so on (2) 'ax i^ + a" Multiplying (1), (2), (3),... together, we have PB' ■ PC' • PI? .•• to n terms ^zfx' - 2ax cos " + aNx" - 2 = a:2« _ 2a''a;» cos « + a^" . [by (4) of Art. 169] = OP«»-20P"-OB"cos7i^ + OB'" . . . (3) which proves the proposition. ip' , !■' M'i m ! "J !'l I!! 'I? i'! m'i I I'll ft ;i! 248 PLANE TRIGONOMETRY. 171. Cote's Properties of the Circle. — These are particu- lar cases of De Moivre's property of the circle. (1) Let OP, produced if necessary, meet the circle at A, aud let AB = BC = CD, etc., = -'!'; n then n$ is a multiple of 2 v. Hence we have from (3) of Art. 170, after taking the square root of both members, OB" - OP" = PB. PC. PD... ton factors. . . I. (2) Let the arcs AB, BC, "-be bisected in the points a, 6, ••• ; then we have, by (1), OB'" - OP*" = Pa . PB . P6 . PC . Pc ... to 2?i factors. Hence, by division. IL OB" 4- OP" = Pa . P6 . Pc . . . to ?i factors . . Cor. If the arcs AB, BC, ... be trisected in the points a,, a2, 6i, 62, ••., then we have OB'"+OB" . 0P''+0P2''=Pax . Paj, • Vb^ . P62 ... to 2»i factors. 172. Resolve sin 6 into Factors. (1) Put a;= 1 ; then we get from (3) of Art. 1G9 2(l-cos^)=2"/'l-cos^/'l-cos^±^Yl-cos^-^±^Y.. '.'(1 Put tf = 2?i<^ in (1), and let 2n« = 7r. n j\ (1) 1 — co8^ = 1 — cos2w<^ = 2sin''n<^; then extracting the square root, we have Bin?}<^ = 2"-'sin<^ • sin («^ 4- 2rt)sin (<^ -f 4«) x ... X sin (-{- 2n« — 2«) (2) I ^ RESOLUTION INTO FACTORS. 249 •ticu- vB U root I. . (2) But sin (tf) + 2n« — 2«) = sin(<^ + 7r — 2«) = sin(2rt — <;^), sin {), and so on. « Hence, when n is odd, multiplying together the second factor and the last, the third and the last but one, and so on, we have sinnt^ =2"~'sin<^sin(2rt+<^)sin(2«— <^)sin(4«-f<^)sin(4«— «^) • •• X sin[(w — 1)« + <^] sin [{n — l)u — <^]. But sin (2«-|-<^) sin (2 a— <^.) = sin^2« — sin^<^, and so on. .*. sinn^=2''~'sin <^(sin''2« — sin*«^)(siu^4« — sin-<^) X ••• ..• X [sin-(n — 1)« — sin'-'i/)] .... (3) Divide both members of (3) by sin <^, and then diminish <^ indefinitely. Since the limit of sin n -4- sin is n, we get n = 2""*sin^2rt sin^4« sin'^Ga x ••• X sin'^(w — 1)« (4) Divide (3) by (4) ; thus sm 7n^ = n sm {l — ^-„-„^ 1 - . .,/ X • • • (5) Put n0 anJ o, •Intf ia +, und every fitctor in the •ecoi nber of (0) ii poitltive; when 0>ir and <2ir, ■in0 \* -, und only the leconU '• U negative; wlieu 9>2irand <3;r, botli menibem are positive, Hinee only tht >ad and third factors are negative; and so on. Hence the + sign was talten in lAiract- ing the square root of (1), , comes Cor. Let 6 = "^, then sin - = 1, and - = ^. Hence (6) be- ^=i^ ^ / \ ^ * ^ J TT 1-3 a. 5 5-7 ; -_ • I I a ■ ■■ - . • • • • 2 2^ 4^' & 22 42 62 82 2 1.3 3.5 5.7 7.9 which is Wallis^s expression for tt. 173. Resolve cos^ into Factors. — In (2) of Art. 172, change into + «, then n<\> becomes n<\>-\- nu, i.e., n + «) sin (<^ 4- 3 «) sin (<^ + 5 «) X ••• sin[ + (2?i — l)rt] (1) But sin(<^ -f 2?i« — a) = sin(<^ -j- tt — «) = sin(« — <{>), sin(<^ + 2nH — 3«) = sin(3« — <^), and so on. Hence when n is even we have from (1) cosn<^=2"'* sin(rt-f <^) sin(« — 6) sin(3« + )sin(3rt — ^) X ••• X siu[(n — 1)« -f <^]sin[(?i — !)« — ] = 2"-'(sin« rt - s'm^) (sin*3« - sin^) X ••• X[sin''(n-l)«-sin*(^] (2) Therefore, putting n = 6, as in Art. 172, we obtain cos H'-'-S)H^ttQ (3) NoTK. — For an alternative proof of the propositions of Arts. 172 and 173, see Locli's Higher Trigonometry, pp. 92-95. SUMMATION OF SERIES. 251 EXAMPLES. IT 4n' 1. If rt = ."-, prove that sin rt sin 5 a sin 9 « • • • sin (4 n — li) « = 2 "+*. 2. Show that 16cosdcos(72°-d)cos(72°4-e)cos(144°-d)cos(144° + ^) = cos 5 $. SUMMATION OF TRIGONOMETRIC SERIES. 174. Sam the Series sin It + sin(rt + /3) + sin(« + 2/3) + ••• + si" ['« + (« - 1)^]- We have 2 sin « sin ^ j3 = cos fa - ^ -cos fu + | j (Art. 45) 2 sin (rt + i3) sin^ j3 = cos fa + ^- cos (rt + f ^), 2 sin (rt -1-2/3) sin ^ /? = cos (rt + 1)3) - cos (rt -\- f ^), etc. = etc. 2sin[rt-f(H-l)i3]sini^ Therefore, if S„ denote the sum of n terms, we have, by addition, 2S„ sin i/3 = cos (rt - ^/3) - cosU + -^^P = 2 sinfrt 4- -■^- iSl sin i wjS . .^ (Art. 45) sinfrt-i-'—^iS'lsin^niS sini/3 252 PLANE TRIGONOMETRY. ''*^ii tidiii; |i' t Mi ill 'h:;; 175. Sum the Series cos «4-co8 (rt4-/3) + eo8 (« + 2/8) H +cos[rt4-(n— 1)/3]. We have 2 oos « sin ^/3 = sin (« + ^/3) — sin (« — ^y3), 2 cos (rt + /3) sin ^/3 = sin (a + |/3) - sin (« + ^y3), etc. = etc. 2cos[rt + (?i-l)^]sin^/3 = 8m a + — — ^ _sin « + — ^^ . Denoting the sura of n terms by S„, and adding, we get 2S„sini/3 = sinr« + ^^Y^/3l-sin(«-i)3). cos I « + — ■"- /sl sin ^ ?i/3 .-. S„ sin^yS Bern. — The sum of the series in this article may be deduced from that in Art. 174 by putting a + - for a. The sums of these two series are often useful;* and the ■tudent is advised to commit them to memory. iir Cor. If we put /3 = — , then sin ^nft = sin tt = 0. Hence we have from Arts. 174 and 175 sin cos rt+sinr«+— '^j + sin^<+y^j4-...sin fi+ -^^ — -^ =0. « + C08[ — ] + COS( U-\ ) + ••• + oosfrt+ '"') + ••• cosr«4-^^^^-^'rl=0. Note. — These two results are very important, and the student sliould carefully notice them. 176. Supa the Series sin'"«4-sin'"(«4-/3)+3in'»(«+2/3)H l-sin- [«+(«- l)i3]. This may be done by the aid of Art. 159 or Art. 160. * See Thompson's Dynamo- Electric Machinery. 3d ed., pp. 345, 346. ■ SUMMATION OF SERIES. 258 Thus, if m is even, we liave from Art. 159 2m-i gin- rt = (-1)2 [cos m«-m COS (wi-2)«4-...] . (1) 2*-^ sin"* (« + (3) m = (— 1)" [cos m («+/3) — w cos (w-2) (rt4-/3) + •••] (2) a,nd so on ; and the required sum may be obtained from the known sum of the series [cos ma + cos m (« -f- )3) + cos m (a + 2/3) H ] and I cos (m — 2) « + cos [ (m — 2) (u + /8) ] • +cos[(m-2)(« + 2y3)] + -..|, etc. We may find the sum of the scries cos"* n + cos"* (a + iS) + cos"* (« + 2 /8) + etc. to n terms in a similar manner by the aid of Art. 168. Hence ■> 0. EXAMPLES. 1. Sum to n terms the series sin* rt 4- sin=» (« + /8) + sin* (a + 2 /3) + ... We have 2sin2a = -(cos2rt-l)by (1), 2 8in2(« + i3) = - [cos 2 (« + ^) - 1] by (2), 2 sin* (a + 2 ^) = - [cos 2 (« + 2 y8) - 1], and so on. Hence 2S„=n— [co8 2rt+co8 2(rt+/8)+cos2(«+2/3) + ...] ^^^cos[2« + (n-l)^]sinrt^ (Art. 175) sin /3 ^ '^ • S —^ cos[2ot + (n — l)j8]sinn)3 " " 2 2sin)3 254 PLANE TRIGONOMETRY. 2. Sum to n terms the series cos' u 4- cos"* 2u + cos' 3 « + . 2 cos ['A /< -f I (>t — 1) 3 rt"| sin f nu 8 sin J « (i 008 rt H — « •> sm nit 8 sin ^ a 177. Sum the Series sinu — sin («; -f /3)4-sin (« + 2/3) to 7i terms Change ft into P + n, and (1) becomes 8in« + 8in(« + tt -f ;3) + sin(rt + 'Jir + 2/3) + ...". Therefore we liave from Art. 174 + P) sin S.= , -^ J - sin Z+1 2 raB • • • Similarly, cos« — co8(«4-/3) + co8(«-f2/3) to n cosr« + (!LzlIK5L±^"] sin !iOL±^ " 8iu2L±i:" 2 17a Sum the Series cosec 6 4- cosec 2 6^ -f- cosec 46 -\- •" to n terms. We have cosec tf = cot - — cot $, 40 coser "Jtf = cot tf — cot 2$, etc. = etc. cosac 2" -'d = cot 2*-''$ - cot 2="-'^. Therefore, by addition, as in Art. 174, S^ = cot|d-cot2'^-"tf. (1) (2) • • • (3) (4) SUMMATION OF SERIES. 255 Note. — Tho nrtlRce employed in this Art., of reBolvInK each term Into the dif- ference of tM'o other*, in exlenaivcly iiMeii In the niimninlion of Kerie*. Practice alone will give the student readineitii in eHeotinK nuvh traniiformationa. If be cannot diicover the mode of redoltitlon In any example, he will often easily recognize It when ho aeea the result of summation . The Htudent, however, la advised to resort to this method of solution only as a last resoarce. (•^) . (4) 179. Sum the Series 6 9 tan 6+ ^ tan - + \ tan -f ... to n terms. J 4 We have tan tf = i;ot tf — 2 cot 2 0, $ 6 ^ tan -= \ cot^ — cot $, . Id itan-=:|cot--^cot-, etc =s etc. \ . B 1 . ^ - — ; tan — ■ = — - cot — - 2«-i 2»-> 2"~' 2""' 1 , ^ cot on-a •>»-a ••• S. = 2^,cot^i-,-2cot2#. 180. Sum the Series 8in«-f a;8in(«4-/3)-f jc'8in(«-|-2/8)H — af"'sin[«-}-(n— 1)^]. Denote the sum by S„, and substitute for the sines their exponential values (Art. 161). Thus, 2 tS,. = (e<« - -?-<») -f 3;(e<(»+^) _ e-*(»+P)) g<« _ a;"e<(« + np) er <« — aj"c- *(" + »'') 1-xcO l-(Be-<^ [Alg. (3) Art. 163] "^ 1 - a?( e<^ + e-'*P) + .r" J^ 256 PLANE TRIGONOMETRY. .'. S.= U sin«— X3iu(rt— ^)--j;'*8in( « -f-n/3)-i-af *^'sin[«+(n— 1)/3] .^>. 1 — 2xco8/8 + «* Cor. If aj < 1, and n be indefinitely increased, S _ si" a — X sin (« — ^) l — 2x cos /3 + «* (2; 5cA. Similarly, cos« -|- a; cos (« + )3) + a''cos (« + 2)3)4- ••• to n terms = cosft— a;co8(«— /3)— a;"co8(«-f n/8)-f a^+'co8[rt+( ^ 7i — 1)/8] 1 — 2a; cos /3 -j-ar* We may obtain (3) from (1) by changing « to « -f-^. {^) Also g^^ co8«-a;c j)sl«-|) .... (4) l-2«cos/8 4- X* ^ 181. Sum the Infinite Series (Bsin (« + ^) + i^sin (« + 2y3) + ^sin (« + 3)8) -f- •••, If l^ and a;co8(« + )3) + ^co8(« + 2/3) + ^cos(« + 3)3)+ •••. Let S denote the former series, and C the latter. Then C + tS = a;e<(<' + '') + -c*(» + 2^) . ^g^Ca+sp)^. ... [2 [3 \ ^\2 ^|3 ^ ; = e*«(e*«*^-l) . . [by (3) of Art. 129] = e»«(gx(co.^ + i«inp)_i) . . . (Art 161) — . gc COS Pgi(a. + X sin fi) gta _ gx coBflj-cos (rt 4-a; sin )3) -f i sin (« +a5sin )3) ] — (cos « + 1 sin «) .... (Art. 161) m . (4) EXAMPLES. Equating real and imaginary parts, we have C = e-^co*^ cos (rt 4- X sin P) — cos «, S = e*<=°"^ sin (« + x sin ft) — sin a. EXAMPLES. Prove the following statements : 1. The two values of (cos 4 ^ + V— 1 sin 4 tf) - are 257 ±(eos2^ + VI^sin2d) . . . (Art. 154) 2. The three values of (cos ^ -f V— 1 sin d) i are cos^+V^^sin^, cos^^-HV-isin?^^^, a 3. The three values of (—1)^ are 1_|_V~3 . 1-V^^ 3 (Art. 154) 2 ' ' 2 4. The six values of (—1)* are contained in v cos i^'' + ^)'' ± y/-iri sin ^^^-±-^- , where r = 0, 1, or 2. 5. The three values of (1 H-V— 1)» are contained in 2* ["cos I + V^^ sin n where = -, f rr, or V «•■ B. The three values of (3 + 4 V— 1) ' are contained in cos —J 1- V — 1 sin — — -! — , where r = 0, 1, or 2. 7. cos 6 ^ = cos" ^ — 15 cos* 6 sin* ^4-15 cos' sin* ^-sin« d. 8. sin 9$-d COS* tf sin ^ - 84 cos" $ si n"" ^ -f 1 2() cos* ^ si n» tf -36co8«^8iu^d-t-iiiu"^, V.) 258 PLANE TRIGONOMETRY. 9. tan n6 II I n tan ^ - n (ti_--l ) ( u - 2) , 10. Given = -— ^: show that 6 is nearly the circular e 2166 -^ measure of 3°. Prove the following : 11. - 64 sin^ 6 =■• sin 7 6-7 sin 5 6 -f- 21 sin 3 - 35 sin 0. 12. -.2»sin'"^=cos 10 ^Si-lOcos 8 O+in cos 6^-120 cmAe + 210 cos 2 (9 -126. 13. 2« (cos« d + sin" 0) = cos 8 ^ + 28 cos -l d + 35. 14. cos«0-f siii''e= J(54-3co84tf). 16. Expand (sin 0)*"+" in terms of cosines of multiples ote. 16. Expand (sin 9)*"-*-^ in terms of sines of multiples of $, 17. Expand (cos ^)^" in terms of cosines of multiples of 0. Use the exponential values of the sine and cosine to prove the following : iQ sin ^ .$ 18. - = cot-. 1 - cos $ 2 19. If log(.T4-y V— i) = rt-|-/3V-l, prove that ar* + y- = e'-", and y = x tan p. 20. If sin (rt + /9V— 1)= a; 4- yV— 1, prove that a;* co8e(;''' u — if sec* « = 1. 21. 2 co8(h cos-'j;) = (.u + \A^ Vl — ^Y EXAMPLES. 269 -X-*)*. 22. (V^iy-' = e'". _ pV2( C 23, e''(cos + V- 1 sin ^) = eVal cos ^ 4. V- 1 sin " 24. The coefficients of x" in the expansion, (1) of e" cos bx, and (2) of e"' sin bx, in powers of x, are l^M:J!!)?eosn^and(^±^siun^. [n l« 25, The coefficient of x" in the expansion of e" cos x in . . 2^ ?l7r powers of a; is — cos |« 4 26, If the sides of a right triangle are 49 and 51, then the angles opposite them are 43° 51' 15" and 46° 8' 45" nearly, 27. If « and b be the sides of a triangle, A and B the opposite angles, then will log b — log a =cos 2 A — cos 2 B + ^ (cos 4 A — cos 4 B) + ^ (cos 6 A — cos 6 B) + . .'.! 28. If A 4- tB = log(m -f- in), then tan B = --, and 2 A = log (n^ + m^) . m 29. cos(e-f j'<^) = cos^| ''e-* -f e*\ 4- i sin <9 '^e-'p — 30, s'm{e-\-ii>)=sine(^-^^^^]-ico8e 31. 2 cos(« -f- ift) = cos «(e'' + e-'^) - i sin rt(e^ - e-^). 32. (a4-i6)(»+'^> = r"e -^»*[cos(j3 log r 4- rtr) 4- i sin (/? log r 4- «r) ], where a+ ib= r(cos r 4 i sin r). 33. log(a + t6) = ilog(a='4^')4-t"tan->-. ■-■' ! ft. 260 PLANE TRIGONOMETRY. ■ffljlBk' 34. [8in(«-d)4-ct'«8intf]'' =:sin"~'«[sin(« — H^) + ei*«sin n$]. 35. ^ = JL_ + J_4-_L_4..... 8 1.3 5-7 9.11 36. Write down the quadratic factors of ic" — 1. Ans. {x — l)[x' — 2xcos^(2rir)'\-l], six factors, putting r = 1, 2, 3, 4, 6, 6. 37. Solve the equation aJ* — 1 = 0. Ans. {x'-l){x'-x + l)(a^-^x-\-l) = 0. 38. Give the general quadratic factor of ic* — a*. Ans. mP — 2 ax cos i^^^ (rTr) -f a^ 39. Find all the values of ^/l. Ans. cos j^(r7r)4- tsin ^(7-7r), r having each integral value from to 11. 40. Write down the quadratic factors of afi + 1. Ans. (x' - V3 ;c 4- 1) («* + l)(a^ 4- V3a; + 1). 41. Write down the general quadratic factor of a^ + 1. Ans. a^ - 2 a; cos (1 4- 2 r) 9" + 1. 42. Find the factors of a;*^ 4. 1 = 0. Ans. (x-f 1) [ar* — 2 a; cos 3Jjj(7r + 2 rTr) 4- 1], seven fac- tors in all. 43; Find a general expression for all the values of >/— 1. Ans. co8^^-^t — - 4- It sin — "*" "" ^^ , where r may have n n any integral value. 44. Solve a;" _ 2 .V* cos ^| TT -I- 1 = 0. Ans. a* — 2 a? cos ^ (3 >*«■ 4- tt) 4- 1 Hi 0, six quadratics. i^];.' EXAMPLES. 261 45. Solve »•» 4- V3 ar* + 1 = 0. Ana. x^ + 2x cos {r x 72° + 0°) + 1 = 0, five quadratics. 46. Write down the quadratic factors of cc*" — 2 x^y" cos « -I- 2/*". Ans. a^-2xycos 'L±^^J[ + y*, n factors. n Prove the following : 47. tan :, 61. 1 + 1+14.1 + ...=='^. .. (i:9l|,l_i_l|li t' iii + 3i + 5!i + 72+- = 8- - - « « '.i^ ;.o o 36 144 324 676 . 35 143 323 676 ; V 64 !!:_,2.2.4.4.6.6.8.8... " ' '^ 2 1.3.3.5.5.7-7.9... 4.36.100.196.324... 55. V2 = 66. ^V3 = 3.35. 99.195.323... 8^80.224.440... 9. 81. 225. 441 ..^ tV- »' II) 262 PLANE TRIGONOMETRY. 67. cos X -f tan ^ sin a; = 68. cos X — cot ^ sin x = 2 V yA 2rr-;yA ^Tr+yA ^Tr-yA 4^+; ' +yj*" sin 2^ sin 6) deduce the value for cos d obtained in Art. 173 69. By aid of the formula cos $ = ^*" "' " and Art. 172, GO. By expanding both sides of Ex. 57 in powers of x and equating the coefficients of x, prove that tan f = .V__2 2_ + o 2 ^ 2 2 v—y ir+y Sir—y 37r+y Sw— y 57r+y 61. Prove in like manner from Ex. 68 that + cotV = 2 2_ + _J_ 2_^_A__ 2 y 27r-y 27r + y 47r-y 4n--fy 63, Prove ^-^^=l-»4-l-l+f3-j\ + A 64. Prove that »+ * siny 1 ^ +. 1 y TT— y 27r— y 7r+y 27r+y Srr—y Av-y Sn+y Sum the following series to n terras : + *^6. sina+sin2«4-sin3aH |-sin7ia = . »» 4- 1 . nrt 8in ^ a sin 2 2 sin « EXAMPLES. 263 66. cos « 4- cos 2 rt + cos 3 « H 1- cos nu = • u sm- o 4t 67. 8iuu + sino«-fsiu5«-| — = sin a nn 68. co8«4-cos3rt + cos5rtH — = sin2na 2 sin a nn • 2 . • 'o . • !!o . wsinrt— 8inn«C08(M4-t)« 69. smV-+sin-2«-f siir3«H — = —-. s.:^ ' . 2 8in« TA «^„« I ^ .,»o I «^„2'j I n8in«-l-cos(n-f-l)«8iuw« 70. co8*rt4-co8-2«-f-co8't>«4-"*= - — ^ J — / _ . 2aiu/« 71. sin'rt + 8iu='(rt + /3) + sin3(« -{.'2p)-\--" ^ .4—^^Jsm^ ^ s.n(^.^.4- -^3^J8in-^/ ~4 sin^/3 4 siii | /i 72. sm''rt4-8in=»2« + siii^']M + sill 8111 ( — ' ]u sin sin ~ v - ' — '— 3 2 V 2 y 2 2 sm 2 . • 3 u 4 8m ^. 73. 8ina8m2« + sin2rt8iu3« + sin3« 8in4« + ••• • — Msin KCosK — si n n« cos ( n + 2>« 2sin« 74. tan « + 2 tan 2 « + 2« tan 2" « + ••• = cot « - 2" cot 2" u. 75. (tan « 4- cot a) + (tan 2 a + cot 2 a) + (tan 2*« -f- cot 2^ a) H =2 cot « — 2 cot 2" «. 76. secasec 2rt 4- sec 2«sec3rt 4- ••• = coseu u [tail (71 -f 1) « — tan wj . 264 PLANE TRIOONOMETRT. :*' I 77. cosec a cosec 2 a + coseo U a cosec 3 « -|- ■ . . = coseo a [cot « — cot (n + 1 ) «]. 78 8in2tf , sin 4 $ ....__ 8ec(2n + l)^-8ec ^ cos $ cos 3 cos 3 6 cos 5 ^ 2 sin 79. cos* a + 008* (rt + j3) + cos* (« + 2 /?) -f ... = f 7i + C08[2ft+(n-l)/3]8in7i^ cos [4« + (»-l)2^]sin 2n/8 2sin/3 + 8 8in2y3 80. tanwtf = sin ^ 4- sin 3 ^ + sin 5 tf + • • • to n terms coH 6 + COS 3 ^ -f 008 CtO -\- •'•to n terms 81. COS $ cos {6 + a) + cos (d + «) cos (d + 2 «) + cos(tf H- 2 rt)cos(e -I- 3 «) H — !^ cos « 4- P08(2^ + ^«)^"'^« 2 2 sin n on sin 0— sin 2d4-8in 3tf to n terms . n + 1. , a\ 82. — = tan — (Tr-ft'). cos tf — cos 2 tf -j-oos 3 $~ ... to n terms 2 83. sin(p + l)tfcosd + sin(/) + 2)dcos2e4---- __ n sinjj^ , sin (jj 4- 1 + n ) 6 sin n $^ "2 2sind 84. sin 3 6 sin d + sin G tf sin 2 d + sin 12 ^ sin 4 ^ + • • • = ^(oos2^-eos2"+>d). 8* sintf fs'm ^Y+ 2 sin -("sin -Y+ 4 sin e = 2"-- sin --"^ - V sin 2 ^. BOBS 9 86. tan sec^ + tan-sec- 4- tan-sec -f...=tand— tan— . 2 4 2 8 4 2" 87. cot 6 cosec ^+2 cot 2 tf cosec 2 ^+2" cot 2^6 cosec 2''^+ ... 1 On-l 2 sin 2^ sin=*2"-^d 2 EX A MPLES. 265 tan I 2"' 88. sin _1 + ____! . 1 . e sin 2 $ sill 2 6 sin ;J $ sin 3 ^ sin 4 ^ --v,(cot3tf-cot4^). sm ^ 89. TTTTi-^ZT. sin tf cos 2 cos 2 tf sin 3 tf sin 3 ^ cos 4 6 = cosec(d + ^ ) tan(n4-l)(^ + ^Vtai/^ + ^y 90. tau-i ^ 1— -- + tan-' 1 + tan-» ^ 1 + 1 + 1' 1 + 2 + 2"^ l+3 + 3« + ... = ^--tan-'-i~. 4 w + 1 91. tan-' X + tan"' X l + 1.2-a:« + tan"' X ; + 1 + 2 . 3 . a?« = tan-' nx. 92. sin « sin 3 « + sin" sin '"^^ + sin " sin— + ... 2 2 2* 2* = ( cos COS 4 2 1 2" =» "> 93. ; + + + cos ^ + COS 3 ^ cos ^ + cos 5 cos ^ + cos 7 tf = ^ cosec ^[tan(n + 1)0 — ta)i 6]. 94. -sec^ + -sec^sec2^ + ,^-8ec^sec2esec22^+ ... = sin^(cot^-cot2"^). 95. ^log tan 2 tf + |^log tan 2^^ + 1 log tan 2«^ + ... = log 2 sin 2$-- log 2 sin 2"^ ' tf. Sum the following series to infinity : 96. cos^+«"^-^cos2^+5^^cos3^+^«fi^cos4tf+... 1 [L [3 — eco.»« cos(^ + sin $ cos $). IMAGE EVALUATION TEST TARGET (MT-3) 1.0 I.I 845 (50 iiillM IIIIIU m iM 22 1.8 1.25 1.4 1.6 -* 6" — ► Photographic Sciences Corporation 23 WEST MAIN STREET WEBSTER, NY. 14580 (716) 872-4503 s. \ ^ cF 'Q^' \ \ %^ O^ % ^^ 'ib" C<'x I I IP' 266 PLANE TRIGONOMETRY. 9/. sin^ — 1 — — = e-cos* sin(sin^). 98. 1 _ ^J^^ + 22ii^ = icos(cos 6) (e«i"« + e-«'°»). 99. 2cos^ + fcos2^-hf cos3e + fcos^^ + ... ;V-, ' cos^ 1 — COS — log(l — cos^). mn r,- a a , sin2^cos^^ , sinS^cos^^ , 100. sin d COS Q H ■ 1 , 1- I? [3 _ eco8«0sin(sin^cos^). 101. cos^ + ?H^cos26 + ^^cos3^4--" . ,1. [2 =-e8iQflco8 cos(^ + sin^e). 102. sin^+-— sin2^ + ^-^'^-sin3^4- < \ = e«'n^ <=»''« sin (^ + sin2^). 103. cos^-|cos2^4-^cos3^ =log/'2cos^Y . 104. cos 2^+^ cos 65 + I cos 10^4- ••• = |log cot-. M\r -a a^sin2^ , ar'sin3^ ..i/cosec^ , .A 105. ifsm^ 1 = cot '( f-cot^l- 2 3 \ X ) T Oj or 106. a; cos 6 — '— cos '26 -\ — cos 3$ cos 4 ^ + • • • 2 3 4 . f = log(l + 2a; cos ^ + ar'). 107. sin^ sin2^ hsin3^ ••• 1 2 3 = cot-' (1 + cot^ ^ 4- cot ^) . 1 1 1 1 -4 1* 2* 3^ 44 90 14-^-34-^54-^74-1- 9g ,-Blnfl> -COS^). / ■ / OcosB). l-siii^e). f siii^^). +cotd )S^4-arO- ? + cotd). PART IT. SPHERICAL TRIGONOMETRY. CHAPTER X. rORMTJLU KELATIVE TO SPHERICAL TRIANGLES. ■ : 182. Spherical Trigonometry has for its object the sohi- tion of spherical triaiigles. . ' ' A spherical triangle is the figure formed by joining any three points on the surface of a sphere by arcs of great circles. The three points are called the vertices of the triangle ; the three arcs are called the sides of the triangle. Any two points on the surface of a sphere can be joined by two distinct arcs, which together make up a great circle passing through the points. Hence, when the points are not diametrically opposite, these arcs are unequal, one of them being less, the other greater, than 180°. It is not necessary to consider triangles in which a side is greater than 180°, since we may always replace such a side by the remaining arc of the great circle to which it belongs. 183. Geometric Principles. — It is shown in geometry (Art. 702), that if the vertex of a triedral angle is made the centre of a sphere, then the planes which form the triedral angle will cut the surface of the sphere in three arcs of great circles, forming a spherical triangle. Thus, let be the vertex of a triedral angle, and AOB, BOC, COA its face-angles. We may construct a sphere with its centre at O, and with any radius OA. Let AB, 267 268 SPHERICAL TRIGONOMETRY, BC, CA be the arcs of great circles in which the planes of the face-angles AOB, BOC, COA cut the surface of this sphere; then ABC is a spherical triangle, and the arcs AB, BC, CA are its sides. Now it is shown in geometry that the three face-angles AOB, BOC, COA are measured by the sides AB, BC, CA, re- spectively, of the spherical triangle, and that the diedral angles OA, OB, OC are equal to the angles A, B, C, respect- ively, of the spherical triangle ABC, and also that a diedral angle is measured by its plane angle. There is then a correspondence between the triedral angle 0-ABC and the spherical triangle ABC : the six parts of the triedral angle are represented by the corre- sponding six parts of the spherical triangle, and all the relations among the parts of the former are the same as the relations among the corresponding parts of the latter. 184. Fundamental Definitions and Properties. — The fol- lowing definitions and properties are from Geometry, Book VIII. : In every spherical triangle Each side is less than the sum of the other two. The sum of the three sides lies between 0° and 360°. The sum of the three angles lies between 180° and 540°. Each angle is greater than the difference between 180° and the sum of the other two. If two sides are equal, the angles opposite them are equal ; and conversely. If two sides are unequal, the greater side lies opposite the greater angle ; and conversely. The perpendicular from the vertex to the base of an isosceles triangle bisects both the vertical angle and the base. / DEFINITIONS AND PROPERTIES. 269 , CA, re- e diedral ), respect- a diedral ^ triedral : the six the corre- d all the 3 same as 3 latter. -The fol- try, Book 360°. lud 540°. ween 180° them are s opposite ase of an e and the ^ The axis of a circle is the diameter of the sphere perpen- dicular to the plane of the circle. The poles of a circle are the two points in which its axis meets the surface of the sphere. One spherical triangle is called the polar triangle of a second spherical triangle when the sides of the first triangle have their poles at the vertices of the second. If the first of two spherical triangles is the polar triangle of the second, then the second is the polar triangle of the first. Two such triangles are said to be polar with respect to each other. Thus: If A'B'C is the polar triangle of ABC, then ABC is the polar triangle of A'B'C. In two polar triangles, each angle of one is measured by the supplement of the corresponding side of the other. Thus : A = 180° - a', B = 180° - 6', C = 180° - c', a = 180° -A', 6 = 180°-B', c = 180°-C'. This result is of great importance ; for if any general equation be established betAveen the sides and angles of a spherical triangle, it holds of course for the polar triangle also. Hence, hy means of the above formidoe any theorem of a spherical triangle may he at once transformed into another theorem hy suhstituting for each side and angle respectively the supplements of its opposite angle and side. If a spherical triangle has one right angle, it is called a right triangle ; if it has two right angles, it is called a bi- rectaMgular trisingle ; and if it has three right angles,* it is called a tri-rectangular triangle. If it has one side equal to a quadrant, it is called a quadrantal triangle ; and if it has two sides equal to a quadrant, it is called a bi-quadrantal triangle. ^ ^^ ? i Ill ^f I' "J I liiil' r 270 SPHERICAL TRIGONOMETRY. Note. — It is shown in geometry that a spherical trianirle mny, in general, be constructed when any three of its six parts are given (not excepting the case in which the given parts are the three angles). In spherical trigonometry we investi- gate the methods by which the unknown parts of a spherical triangle may bo com- jfutcd from the above data. EXAMPLES. 1. In the spherical triangle whose angles are A, B^ C, prove B+C-A<^ (1) C+A-B<7r (2) A + B-C<7r (3) 2. If C is a right angle, prove A + B 90°. The second follows from the equation cos c = cos a cos b. 188. Ambiguous Solution. — When the given parts of a right triangle are a side and its opposite angle, the triangle cannot be determined. For two right spherical triangles ABC, X.'BC, right angled at C, may always be found, having the angles A and A' equal, and BC, the side opposite these angles, the same in both triangles, but the remaining sides, AB, AO, and the remaining angle ABC of the one triangle are the supplements of the re- maining sides A'B, A'C, and the remaining angle A'BC of the other triangle. It is therefore ambiguous whether ABC or A'BC be the triangle required. This ambiguity will aldO be found to exist, if it be attempted to determine the triangle by the equation sin 6 = tan a cot A, since it cannot be determined from this equation whether the side AC is to be taken or its supplement A'C. 189. Quadrantal Triangles. — The polar triangle of a right triangle has one side a quadrant, and is therefore a quadrantal triangle (Art. 184). The formukc for quad- rantal triangles may be obtained by applying the ten formulae of Art. 185 to the polar triangle. They are as follows, c being the quadrantal side : cos C = — cos A cos B (1) sin B = sin 6 sin C (2) (1) (2) EXAMPLES. 275 sin A = sin a sin C /3\ cos 6 = — tan A cote (4\ cos a = — tan B cot C . . . , /5\ sin A = tan B cot 6 /g\ sin B = tan A cot a /j\ cosC = — cotacot& /g\ cos b = cos B sin a ...... . .... (9) cos a = cos A sin 6 /jq\ EXAMPLES. In the right triangle ABC in which the angle C is the right angle, prove the following relations : 1. sin" a + sin^ b — sin- c = sin^a sin^b. 2. cos^A sin^c = sin^c — sin^a. 3. sin^ A cos^c = sin^ A — sin^a. 4. sin^ A cos^d sin^c = sin^c — sin^ft. 5. 2cosc = cos(a + &) + cos(a — 6). 6. tanj(c + a)tai4(c-a)=tan2^6. 7. sin2^ = sin2^cos«^ + cos2-sin''^. 2 2 2 2 2 8. sin(c-6) = tan='Asin(c-H6). 9. If & = c = ^, prove cos a = cos A. 10. If a = 6 = c, prove sec A = 1 + sec a. 11. If c < 90°, show that a and b are of the same species. 12. If c > 90°, a and 6 are of different species. 13. A side and the hypotenuse are of the same or oppo- site species, according as the included angle <, or > -. 2 wr ^E [■iih 276 SPHERICAL TRIGONOMETRY. OBLIQUE SPHERICAL TRIANGLES. 190. Law of Sines. — In any spherical triangle the sines of the sides are proportional to the sines of the opposite angles. Let ABC be a spherical triangle, the centre of the sphere ; and let a, b, c denote the sides of the triangle opposite the angles A, B, C, respectively. Then a, b, and c are the measures of the angles BOG, COA, and AOB. From any point D in OA draw DG ± to the plane BOC, and from G draw GE, GF ± to OB, OC. Join DE, DF, and GO. Then DG is ± to GE, GF, and GO (Geom. Art. 487). Hence, DE is ± to OB, and DF ± to OC (Geom. Art. 507). .-. Z DEG = Z B, and Z DFG = Z C . . (Art. 183) In the right plane triangles DGE, DGF, ODE, ODF, DG = DE sin B = OD sin DOE sin B = OD sin c sin B, DG = DF sin C = OD sin DOF sin C = OD sin b sin C. .'. sin c sin B = sin b sin C ; or sin 6 : sin c : : sin B : sin C. Similarly, it may be shown that sin a : sin c : : sin A : sin C. sin a _ sin 6 _ sin c sin A sin B sin C Note. — The common value of theae three ratios is called the modulus of the spherical triangle. Sch. In the figure, B, C, 6, c are each less than a right angle ; but it will be found on examination that the proof will hold when the figure is modified to meet any case which can occur. For example, if B alone is greater than LAW OF COSINES. 277 90°, the point G v/ill fall outside of OB instead of between OB and OC. Then DEG will be the supplement of B, and thus we shall still have sin DEG = sin B. 191. Law of Cosines. — In any spherical triangle, the cosine of each side is equal to the product of the cosines of the other two sides, plus the product of the sines of those sides into the cosine of their included angle. Let ABC be a spherical triangle, the centre of the sphere, and a, h, c the sides of the triangle opposite the angles A, B, C, respectively. Then a == Z BOG, b = Z COA, c = Z AOB. From any point D in OA draw, in the planes AOB, AOC, respectively, the lines DE, DF ± to OA. Then ZEDF = ZA . (Art. 183) Join EF; then in the plane triangles EOF, EDF, we have EF''=:OE' + OF'-2 0E.OFcosEOF . r>2 EF' = DE' H- DF' - 2 DE . DF cos EDF also in the right triangles EOD, FOD, we have tTt]^2 FC2 0E'-DE' = 0D; and OF - DF = OD (1) (2) (3) Subtracting (2) from (1), and reducing by (3), and transposing, we get or 2 OE . OF cos EOF = 2 OD' + 2 DE • DF cos EDF. OD OD , DF DE ^^.j. .'. cos EOF = — — • — — -f — — . — — cos EDF, OF OE OF OE ' cos a = cos 6 cos c -f sin & sin c cos A (4) \ I ii f ■ *'S 278 SPHERICAL TRIGONOMETRY. By treating the other edges in order iii the same way, or by advancing letters (see Note, Art. 96) we get cos h = cos c cos a + sin c sin a cos B cos c = cos a cos h -\- sin a sin 6 cos C (5) (6) ^c^. Formula (4) has been proved only for the case in which the sides h and c are less than quadrants; but it may be shown to be true when these sides are not less than quadrants, as follows : (1) Suppose c is greater a, — <:~- ~*f.^_^ than 90°. Produce B A, BC B ^^^ ^* ~^ to meet in B', and put AB'=c', CB'=a'. >B' '0' Then, from the triangle AB'C, we have by (4) cos a' = cos h cps c' + sin b sin c' cos B'AC, or cos(7r— a)=:cos6cos(7r— c) + sin 6 sin(7r — c)cos(7r— A). .*. cos a = cos b cos c -f- sin 6 sin c cos A. (2) Suppose both 6 and c to be greater than 90°. Produce .' , AB, AC to meet in A', and put A'B = c', A'C = b'. Then, from the triangle A'BC, we have by (4) cos a = cos y cos c' + sin V sin c' cos A' ; but 6' = TT — 6, c' = TT — c, A' = A. .*. cos a = cos b cos c + sin b sin c cos A. The triangle AB'C is called the colunar triangle of ABC. 192. Relation between a Side and the Three Angles. - In any spherical triangle ABC, cos A = — cos B cos C + sin B sin C cos a. -B' RELATION BETWEEN SIDE AND ANGLES. 279 Let A'B'C be the polar triangle of ABC, and denote its angles and sides by A', B', C, a', b', c' ; then we have by (4) of Art. 191 cos a' = cos b' cos c' + sin b' sin c' cos A' ; but a' = TT - A, 6' = TT - B, c' = TT - C, etc. . (Art. 184) Hence, substituting, we get v cos A = — cos B cos C + sin B sin C cos a . . . . (1) Similarly, -- - cos B = — cos C cos A + sin C sin A cos b cos C = — cos A cos B + sin A sin B cos c (2) Hem. — This process is called " applying the formula to the polar triangle." By means of the polar triangle, any formula of a spherical triangle may be immediately transformed into another, in which angles take the place of side?, and sides of angles. 193. To show that in a spherical triangle ABC, cot a sin 6 = cot A sin C + cos C cos b. ' ' Multiply (6) of Art. 191 by cos b, and substitute the result in (4) of Art. 191, and we get cos a = cos a cos'^ b + sin a sin 6 cos b cos C + sin b sin c cos A. Transpose cos a cos^ b, and divide by sin a sin 6 ; thus, . . , , r\ , sin c cos A cot a smb~ cos b cos C H sin a = cos b cos C + cot A sin C . (by Art. 190) By interchanging the letters, we obtain five other formulae like the preceding one. The six formul&e are as follows : cot a sin & = cot A sin C + cos C cos 6 . ... (1) cot a sin c = cot A sin B + cos B cos c cot b sin a = cot B sin C + cos C cos a cot 6 sin c = cot B sin A + cos A cos c cot c sin a = cot C sin B -|- cos B cos a cot c sin 6 = cot C sin A + cos A cos b (2) (•^) (4) (r») (6) i 9^ I. ri* 280 SPHERICAL TRIGONOMETRY. EXAMPLES. 1. If a, 6, c be the sides of a spherical triangle, a', &', c' the sides of its polar triangle, prove sin a : sin 6 : sin c s= sin a' : sin &' : sin c'. 2. If the bisector AD of the angle A of a spherical triangle divide the side BC into the segments CD = 6', BD = c', prove sin & : sin c = sin 6' : sin c'. 3. If D be any point of the side BC, prove that cot AB sin DAC + cot AC sin DAB = cot AD sin BAC. cot ABC sin DC + cot ACB sin BD = cot ADB sin BC. 4. If a, /3, y be the perpendiculars of a triangle, prove that sin a sin a = sin b sin /? = sin c sin y. 5. In Ex. 4 prove that sin a cos a = Vcos'* b -\- cos^ c — 2 cos a cos b cos c. 194. Useful FonnulfiB. — Several other groups of useful formulae are easily obtained from those of Art. 191 ; the following are left as exercises for the student : sin a cos B = cos b sin c — sin b cos c cos A sin a cos C = sin & cos c — cos b sin c cos A sin b cos A = cos a sin c — sin a cos c cos B sin b cos C = sin a cos c — cos a sin c cos B sin c cos A = cos a sin 6 — sin a cos b cos C sin c cos B = sin a cos b — cos a sin b cos C (1) (2) i^) (4) (5) (6) FORMULA FOR THE HALF ANGLES. 281 .t /.» (1) (2) {^) (4) (5) (6) Applying these six formulae to the polar triangle, we obtain the following six : sin A cos b = cos B sin C + sin B cos C cos a sin A cos c = sin B cos C + cos B sin C cos a sin B cos a = cos A sin C 4- sin A cos C cos b sin B cos c = sin A cos C + cos A sin C cos b sin C cos a = cos A sin B 4- sin A cos B cos c sin C cos b = sin A cos B + cos A sin B cos c (7) (8) (9) (10) (11) (12) 195. FormulsB for the Half Angles. — To express the sine, cosine, and tangent of half an angle of a spherical triangle in terms of the sides. I. By (4) of Art. 191 we have . cos a — cos & cos C ^ o • 2^. / A 4. An\ COS A = — = 1 — 2 sin^ - (Art. 49) sin b sin c 2 o • 2 A ^ COS a — COS b cos c 2 Sin & sine v,;', cos (6 — c) — cos a sm sin c sin 2 A = sin^(a+&-c)sin|(a-& + c) ^^^^ ^^^ 2 sin b sin c Let 2s = a + 64-c; so that s is half the sum of the sides of the triangle ; then a + b — c = 2{s — c), and a — b-{-c = 2{s — b). • 2 A sin (s — &) sin (s — c) 2 sin & sin c ^ .-. sin ^ =x^^"i'' ~ ^^ ^^" il".'^) 2 \ sin b sin c (1) m\ \ 282 SPHERICAL TUIGONOMETRY. 11 V 1 Advancing letters, sm — c) sin (s — g ) B _ /sin (.s — c) si 2 >/ sin c si sma . . . (2) II. . C /sin (.s — a) sin (s — 6) ,o\ sin — =\/ \__— ^_-— ^ L ... (3) 2 \ sin a sin 6 J 2 CDs'* — = 1 -f- cos A (Art. 49) = 1 + COS a — cos h cos c sin& sine ,*. cos __ cos a — cos (6 4- c) sin h sin c 2A_ sin \{a + 64-0) sin ^(6 4- c — a) sin 6 sin c _ sin s sin (g — ») sin 6 sine cos -^ sin 3 sin (8 — a) sin h sin e (4) Advancing letters, B /sinssin(s — ^) cos — =\/ : ^ '- 2 \ sm e sin a • • (5) cos C _ Isin s sin (&• — e) ,n.^ 2 Ai sin a sin 6 III. By division, we obtain tan tan A _ /sin {h — b) sin (s — c) 2 >/ sin s sin (.s — a) ]^ — / sill ('** — c) sin (■*» — «) 2 Al sins sin (s — i^) (7) (8) tang=J'^^:^'^"^'~^ ... (9) 2 \ sinssinf.s — c) ^ . (4) . (5) . (6) . (7) . (8) • (9) FORMULA OR THE HALF ANGLES. 283 Sch. The positive sign jnusfc be given to the radicals in each case in this aj'ticle, because |-A, ^B, ^C are each less than 90°. Cor.l. tan^tan? = ^^^-^^ .... .(10) 2 2 sins ^ ^ tan — tan = — >^ ^ (11) 2 2 sins ^ ^ . C. A sin(.s' — ?/) ,^fv, tan - tan— = — ^ '- (12) 2 2 sins ^ ' A A Cor. 2. Since sin A = 2 sin — cos — , . A 2Vsins sin(s — «) sin(s — />) sin(s — c) .^o\ .-. sin A = —-^ > i — r^^ ^ >^ ^ , (1,>) sni6 sine - 2w .... — • •••••••••«»l ATI: J Sin 6 sin c where «^ = sinssin(s — a) sin(s — 6) siii(s — c). EXAMPLES. M -n • 2A 1 — cos^a— ('os-6— cos^c4-2cosacos&cosc 1. Prove siirA= — r-^ — r— ^ - snrb sure _ 4ri^ sin^& sin^c where 4 w* = 1 — cos^ a — cos^ 6 — j3os^ c + 2 cos a cos b cos c. C C 2. Prove cose = cos(a+ &)sin^ — h cos (a — &) cos^* - • o T3 . A . B . C sin(s— a) sin(s — &)sin(s — c) o. Prove sin — sin— sin - = — -^^ ^ -^^ ^ ^ -. 2 2 2 sin a sin 6 sm e 4. Prove cos ^ 4-cosB ^sin ( » + &) . 1 — cosC sine e T> ^cosA4 cosB • / ,x . A 5. Prove 2 sin (a — b) sine = 0. 1 — cui U ^^*v ^ -o cos A — cos B sin (a ~ b) 6. Prove r^ ^ -' 1 + cosC sine II vm? 284 SPHERICAL TRIGONOMETRY. 196. FormulflB for the Half Sides. — To express the sine, cosine, and tangent of half a side of a spherical triangle in terms of the angles. By (1) of Art. 192, we have c^sA + cosBcosC^^_23i^.a (Art. 49) smBsiiiC 2 ^ ^ .-. 2 3111=^^= cosA4-cos(B + C ) 2 sin B sill C . sin2^ = - cosi(A + B+C)cos^(B+C-A) .^ .^. 2 sin B sin C ^ * ^ Let 2S = A + B + C; then B + C - A = 2 (S - A). Proceeding in the same way as in Art. 195, we find the following expressions for the sides, in terras of the three angles : sin« = -.CcosSc^3S^A) .^. 2 \ sin B sin ^ . b I cos S cos (S — B) ,o\ sin- = ^/ ; — —\ — - — ^ (2) 2 A/ sinCsinA ^ ^ 2 \ sinAsinB ^ ^ cos - = ^ / co«(S-B)cos(S-C) ... 2 \ sinBsinC ..... v / 2 \ smCsinA ..... vu; cos^ = J^^^S:^S«pISEl} (6) 2 \ SinAsinB ^ ^ — tan ^ = J- -^^'^'^ S ^os ( S - Al_ .... (7) 2 \ cos (S - B) cos (S - C) '^ Itill'l! FOHMUL^ FOR THE HALF SIDES. 286 tan H- cQsS cos(S -r B) co8(S--C)cos(S-A) ... . (8) . (1) . (2) . (3) . (4) ■ (5) . (6) . (7) tan£ = J cos S cos (S - C) 2 \ cos (S- A) cos (S- (S - A) cos (S - B) • • • • V / Sch. 1. These formulae may also be obtained immediately from those of Art. 195 by means of the polar triangle. Sch. 2. The positive sign must be given to the above radicals, because -, -, -, are each less than 90°. Sch. 3. These values of the sines, cosines, and tangents of the half sides are always real. For S is > 90° and < 270° (Art. 184), so that cos S is always negative. Also, in the polar triangle, any side is less than the sum of the other two (Art. 184). .-. TT — A<7r — B+TT— C. .-. B + C-A<7r. .'. cos (S — A) is positive. Similarly, cos (S — B) and cos (S — C) are positive. Cor. Since sin a = 2 sin - cos % smct: .2V-cosScos(S-A)cos(S-B)cos(S-C) sin B sin C (10) 2N sin B sin C where N= V— cos S cos (S -- A) cos (S — B) cos (S — C). m ''-'^B- 286 SPHERICAL TRIGONOMETRY. EXAMPLBS. 1. Prove cosC=— cos(A + B)cos^^ — cos(A— B) sin^-- a b .:. C 2. Prove sin - sin - sin = — 9. O. 9 si — N cos S 2 sin A sin B sin C where N = V— cos S cos (S — A) cos (S — B) cos (S — C). 197. Napier's Analogies. Let m or sin A _ sinB sin a sin b sin A + sinB sin a + sin 6 sin A — sin B (Art. 190) (1) (Algebra) (2) ... (3) sin a — sin b cos A + cos B cos C = sin B sin C cos a (Art. 192) = m sin C sin b cos a, by (1 ) (4) and cos B + cos C cos A = sin C sin A cos b = m sin C sin a cos & . . (5) .-. (cos A + cosB)(l 4-cosC) = msinC sin (a + 6), (6) Dividing (2) by (6), sin A 4- sin B from (4) and (5) sin a 4- sin 6 1+ cos C cos A + cos B sin (a + b) sin C .-. tani(A-j-B) = cosi(a-6)^^^C cos ^ (a + 6) 2 (Arts. 46, 46, and 49) (7) Similarly, tan^ (A - B) = -"^^^i (^ - ^) cot^ . . (8) ^' ^^ ' sini(a + &) 2 ^ ' V DELAMBRE'8 ANALOGIES. 287 I ♦ -C). 30) (1) ra) (2) . (3) rt. 192) Kl)(4) . (5) (6) (5) osC _ — -I — ■■ • (7) 49) Writing 71 — A for a, etc., by Art. 184, we obtain from (7) and (8) ScL The formulae (7), (8), (9), (10) are known as Napier's Analogies, after their discoverer. The last two may be proved without the polar triangle by starting with the formulae of Art. 191. Cor. Ill any spherical triangle ivhose parts are positive, and less than 180°, the half-sum of any two sides and the half- sum of their opposite angles are of the same species. C For, since cos ^ (a — 6) and cot — are necessarily positive, therefore by (7) tan^(A4-B) and cos ^ (a + 6) are both positive or both negative. .-. |(A + B) and ^ (a + 6) are both > or both < or both = 90°. 198. Delambre's (or Gauss's) Analogies. sini(A4-B) A B , A . B = sm 2 ^^^ 2 + ^°^ 2 ^^^ 2 V sin {s — b) sin (.s — c) /sin s sin (s — 6) sin 6 sin c ^ sin c sin a + . (8) / sin8sin(.t-a) / sin (6--c) sin (s - ct) ,^^^^ jg^^ \ sin 6 sin c Af sin c sin a _ sin (s — &) + sin {s — a) / sin s sin {s — c) sin c A( sin a sin b ^cos^{a-b)G (Art.. 45 and 195) • 1'' cos 2 k 288 SPHERICAL TEW ONOMETR Y. .: sin^ (A + ]i) cos- — oos^ (d— 6) cos - . . . (1) Similarly, we obtain the following three equations : c C sin^ (A — B) sin- = sin^ (a — 6)co8 - ... (2) c C cos |( A 4- B) cos - = cos I (a + 6) sin - . . . (3) c • C cos i (A — B) sin - = sin ^ (a + 6) sin — (4) 8c\i. 1. When the sides and angles are all less than 180", both members of these equations are positive. Sell. 2. Napier's analogies may be obtained from De- lambre's by division. Note. — Delarabre's analogiea were discovered by him in 1807, and publiBhed in the Connaissance des Temps for 1809, p. 44:J. They were subsequently discovered independently by Gauss, and published by bim, and arc sometimes improperly called Gauias's equutious. Both systems may be proved geometrically. The geometric proof is the one originally given by Delambre. It was rediscovered by Professor Crofton in 1869, and published in the Proceedings of the Loudon Mathe- matical Society, Vol. HI. [Casey '■ Trigonometry, p. 41], EXAMPLES. In the right triangle ABC, in which C is the right angle, prove the following relations in Exs. 1-45 : 1. sin^acos^& = sin (c + &) sin (c — &). 2. tan* a : tan* 6 = sin*c — sin*& : sin*c — sin* a. 3. cos' a cos* B = sin* A — sin* a. 4. cos* A + cos* c = cos* A cos* c + cos* a. 5. sin* A — cos* B = sin* a sin* B. 6. If one of the sides of a right triangle be equal to the opposite angle, the remaining parts are each equal to 90°. EXAMPLES. 289 . (1) . (2) . (3) . (4) Ml 180°, om De- mbllshed in discovered improperly :ally. The Bcovered 'oy idon Matbe- ;ht angle, 7. Tf the angle A of a right triangle be acute, show that the difference of the sides which contain it is less than 90°. 8. Prove . B sin (s — a) tan = \ ^« 2 sins 9. Prove (1) 2n = sinasin&; (2) 2N = sinasinB. 10. Prove sin^ a sin^ b = sin* a + sin^ b — sin' c. 11. Prove 12. Prove ^^^,A^ sin(c-6) ^ 2 sin(c + 6) y • • ' 2 sin2^ = sin2^(a + b) + sinH(a - b) . 4W 13. In a spherical triangle, if c = 90°, prove that tan a tan b + sec C = 0. 14. In a spherical triangle, if c = 90°, prove that sin^p = cot ^ cot <^, , where p is the perpendicular on c, and and <^ are the seg- ments of the vertical angle. ,;: , 15. Show that the ratio of the cosines of the segments of the base made by the perpendicular from the vertex is equal to the ratio of the cosines of the sides. 16. If B be the bisector of the hypotenuse, show that : sin^'a. 4- sink's ': v' sin'B = 4 cos 2C Lial to the L to 90°. 17. Prove tanS = cot- cot-. 18. Construct a triangle, being given the hypotenuse and (1) the sum of the base angles, and (2) the difference of the base angles. 19. Given the hypotenuse and the sum or difference of the sides : construct the triangle. 290 SPHEliICA L TlilGONOMETU Y. 20. Given the sum of the sides a and h, and the sum of the base angles : solve the triangle. oi c!u i.u 1. • A Vsinc-f sm u + Vsinc — sin M 21. Shovir that sin — = -- ^ 9 2 Vsin c 22. sin ' \2C08/ h sin c 23. cos V 2 cos 6 sine 24. sin (a + 6) tan ^ ( A. + H) = sin (a — 6) cot ^ ^ - B). cos a -\- cos h 2o. sin(A + B): 26. sin(A-B) 27. cos(A + l.) = - 28. cos(A-B) = — 1 + cos a cos h cos ft — cos a 1 — COS a cos 6 sin a sin 6 1 4- cos a cos 6 sin rt sin 6 cos a cos ft 29. sin2 '^ = sin^ ^ cos'' - + cos" ^ sin'' -. 2 2 2 2 2 30. sin (c - 6) = sin (c + ft) tan* _. A B 31. sin (a — ft) = sin a tan sin ft tan ^ ^ 2 2 32. sin (c — a) = cos a sin 6 tan B 33. If ABC is a spherical triangle, right-angled at C, and cos A = cos" a, show that if A be not a right angle, ft-fc = ^7r or fTT, according as ft and c are both < or both > |. EXAMPLES. 291 B). 34. If «, /8 be the arcs drawn from the riyht angle respectively perpendicular to and bisecting the hypotenuse c, show that 8in=*-^(l + sin2«) = sin»/?. 35. In a triangle, if C be a right angle and D the middle point of AB, show that c 4 cos''- sin^ CD = sin'* a + sin* 6. In a right triangle, if p be the length of the arc drawn from the right angle C perpendicular to the hypotenuse AB, prove : 36. cot*^7 = cot- a 4- cot^ 6. 37. cos^i) = cos^ A + cos'^B. 38. tan- a = ^ tan a' tan c. 39. tan'* b = ± tan b' tan c. 40. tan'* a : tan'* 6 = tana' : tan b'. 41. sin^/) = sin a'sin 6'. 42. sin/) sin c= sin a sin 6. 43. tan a tan b = tan c sin p. 44. tan'* a + tan ^6 = tan^c cos'^p. 45. cot A : cot B = sin a' : sin b'. In the oblique triangle ABC, prove the following : 46. If the difference between any two angles of a tri- angle is 90°, the remaining angle is less than 90°. 47. If a triangle is equilateral or isosceles, its polar tri- angle is equilateral or isosceles. 48. If the sides of a triangle are each -, find the sides of the polar triangle. 41 ^92 SPHERICAL TRIGONOMETRY. 40. If in a triangle the side a = 90°, show that cos A 4- cos B cos C = 0. 50. If and 6* are the angles which the internal and ex- ternal bisectors of the vertical angle of a triangle make with the base, show that cosO = cos A ~ cos B 2 cos C ■J and cos 6' = cos A + cos B 2sin^ oos A. 51. Given the base c and - — — = — cos C : find the locus of the vertex. 52. Prove 4N^ = 1 — cos'^A — cos^B — cos^C — 2 cos A cos B cos C. 53. If p, q, r be the perpendiculars from the vertices on the opposite sides, show that (1) sin a sin J) = sin 6 sin 7 = sin c sin r = 2n. (2) sinAsinj9 = sinB sin 7 = sinCsinr = 2N. 54. Prove 8 n'' = sin^ a sin^ b sin^ c sin A sin B sin C. sin'^ A + sin'^ B -f- sin*^ C 1 + cos A cos B 00s C 55. Prove sin^ a -|- sin^ 6 + sin^ c 1 — cos a cos & cos c 56. If I be the length of the arc joining the middle point of the base to the vertex, find an expression for its length in terms of the sides. Ans. cosh cos (I -f- cos h 2 cos 2 57. If CD, CD' are the internal and external bisectors of the angle C of a triangle, prove that ^ /-,T^ cx)t a "f- cot b - ^ / , T^. cot a ~ cot & cot CD = — , and cot CD' = 2cos^ 2 2sin^ EXAMPLES. 298 58. Show that the angles $ and 6', made by the bisectors of the angle C in Ex, 55 with the opposite side c, are thus given : cota — cot6 . ^-p, _ .. cot^ = — sin CD, ! 2 sin C .,f. .,■ cotrt + cotft . cot B = — sin CD'. cos- C 59. Show that the arc intercepted on the base by the bisectors in Ex. 55 is thus given : cotDD' 60. Prove that cos^6 — cos^c sin'^A — sin^B 2 sin A sinB siiiC cos^c — cos^a cos 6 cot B — cos c cot C cos c cot C — cos a cot A cos* a — cos* 6 cos a cot A — cos h cos B 61. If s and s' are the segments of the base made by the perpendicular from the vertex, and m and m' those made bj- the bisector of the vertical angle, show that tan- .S'' m — )/ .,a — h tan = tair 2 2 2 62. Prove sin 6 sin c + cos h cos c cos A = sin B sin C — cos B cos C cos a. 63. Show that the arc I joining the middle points of the two sides a and 6 of a triangle is thus given : cosZ = 1 + cos a 4- cos h ■\- cos c ■ ■ — ■ - » 4 cos cos- 2 2 SPHERIC A L TRIGONOMETR Y. 64. If the side c of a triangle be 90°, and S the arc drawn at right angles to it from the opposite vertex, show that COt^SrzzCOt^A + COt^B. 65. Prove that the angle (f> between the perpendicular from the vertex on the base and the bisector of the vertical angle is thus given : tan = «^« ii'^^^^ltan ^ ( A - B). ^ cos I (a + 6) 66. In an isosceles triangle, if each of the base angles be double the vertical angle, prove that cosacos- = cosf c + -V 67. If a side c of a triangle be 90°, show that (1) cotacot& + cos C = 0. (2) cos S cos (S - C) + cos (S - A) cos (S - B) = 0. 68. In any triangle prove cos a — cos b sin (A — B) _ ^ 1 — cos c sin C 69. tan|(A+B):tan|(A-B) = tan ^ (a + 6) : tan ^ (a — 6). 70. tan ^ (A -f- a) : tan |( A — a) = tan |(B + 6) : tan ^ (B - &). 71. If the bisector of the exterior angle, formed by pro- ducing BA through A, meet the base BC in D', and if BD = c", CD' = b", prove sin ft : sin c = sin ft" : sin c". 72. If D be any point in the side BC of a triangle, prove sinBI) _sinBAD sinC sin CD sin cad" sin b' EXAMPLES. 295 0. 73. If A = a, show that B and b are either equal or supplemental, as also C and c. 74. If A = B + C, and D be the middle point of a, show that a = 2 AD. 75. When does the polar triangle coincide with the primitive triangle? 76. If D be the middle point of c, show that cos a -f- COS 6 = 2 cos cos (^I). 2 \,,,^ ■, 77. In an equilateral triangle show that (1) 2 cos- sin — = 1. ^ ^ 2 2 (2) tan2^ + 2cosA=l. 78. If & 4- c = TT, show that sin 2 B + sin 2 C = 0. 79. Show that sin b sin c + cos & cos c cos A = sin B sin C — cos B cos C cos a. 80. If T> be any point in tlie side BC of a triangle, show that cos AD sin a = cos c sin DC + cos b sin BD. 81. Prove cos^ ==cos''|^(a+6)sin'' -|-t^os^|(a— 6)cos*— • 82. '' sin2^=sin4(a+&;sin2^+sin2^(a-6)cos2p. a Z ^ 83. " sin s sin (s — a) sin (s — b') sin (s — c) = ^(1 — cos^a — cos^6 — cos^c 4- 2 cos a cos & cose). 84. If AD be the bisec^'^r of the angle A, prove that (1) cos B 4- cos ^ ^ sin - sin ADB cos AD. (2) cos C - cos B = 2 cos ^ cos ADB. 296 SPHERICAL TRIGONOMETRY. 85. Prove cos a sin 6 = sin a cos 6 cos C + cos A sin c. 86. " sin C cos a = cos A sin B + sin A cos B cos C. 87. In a triangle if A = ^, B = ^, C = ^, show that O ij M ir a + 6 -|-c= - 88. Prove sin(S-A): 1 + cos a — cos b — cos c . a • b ■ c 4 cos - sin - sin - 89. If 8 be the length of the arc from the vertex of an isosceles triangle, dividing the base into segments a and /3, prove that tan - tan " = tan - tan —^ — 90. If b = c, show that . a A sin - cos — 2 2 sin b = r» and sin B = . A a sin — cos - 2 2 91. If AB, AC be produced to B', C, so that BB', CC shall be the semi-supplements of AB, AC respectively, prove that the arc B'C will subtend an angle at the centre of the sphere equal to the angle between the chords of AB, AC. PRELIMINAHY OBSERVATIONS. 297 ■.) f. CHAPTER XI. SOLUTION OF SPHERIOAL TEIANGLES. / ;■•'.. 199. Preliminary Observations. — In every spherical tri- angle there are six elements, the three sides and the three angles, besides the radius of the sphere, which is supposed constant. The solution of spherical triangles is the process by which, when the values of any three elements are given, we calculate the values of the remaining three (Art. 184, Note). In making the calculations, attention must be paid to the algebraic signs of the functions. When angles greater than 90° occur in calculation, we replace them by their supple- ments ; and if the functions of such angles be either cosine^ tangent, cotangent, or secant, we take account of the change of sign. It is necessary to avoid the calculation of very small angles by their cosines, or of angles near 90° by their sines, for their tabular differences vary too slowly (Art. 81). It is better to determine such angles, for example, by means of their tangents. We shall begin with the right triangle; here two ele- ments, in addition to the right angle, will be supposed known. "'U SOLUTION OF RIGHT SPHERICAL TRIANGLES. 200. The Solution of Right Spherical Triangles presents Six Cases, which may be solved by the formulae of Art. 185. If the formula required for any case be not remem- bered, it is always easy to find it by Napier's Kules (Art, 298 SPHERICAL TRIGONOMETRY. 186). In applying these rules, we must choose the middle part as follows : When the three parts considered are all adjacent, the one between is, of course, the middle part. When only two are adjacent, the other one is the middle part. Let ABC be a spherical triangle, right-angled at C, and let a, b, c denote the sides opposite the angles A, B, vJ, respectively. W(^ shall assume that the parts are aU positive and less than 180° (Art. 182). 201. Case I. — Given the hypotenuse c and an angle A ; to find a, b, B. By (•^). (5), and (8) of Art. 185, or by Napier's Rules, we have sin a = sin c sin A, tan b = tan c cos A, cot B = cos c tan A. Since a is found by its sine, it would be ambiguous, but the ambiguity is removed because a and A are of the same species [Art. 187, (1)]. B and b are determined imme- diately without ambiguity. If a be very near 90°, w^e commence by calculating the values of b and B, and then determine a by either of the formulae tan a = sin b tan A, tan a = tan c cos B. Check. — As a final step, in order to guard a. r, b, c. 89° 32' 29". 207. Qnadrantal and Isosceles Triangles. — Since the polar triangle of a quadrantal triangle is a right triangle (Art. 184), we have only to solve the polar triangle by the I'orniulse of Art. 185, and take the supplements of the parts thus found for the required parts of the given triangle ; or we can solve the quadrantal triangle immediately by the formulae of Art. 189.* A biquadrantal triangle is indeterminate unless either the base or the vertical angle be given. An isosceles triangle is easily solved by dividing it into two equal right triangles by drawing an arc from the vertex to the middle of the base. The solution of triangles in which a + & = tt, or A + B = ir, can be made to depend on the solution of right triangles. Thus (see the second figure of Art. 191) the triangle K'AC has the two equal sides, «' and b, given, or the two equal angles, A and B', given, according as a + 6 = tt or A -f B = tt in the triangle ABC. EXAMPLES. •; . Solve the following right triangles : ' 1. 3. 5. Given c=32°34', find a=22°15'43". Given c= 69° 25' 11", find a=50° 0' 0", Given c=55° 9' 32", find 6 =51° 53', Given c=127°12', find &=39° 6' 25", Given a.= 118° 54', find A =95° 55' 2", a = 44° 44'; &=24°24'19", A=54°54'42"; 6 =56° 50' 49", «=22°15' 7"; A=27°28'37'.5, a=141°ll'; A =128° 5' 54", B = 12°19'; 6= 10° 49' 17", B=50°8'21". B = 63°25'4". B=73°27'll".l. B = 52° 21 '49". c= 118° 20' 20". * Quadrantal triangles are generally avoided in practice, but when unavoidable, they are readily aolved by either of these method*. I'' 304 0. SPHERIC A L TRW ON OMETR Y. 7. 8. 9. 10. 11. 12. Given a =29° 46' 8", find A= 54° 1'16", Given a= 77° 21' 50", find 6=28°14'31".l, or //=151°45'28".9, Given a =68°, find 6=25°52'33".5, or 6'=154°7'26".5, Given a = 144° 27' 3", find A = 126° 40' 24", Given a =36° 27', find A = 46°59'43".3, Given A = 63° 15' 12", find a. =49° 59' 56", Given A = 67° 54' 47", B = 137°24'2r'; 6=155°27'64". A =83° 56' 40"; c= 78° 53' 20", c'=101°6'40", A = 80°; c=70°18'10".2, c'=109°41'49".8, &=32°8'56"; B=47°13'43", 6=43°32'31"; B=57°59'19".2, B = 135° 33' 39"; 6=143°5'12", B = 99° 57' 35"; &=100°45', c=142°9'13". B =28°49'57".4, B'=151°10'2".6. B =27°36'58".7, B'=152°23'l".3. c= 133° 32' 26". c=54°20'. c=120°55'34". c=94°5'. find a= 67° 33' 27", 13. Solve the quadrantal triangle in which c = 90°, A = 42°l', B = 121°20'. Ans. C = 67°16'22", 6 = 112°10'20", a = 46°31'30". 14. Solve the quadrantal triangle in which a = 174° 12' 49".l, h = 94° 8' 20", c = 90°. Arts. A = 175°57'10", B = 135°42'55", C = 135°34'8". SOLUTION OF OBLIQUE SPHERICAL TRIANGLES. 208. The Solution of Oblique bpherical Triangles presents Six Cases ; as follows : - I. Given two sides and the included angle, a, b, C. II. Gfiven two angles and the included side, A, B, c. III. Given two sides and an angle opposite one of them, a, 6, A. CASE I. 306 2° 9' 13". i°49'57".4, 51°10'2".0. 7°36'58".7, 52''23'1".3. L33°32'26". 54° 20'. 120° 55' 34". p94°5'. = 46° 31* 3b". = 135° 34' 8". NGLES. ngles presents b, C. B, c. ; one of them, IV. Given two angles and a side opposite one of them, A, B, a. V. Given the three sides, a, b, c. VI. Given the three angles, A, B, C. These six cases are immediately resolved into three pairs of cases by the aid of the polar triangle (Art. 184). For when two sides and the included angle are given, and the remaining parts are required, the application of the data to the polar triangle transforms the problem into the supplemental problem : given two angles and the included side, to find the remaining parts. Similarly, cases III. and IV. are supplemental, also V. and VI. The parts are all positive and less than 180° (Art. 182). The attention of the student is called to Art. 199. 209. Case I. — Oive i two sides, a, b, and the included angle C ; to find A, B, c. By Napier's Analogies, (7) and (8) of Art. 197, tan ^ (A + B) = - ^^" ^ (« - ft) cot cos ^ (a + 6) C _— • 2 , i/A T>\ sini^(a — &) ,C tan i (A - B) = . \) ,( cot-, sin ^ (a + 6) 2 These determine ^(A + B) and ^(A — B), and there- fore A and B ; then c can be found by Art. 190, or by one of Gauss's equations (Art. 198). Since c is found from its sine in Art. 190, it may be uncertain which of two values is to be given to it : if we determine c from one of Gauss's equations, it is free from ambiguity. We may therefore find c from (3) of Art. 198. Thus cos ^ (A -)- B) cos- = cos i (a + &) sin-- 2 ^ lii I It- 1 306 SPHERICAL TRIGONOMETRY. 1 I M 11 Check, tan ^ (a + &) cos ^ (A + B) = cos ^ (A — B) tan -• There is no ambiguity in this case. Ex.1. Given a = 43° 18', & = 19°24', C = 74°22'; find A, B, c. Solution. i {a 4- h) = 31° 21', ^ (ct - 6) = 11° 57', \ C •- 37° 11'. logcosH«-&)= 9-9904848 log£eci(a + &)= 0.0685395 log cot 5:^10.1199969 ^ 2 log tan^(A+B) =10.1790212 .-. ^(A4-B) = 56°29'17" i(A-B) = 27°41' 0".5 .-. A = 84°10',17".5 B = 28° 48' 16".5. c = 41°35'48".5. log sin |(a - 6) = 9.3160921 iGgcosec^(a+6)= 0.2837757 logcot- =10.1199969 ^ 2 log tan \ ( A -B) = 9.7198647 .-. i(A-B)=27°41'0".5. log cos \{a + h)= 9,9314605 log sec i(A 4- B) = 0.2579737 log sin -= 9.7813010 ^ 2 log cos -= 9.9707352 ^ 2 ^ = 20° 47' 54".25. 2 Othei'wise tkus: Let fall the perpendicular BD, dividing the triangle ABC into two right tri- angles, BDA, BDC. Denote AD by m, the angle ABD by <^, and A-^^ BDhyp. Then by Napier's Rules, b we have cos C = tan (6 — m) cot a ; sin (& — m) = cot C tanp ; sin m = cot A tanp. .'. tan(6 — m) = tanacosC (1) and tan A sin m = tan C sin (& — m) (2) CASE 11. 307 ban '; find 11'. 3160921 2837757 1109969 ,7198647 '0".5. ,9314605 1.2579737 1.7813010 ).9707352 np. . . (1) From (1) m is determined, and from (2) A is determined. In a similar manner B may be found. Also, from the same triangles, we have by Napier's Rules cosa"= cos (6 — m) cosp; cos c = cos m cosp. .•. cosccos (6 — m) = cosmcosa, from which c is found. Note. — This method has the advantage that, in using it, nothing need be remem- bered except Napier's Rules. If only the side c is wanted, it may be found from (4) of Art. 191, without pre- vi3usly determining A and B. This formula may be adapted to logarithms by the nae of a subsidiary angle (Art. 90). Ex.2, Given 6=120°30'30", c=70°20'20", A=50°10'10"; find B, C, a. A71S. B =: 135° 5' 28".8, C = 50° 30' 8".4, a = 69° 34' 56". 210. Case 11. — Given two angles, A, B, and the included side c ; to find a, b, C. By Napier's Analogies (9) and (10) of Art. 197, cos^(A + B) 2 : tanH«-&)= '^"^^^~?^ -ian£, ' ^^ ^ sin^(A + B) 2 from which a and b are found. The remaining part C may be found by (2) of Art. 198. sin ^ (a — &) cos - = sin ^ (A — B) sin -• n Check, 3s ^ (a — b) cot — = cos ^ (a -f b) tan ^ (A + B). There is no ambiguity in this case. ti m,f ■•■•V- ■:,;■■ t' 308 SPHEltlCAL TRIGONOMETRY. Ex. 1. Given A=68°40', B = 56°20', c=84°30'; find a, b, C. Solution. i (A + B) = 62° 30', i (A - B) = 6° 10', ^ = 42° 15'. logcosi(A-B)= 9.9974797 logseci(A+B)= 0.3355944 log tan ?= 9.9582465 ^ 2 . . log tan |(a + 6) = 10.2913206 .-. i(a. + &) = 62°55' 9" ^(a-b)= 6° 16' 39" a = 69° 11' 48" b = 56° 38' 30". C=::97°19' 3".5. log sin ^(A - B) =9.0310890 logcoseci(A + B) =0.0520711 logtan- =9.9582465 ^ 2 log tan |(a - &) =9.0414066 .-. ^(a- 6) =6° 16' 39". log sin |( A - B ) = 9.0310890 log cosec ^(a - b) =0.9612050 log sin £=9.8276063 ^ 2 log cos ^=9.8199003 . § = 48°39'31f". Otherwise thus : Let fall the perpendicular BD (see last figure). Denote, as before, AD by w, the angle ABD by , and BD by p. Then by Napier's Rules, we have cos c = cot ^ cot A ; cos (^ = cote tan ^; cos(B — <^) = cotatanp. .*. cot <^ = tan A cos c (1) tan a cos (B — )= cos (^ tan T (2) From (1) <^ is determined, and from (2) a is found. Similarly 6 may be found. Also, from the same triangles, we have cos C sin (f> = cos A sin(B — ), from which C is found. CASE III. 309 a, b, C. 15'. )310890 0520711 9582465 0414066 ' 16' 39". ,0310890 .9612050 .8276063 1.8199003 If". (see last BD by , IP- . (1) • (2) IS found. Ex. 2. Given A = 135° 5' 28".G, C = 50° 30' 8".6, b = 69° 34' 56".2 ; find a, c, B. Ans. a = 120° 30' 30", c = 70° 20' 20", B = 50° 10' 10". 21L Case III. — Given two sides, a, b, and the angle A opposite one of them ; to find B, C, c. The angle B is found from the formula, sinB = ^HL^sinA (Art. 190) (1) sin a Then C and c are found from Napier's Analogies, ' tang = ^^"i(^-^) cotHA-B) . . . 2 sini(a + 6) ^^ ^ • : ' c sini( A + B) , , / ,v tan- = -^ — " ^ ^ — ^tanH« — ^)- • • 2 sini-(A-B) ^^ ' (2) (3) Check, sin A sin a sinB sin b sinC sin c Since B is found from its sine in (1), it will have two values, if sin A sin 6 < sin «, and the triangle, in general, will admit of two solutions. When sin A sin b > sin a, there will be no solution, for then sin B > 1. In order that either of these values found for B may be admissible, it is necessary and sufficient that, when sub- stituted in (2) and (3), they give positive values for C c tan— and tan-, or which is the same thing, that A — B and Z Li a — b have the su,me sign. Hence we have the following Rule. — If both values of B obtained from (1) be such as that A — B and a — b have like signs, there are tivo complete solutions. If only one of the values of B satisfies this condi- tion, there is only one triangle' that satisfies the problem, since ]'i-m 810 SPHERICAL TRIGONOMETRY. '^i'^- in this case C, or c > 180°. If neither of the values of B makes A — B and a — b of the same signs, the problem is impossible. This case is known as the ambiguous case, and is like the analogous ambiguity in Plane Trigonometry (Art. 116), though it is somewhat more complex. For a complete discussion of the Ambiguous Case, the student is referred to Todhunter's Spherical Trigonometry, pp. 53-58 ; McCol- lend and Preston's Spherical Trigonometry, pp. 137-143; Serret's Trigonometry, pp. 191-195, etc. Ex. 1. Given a=42°45', &=47°15', A=56°30'; findB, C,c. Solution. log sin 6 = 9.8658868 colog sin a = 0.1682577 log sin A = 9.9211066 log sin B =9.9552511 .-. B = 64°26'4", B' = 115° 33' 56". ^{a-hb) = 45° 0' 0". ^(a-b) =- 2° 15' 0". ■^(A + B) = 60° 28' 2". |(A _ B) = - 3° 58' 2". |(A + B')= 86° 1'58". |(A-B') = -29°31'58". Since both values of B are such that A — B, A — B', and a — b, are all negative, there are two solutions, by the above Rule. (1) TT/iew B = 64° 26' 4". logsini(a-6) =8.5939483- cologsin^(a+&) =0.1505150 log cot ^( A - B) = 1.158941.'.- log tan ^=9.9034046 ® 2 C = 38° 40' 48". .-. C = 77° 21' 36". log sin i( A + B) =9.9395560 cologsin^(A-B)=1.1599832- log tan i{a-b) = S.5942832 - log tan -=9.6938224 ^ 2 .-. -=26° 17' 40". 2 .-. c =53° 35' 20". CASE III. 311 es of B Mem is like the t. IIC), jomplete referred ; McCol- 137-143; idB,C,c. r 0' 0". 2° 15' 0". 3° 28' 2". 3° 58' 2". G° 1'58". 9° 31' 58". — B', and the above 9.93955G0 1.1599832- 8.5942832- =9.6938224 =2G°17'40". = 53° 35' 20". (2) When B' = 115° 33' 56". log sin ^(a - b) =8.5939483- cologsin|(rt+ft) =0.1505150 log cot ^(A-B') =0.2467784- log tan ^-=8.9912417 ^ 2 C' 2 5°35'50i". •. C'=11°11'40^". logsin|(A + B') =9.9989581 colog sin |( A -B') =0.3072223- log tan |(a - ^;) = 8.5942832- log tan ^^'=8.9004636 ^ 2 .... ^' = 4° 32' 47i". 2 * .-. c'=9° 5' 34V'. Ans. B = 64° 26' 4", C =77° 21' 36", c =53° 35' 20"; B' = 115° 35' 56", C' = 11°11'40V', c'= 9° 5' S^". Otherwise thus: Let fall the ' perpendicular CD; denote AD by m, the angle ACD by . . •. cot (f} = cos b tan A Again, . ,^ cos a = cos (c — m) cosp ; cos 6 = cos m cosjp. .•. cos (c — m) = cos a cos 7Ji -i-b . ' . Also, cos (C — <^) = cot a tsaip ; cos <^ = cot b tajip. .'. cos (C — <^) = cot a tan 6 cos . . (2) (3) Lastly, . T3 sin 6 . . sin B = -^ — sin A sin a (4) (5) The required parts are given by (1), (2), (3), (4), (5). Ex. 2. Given a = 73°49'38", ?;=120°53'35", A=88°52'42": find B, C, c. Ans. B = 116° 44' 48", C = 1 16° 44' 48", c = 120° 55' 35". :) = 9.7013681 tan -=9.8194553 2 •. ? =33° 25' 10". 2 B =66° 50' 20". tanr= 9.5208234 sin(s-c)= 9.4460251 tan ^=10.0747983 C :49°54'35 » QK" C =99° 49' 10". .-<. Ex.2. Given a = 100°, & = 37° 18', c = 62°46'; find A, B, C. Ans. A=176°15'46".56, B = 2°17'55".08, C=3°22'25".46. 214. Case VI. — Given the three angles, A, B, C ; to find the sides. As in Art. 213, the formulae for the tangents are to be preferred. Putting tan R =^ ^^^ ^^ _ ^^ ,," |i:|y^^^> we have, from (7), (8), (9) of Art. 196, tan - = tan R cos (S — A), 2t CASE VI. 315 tan- = tan R cos (S — B), tan I = tan R cos (S — C), by which the three sides may be found. rn 7 sin A sin B sin C VhecK, = = • sin a sm h sin c Ex. 1. Given A = 68° 30', B = 74° 20', C = 83° 10'; find Solution. a, b, c A= 68° 30' B= 74° 20' C= 83° 10' 28 = 226° 0' log (- cos S) = 9.5918780 log cos (S - A) = 9.8532421 log cos (S - B) = 9.8925365 log cos (S - C) = 9. 9382576 log tan2R = 9.9078418 log tan R = 9.9539209.* S = 113° 0' S _ A = 44° 30' . S-B= 38° 40' S - C = 29° 50' lt)g tan - = 9.8071630 ^ 2 log tan - = 9.8464574 ^ 2 log tan - = 9.8921785 ^ 2 a = 65° 21' 22^" 6 = 70° 9' 9^', c = 75°55' 9". Check, sin a _ sin b _ sine ^ sin A sin B sin C Ex.2. Given A=59°55'10", B=85°36'50°, C=59°55'10"; find a, b, c. Ans. a = 129° 11' 40", b = 63° 15' 12", c = 129° 11' 40". * Tlio neceBsary additions may be conveniently performed by writing log tan R on a slip of paper, and holding it Buccessively over log cos (8 -A), log cos (S — B), and l0gC08(S-C). t»l .' ••♦ iW' 316 SPHERICAL TRIGONOMETRY. EXAMPLES. Solve the following right triangles : Given c= 84° 20', find a= 35° 13' 4", Given c= G7°54', find a= 39°35'i)l", 3. Given c= 22° 18' 30", find a= 1G°17'41", 4. Given c=145°, find a= 13° 12' 12", 6. Given c= 98° 6' 43", find a =137° 6', 6. Given c= 4G°40'i2", find a= 26°27'23".8, 7. Given c= 7C°42', find b= 70° 10' 13", 8. Given c= 91° 18', find b= 94°18'53".8, Given c= 86° 51', find a= 86° 41' 14", 10. 11. 12. 13 14 Given c= 23° 49' 51", find b= 19° 17', Given c= 97° 13' 4", find 6= 79°13'38".2, Given c= 37° 40' 20", find 6= 0°2G'37".2, A= 35° 25'; 6= 83° 3' 29", B= 85° 59' 1". \ 15 A= 43° 28'- b= 60°46'25f', B= 70° 22' 21". A= 47° 39' 36"; 16. b= 15° 26' 53", B= 44°33'53".4, A= 23° 28'; 17. 6= 147° 17' 15", B = 109° 34' 33". A=138°27'18"; 18. 6= 77° 51', B= 80° 55' 27". A= 37° 46' 9"; 10. b= 39°57'41".4, B= 62° 0' 4". a= 47° 18'; A= 49° 2'24".5, B= 75°9'24".7o. 20. a= 72° 27'; A= 72''-'29'48", B= 94° 6'53".3, b= 18° 1'50"; 21. A= 88° 58' 25", B= 18° 3' 32". a= 14° 16' 35"; A= 37°36'49".3, B= 54°49'23".3. 22. a= 132° 14' 12"; A =131° 43' 50", B= 81°58'53".;5. 23. «:= 37° 40' 12"; A= 89° 25' 37", B= 0°43'33". tp EXAMPLES. 317 13. Given a= 82° 6', B= 43° 28'; find A= 84°34'28", 6= 43° 11' 38", c= 84° 14' 57". 14. Given a= 42°30'30", B= 53° 10' 30"; find A= 53° 50' 12", b= 42° 3' 47", c= 56° 49' 8". 85° 59' 1". 1 15. Given a= 20°20'20", B= 38° 10' 10"; find A = 54°35'16".7, 6= 15°16'50".4, c= 25°14'38".2. 70° 22' 21". 16. Given a= 92° 47' 32"; B= 50° 2' 1"; . find A= 92° 8' 23", 6= 50°, c= 91°47'40". 44°33'53".4. 17. Given b= 54° 30', A= 35° 30'; \. ■ 109° 34' 33". find B= 70°17'35", a= 30° 8'39".2, c= 59°51'20". 18. Given 6=155°46'42".7, A= 80° 10' 30"; : 80° 65' 27". find B=153°58'24".5, a= 67° 6'52".6, c=110°46'20". 19. Given a= 35° 44', A= 37° 28'; b 62° 0' 4". find b= 69° 50' 24", or 6'=110° 9'36", c= 73°45'15", c'=rl06°14'45". B= 77° 54', B=102° 6'. : 75°9'24".7o. 20. Given rt=129°33', find b= 18° 54' 38", A = 104° 59'; c=127° 2' 27", B= 23° 57' 19", = 94° 6'53".3. or &'=161° 5'22", c'= 52°57'33", B'=156° 2'41". 21. Given a= 21° 39', A= 42° 10' 10"; = 18° 3' 32". find b= 25° 59' 27 ".8, or 6'= 154° 0'32".2, c= 33°20'13".4, c'=146°39'46".6. B= 62° 23' 2".8, B'=127°36'57".2. = 54° 49' 23".;]. 22. Given a= 42°18'45", find b= 60° 36' 10", A= 46° 15' 25"; c= 68° 42' 59", B= 69° 13' 47", or 6' =119° 23' 50", c'=lll°17' 1", B'=110°46'13". = 81°58'53".;J 23. Given &=160°, B = 150°; 1 find a= 39° 4'50".7, c=136°50'23".3. A= 67° 9'42".7, = 0°43'33". 1 or a'=140°55' 9".3, c'= 43° 9' 36". 7, A'=112°50'17".3. s R 1 318 24. 25. 26. 27. 28. ^o» 30. 31. 32. 33. 34. 35. 36. SPUE RICA L TRIG ONOMETR F. Given a= 25° 18' 45", A= 15° 58' 15". Ana. Impossible ; why? Given a= 32° 9' 17", b= 32° 41'; find A = 49° 20' 17", B= 50° 19' 16", c= 44°33'17". Given a= 55° 18', b= 39° 27'; find A= 66° 15' 6", B= 45° 1'31", c= 63°55'21". Given a= 56° 20', b= 78°40'; find A= 56° 51' 7", B= 80° 31 '48", c= 83°44'44i". Given a= 8'3°40', 6= 32° 40'; find A = 88°11'57".8, B= 32°42'37".8, c= 87°11'39".S. Given a— 37° 48' 12", b= 59° 44' 16"; find A= 41° 55' 45", B= 70° 19' 15", c= 66° 32' 6". Given a= 116°, b= 16°; find A = 97°39'24".4, B= 17°41'39".9, c=114°55'20".4. Given A = 52° 26', B= 49° 15'; find a= 36°24'34".5, 6= 34° 33' 40", c= 48° 29' 20". Given A = 64° 15', B= 48° 24'; find a= 54° 28' 53", 6= 42° 30' 47", c= 64° 38' 38". Given A = 54° 1'15", B = 137°24'21"; find a= 29° 46' 8", 6=155°27'55", c=142° 9' 12". Given A= 46° 59' 42", B= 57° 59' 17"; find a= 36° 27', 6= 43° 32' 37", c= 54° 20' 3". Given A = 55° 32' 45", B=101°47'56"; find a= 54°41'35", 6= 104° 21' 28", c= 98° 14' 24". Given A = 60°27'24".3, B= 57°16'20".2; find a= 54°32'32".l, b= 51°43'36".l, c= 68° 56' 28". 9. 4° 33' 17". >3°55'21". 33°44'44i". 87m'39".S. 66° 32' 6". 14°55'20".4. 48° 29' 20". 64° 38' 38". 42° 9' 12". 54° 20' 3". «)8°14'24". 68°56'28".9. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. EXAMPLES. ijiy Solve the following quaili •antal triangles : Given B= 74° 45', a= 18° 12', c= 90°; find 6^ 86°17'15".6, A= 17° 34' 2", C = 104° 31' 13". Given A = l 10° 47'50", B=135°35'34''.5, c= 90°; find a =104° 53' 0".8, 6=133°39'47".7, C = 104°41'37".2. Solve the following oblique triangles: Given a= 73° 58', b= 38° 45', C= 46° 33' 39"; find A=116° 8' 28", B= 35° 46' 39", c= 51° I'll". Given a= 96° 24' 30", 6= 68° 27' 26", C= 84° 46' 40"; find A= 97°53'0i", B= 67°59'39i", c= 87° 31' 37". Given a= 76° 24' 40", b= 58° 18' 36", C = 116° 30' 28"; find A= 63°48'35V', B= 51°46'12i", c= 104° 13' 27". Given a= 86°18'40", b= 45° 36' 20", C = 120° 46' 30"; find A= 64°48'53|", B= 40°23'15|", c=108°39'llV'. Given a= 88° 24', b= 56° 48', C = 128°16'; find A= 65° 13' 3^', B= 49° 27' 51", c= 120° 10' 52". Given a= 68°20'25", 6= 52° 18' 15", C=117°12'20"; find A= 56° 16' 15", B= 45° 4'41", c= 96°20'44". Given a= 88°12'20", 6=124° 7'17", C= 50° 2' 1"; find A= 63° 15' 12", B= 132° 17' 59", c= 59° 4' 25". Given a= 32° 23' 57", 6= 32° 23' 57", C= 66°49'17"; find A= 60° 53' 2", B= 60° 53' 2", c= 34° 19' 11". Given b= 99° 40' 48", c=100°49'30". A= 65° 33' 10"; find B= 95° 38' 4", C= 97°26'29".l, a= 64°23'15".l. Given A = 31° 34' 26", B= 30° 28' 12", c= 70° 2' 3"; find «= 40° 1' 5\", b= 38° 31' 3^", C=130° 3' 50". ■ 320 49. 50. 51. 52. * 53. 54. 65. 56. 57. SPHERICAL TRIGONOMETRY. 58. 59. 60. Given A = find a= Given A = find a= Given A = find a= Given Ax find «= Given A = find a- Given A: find a- GivenA: find a- GivenA: find a: Given «= find B = or B': Given a- find B: or B': Given a-. find B- or B': Given a-. find B: or B': :130°5'22".4, : 84° 14' 29", : 96° 46' 30", = 102° 21 '42", : 84° 30' 20", : 94° 34' 52V', = 107° 47' 7", = 70° 20' 50", = 128° 41' 49", = 125° 44' 44", = 129°58'30", = 85° 59', = 95° 38' 4", = 99°40'48", = 70°, = 57^ 56' 53", = 62° 15' 24", = 62°24'24".8, = 117°35'35".2, = 52° 45' 20", = 59°24'15|", = 120° 35' 44 1", = 48° 45' 40", = 55° 39' 57", = 124° 20' 3", = 46° 20' 45", ' 6= = 54° 6' 19", \ C = = 125°53'41", C'= B = h-. B= 6= B = B= h-. B = h-. B= 6= C: C: B: 6: h: C: C: 6: C: C: h: C: ! nf C'= 32°26'6".41, 44° 13' 45", 84° 30' 20", 78° 17' 2", 76° 20' 40", 76° 40' 48 V', 38° 58' 27", 38° 27' 59", 107° 33' 20", 82° 47' 35", 34° 29' 30", 47° 29' 20", 97° 26' 29", 100° 49' 30", 131° 18', 137° 20' 33", 103° 18' 47", 155°43'11".3, 59° 6'10".6, 71°12'40", 115° 39' 55V', 26°59'55".2, 67° 12' 20", 116° 34' 18", 24° 32' 15", 65° 18' 15", 116° 55' 26", 24°12'53".3, c= C = c= C = c= C = c- C = c= C = C-- C = h: B: C- C = A = c'= A= c'= A = c= c'= A = c= c'= : 51° 6 : 36° 45 : 126° 46 =125° 28 130° 46 : 130° 51 : 51° 41 : 52° 29 : 124° 12 =127° 22 : 50° 6 : 36° 6 : 64° 23 : 65° 33 =116°; : 94° 48 53° 42 153° 9 70° 25 46° 22 97° 33 29° 57 42° 20 93° 8 27° 37 40° 10 90° 31 27° 23 ll"/); 26". 13V'. 33V'. 14"; 45". 31"; 7". 20"; 50". 15"; 10". 12". 38"; 26". 10"; 18".8, io".r,. 30"; 9".(i, 20". 30" ; 46", 14". 19. EXAMPLES. 321 ,1° 6'11".0; I 61. 16° 45' 26". >6°46';^ 25°28'13|". ^0°46'; 30° 51' 33 V- 24° 12' 31"; .27° 22' 7". 50° 6' 20"; 36° 6' 50". 46° 22' 10"; : 97°33'18".8, L'0°57'10".r). 62. 51° 41' 14"; 163. 52° 29' 45". 64. 64° 23' 15"; |g5 65° 33' 10". 116° ; 94° 48' 12". 53° 42' 38"; 153° 9' 35V', 70° 25' 26". 18. 42° 20' 30"; : 93° 8' 9".<'», = 27° 37' 20". = 40° 10' 30"; |9 = 90° 31' 46", = 27" 23' 14". Given a=150°57' 5", find B = 120° 47' 44", or B'= 59° 12' 16", Given a= 50° 45' 20", find B= 57°34'51".4, or B'=122°25' 8". 6, Given a= 40° 5'25".6, find B= 42° 37' 17 ".5, or B'=137°22'42".5, Given a= 99° 40' 48", find B= 65° 33' 10", (No ambiguity ; why?) Given A = 79° 30' 45", find b= 36° 5'34f' , (No ambiguity ; why?) Given A = 73° 11' 18", find b= 41°52'34|", (Only one solution ; why?) Given A = 46° 30' 40", find b= 33°18'47i", (Only one solution ; why?) Given A = 61° 29' 30", find b= 15°30'30".5, (Only one solution ; why?) Given A = 36° 20' 20", find 6= 55° 25' 2^", or 6' =124° 34' 57^", b-. C = C'= 6: C: 134° 15' 54", : 97° 42' 55", : 29° 9' 9", : 69° 12' 40", :115°57'50".6, : 25°44'31".6, : 118° 22' 7".3, :160° 1'24".4, b C C'= 50°18'55".2, c'= c'= A = c= c'= A = 6= 64° 23' 15". C= 97° 26' 29", B = C-- B: C: B: B: V.= 46° 15' 15", 50° 24' 57", 61° 18' 12", 41° 35' 4", 36" 20' 20", 60° 32' 6", 24° 30' 30", 39° 33' 52", 46° 30' 40", 81° 27' 26 V', A = 144° 22' 42"; 55° 42' 8", 23° 57' 29". 44° 22' 10"; 95n8'16".4, 28° 45' 5".2. 29°42'33".8; 153°38'42".4, 90° 5'41".0. : 95° 38' 4"; = 100° 49' 30". a= 53°18'20"; C= 70° 55' 35". a= 46° 45' 30"; C= 60°42'46".5. a= 42° 15' 20"; C = 110° 3'14".6. a= 34° 30"; C= 98'48'58".5. a= 42° 15' 20"; C = 119°22'27i", c'=162°34'27", C'=164°41'56". 322 SPHERICAL THIGONOMETRY. ■ i 1 , 1 70. Given A = 52° 50' 20", B= 66° 7' 20", «= 59° 28' 27"; v; : 1 find b= 81° 15' 15", c= 110° 10' 50^", 0=119° 43' 48", ipf^ 71. or b'= 98° 44' 45", Given A =115° 36' 45", c'=138°45'26", B= 80° 19' 12", C'=142°24'59". b= 84°21'56"; . ■ ' ■ 1 find a =114° 26' 50", c= 82° 33' 31", C= 79° 10' 30". I ■ 72. Given A = 61° 37' 52". 7, B = 139°54'34".4, 6=150°17'26".2; 1 find a= 42°37'17".5, 0=129° 41' 4".8, C= 89°54'19".0, ■v . . ^ or a'=137°22'42".5. c'= 19°58'35".6, Q/1 C'= 26°21'17".6. _Jh: iJ. iS |! 73. Given A = 70°, B = 120°, &= 80°. :9H|lBn Ans. Impossible ; why ? cS7 ' ' ^^HHI 1 74. Given a= 108° 14', b= 75° 29', c= 56° 37'; ! find A = 123° 53' 47", B= 57° 46' 56", C= 46°51'51".5. ^"'^• 'In 1 75. Given a= 57° 17', b= 20° 39', c= 76° 22'; find A= 21° 1' 2", B= 8° 38' 46", C = 155°31'36".5. ^^■ ■■fe^^^^ ,2-^„LjM 76. Given a= 68° 45', b= 53° 15', c= 46° 30'; find A= 94° 52' 40", B= 58° 5' 10", C= 50°50'52|". c) H 1 • 77. Given a= 63° 54', &= 47° 18', c= 53° 26'; 1 " find A= 86° 30' 40", B= 54° 46' 14", C= 63°12'55i". 78. Given a= 70° 14' 20", 6= 49°24'10", c= 38° 46' 10"; ! find A= 110° 51 '16", B= 48° 56'. .4", C= 38°26'48". 9 c= 97°12'25"; 79. Given a=124°12'31". b= 54° 18' 16", ''^^HII find A=127°22' 7", B= 51° 18' 11", C= 72° 26' 40". H| 80. Given a= 60° 12' 4", ?>=116°44'48", c=129°ll'42"; find A= 59° 4' 26", B= 94° 23' 10", C = 120° 4' 50". HI ^^' Given a =100°, b= 50°, c= 60°; ),*j find A=138°15'46".4, 'b= 31°11'14".0, C= 35°49'68".2. ^H ^^' Given A = 86° 20', B= 76° 30', C= 94° 40'; ,'^] 1 find a« 87° 20' 28", bm 76''44'2i", c« 93° 66' 31". lat EXAMPLES. 59° 28' 27"-, 19° 43' 48", 42° 24' 59". 84° 21' 56"; 79° 10' 30". L50°17'26".2; 89°54'19".0, 26°21'17".6. 80°. 56° 37'; 46°51'51".5. 76° 22'; :155°31'36".5. : 46° 30'; : 50°50'52i". : 53° 26'; = 63° 12' 551". = 38° 46' 10"; = 38° 26' 48". = 97° 12' 25"; = 72° 26' 40". = 129° 11' 42"; = 120° 4' 50". = 60°; = 35°49'58".2, = 94° 40'; =3 93° 56' 31". 83.! Given A = 96° 45', find a= 88° 27' 49", 184. Given A = 78° 30', find a= 74° 57 '46", 1 85. Given- A = 57° 50', find a= 58° 8' 19", 1 86. Given A = 129° 5' 28", find a =135° 49' 20", |87. Given A = 138° 15' 50", find a=100° 0' 8". 4, |88. Given A= 102° 14' 12", find a=104°25' 8", 89. Given A = 20° 9' 56", find a= 20° 16' 38", B = 108°30', 6=107° 19'52", B = 118°40', 6=120° 8' 49", B= 98° 20', b= 83° 5' 36", B = 142° 12' 42", 6 =144° 37 '15", B= 31° 11' 10", 6= 49°59'56".4, B= 54° 32' 24", b= 53° 49' 25", B= 55° 52' 32", b= 56°19'41", 323 C=116°15'; c=115°28'13V'. C= 93° 20'; c=100°18'llf". C= 63° 40'; c= 64° 3' 20". C = 105° 8' 10"; c= 60° 4' 54". C= 35° 50'; c= 60° 0'11".2. C= 89° 5' 46"; c"= 97° 44' 18". C = 114° 20' 14"; c= 66° 20' 43". ^i 90. If a, b, c are each < ^, show that the greater angle may txceed -• 2 91. If a alone >^7r, show that A must exceed ^• 92. If a and b are each >^7r, and c < ^tt, prove that : (1) The greatest angle A must be >i'n-; (2) B may be > j^ir; (3) C may or may not be < ^tt. 93. If cos a, cos b, cos c are all negative, prove that cos A, )s B, cos C are all necessarily negative. 94. In a spherical triangle, of the five products, cos a cos A, IS h cos B, cos c cos C, cos a cos b cos c, ~ cos A cos B cos C, show lat one is negative, the other four being positive. I ] 324 SPHERICAL TRIGONOMETR F. A ■ I ■ CHAPTER XII. THE IN-OIROLES AND EX-OIEOLES. — AREAS. 215. The In-Circle (Inscribed Circle). — To find the angular radius of the in-circle of a triangle. Let ABC be the triangle; bisect the angles A and B by the arcs AG, BO; from draw OD, OE, OF perpendicular to the sides. Then it may be shown that is the in-centre^ and that the per- pendiculars OD, OE, OF are each equal to the required angular radius. Let 2 s = the sum of the sides of the triangle ABC. The right triangles OAE, OAF are equal. .-. AF = AE. Similarly, BD = BF, and CD = CE. .-. BC + AF = AC + BF = s. .-. AF = s-BC = 8-a. Now tan OF = tan OAF sin AF . (Art. 186) or, denoting the radius OF by r, we have tan r = tan — sin (s — a) 2 ^ ^ (1) or tan r V sin (s — g) sin {s — b) sin {s — c) sins n sins (Art. 195) (2) THE ESCEIBED CIRCLES. 325 a) Also, sin(s— a) =sin ^(6 + c) cos -|-a — cos ^(6 + c) sin ^a sin ^ a cos ^ a . A sin — 2 [cos |(B - C) - cos i(B + C)] (Art. 198) _ sinct s in^B si n^^C sin^A which in (1) gives . B . -C sin — sill - , 2 2. tan r = :— : — sm a cos ^ A N , . . . (3) (Art. 196) (4) 2 cos ^ A cos ^B cos|-C an equation which is equivalent to the following: cot7' = ^— [cosS+cos(S-A)+cos(S-B)+cos(S-C)](5) ^ JN 216. The Ex-Circles. — To find the dtngular radii of the ex-circles of a triangle. A circle which touches one side of a triangle and the other two sides produced, is called an escribed circle, or ex-circle, of the triangle. It is clear that the three rx-circles of any triangle are the in-circles of its colunar triangles (Art. 191, Sch.). Since the circle escribed to the side a of the triangle ABC is the in-circle of the colunar triangle A'BC, the parts of which are a, ir — b, tt — c, A, TT — B, IT — C, the problem becomes identical with thiit A<^ I Y ^^A of Art. 215 ; and we obtain the value for the in-radins of the colunar triangle A'BC, by substituting for b, c, B, C, their supplements in the five equations of that article. JiT ' 326 SPHERICAL TEIGONOMETJiY. Hence, denoting the radius by i\, we get tan ?•„ = tan I A sin s (1) =^- — (2) sm(s — a) cos 4 B cos 4 C ■ /o\ = 2 2 — gin a (3) cos \ A ^ NL (4) 2cos^Asin^Bsin^C ' '■ ' ' ^ ' cotr,=--[-cosS-cos(S-A)+cos(S-B)+cos(S-C)](5) These formulae may also be found independently by methods similar to those employed in Art. 215, for the in-circle, as the student may show. Sch. Similarly, another triangle may be formed by pro- ducing BC, BA to meet again, and another by producing CA, CB to meet again. The colunar triangles on the sides b and c have each two parts, b and B, c and C, equal to parts of the primitive triangle, while their remaining parts are the supplements in the former case of a, c. A, C, and in the latter, of a, b, A, B. The values for the radii r^ and r<. are therefore found in the same way as the above values for r„ ; or they may be obtained from the values of ?•„ by advancing the letters. Thus, tan Vi, = tan I B sin s = , etc., sin(.s — 6) and tan r^ = tan i C sin s = -r-- , etc. sin (s—c) 217. The Circumcirde. — To find the angular radius of the circumcirde of a tnanyle. The small circle passing through the vertices of a spheri- cal triangle is called the circumscribing circle, or circumcirde, of the triangle. (1) (2) (3) . . (4) -C)] (5) ently by , for the d by pro- producing the sides equal to ling parts C, and in found in )y may be etters. itC., tc. • radius of »f a spheri- ircumcircle, THE CIRCUMCIRCLE. 327 Let ABC be the triangle; bisect the sides CB, CA at D, E, and let be the intersection of perpendiculars to CB, CA, at D, E; then is the circuiu- centre. For, join OA, OB, OC; then (Art. 186) cos OB = cos BD cos OD, cos OC = cos DC cos OD. .-. OB = OC. Similarly, OC = OA. Now the angle OAB = OBA, OBC = OCB, OCA = OAC. .-. OCB + A=|(A + B + C)=S. .-. OCB = S-A. Let OC = R ; then, in the triangle ODC, we have cos OCD = tan CD cot CO = tan | a cot R . (Art. 18G) tan 4 a .-. tan R = or tan R = — cos(S- A) cosS N . . . . (1) (Art. 196) (2) Also cos(S - A)= cosi[(B + C) - A] = cos|(B + C) cos|A + sin|(B + C) sin|A = sin^^?iA[-cos|(64.c)4.cosK&-c)] (Art. 198) COS ■«- ct sin A cos^a cos^&cos|c, which in (1) gives tanR = sin^g (3) sin A cos 1 6 cos ^c 2sin4a sin46sin4c ,. , ^..j,, ,.. = ^ 2 2_ . (Art. 195) (4) II i! £ i:- M; f , &28 BPHEBICAL TttlGONOMETHY. which may be reduced to the following : tan R = — [8in(s — a) + sin(s — h)-\- sin(s — c) — sin s] (5) 2n 218. Circumcircles of Colunar Triangles — To find the angular radii of the circumcircles of the three colunar triangles. Let Rj, Rgj ^3 ^6 the angular radii of the circumcircles of the colunar triangles on the sides a, 6, c, respectively. Then, since Ri is the circumradius of the triangle A'BC whose parts are a, v — b, tt — c, A, tt — B, tt — C, we have, from Art. 217, tanRi = -*?5i^ (1) cosS tanR,=: "^^(^-^) (2) tan Ri = -— -4^^i^^-- (3) sin Asin^o sm^c tan R, = ^ ^"^ ^ g cos ^ 6 cos ^ c ^ ,^. tan Rj = — - [sins— sin(s— a)+sin(s— 6)+sin(s— c)] (6) Similarly, tan R, = - *^5Li^ = ^^^i^Jni^ = etc., cosS N ' and tan R, = - ^-^^^ = ^i^-Cl ^ ' cos S N EXAMPLES. Prove the following : 1. cos 8 4- cos (s — a) + cos {s — b)-\- cos (s — c) = 4 cos ^ a cos ^ 6 cos ^ c. 2. cos(s — 6)-f cos(s — c) — cos(8 — a) — COSS = 4 cos ^ a sin ^ b sin J^ c. ^ PROBLEM. 329 3. tann = 52iiCc^si„^,^ N cos^B 2 cos ^ 13 sin ^C sin ^ A 4. tanr,=.^^^iA^iMsinc = — — ^^^ cosiC 2 cos I C sin ^ A sin ^B* 6. cot r : cot Vi : cot i-g : cot r^ = sin s : sin (s - a) : sin (s - b) : sin (s - c). 6. tan r tan r^ tan ?-2 tan r^ = nl 7. cot r tan rj tan ?'a tan r^ = sin^s. 8. tanR2 = 2£2^i5L^li£2i±f. n 9. tanR3 = ^cosiacosi6^inic_ 10. tanRi:tanR2:tanR3r^cos(S-A):cos(S-B):cos(S-C). 11. cot R cot Rj cot Eg cot R3 = N". 12. tan R cot Rj cot R2 cot R3 = cos" S. i AREAS OF TRIANGLES. 219. Problem. — Tojind the area of a spherical triangle, having given the three angles. Let r = the radius of the sphere. E = the spherical excess = A + B -f C — 180°. K = area of triangle ABC. j-t is shown in Geometry (Art. 738) that the absolute area of a spherical triangle is to that of the surface of the sphere as its spherical excess, in degrees, is to 720°. .-. K:47rt-2=E:720°. ... K = -^-,rr2 180° • • (1) 330 SPUEliWAL TRIGONOMETRY. Cor. The areas of the colunar triangles are (^2A-E) , 1215- E), (2C_-E)^ 180° ^' 180° " ' 180° i! 4 220. Problem. — To jind the area of a tnangle, having given the three sides. Here the object is to express E in terms of the sides. I. CagnolVs Theorem. sin IE = sin ^(A + B + C - tt) = sin ^(A + B) sin ^C — cos ^(A + B) cos ^C = 5^5LK^^-[oosi(a-?/)-cosi(« + 6)] • (Art. 198) cos^c _ sin ^ g sin ^6 sin C _ sin ^ a sin ^b 2n / a .f 1Q''\/'1N cos ^ c cos ^ c sin « sin 6 .-. sin^E = ''^ 2 cos ^a cos ^6 cos ^c II. Lhuilier^s Theorem. ^ cosKA+B + C-tt) (2) ^ sini(A + B)-sinK ^-C) (j^j., ak>. cos^(A4-B) + cos|(7r-C) ^ ' ^ _ sin|(A + B) — cos|C ~ cos ^( A + B) + sin I C cos ^{a — b) — cos ^c cos^C cos ^(a + 6) + cos ^c sin^C , . . . (Art. li)6) . . . . . (Art. 4.5) = Vtan|.stan|(s-a)tan^(.s-6)tan|(s-c) (Art. 195) (3) ^ sin i(s - b) sin ^(s - a) ^^^ ^ ^ COSi^SCOS^(.S — c) ASEAS OF TlilANOLES. 331 221. Problem. — To find the area of a triangle, having given two sides and the included angle. co8^E= cos [^ (A -f- B) - (^TT - :|C)] = cos^( A + B) sin^C + sin^(A -|- B) cos^C = co8^{a + b)fim^C + cosi{a-b)cm^C (Art. 19 + sin|asin^&cosG]sec^c . . (1) Dividing (1) of Art. 220 by this equation, and reducing, we have tanjE- tanjatan^fesinC 1 + tan 4 a tan ^ h cos C ' ' ' \ ) EXAMPLES. 1. Given a = 113° 2' 56".64, b = 82° 39' 28".4, c = 74° 54' 31 ".06; find the area of the triangle, the radius of the sphere being r. By formula (3) of Art. 220, a =113° 2'56".64 6= 82°39'28".40 c= 74° 54' 31 ".06 2s = 270°36'56".10 s = 135°18'28".05 a^. 22°15'31".41, b= 52°38'59".65, s-c= 60°23'56".99. 8 8 K = mm X TT^-^ . = IMMf»'' = area. is = 67°39'14".025 ^(.s--a) = ll° 7'45".705 i(s-6) = 26°19'29".825 |(s-c) = 30°ll'58".495 logtan ^8 = 3860840 log tan ^(.s - a) = y.2938583 log tan l(s -b) = 9.6944058 log tan i (s - c) = 9. 7649261 logtan2^E = 9.1392742 log tan ^E = 9.5696371. iE = 20°21'58".25. E = 81° 27' 53" = 293273". . . [(1) of Art. 219] ill ^i! :■! » 832 SPHERICAL TRIGONOMETRY. 2. Given A = 84° 20' 19", 13 = 27° 22' 40", C = 75°33'; find E = 7°15'59". 3. Given a = 40° 24', 6 = 07° 14', c = 81°12'; find K = 4. Given a =108° 14', 6 = 75° 29', c = 56°37'; find E = 48° 32' 34".5. 5, Prove cos4E = 1 + ""^" + ^ ^^ ^- + ""^^ 4 cos ^ a cos ^ 6 cos ^ c __ cos'^g + cos'^6 -f- cos^^c — 1 _ 2cos^acos^6cos|c 7. 6. « ainiE=v P^^^"^^^~"^^"^^^^~^)^"^^(^~^) \ cosia cosi^cosAc. cos^a cos 1^6 cos^^c. « C03lE=v fc^^^^"^^^^~^^^"^^^^~^^^"^^^^~^^ \ cos^acos^ftcos^c 8. " eottE^: ^"^^^^"^^^'^^^^^ ^ sin C _ eot|6cot^c + C0S A sin A cot^ccot^g-t-cosB sinB EXAMPLES. Prove the following : 1. sin (s — a) + sin (s — &) + sin (s — c) — sin s = 4 sin ^ a sin ^6 sin^c. 2. sin s + sin (s — 6) + sin (s — c) — sin (s — a) = 4sin^a cos^6 cos^c. EXAMPLES. 333 3. sin(s — 6)sin(.s — o) -f siii(.'} — c)8iu(s — a) + sin (a — a)8in(s — 6) + sins sin (s — a) + sill s sin (.9 — b)-\- sin 8 sin (a — c) = sin b sin c + sin c sin a -f sin a sin 6. 4. sin(.«» — 6)sih(s — c) + sin(s — c)sin(.s — a) — sin(s — a)sin(.s — 6) + sins 8in(s— (;) 4- sin 8 yin (.s — b)~ sin s sin (.s — c) = sin 6 sin c + sin c sin a — sin a sin b. 5. sin's 4- 8in*(s — «) + sin-(s — 6) + 8in^(s ~ c) = 2(1 — cos rt cos & cose). 6. sin's + sin''(s — a) — sin'(8 — 6) — sin'(s — c) = 2 cos a sin 6 sine. 7. cos's 4- cos'(s — a) + cos'(s — &) -f cos'(s — e) = 2(1 -}- cos a cos & cose). 8. cos's + cos'(s — a) — cos'(« — 6) — cos'(s — c) = — 2 cos a sin 6 sin e. 9. tan r cot Vi tan rj tan rg = sin' ( s — a) . 10. tan r tan Vi cot rj tan rg = sin' (s — 6) . 11. tan r tan rj tan j'a cot 9-3 = sin'(s — c) . 12. cotr sins = cot^Acot^B cot^C. 13. tan Vi + tan r^ -f- tan r^ — tan r = 4NsinS sin A sin B sin C '" Hj . . , . . ■ 4sin4^asini^6sin4c 14. cot r, + cot r^ + cot ? 3 — cot r = ^ — ^ — 15. tan r, : tan n : tan r. sma sin & sin c 1 4- cos A 1 4- cos B 1 + cos C ^e tanr,+tanr2 4-tan rs— tan 7- i,^ • „^„„ • „^„i. . „..„„\ 16. — ~ = *(14-cosa4-coso4-cosc). cot Vi 4- cot rj 4- cot r^ — cot r :r]^ ■:1 334 SPHERICAL TRIGONOMETRY. n' 17. cotV. + cot'^r, + cot=r3 + cot^r = 2(l-cosacq s6^osc). 18. 19. _J_,_^ 1 !_ sin'^r siirri siu-rj siii^rs 2 cos g sin 6 s i n c »2 cot r-i cot '/'g + cot j'a cot rj + cot i\ cot 7', + cot r(cot r, + cot rg + cot r^ _ sin b sin c + sin o sin a + sin a sin h _ _ 20. tan »*2 tan r^ + tan r^ tan j'l -f tan r, tan rj 4- tan r(tan )\ + tan ro -f tan r^) = sin b sin c + sin c sin a + sin a sin 6. 21. cot R tan R, cot Rg ^ot R., = cos- (S — A) . 22. cot R cot Ri tan R^ cot Kg = cos- (S - li). 23. cot R cot Ri cot Rg tan Rg = cos^ ( S - C) . 24. tan Ri -f tan Rg = cot r + cot r^. 25. tan R, + tan Rj -f tan Rg — tan R = 2 cot r. 26. tan II — tan lli + tan R, + tan Rg = 2 cot i\. 27. tan R + tan Rj — tan Rg -|- tan Rg = 2 cot r^ 28. tan R + tan R, + tan Rj — tan Rg = 2 cot r^. 29. cot Vy + cot Ti 4- cot rg — cot r = 2 tan R. 30. cot r — cot )\ + cot r^ + cot rg = 2 tan Rj. 31. cot )• 4- cot r, — cot r^ -|- cot rg = 2 tan Rj. 32. cot r + cot ri 4- cot rj — cot r^ = 2 tan R,. 33. tan R 4- cot r = tan R, 4- cot r, = etc., = \ (cot r 4- cot r, 4- cot rj 4- cot i\). 34. tan^ R 4- tan* R, 4- tan'^ Kj 4- tan- R., 2(1 4-<'os A cosB cosC) T?'^ 35. 36. EXAMPLES. 335 tan' R + tan' R, + tan" R^ + tan^ R, _ . cot^- + cot2ri + cot=*?-2 + cot'r3 ~ * tan' R + tan' R^ - tan' B^ - tan' R^ _ 2(cos AsinBsinC ) 37. 38. 39. 40. 41. cot' r + cot' I'l — cot' 7-2 — cot' Va = N' 2 cos a sin b sin c ji^" tan' R + tan' R, - tan' Rg - tan' R., _ _ cos A cot' r + cot' Vi — cot' rg - cot' r^ ~ cos a* tan R cot R, = tan ^ 6 tan ^c. (cot r 4- tan R)'-f 1 = /'»i"Cf' + si"^> + sincV (cot r, - tan R)'+ 1 = A'^" ^ + sin o - sin gV^ 42. tan J A sin (s — a) = 2 cos ^ A cos ^B cos \Q 43. 44. 45. 46. 47. 48. tan r _ cos(S - A) cos(S - B) cos(S - C) tanR 2 cos ^ A cos ^B cos ^C cot(s — 6) cot(s — c) + coL(.s — r')cot(.s — a) + cot (.9 — a) cot(s — 6) = cosec'r. cot(8-6)cot(s— c) - cotscot(s— 6) — cot8cot(s-c) = cosec'r,- cot(s— c)cot(«— a) — cot«cot(s— c) — cot«cot(«— a) = cosec'r2. cot(s— a)cot(«— 6) — cot 5 cot (s— a) — cots cot (s — 6) = cosec'^7'3. cot(s — g) cot(jt — 6) cot(« — c) 2 cot a sin'r, sin'rj sin'r, sin'r "s 3cot(» — a) cot(8 — 6)cot(5 — c). 336 8PHERWAL TRIGONOMETRY. Wl- '■ Si V 49. " cosec'ri + cosec'/'a + cosec';'3 — cosec'r = — 2 cot s[cot(s — a) + cot(s — &) + cot(s — c)]. 50. 1 +^+^ + - sin^r sin^ri sin'^7*2 sin''? 3 51. cot R — cot El — cot R2 — cot R3 = — 2 S tan {s — ft) tan {s — h) tan s tan (s — a) tan (s — b) tan (s — c) 2 N cos 8 n 52. sin(A-iE) = - n 2 cos ^a sin ^ftsin^c 53. sin(B-iE) = - n 54. sin(C-^E) 2sin^acos^6sini^c n 2 sin ^ a sin ^6 cos ^c 55. cos(A-:^E) = — --5— -i; — .^. , . / ♦ 2co8^ttsin^o sin J^c 56. cos(B-iE) sin'^c + sin'^^ft— sin^^ft 2sin^acos^6sin ^c 57. co8(C - i E) = '"^^"'^^ + ^^"'i^ - ^"^H ^. 2 sin 1^^ a sin ^ &cos^c 58. cot(A-^E) = "^H^tani&-cosC ^ ^ ^ sinC _ t an ^ 6 tan ^ c + cos A _ tan ^ a cot 1 6 — cos B sin A sin B 59. tan^(A— ^E) = Vcot^scotJ^(s— a)tan^(8— 6)tan^(s— c). 60. tan^(B — .J^E) = Vcotistan^(8— a)cot|(,s— 6)tanJ^(.s— c). 61. tan^(C— J^E)= Vcot^«tan|(«— a)tan|(s— 6)cot^(s— c). 62. If S, Si, Sj, S^denotc the sums of the angles of a triangle and its three colunars, prove that S + S, + Sj + Sg = 3 tt. 63. In an equilateral triangle, tan R = 2 tanr. EX A HPLES. 337 64. If El, Ej, Eg denote the spherical excesses of the colunars on a, h, c, respectively, show that E + Ei + Eg + E3 = 27r ; and therefore the sum of the areas of any triangle and its colunars is half the area of the sphere. 66. Given a = 108° 14', 6 = 75° 29', c = 56°37'; find E = 48° 32' 34".5. 66. Given a = 63° 54', & = 47°18', c = 53°26'; find E = 24° 29' 49^". 67. Given a = 69° 15' 6", 6 = 120° 42' 47", c = 159° 18' 33"; find E = 216° 40' 23". 68. Given a = 33°i'45", & = 155° 5' 18", C = 110°10'; find E = 133° 48' 55". 69. Given a=.l=c = 1", on the earth's surface ; find E = 27".21. 70. Given a = b = c = 60°, on a sphere of 6 inches radius ; find the area of the triangle. Ans. 19.845 square inches. 71. If a=6=^, and c=^, prove sin^E=^, and cosE=|. 72. If C = ^, prove sin ^ E = sin ^ a sin ^ 6 sec ^ c, and ill cos^^E = cos^acos^fi sec^c. 73. If a = 6, and C = -, prove tan E = ^ tan a sec a. 74. If A4-B4-C = 27r, prove cos''^a4-cos''^6 + cos2^c=l. 75. If a + 6 = TT, prove that E = C ; and if E' denote the spherical excess of the polar triangle, prove that sin^E' = sinaco8^C. ; 37r. nc -o -211? VsmTEsinlEiSinAEsSin^Ea 76. Prove 8mHE = ' ^ . / , ., W ' cot \ a cot ^ cot i c 838 SPHERICAL TRIG ON O METR Y. Uy\ CHAPTER XIII. APPLIOATIONS OF SPHEEIOAL TRIGONOMETEY. SPHERICAL ASTRONOMY. ■t i * 222. Astronomical Definitions, The celestial S2)here is the imaginary concave surface of the visible lieavens in which all the heavenly bodies appear to be situated. The sensible horizon of a place is the circle in which a plane tangent to the earth's surface at the place meets the celestial sphere. The rational horizon is the great circle in which a plane through the centre of the earth parallel to the sensible horizon meets the celestial sphere. I^ecause the radius of the celestial sphere is so great, in comparison with the radius of the earth, these two horizons will sensibly coincide, and form a great circle called the celestial hori- zon. The zenith of a place is that pole of the horizon which is exactly overhead ; the other pole of the horizon directly ur.derneath is called the nadir. Vertical circles are great circles passing through the zenith and nadir. The two principal vertical circles are the celestial meridian and the prime vertical. The celestial meridian of a place is the great circle in which the plane of the terrestrial meridian meets the celestial sphere ; the points in which it cuts the horizon are called the north and south points. SPHERICAL ASTRONOMY. 339 The prime vertical is the vertical circle which is per- pendicular to the meridian ; the points in which it cuts the horizon are called the east and ivest points. The axis of the earth or of the celestial sphere is the imaginary line about which the earth rotates. The celestial equator, or equinoctial, is the great circle in which the plane of the earth's equator intersects the celestial sphere. The poles of the equinoctial are the points in which the axis pierces the celestial sphere. Hour circles, or circles of declinathn, are great circles passing through the poles of the equinoctial. The ecliptic is a great circle of the celestial sphere, and the apparent path of the sun due to the real motion of the earth round the sun. The equinoxes are the points in which the ecliptic cuts the equinoctial. There are two, called the vernal and the autumnal equinox, which the sun passes on March 20 and September 22. The obliquity of the ecliptic is the angle between the planes of the ecliptic and equator, and is about 23° 27'. Circles of latitude are great circles passing through the poles of the ecliptic. which irectly 223. Spherical Coordinates. — The position of a point on the celestial sphere may be denoted by any one of three systems. In each system two great circles are taken as standards of reference, and the point is determined by means of these circles, which are called its spherical, coordinates, as follows: I. The horizon and the celestial meridian of the place. The azimuth of a star is the arc of the horizon inter- cepted between the south point and the vertical circle if ^t > i& 340 SPHERICAL TRIGONOMETRY. passing through the star; it is generally reckoned from the south point of the horizon round by the west, from 0" to 360". The altitude of a star is its angular distance above the horizon, measured on a vertical circle. The complement of the altitude is called the zenith distance. II. Tlie equinoctial and the hour circle through the vernal equinox. The right ascension of a star is the arc of the equinoctial included between the vernal equinox and the hour circle passing through the star; it is reckoned eastward from 0° to 360°, or from O** to 24\ The angle at the pole between the hour circle of the star and the meridian of the place is called the hour angle of the star. The declination of a star is its distance from the equinoc- tial, measured on its hour circle ; it may be north or south, and is usually reckoned from 0° to 90°. It corresponds to terrestrial latitude. The poh " distance of a star is its distance from the pole, and is the complement of its declination. The right ascen- sion and declination of celestial bodies are given in nautical almanacs. III. The ecliptic and the circle of latitude through the vernal equinox. The latitude of a star is its angular distance from the ecliptic measured on a circle of latitude ; it may be north or south, and is reckoned from 0° to 90°. The longitude of a star is the arc of the ecliptic inter- cepted between the vernal equinox and the circle of latitude passing through the star. ] )ernal the north inter- ,titude 8PBEHICAL COORDINATES. 341 ftt» 2.^4. Qraphic Representation of the Spherical Coordi- nates. — The figure will serve to illustrate the pre- ceding definitions. is the earth, PHP'R is the meridian, P the north pole, HR the horizon, EQ the equinoctial, Z the zenith. Then, of a place whose ^nith is Z, QZ is the ter- strial latitude ; and since QZ = PE, .*. PR = the latitude. But PR is the elevation of the pole above the horizon. Hence the elevation of the pole above the horizon is equal to the latitude. Let V be the vernal equinox, and let S be any heavenly body, such as the sun or a star ; then its position is denoted as follows : ^©V-^1 A\ VK = right ascension of the t )od3 f = a, KS = declination " (I « = 8, ZPS or QK = hour angle « u (( = t, PS = north polar distance " (( (( =i>» HT = azimuth « « i( = a, TS = altitude ' « « u = A, ZS ~ zenith distance " « ii = 2, QZ H = latitude of the observe ;r ^4,. The triangle ZPS is called the astronomical triangle; ZP = 90° — <^ = co-latitude of the observer, PS = 90°-8, SZ = 90°-A. :i:' !■, 342 SPHF.niCA L TlilGONOMETIt Y. 7 Let the small circle MM', passing througli S, and parallel to the equinoctial, represent the apparent diurnal iiiotion of the heavenly body S (the declination being supposed con- stant); then the body S will appear to rise at A (if we sup- pose the Eastern heiulsphere is represented in the diagram). It will be at B at 6 o'clock in the morning, at M at noon, at M' at midnight, and at <» it will be east. 225. Problems. — By means of the foregoing definitions and diagram we may solve several astronomical problems of an elementary characster as follows : (1) Given the IrUtude of a place and the declination of a star; to find the time of its rising. Let A be the position of the star in the horizon. Then in the triangle APB, right angled at It, we liave cos RPA = - cos ZPA = tan KP cot AP. .*. cos t = — tan <^ tan S . . from which the hour angle is found. (1) Since the hourly rate at which a heavenly body appears to move from east to west is 15", if the hour angle be divided by 15 the time will be found. In the case of the sun, formula (1) gives the time from sunrise to noon, and hence the length of the day. Ex. Required the apparent time of sunrise at a place whose latitude is 40° 3G' 23".9, on July 4, 1881, when the sun's declination is 22° 52' 1". ' <^ = 40°36'23".9, 8 = 22° 52' 1". log tan <^ = 9.0.331352 log tan 8 = 9.6250362 log cos t = 9.5581714- .-. t =111° 11' 44" _ 71. 24"' 47», nearly, t \ (1) PROBLEMS OF SPIIEHWAL AnTllONOMY. o48 which taken from 12'', the time of apparent noon, gives 4'' 35"' 13", the time of apparent sunrise.* (2) Oiven the latitude of a place and the declination of a star; to find its azimuth from the north at rising. Let A = the azimuth = All. Then in the triangle APR we have sin AP = cos AK cos PR, or sin 8 = cos A cos . .*. cos A=sindsec<^ (2) Ex. Required the hour angle and azimuth of Arcturus when it rises to an observer in New York, lat. 40° 42' N., the declination being 19° 57' N. Ans. 7" 12"" 46'.3 ; N. G3° 15' 11" E; (3) Oiven the latitude of the observer and the hour angle and declination of a star; to find its azimuth and altitude. Here we have given, in the triangle ZPS, two sides and the included angle; that is, PZ = 90°-) cos ^ t, sm^{p — A) sin ^« = sin ^(8 — <^)cos^f, cos^(2)-f- A) cos -^2 = sin 1(8 + ^)sin^^, cos^(2> — A) sin ^z = cos|(8 + ^) sin^f. Hence, when <^, t, and 8 are given, that is, the latitude of the place, and the hour angle and the declination of a heavenly body. A, z, and p can be found. In a similar manner may be solved the converse problem : Given the latitude of the observer and the azimuth and altitude of a star; to find its hour angle and declination. * In these examplea no corrections arc applied tdt refraction, semi-diameter of the sun, change in declination from noon, etc. 44 SPHERICAL TRIGONOMETRY. ^4) Given the right ascensioiis and declinations of two stars; to find the distance between them. Let P be the pole, S and S' the two stars, p^a-ar Let « and «' be the right ascensions of the stars ; 8 and 8' their declinations ; and d the required distance. Then we have given, in the triangle PSS', two sides and the included angle ; that is, PS = 90°-8=i), PS' = 90°-8'=i)', and P = « - a'. This may be solved by Art. 198, or by the second method of Art. 209, as follows : Draw SD perpendicular to PS' produced; let PD Then cosP = tanPDcotPS. m. .*. tan m = cos P tan p. Also cos SS' = cos PS cos S'D sec PD .•. cos d = cosp cos S'D sec m. (Art. 209) Ex. Required the distance between Sirius and Aldebaran, the right ascensions being 6" 38"" 37'.6 and 4'' 27"" 25".9, and the declinations 16° 31' 2" S. and 16° 12' 27" N., respectively. Here P = 2'' 11« 11'.7 = 32° 47' 65" j9 = 106°31' 2" m = 109° 25' 55" p'= 73° 47' 33" S'D= 35° 38' 22" j> = 106°3r 2", m = 109° 25155", SS' = d= 46° 0'44". log cos P = 9.9245789 log tan p = 0.5279161 - log tan m = 0.4524950- .-. m = 109° 25' 55". log cos S'D = 9.9099302 log cos p^- 9. 4537823- colog cos m = 0.4779643 - log cos cZ = 9.8416768 PROBLEMS OF SPHERICAL ASTRONOMY. 346 \> (5) Given the right ascension and declination of a star; to find its latitude and longitude. Let V be the vernal equinox, S the star, VD, VL the the equator and the ecliptic, SI), SL perpendicular to VD, VL. Then VD = right ascension = «, SD = dec- lination = 8, VL = longitude = X, SL = latitude = I. Denote the ob- liquity of the ecliptic DVL by w, and the angle DVS by 0. From the right triangles SVD, SVL we get cot ^ = sin rt cot 8 (1) tanX = cos (^ — o)) tanasec^ (2) sin I = sin {6 — w) sin 8 cosec 6 (3) From (1), $ is determined; and from (2) and (3), X and / are determined. Ex. Given the right ascension of a star 5'* 6"" 42'.01, and its declination 45° 51' 20". 1 N. ; to find its longitude and latitude, the obliquity of the ecliptic being 23° 27' 19". 45. « = 76°40'30".15 8 = 45° 51' 20".l ^ = 46° 38' 11 ".8 23° 27' 19".45 tf-a) = 23°10'52".35 A = 79° 68' 3".44. Z=22°51'48".4. log sin «= 9.9881479 log cot 8 = 9.9870277 log cot d = 9.9751756 log cos (^ - o>) = 9.9634401 log tan a = 10.6255266 cologcos^= 0.1632816 log tan\ = 10.7522483 log sin (^ - 0)) = 9.5950996 log sin 8 = 9.8558743 cologsin^= 0.1384575 log sin/ = 9.5894314 346 SPIJElilCA L TltlOONOMETlt V. EXAMPLES. 1. Find the apparent time of sunrise at a place whose latitude is 40" 42', when the sun's declination is 17° 49' N. Ana. 4^ CC". 2. Given the latitude of a place = 40° 36' 23".9, the hour angle of a star=4()°40'4".o, and its declination = 2.'{° 4' 24".3 ; to lind its azimuth and altitude. An8. Azimuth = 80° 23' 4".47, altitude = 47° 15' 18".3. 3. Find the altitude and azimuth of a star to an observer in latitude 38° 53' N., when the hour angle of the star is 3h 15... 20« w., and the declination is 12° 42' N. Ana. Altitude = 39° 38' 0"; azimuth = S. 72° 28' 14" W. 4. Given the latitudes of New York City and Liverpool 40° 42' 44" N. and 53° 25' N., respectively, and their longi- tudes 74°0'24"W. and 3° W., respectively; to find the shortest distance on the earth's surface between them in miles, considering the earth as a perfect sphere whose radius is 3950 miles. NoTB. — This ia evidently a case of (4) where two ■ides and the included angle are given, to tind the third aide. Ana. 3305 miles. 6. The latitudes of Paris and Pekin are 48° 50' 14" N. and 39° 54' 13" N., and their difference of longitude is 114° 7' 30"; find the distance between them in degrees. ^ Ana. 78° 56' 40". GEODESY. 226. The Chord al Triangle. — Given ttvo aidea and the included angle of a apherical triangle; to find the correspond- ing angle of the chordal triangle. GEODESY. 847 14" N. ade is s. ^ ;G'40". The chordal triangle is the triiiuglo formed by the chords of tlie sides of a spherical triangle. Let ABC be a spherical triangle, the centre of the sphere, A'BC the cohinar triangle, and M, N the middle points of the ares A'B, A'C. Then the chord AB is parallel to the radius ()M, since they are both perpendicular to the chord A'P>. Simi- larly, AC is parallel to ON. In the spherical triangle A'MN, we have cosMNsscosA'NcosA'M + sinA'NsiuA'McosA' (Art. 191) Denote the angle BAC of the chordal triangle by A^ Then arc MN or angle MON = A„ A'N = ^(7r-6), A'M = Ht - c), and A' = A. .'. cos A, = sin^i sin|^c-f- cos^6 cosl^ccos A . (1) with similar values for cos B^ and cos C^. Cor. 1. If the sides h and c are small compared with the radius of the sphere, Ai will not differ much from A. Let Ai = A - ^ ; then cos A, = cos A + ^ sin A, nearly. But sin ^6 sin^c = sin^|(6 -f c) — sin'^ft — c), and cos ^Z> cos ^c = (iOii^\{h + f, — sin'-* \{h — c). Substituting in (1) and reducing, we get ■e = tan|Asin2|(i>4-c)-cot^A8in''^(&-c) . (2) which is the circular measure of the excess of an angle of the spherical triangle over the corresponding angle of the chordal triangle. The value in seconds is obtained by dividing tlie circular measure by the circular measure of one se(H)nd, or, approxi- mately, by the sine of one second. I H 848 SPHERICAL TRWONOMKTJiY. Coi'. 2. The anijlex of the chordal triangle are, reftpectively, eqnai to the arcs joining the middle points of the sides of the colunar triang^^s. 22n. Legendre's Theorem. — If the sidps of a spherical triangle be small compared with the railius of the sphere, then each angle of the spherical triangle exceeds by one-third of the spherical excess the corresponding angle of the plane triangle, the sides of which are of the same length as the arcs of the spherical triangle. Let a, b, c be the lengths of the sides of the spherical triangle, and r the radius of the sphere; then the circular measures * of the sides are respectively -, - , — Hence, '/• r r uegleci.ing powers of - above the fourth, r C08- — r K ■ cos COS r r sin b . c -sin- r r C08A = ; (Art. 191) Y pT^ pr \^^^- ^^^) be. bc\ 2»-» 24<«-- A ^^ J \ 2bc.r 246a-'. A *>»*/ = y + c'-q' _ 26«c' + 2c'a- 4-2a'ft«-a«- 6«- c« 26c 246cr» (1) *The u 1 ? is tli>> "'rrular meaaiire of the angle wbicb tbe arc u aubU ida al tb« r he centre of tuc scbere; and aiaiilarly fur - nud .. r r GEODESY. 849 lively, of the kerical e, then of the iangle, of the herical ircular Hence, t. 191) rt.l56) . (1) t( idi »l the Now if A', B', C denote the angles of the plane triangle whose sides are a, b, c, respectively, we have cosA' = ^li^' .... (Art.%) and .;»» A > - 2&V 4- 2c'a« •:■ 2a«ft' -a^-b*-c* 46V (Art. 100) Therefore (1) becomes cosA==cosA'-^l^A' .... (2) Let A =. A' 4- ^, where is a very small quantity ; then cos A = (!08 X' — 6 sin A', neprly. . /J 6c sin A' A • ^'^~~7n^'^3? (Art. 101) where A denotes the area of the plane triangle whose sides are a, 6, c. We have therefore A = A' + A 3r» A Similarly 13=:H' + A, C = C'4-A. .-. A + Ii-fC-A' ^B'-C' = -; or A-fB + C — ir=— = spherical excess (Art. 210) .-. A — A'=B — B'=C— ^'= --^= J spherical excess. Cor. 1. If the sides of a spherical triangle be very small compared xnth the radius of the sphere, the area of the spherical triangle is approximoMy equal to aA4-«^-+A^ + '^. \ 24r« ; i 350 SPHERICAL TliiaONOMETliY. For, tau \ E = x/tau — tan -— tan '^ tan '^^^ (Art. 220) ' ^ \ 2r 2r 2r 2r ^ ' and tan A = ^[1+ «n. tan^=^ = ^-«ri+ (^^1 etc. 2r 2rL 12j'2j' 2r 2r [ lii''' J (Art. 156) * \2r 2r 2/ 2r [ 12r'JL 12r J .-. iE = A /i + Z±llZliLt±l^ZliOl±Vii£): (Arts. 101 and 156) \i\ Vlr' J 4i\ 24:7^ J ... E,- = Afl+^^Y That is : ^/le area of the sphei'kal triangle exceeds the area a^ 4. lyi 4. c- 0/ f /ie j?i(t/ie triangle by — ^>l~? — i^^'*^ ''Z ^'^^ Ztt^^er. 24 7- 1 Cor. 2. If we omit terms of the second degree in -, we have • r E».2^^. Hence, if the sides of a spherical triangle be very small compared tmth the radius of the sphere, its area is approxi- mately equal to the area of the plane triangle having sides of the same length. 228. Roy's Rule. — The are 'j ^f a spherical triangle .;n the Earth's surf((ce being known, to establish a formula for com- puting t' 8 spherical excess in seconds. Let A be the area of the triangle in square feet, a id n the number of seconds in the spherical excess. Then we have ROY'S RULE. 351 A = 180° X 60 X 60 ni* .... (Art. 210) 180x60x60 206265 ^^^ Now, the length. of a degree on tlie Earth's surface is found by actual measurement to be 365155 feet. ... ^=365155. .,,^180x365155. 180° TT Substituting this value of r in (1), and reducing, we get log n = log A -9.3267737 (2) This formula is called General Roy's rule, as it was used by him in the Trigonometric Survey of the British Isles. He gave it in the following form : From the logarithm of the area of the triangle^ taken as a j^lane triangle, in .square feet, subtract the constant logarithm 9.3267737; and the re- mainder is the logarithm of the excess above 180°, in seconds, nearly. Ex. If the observed angles of a spherical triangle .are 42° 2' 32", 67° 55' 39", 70°1'48", and the side opposite the angle A is 27404.2 feet, required the number of seconds in the sum of the errors made in observing the three angles. Here the ajyparent spherical excess is A + B + C - 180° = - 1". The area of the triangle is calculated from the expression a^ sin B sin 2 sin A (Art. 101) and by Roy's Rule the computed spherical excess is found to be .23". Now since the computed spherical excess may be suppose _i to be the real spherical excess, the sum of the obser/ed angles ought to have been 180° -f- .23". Hence it appears thnt the sum of the errors of the obser- vations is .23" — (— 1") = 1".23, which the observer must m ! I «^AM*MUM»ie 352 SPHERICAL TRIGONOMETRY. add to the thiee observed, angles, in such proportions as his judgment may direct. One way is to increase each of the observed angles by one-third of 1".23, and take the angles thus corrected for the true* angles. •JV 229. Reduction of an Ang^le to the Horizon. — Given the angles of elevation or dejrreKsion ofttvo objects, which are at a small angular distance from the horizon, and the angle which the objects subtend, to find the horizontal angle between them. Let a, b be the two objects, the angular distance between which is ireasured by an observer at ; let ()Z be the direction at right angles to the observer's horizon. De- scribe a sphere round as a centre, and let vertical planes through Oa, 06, meet the horizon at OA, OB, re- spectively ; then the horizontal angle AOB, or AB, is required. Let ab = 0, AI> = ^ -f a;. Art = h, B6 = k. Then in the triangle aZb we have cos AB = cos aZb = cos ab — cos aZ cos 6Z or ,a, V cos^ COS {$ + a;) = - sin aZ sin 6Z s'lnh sink cos h cos k This gives the exact value of AB ; by approximation we obtain, where x is essentially small, y, . /I cos 6 — hk .*. a; sin ^ = hk — | {h^ + fc*) cos $, nearly. e 2 hk - {h' + k') ( cos^ ;^ - sin2 x = 2sin0 = \i{h 4 A;)' tan ^ e - {h - ky cot^ ^]. SMALL VAUIATlOys IN PA UTS OF TRIANGLES. 353 1 EXAMPLES. 1. Prove that the angles subtended by the sides of a spherio'.l triangle at the pole of us circunicircle are respec- tively double the corresponding angles of its chordal tri- angle. 2. If Ai, B|, Ci ; Aj}, B^ ^2 > A.,, B3, C;j ; be the angles of the chordal triangles of the colunars, prove that , co8Aj=co8ja8inS, co8B, = 8in J/>8iu(iS— C), co8C,=8liUc8in(S— B), C08A,=8iiUa8m(S — C),co8B,=co8.j68iaS, co8C,=8in Jc8in(S- A), co8A3=siniasin(S— B), co8B3=8inA^8in(S— A),co8C3=:co8ic8inS. 3. Prove Legendre's Theorem from either of the formulae for sin ^ A, cos ^ A, tan ^ A, respectively, in terras of the sides. 4. If C = A -f B, prove cos C = — tan ^ a tan ^ 6. 230. Small Variations in the Farts of a Spherical Tri- angle. It is sometimes important in Geodesy and Astronomy to determine the error introduced into one of the computed parts of a triangle from any small error in the given parts. If two parts of a spherical triangle remain constant, to deter- mine the relation between the small variations of any other two parts. Suppose C and c to remain constant. (1) Required the relation between the small variations of a side and the opposite angle (a, A). Take the equation sin A sine = sin C sin a (1) We suppose a and A to receive very small increments da and dA; then we require the ratio of da and dA when bo'h are extremely small. Thus sin (A 4- dA) sin c = sin C sin (a + da), IM^4 864 SPHERICAL TRIGONOMETRY. or (sin A cos dA + cos A sin t/A) sin c = sin C (sin a cos iki -f cos a sin da) (2) Because the arcs dk. and da are extremely small, tlieir sines are equal to the arcs themselves and their cosines ccjual 1 : therefore (2) may be written sin A sin c -f cos A sin cdX = sin C sin a 4- sin C cos ada (3) Subtracting (1) from (3), we have cos A sin cdA = sin C cos ada. da _ c os A sin c _ tan a dA. sin C cos a tan A (2) Required the relation between the small variations of the other sides («, b). We have cos c = c»s a cos 6 + sin a sin 6 cos C (1) .*. co8c = cos(a+da)cos(6+(/6) +sin(a+— cosfeco sC) ,. sin a sin b n cot B sin C da , cotAsinCdft ,» , ^nox = : 1 ; .... (Art. I'Jo) Sin a sm b da__ cos A db cos B (',<) Required the relation between the small variations of the other angles (A, B). SMALL VARIATIONS IN PARTS OF TRIANGLES. 365 . (2) I, their cosines a (3) nations . (1) lb)cos C, D. (2) product !0s C) dh, A.rt. 193) variations By means of the polar triangle, we may deduce from the result just found, that dB cos^ COS b (4) Required the relation between the small variations of a side ami the adjacent angle {b, A). We have cot c sin 6 = cot C sin A + cos 6 cos A . . . (Art. 103) Giving to b and A very small increments, and subtracting, as before, we get cote cos 6r7/> = cot C cos Ad A — sinftcos Art6 — cosftsin AdA. (cote cos b 4- sin b cos \)db== (cot C cos A — cos b sin A)dX. cos a ,, cos B , .'. — : — db = da sine sinC db cos B sin 6 dA cos a sin B EXAMPLES (Arts. 191 and 192) sin 6cotB cos a 1. If A and c are constant, prove the following relations between the small variations of any two parts of the other elements : da dC dh dC tana tanC tana sma db dB sin C dC dB cos a. sinC ' 2. If B and C remain constant, prove the following: db tan b dA do tan c ' dA • A i. I. — — = sin A tan b ; do da da do = sin B sin C. sma sin c cos b I 356 SPHERICAL TRIGONOMETRY. POLYEDRONS. 231. To find the Inclination of Two Adjacent Faces of a Eegnlar Folyedron. Let C and D be the centres of the cir- cles inscribed in the two adjacent faces whose common edge is AB ; bisect AB in E, and join CE and DE ; CE and DE will be perpendicular to A]i. .•. Z CED is the inclination of the two adjacent faces, which denote by I. In the plane CED draw CO and DO at right angles to CE and DE, respectively, and meeting in 0. Join OA, OE, OB, and from as centre describe a sphere, cutting OA, OC, OE at a, c, e, respec- tively ; then ace is a spherical triangle. Since AB is per- pendicular to CE and I'^E, it is perpendicular to the j)lane CED; therefore the plane AOB, in which AB lies, is per- pendicular to the plane CED. .•. Z aec is a right angle. Let m be the number Ox sides in each face, and n the number of plane angles in each solid angle. Then Zace=ZACE = -?^ = -51. 2m m and /.cae = ^Z of the planes OAC and OAD. .*. Z cae = 2Tr 2n IT n In the right triangle cae we have cos cae = cos ce sin ace. But ^os ce = cos COE = cos ( ^ — - IT • T • ir .•. cos- = sm-sin— n 2 m .'. 8in- = co8-cosec— • 2 n m = sin -• VOLUME OF A PAIiAlLELOPlPED. 357 Cor. 1. Ifrbe the radius of the inscribed sphere, and a be a side of one of the faces, then r = -cot— tan — 2 m 2 For, r = OC=CEtanCEO=AEcotACEtanOEO = „ cot tan — 2 m 2 Cor. 2. IfRbe the circumradius of the polyedron, then R = - tan-tan • 2 n 2 For, r = C) A cos aoc = K cot eca cot eac = R cot "^ cot - • »u n .'. R s= - tan- tan-- 2 2 Cor. 3. The surface of a regtdar polyedron, F being the number of faces, — '^^i^i-i. cot — • 4 m For, the area of one face = " w cot—- .-. etc. 4 m Cor. 4. The volume of a regular polyedron ma'rF . rr ^ cot— • 12 m For, the volume of the pyramid which has one face of the polyedron for base and O for vertex ~o*~T~^ot — - .'. etc. o 4 m 232. Volume of a Parallelopiped. — To find the volume of a parallelopiped in terms of its edges and their incHnations to one another. 358 SPHERICAL THWONOMETRY. Q Let the edges be OA =■ a, OB =6, OC = c, and let the inclinations l)e BOC = «, COA = /3, AOB = y. Draw CH perpendicular to the face AOBE. Describe a Sphere round O as centre, meeting OA, OB, 00, OE, in a, b, c, e, respectively. The volume of the parallelo- piped is equal to the area of tlie base OAEB multiplied by the altitude CH ; that is, volume = ab sin y • CH = abc sin y sin ce where ce is the perpendicular arc from c on ab. .'. volume = a6c sin y sin etc sin 6ac . . . (Art. 186) 2n = a6c sin y sin ^ sin /3 sin y (Art. 195) = abc Vl — cos' « — cos" /8 — (los" y -H 2 008 a cos ft cos y. Cor. 1. The surface of a paralklopiped — 2 {be sin « 4- ca sin /3 + at sin y) . Cor. 2. The volume of a tetraedron = ^ abc Vl — cos' a — cos* 0— cos' y + 2 cos « cos ft cos y. For, a tetraedron is one-sixth of a parallelopiped which has the same altitude and its base double that of the tetraedron. 233. Diagonal of a Parallelopiped. — To find the diagonal of a parallelopiped in terms of its edges, and their mutual inclinaiions. Let OD (figure of Art. 232) be a parallelopiped, whose edges OA = a, OB = b, OC = c, and their inclinations BOC = «, COA = ^, AOB = y ; let OD be the diagonal required. TABLE OF FORMULA. 859 and OE the diagonal of tho face OAB. Then the triangle OKD gives OD* « UE' -f ED* 4- 2 OE . ED cos COE « rt» -f &» -H 2 o6 cos y 4- e' -\-2c' OE cos COE (1) Now, it is clear that OE cos COE is the projection of OE on the line OC, and therefore it mnst l)e equal to the sum of the projections of OB and BE (or of OB and OA), on the same line.* .-. OE cos COE = b cos « -|- a cos /8, which in (1) gives OD' = a' + Ir 4- c* 4- 26c cos« 4- 2ca cos ft 4- 2a6 cos y . (2) 234. Table of FormulsB in Spherical Trigonometry. — For the convenience of the student, uiany of the preceding formulae are summed up in the following table : 1. cose —cos a cos 6 (Art. 185) 2. sin 6 = sin B sine. 3. sin a = sin A sine. 4. cosC = -cos AcosB (Art. 189) 5. sin B = sin b sin C. 6. sin A = sin a sin C. - sin a ^\n b sin c (Art. 190) (Art. 191) sin A [sin B sin C 8. cos a =eos 6 cose 4- sin 6 sine cos A . 9. cost = cos c cos a 4- sine sin a cos B. 10. cos c = cos a cos 6 4- sin a sin b cos C. 11. cos A = —cos B cos 4- sin B sin C cos a (Art. 192) 12. cos B = — cos C cos A 4- sin C sin A cos b. * From the nature of projecUona (Plane and Solid Qeoiii., Art. 3'26). I -,"?>. 1% i>>> ^^^. IMAGE EVALUATION TEST TARGET (MT-3) 1.0 I.I 1.25 IIIIIM |45 Ik tii Ui M 12.0 R 1116 il V <^ /i c^: ^ :> ? Photographic Sciences Corporation 23 WEST MAIN STREET WEBSTER, N.Y. 14580 (716) 872-4503 <^ f\ iV \\ ■4"^ O ^ n.^ "l-(S-rMc5?.(S:rC). ^ \ sin B sin C / il rt. 193) .rt. 194) \.it. 195) s — c) Art. 196) TABLE OF FORMULA. 361 32. tan i«=V- CO S 8 cos ( S — A) cos (S - B) cos (S - C) 33 sin a = 2 V-co38cos(8-A) co s(S-^)cos(S-0) sin B sin C 34. tani(A + B)^52iM^L=l|lcot|C . . (Art. 197) • cos ^ (a + 6) ^ 35. tanKA-B) = ^^"^^"~ -^cot^C. ^^ ^ sin^(a + 6) ^ 36. tan A (a + 6) = -— ^^^ ~^U an Ac. ^^ ^ cos^A + B) ^ 37. tan Ha -6) ^ gl!li(A - B) ^^^ . 38. sinHA+B)oos^c = cos|(a-6)cos|C (Art. 198) 39. sin ^A — B) sin ^c= sin ^{a — b) cos|C. 40. cos ^A 4- B) cos ^c = cos ^(a + 6) sin ^C. 41. cos^A — B) sin |c = sin |(a + 6) sin JC. 42. tan r Vsin (s — a)^in (s — ft) sin (g — c) sins ?i sins (Art. 215) 43. tanR = -— [sin(s— a)+sin(.s— 6)+sin(s— c)— sins] (Art. 217) E 44. K = areaof A = -^7rr2 (Art. 219) loO 45. sin^E n 2 COB ^ a cos ^ & cos ^ c . . . (Art. 220) 46. tan | E = Vtanistan^(s— a) tan^ (••* — &) tan^s— c). 362 SPHERICAL TRIGONOMETRY. EXAMPLES. 1. Find the time of sunrise at a place whose latitude is 42° 33' N., when the sun's declination is 13° 28' N. Alls. 5'^9"'13'. 2. Find the time of sunset at Cincinnati, lat. 39° 6' N., when the sun's declination is 15° 56' S.* Ans. 5"* 6"". 3. Find the time of sunrise at lat. 40° 43' 48" N., in the longest day in the year, the sun's greatest declination being 23° 27' K Ans. 4»> 32™ 16'.4. 4. Find the time of sunrise at Boston, lat. 42°21'N,, when the sun's declination is 8° 47' S. Ans. &" 14°'. 5. Find the length of the longest day at lat. 42° 16' 48".3 N., the sun's greatest declination being 23° 27' N. Ans. IS'' 5" 50*. 6. Find the length of the shortest day at New Bruns- wick, N.J., lat. 40° 29' 52".7 N., the sun's greatest declina- tion being 23° 27' S. 7. Find the hour angle and azimuth of Antares, declina- tion 26° 6' 8., when it sets to an observer at Philadelphia, lat. 39° 57' N. Ans. 4" 23"' 5'.7 ; S. 54° 58' 44" W. 8. Find the hour angle and azimuth of the Nebula of Andromeda, declination 40° 35' N., when it rises to an ob- server at New Brunswick, N. J., lat. 40° 29' 52".7 N. 9. Find the azimuth and altitude of Regulus, declination 16° 13' N., to an observer at New York, lat. 40° 42' N., when the star is three hours east of the meridian. Ans. Azimuth = S. 71° 12' 30" E. ; Altitude = 44° 10' 33". 10. Find the azimuth and altitude of Fomalhaut, dec- lination 30° 25' S., to an observer in lat. 42° 22' N., when the star is 2'' 5"" 36' east of the meridian. Ans. Azimuth = S. 27° 18' 40" E. ; Altitude = 11° 41' 37". EXAMPLES. 363 13' * ^ 11. Find the azimuth and altitude of a star to an observer in lat. 39° 57' N., when the hour angle of the star is 5h ijm 40S east, and the declination is 62° 33' N. Ans. Azimuth = N. 35° 54' E. ; Altitude = 30° 24'. 12. Find the hour angle {t) and declination (S) of a star to an observer in lat. 40° 36' 23",9 N., when the azimuth of the star is 80° 23' 4".47, and the altitude is 47° 15' 18".3. Ans. « = 46° 40' 4".53 ; 8 = 23° 4' 24".33. 13. Find the distance between Regulus and Antares, the right ascensions being 10" 0'" 29'.11 and 16" 20'» 20».35, and the polar distances 77° 18' 41".4 and 116° 5' 55".5. Ans. 99°55'44".9. 14. Find the distance between the sun and moon when the right ascensions are 12" 39'" 3».22 and 6" 55"* 32'. 73, and the declinations 9° 23' 16".7 S. and 22° 50' 21".9 N. Ans. 89°52'55".5. 15. Find the shortest distance on the earth's surface, in miles, from New York, lat. 40° 42' 44" N., long. 74° 0' 24" W., to San Francisco, lat. 37° 48' N., long. 122° 23' W. Ans. 2562 miles. 16. Find the shortest distance on the earth's surface from San Francisco, lat. 37°48'N., long. 122° 23' W., to Port Jackson, lat. 33° 51' S., long. 151° 19' E. Ans. ; 6444 nautical miles. 17. Given the right ascension of a star 10" l" 9'.34, and its declination 12° 37' 36".8 N. ; to find its latitude and longi- tude, the obliquity of the ecliptic being 23° 27' 19".45. Ans. Latitude =s ; Longitude = , 18. Given the obliquity of thei ecliptic w, and the sun's longitude A; to find his right ascension a and decliiiation 8. Ans. tan a = cos — sin 60° tan 75°. 26. Given the sun's declination and the latitude ; show how to find the time when he is due east. 27. If the sun rise northeast in latitude <^, prove that cot hour angle at sunrise = — sin . 28. Given the latitudes and longitudes of two places ; find the sun's declination when he is on the horizon of both at the same instant. 29. Given the sun's declination 8, his altitude h at 6 o'clock, and his altitude h' when due east; prove sin'^ 8 = sin h sin h'. EXAMPLES. 365 18".5, and ascension 5' 33".9 N. '., and the his right 56° T 6".5. )6, and the longitude 58' 34".4 S. >, and his sion (a). ^ 45"' 14».7. a.t noon of sun's dec- md also at tude. ^th of day ;ude; show (ve that cot wo places ; zon of both itude h at ast ; prove 30. Given the declination of a star 30°; find at what latitude its azimuth is 45° at the time of rising. 31. Given the sun's declination 8, and the latitude of the place <^ ; find his altitude when due east. 32. Given the declinations of two stars, and the differ- ence of their altitudes when they are on the prime vertical ; find the latitude of the place. 33. If the difference between the lengths of the longest and shortest day at a given place be six hours, find the latitude. 34. If the radius of the earth be 4000 miles, what is the area of a spherical triangle whose spherical excess is 1° ? 35. If A", B", C" be the chordal angles of the polar triangle of ABC, prove cos A" = sin I A cos [s — a), etc. 36. If the area of a spherical triangle be one-fourth the area of the sphere, show that the bisector of a side is the supplement of half that side. 37. If the area of a spherical triangle be one-fourth the area of the sphere, show that the arcs joining the middle points of its sides are quadrants. 38. Given the base and area ; show that the arc joining the middle points of the sides is constant ; and if it is a quadrant, then the area of the triangle is irr'K 39. Two circles of angular radii, a and (S, intersect orthogonally on a sphere of radius r; find in any manner the area common to the two. 40. If E be the spherical excess of a triangle, prove that |E = tan|a tan^& sinC - |(tania tan^&)2siu2C + etc. 41. Show that the sum of the three arcs joining the middle points of the sides of the colunars is equal to two 366 SPHERICAL TRIGONOMETRY. right angles, the sides of the original triangle being regarded as the bases of the colunars. 42. Prove that cos^^a sin^S + sin»^ ft sin^CS - C) + sin^J-c sin^CS - B) + 2cos |a sin |& sin|c sin S sin(S — B) sin(S — C) = 1. 43. Having given the base and the arc joining the middle points of the colunar on the base, the circumcircle is fixed. 44. Prove sin^6 sin^c sin(S - A) + cos^6 cos^csinS = cos^a. 45. If A 4- B + C = 2 TT, prove that and cos C = — cot | a cot ^ b. 46. Solve the equations, sin & cos c sin Z + sin c cos 6 sin Y = sin a, sin c cos a sin X + sin a cos c sin Z = sin &, sin a cos b sin Y + sin & cos a sin X = sin c, for sin X, sin Y, and sin Z. 47. If 6 and c are constant, prove the following relations between the small variations of any two parts of the other elements of the spherical triangle ABC : ^^tanB. i^ = -sin.xtanC; dC tanC dB da ■ 1, • dA sin A . — = sin B sm c ; "tft " ^~~5 n ' dA dB smBcosC da = — sin a tan B dA sin A .— Olll tt vein A^ 5 - ^^ -f-fc • /-t dC dG cos B sin 48. If A and c remain constant, prove the following : da^sina ^ = cos C. dB tanC db regarded S-B) C)=l. ;he middle e is fixed. ^csinS = cos^a. a 6, QC, g relations f the other lowing : EXAMPLES. 867 49. If B and C remain constant, prove the following db = sin A tan c ; da db sin« sin b cos c 50. If A and a remain constant, prove the following ^ = ^"'^" ^ . ^ _ tan c , dB tan B ' dC~ tan C ' rfft _ _ cosB . db___ sin b dc cos C dC tan B cos c 51. Two equal small circles are drawn touching each other; show that the angle between their planes is twice the complement of their spherical radius. 52. On a sphere whose radius is r a small circle of spher- ical radius 6 is described, and a great circle is described having its pole on the small circle ; show that the length of their common chord is 2r sinO V-cos2^. 53. Given the base c of a triangle, and that tan ^a tan ^ 6 = tan^^ B, B being the bisector of the base, find a — 6 in terms of c. 54. If C = A -I- B, show that 1 — cos a — cos b + cos c = 0, 55. If A denote one of the angles of an equilateral tri- angle, and A' an angle or its polar triangle, show that cos A cos A' r= cos A + cos A'. 56. Show that cos a cos B — c os b cos A cos C -f cos c sin a — sin b sine 57. Prove cos A = ^os g sin 6 — s in a cos b cos sine and cos A + cos B = 2^ (^ + &) sin^O . sine 368 SPHERICAL TltlGONOMETRY. 58. Prove Legeiidre's Theorem by means of the relations si n A _ sinB _ s inC . sin a sin 6 sin c 59. Two places are situated on the same parallel of lati- tude ; iind the difference of the distances sailed over by two ships passing between them, one keeping to the great circle course, the other to the parallel ; the difference of longitude of the places being 2X. Ans. 2r[A.cos <^ — sin~'(co8^sin X)]. 60. If the sides of a triangle be each 60°, show that the circles described, each having a vertex for jwle, and passing through the middle points of the sides which meet at it, have the sides of the supplemental triangle for common tangents. 61. Find the volume and also the inclination of two adjacent faces (1) of a regular tetraedron, (2) of a regular octaedron, (3) of a regular dodecaedron, and (4) of a regular icosaedron, the edge being one inch. Ans. (1) 117.85 cu. in., 70°31'43".4; (2) .4714CU. in., 109°28'16"; (3) 7.663 cu. in., 116° 33' 54"; (4) 2.1817 cu. in., 138° 11'22".6. 62. In the tetraedron, prove (1) that the circumradius is equal to three times its in-radius, and (2) that the radius of the sphere touching its six edges is a mean proportional between the in-radius and circumradius. 63. Prove that th ratio of the in-radius to the circum- radius is the same in the cube and the octaedron, and also in the dodecaedron and icosaedron. relations el of lati- l over by the great ierence of ^sin X)]. f that the itl passing leet at it, r common )n of two : a regular f a regular °31'43".4; °28'16"; i°33'54"; ,° 11' 22".6. rcum radius ■j the radius roportional the circum- n, and also