IN MEMORIAM FLOR1AN CAJORI BOWSER'S MATHEMATICS. ACADEMIC ALGEBRA. With numerous Examples. COLLEGE ALGEBRA. With numerous Examples. PLANE AND SOLID GEOMETRY. With numerous Exer- cises. ELEMENTS OF PLANE AND SPHERICAL TRIGONOME- TRY. With numerous Examples. A TREATISE ON PLANE AND SPHERICAL TRIGONOME- TRY, and its applications to Astronomy and Geodesy. 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 numerous Exam- plea. AN ELEMENTARY TREATISE ON ANALYTIC MECHANICS. With numerous Examples. AN ELEMENTARY TREATISE ON HYDROMECHANICS. With numerous Examples. A TREATISE ON PLANE AND SPHERICAL TRIGONOMETRY, AND ITS APPLICATIONS TO ASTBONOMY AND GEODESY, NUMEROUS EXAMPLES. BY EDWARD A. BOWSER, LL.D., PROFESSOR OF MATHEMATICS AND ENGINEERING IN RUTGERS COLLEGE. BOSTON, U.S.A. : PUBLISHED BY D. C. HEATH & CO. 1892. COPYRIGHT, 1892, BY E. A. BOWSER. CAJORI TYPOGRAPHY BY J. S. GUSHING & Co., BOSTON, U.S.A. PRESSWORK BY BERWICK & SMITH, BOSTON, U.S.A. PEEFACB. 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 911384 IV PREFACE. examples at the ends of the chapters have been care- fully graded, beginning with those which are easy, and extending to those which are more and more diffi- 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 text-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. RUTGERS COLL,KOK, New Brunswick, N. J., April, 181L'. TABLE OF CONTENTS. PART I. PLANE TRIGONOMETRY. CHAPTER I. MEASUREMENT or ANGLES. RT. PAGE 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 0. The Sexagesimal Method 5 7. The Centesimal or Decimal Method 6 8. The Circular Measure 6 9. Comparison of the Sexagesimal and Centesimal Measures . . 8 10. Comparison of the Sexagesimal and Circular Measures 9 11. General Measure of an Angle 11 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 yi CON TENTS. ART. PAGE 18. Functions of Angles of Any Magnitude 23 19. Changes in Sine as the Angle increases from to 360 25 20. Changes in Cosine as the Angle increases from to 360. . . 26 21. Changes in Tangent as the Angle increases from 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 Formulae 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 Complemental 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 35. 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 46. Formulae for transforming Sums into Products 55 46. Useful Formulas 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 60. Functions of 3 x in Terms of the Functions of x 61 61. Functions of Half an Angle 63 62. Double Values of Sine and Cosine of Half an Angle 63 CONTENTS. vii ABT. PAGE 53. Quadruple Values of Sine and Cosine of Half an Angle 65 54. Double Value of Tangent of Half an Angle 66 55. Triple Value of Sine of One-third an Angle 67 56. Find the Values of the Functions of 22 69 57. Find the Sine and Cosine of 18 69 58. Find the Sine and Cosine of 36 70 59. If A + B + C = 180, to find sin A -f sin B + sin C, etc 70 60. Inverse Trigonometric Functions 72 61. Table of Useful Formulae 75 Examples ... 77 CHAPTER IV. LOGARITHMS AND LOGARITHMIC TABLES. TRIGONOMETRIC TABLES. 62. Nature and Use of Logarithms 87 63. Properties of Logarithms 87 64. Common System of Logarithms 01 65. Comparison of Two Systems of Logarithms 93 66. Tables of Logarithms 95 67. Use of Tables of Logarithms of Numbers 98 68. To find the Logarithm of a Given Number 99 69. To find the Number corresponding to a Given Logarithm . . . 102 69a. Arithmetic Complement 103 70. Use of Trigonometric Tables 105 71. Use of Tables of Natural Trigonometric Functions 106 72. To find the Sine of a Given Angle 106 73. To find the Cosine of a Given Angle '. 106 74. To find the Angle whose Sine is Given 108 75. To find the Angle whose Cosine is Given 108 76. Use of Tables of Logarithmic Trigonometric Functions 110 77. To find the Logarithmic Sine of a Given Angle 112 78. To find the Logarithmic Cosine of a Given Angle 112 79. To find the Angle whose Logarithmic Sine is Given 114 80. To find the Angle whose Logarithmic Cosine is Given 115 81. Angles near the Limits of the Quadrant 116 Examples 117 viii CONTENTS. CHAPTER V. SOLUTION OF TRIGONOMETRIC EQUATIONS. ART. PAGE 82. Trigonometric Equations 126 83. To solve m sin = a, m cos = b 128 84. To solve a sin -f b cos = c 129 85. To solve sin ( + x) = m sin 131 86. To solve tan ( + a) = m tan x 132 87. To solve tan ( + x) tan a; = m 133 88. To solve m sin (0 + x) = a, w sin (0 + x} - b 134 89. To solve x cos a -f y sin a = w, sin a y cos , 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, ..., 9, , ... stand for mere numbers. Thus, n- stands for a number, and a number only, viz., 3.14159 ..., but in the expression ' the angle IT,' that is, ' the angle 3.14159 ...,' there must be some unit understood. The unit understood here is a radian, and therefore ' the angle n ' stands for ' n radians' or 3.14159 ... radians, that is, two right angles. Hence, when an angle is referred to, n is a very convenient abbreviation for two right angles. So also ' the angle a or 6 ' means ' a radians or 9 radians.' The units in the three systems, when expressed in terms of one common standard, two right angles, stand thus : 8 PLANE TRIGONOMETRY. The unit in the Sexagesimal Method = of 2 right angles. 180 " " " " Centesimal " = " " 200 " " " " Circular " = - " " " " 7T If D, G, and 6 denote the number of degrees, grades, ad radians respectively in any angle, then JUA.^ m 180 200 TT' because each fraction is the ratio of the angle to two right angles. 9. Comparison of the Sexagesimal and Centesimal Meas- ures of an Angle. Although the centesimal method was never in general use among mathematicians, and is now totally abandoned everywhere, yet it still possesses some interest, as it shows the application of the decimal system to the measurement of angles. From (1) of Art. 8 we have JD_ _G 180 200* .-. D = ^G, andG= D. 10 9 EXAMPLES. 1. Express 49 15' 35" in centesimal measure. First express the angle in degrees and decimals of a degree thus : 60) 35" 60) 15' .583 49.25972 10 9) 492.5972 64*. 733024-.. . .-. 49 15' 35"- 54^ 73 V 30 V \24 .... COMPARISON OF MEASURES. 9 2. Express 87 g 2 V 25 U in degrees, etc. First express the angle in grades and decimals of a grade thus : 872 V 25 u =87 g .0225 78.32025 60 19.215 _ 60 12.9 .-. 87*2 V 25 VV =78 19' 12".9. Find the number of grades, minutes, and seconds in the following angles : 3.' 51 4' 30". Ans. 56* 75 V O x \ 4. 45 33' 3". 50s61 v 20 N \37. 5. 27 15' 46". 30* 29 V 19 V \75 . -. 6. 157 4' 9". 174*52 V 12 V \962.... Find the number of degrees, minutes, and seconds in the following angles : 7. 19* 45 V 95 V \ Ans. 17 30' 48".78. 8. 124^ 5 V 8 V \ 111 38' 44".592. 9. 55 18 V 35 V \ 49 39' 54".54. 10. Comparison of the Sexagesimal and Circular Meas- ures of an Angle. From (1) of Art. 8 we have 180 10 PLANE TRIGONOMETRY. EXAMPLES. 1. Find the number of degrees in the angle whose circu- lar measure is 1. Here = i- 7T 2 7T . *^" ^ * O8 *3ft' "IfWlO - 22 = > 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* .-. = 11=11! w = (.333-.) ir =1.0453 -. 3. Express, in degrees, the angles whose circular measures 7T 7T 7T 7T 2 2' 3' 4' 6' NOTE 1. The student should especially accustom himself to express readily in circular measure an angle which is given in degrees. 4. Express in circular measure the following angles : 60, 22 30', 11 15', 270. Ans. * |, * ^. o o 16 2 5. Express in circular measure 3 12', and find to seconds the angle whose circular measure is .8. ("Take TT = ~\ Ans. ~, 45 49' 5" T 5 T . 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" T 8 T . NOTE 2. Questions in which angles are expressed in different systems of meas- urement are easily solved by expressing each angle in right angles. GENERAL MEASURE OF AN ANGLE. 11 7. The sum of the measure of an angle in degrees and twice its measure in radians is 23f ; find its measure in degrees (TT 2 T 2 ). Let the angle contain x right angles. Then the measure of the angle in degrees = 90 x. " " " " " " " radians = ^ x. .-. 652^=163, .-. x = -> 4 .-. the angle is J of 90 = 8. The difference between two angles is -, and their sum is 56 ; find the angles in degrees. Ans. 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 know 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 arc 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 whole number, positive or negative. Then (1) 2n x 180 + A represents algebraically an angle in the first quadrant. (2) 2n x 180 A represents algebraically an angle in the fourth quadrant. (3) (2n -f- 1) 180 A represents algebraically an angle in the second quadrant. (4) (2n + 1) 180 + A represents algebraically an angle in the third quadrant. In circular measure the corresponding expressions are (1) 2n7r+0, (2) 2wir-0, (3) (2n + l)ir-0, (4) (2n EXAMPLES. State in which quadrant the revolving line will be after describing the following angles : (1) 120, (2) 340, (3) 490, (4) - 100, (5) -380, (6)j7r, (7)10* + ?. 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 remaiMder obtained by subtracting it from two 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 J?r is [ ? *ir i= *ir. The supplement of JTT is (TT JTT) = J EXAMPLES. 13 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. 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. 46* 95\ 9. 63 19' 17". 70*.35 V 70 V \98 .... 10. 103 15' 45". 114 73v 61 w L 11. 19 0'18". 21*11 X 66 V \6. 12. 143 9' 0". 159* 5 X 55 V \5. 13. 300 15' 58". 333* 62 V 90 V \ 1234567890. 14. 27 41' 51", 30 775. 15. 67.4325. 74*.925. 16. 8 15' 27". 9* 17 X 50 N \ 17. 97 5' 15". 107 g 87 V 50 V \ 18. 16 14' 19". 18* 4 V 29 XV .... 19. 132 6'. 146* 77 V 77 v \t. 14 PLANE TRIGONOMETRY. Express the following 11 angles in degrees, minutes, and seconds : 20. 105 g 52 V 75 V \ Ans. 94 58' 29".l. 21. 82 g 9 V 54 V \ 73 53' 9".096. 22. 70*15 V 92 V \ 63 8' 35".808. 23. 15* V 15 V \ 13 30' 4".86. 24. 154* 7 V 24 V \ 138 39' 54".576. 25. 324* 13 V 88 V \7. 291 43' 29".9388. 26. 10* 42 V 50 V \ 9 22' 57". 27. 20* 77 V 50 V \ 18 41' 51". 28. 8* 75\ 7 52' 30". 29. 170M5 V 35 V \ 153 24' 29 ".34. 30. 24* V 25 V \ 21 36' 8".l. Express in circular measure the following angles : 1453 TT 31. 315, 24 13'. Ans. JTT, 32. 95 20', 12 5' 4". 10800 143 w 2719 TT 270' 40500 33. 22i, 1, 57.295. -, -|-, 1 radian. 8 180 34. 120, 45, 270. 2.09439, |, |TT. 35. 360, 3-J- rt. angles. 27r, JTT. Express in degrees, etc., the angles whose circular meas- ures are : 1 on 36. |TT, ITT, . Ans. 112.5, 120, -- degrees. 2 7T 37. -, -, - - degrees, - - degrees, degrees. 463 7T 7T 7T 38. -, .7854. 47 43' 38"^, 45. 6 EXAMPLES. 15 39. 41., ^ 2.504. A,is. 257 49' 43".39, 15, 143.468. 40. .0234, 1.234, ?. 120'27", 70 42' 11", 38 11' 60". o 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-J-feet. Ans. If 42. Find the number of degrees in an angle at the centre of a circle of radius 10 feet, which intercepts an arc of STT feet. Ans. 90. 43. Find the number of right angles in an angle at the centre of a circle of radius 3 T 2 T inches, which intercepts an arc of 2 feet. Ans. 4f . 44. Find the length of the arc subtending an angle of 4| radians at the centre of a circle whose radius is 25 feet. Ans. 112-1- ft. 45. Find the length of an arc of 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 which 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 ffj for TT). Ans. .00004848. 16 PLANE TRIGONOMETRY. CHAPTER II. THE TKIGONOMETKIO FUNCTIONS, 13. Definitions of the Trigonometric Functions. Let RAD 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 B 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 A 6 c opposite these angles by the corresponding small italics, a, b } c* we have the following definitions : a = opposite side ig called ^ ^ Qf &f> ang]e A c hypotenuse - = adjacent side is called the cosine of the angle A. c hypotenuse a = opposite side ig called the ^ of ^ angle A b adjacent side 6 = adjacent side ig called ^ cot nt of the angle A a opposite side c hypotenuse -,-, -, , -, f , i - = J r is called the secant 01 the angle A. 6 adjacent side c hypotenuse -.-. ^ ., T A - = !L -*- is called the cosecant ot the angle A. a opposite side 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 chosen unit of length. TEIGONOMETRIC FUNCTIONS. 17 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 opposite side to the hypotenuse; or sin A = c The cosine of an angle is the ratio of the adjacent side to the hypotenuse; or cosA = c The tangent of an angle is the ratio of the opposite side to the adjacent side; or tan A = - b The cotangent of an angle is the ratio of the adjacent side to the opposite side; or cot A = a The secant of an angle is the ratio of the hypotenuse to the adjacent side; or sec A = - 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 The coversed sine of an angle is unity minus the sine of the angle; or covers A = 1 sin A = 1 c 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 another ; 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 cosec A = - -, or sin A = , sin A cosec A sec A = , or cosA = V^X \j\JiJ J.JL cos A sec A tan A = -, or cotA = j Vyj. V/WL/J.JL. cot A tan A The powers of the Trigonometric functions are expressed as follows : (sin A) 2 is written sin 2 A, (cos A) 3 is written cos 3 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 2 A is an abbreviation for (sin A) 2 , 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- R, pendicular to AB. Take P", any point in AB, and draw P"C" per- ^ , , x pendicular to AD. A c c/ 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. PC = P'C' = P"C" ' 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. 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 b tan A = b cot B, b = a cot A = a tan B = c cos A = c sin B, c = a cosec A = a secB = b sec A = b cosecB. 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, 6, 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 = T 8 T , cos A = T f, tan A = y 8 ^, etc. ; (2) sin A = f , cos A = 9 T , etc. ; (3) sin A = ft, tan A = |f, etc.; (4) sin A = ffy, tan A = - 4 g 8 T -, etc. ; (5) sin A = fff, tanA = f||-, etc. In a right triangle, given : 2. a = Vm 2 -f- n 2 , b = V2 mn ; calculate sin A. Vra 2 + n 2 Ans. - m -\-n 3. a = Vm 2 mn, b = n; calculate sec A. m ~ n - n vnr-\- mn mn+n 2 m 2 - n 2 4. a = Vm 2 + mn, c m -f- n ; calculate tan A. -*/ 5. a = 2 ran, b = ra 2 n 2 ; calculate cos A. ra 2 +n 2 6. sin A = -I, c = 200.5; calculated. 120.3. 7. cos A = .44, c = 30.5 ; calculate 6. 13.42. 8. tan A = JJL, 6 = f T ; calculate c. & Vl30. 20 PLANE TRIGONOMETRY. 15. Functions of Complemental Angles. In the rt. A ABC we have sin A = -, and cosB = -. (Art. 13.) c c .*. sin A = cosB. But B is the complement of A, since their sum is a right angle, or 90 ; i.e., Also, sin A cos A tan A cot A sec A = cosB = cos (90 -A) = sinB = sin (90 A) = cotB = cot (90 -A) = tanB = tan (90 -A) a c b ~j c a b' b > a = cosecB =cosec(90- A) = --, cosecA =secB = sec (90 A) = -, a vers A = covers B = covers (90 A) = 1 -, c covers A = vers B = vers (90 - A) = 1 -. c Therefore the sine, tangent, secant, and versed sine of an angle are equal respectively to the cosine, cotangent, cosecant, and coversed sine of the complement 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. REPRESENTATION OF FUNCTIONS BY LINES. 21 Draw the tangents AT, BT' meeting DP produced to T', and draw PC, PD _L to OA, OB. Then PC was called the sine of- the arc AP. OC AT BT' OT OT' AC BD cosine tangent cotangent secant cosecant versed sine coversed sine 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 0, we have, when A = OP = 1, 22 PLANE TRIGONOMETRY. smO =^ = ~ = ?C, OP 1 , OA 1 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 denned 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. The 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 which 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. " cot A cos A + sin A = cosec A. 3. " (tan A - sin A) 2 + (1 - cos A) 2 = (sec A - I) 2 . 4. " tan A -f cot A = sec A cosec A. 5. " (sin A-|- cos A) -5- (sec A -h cosec A) = sin A cos A. 6. " (1 + tanA) 2 + (1 + cot A) 2 = (sec A + cosec A) 2 . O M POSITIVE AND NEGATIVE LINES. 23 7. Given tan A = cot 2 A ; find A. 8. " sin A = C os3A; find A. 9. " sin A = cos (45 - A) ; find A. 10. " tanA = cot6A; find A. 11. cot A = tan (45+ A) ; find A. 17. Positive and Negative Lines. Let AA' and BB' be two perpendicular right lines intersecting at the point O. Then the position of any point in the line AA' or BB' will be deter- mined if we know the distance of the point from O, and if we know also upon which side of the A" r- 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 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 AA' upwards as positive, and consequently distances measured from BB' towards the left and from AA' downwards as negative. 18. Trigonometric Functions of Angles of Any Magni- tude. In the definitions of the trigonometric functions given in Art. 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. PLANE TRIGONOMETRY. Let AOP be an angle of any mag- nitude formed by OP revolving from an initial position OA. Draw PM J_ to AA'. Consider OP as always positive. Let the angle AOP be denoted by A ; then whatever be the magnitude of the angle A, the defini- tions of the trigonometric functions are MP _.__ OP 7 sin A OP cosec A = I. When A lies in the 1st quadrant, MP is positive because measured from M upwards, OM is positive because measured from towards the right (Art. 17), and OP is positive. Hence in the Jirst quadrant all the func- tions are positive. II. When A lies in the 2d quadrant, as the obtuse angle AOP, MP is positive because measured from M upwards, OM is negative because measured from O towards the left (Art. 17), and OP is positive. Hence in the second quadrant MP sin A = -^ is positive ; cos A = is negative; MP tan A = is negative; and therefore sec A and cot A are negative, and cosec A is positive (Art. 13). FUNCTIONS OF ANGLES. III. When A lies in the 3d quadrant, as the reflex angle AGP, MP is negative because measured from M downwards, OM is negative, and OP is positive. 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 + - + - NOTE. It is apparent from this table that the signs of all the functions in any quadrant are known when those of the sine and 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 to 360. Let A de- note the angle AOP described by the revolution of OP from its initial posi- tion OA through 360. Then, PM being drawn perpendicular to AA', MP sin A = OP' whatever be the magnitude of the angle A. 26 PLANE TRIGONOMETRY. When the angle A is 0, P coincides with A, and MP is zero ; therefore sin = 0. As A increases from 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 positive; 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 360, MP decreases from OP to zero, and is negative; therefore sin 300= 0. Hence in the 4th quadrant sin A is negative, and increases algebraically from 1 to 0. 20. Changes in the Cosine as the Angle increases from to 360. In the figure of Art. 19 When the angle A is 0, P coincides with A, and OM = OP ; therefore cosO= 1. As A increases from to 90, OM decreases from OP to zero and is positive; 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 quadrant 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. 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 to 360. In the figure of Art. 19 When A is 0, MP is zero, and OM = OP; therefore tan 0=0. As A increases from 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 tan 90 =00. Hence in the 1st quadrant tan A is positive, and increases from to oo. 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 oo to 0. When A passes into the 2d quadrant, and is only just greater than 90, tan A changes from -f oo 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 the 3d quadrant tan A is positive, and increases from to oo. As A increases from 270 to 360, MP decreases from OP to zero, and is negative, OM increases from zero to OP, and is positive; therefore tan 360= 0. Hence in the 4th quadrant tan A is negative, and increases algebraically from oo to 0. 28 PLANE TRIGONOMETRY. 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 with those given in the following table. 22. Table giving the Changes of the Trigonometric Functions in the Four Quadrants. QUADRANT. I. II. III. IV. sin varies from -f to 1 + 1 toO to - 1 - 1 to cos " " + 1 toO to - 1 - 1 toO + .Oto 1 tan ". " + to co co to + to co - co toO cot " " + oo toO to co + co to to -co sec " " + 1 to co co to 1 - 1 to - co + CO tO 1 cosec " " + CO tO 1 + 1 to co - CO tO - 1 1 tO CO vers " " + Oto 1 f 1 to 2 + 2 to 1 + 1 toO NOTE 1. The cosecant, secant, and cotangent of an angle A have the same sign as the sine, cosine, and tangent of A respectively. The sine and cosine vary from 1 to 1, passing through the value 0. They are never greater than unity. The secant and cosecant vary from 1 t 1, passing through the value oo. They are never numerically less than unity. The tangent and cotangent are unlimited in value. They have all values from oo to +00. The versed sine and cover sed sine vary from to 2, and are always positive. The trigonometric functions change sign in passing through the values and oo, and through no other values. In the 1st quadrant \\\e functions increase, and the cofunctions decrease. NOTE 2. From the results given in the above table, it will be seen that, if the value of a trigonometric function be given, we cannot fix on one angle to which it belongs exclusively. Thus, if the given value of sin A be \, we know since sin A passes through all values from to 1 as A increases from to 90, that one value of A lies between RELATIONS BETWEEN FUNCTIONS. 29 and 90. But since we also know that the value of sin A passes through all values between 1 and as A increases from 90 to 180, it is evident that there is another value of A between 90 and 180 for which sin A = . 23. Relations between the Trigonometric Functions of the Same Angle. Let the radius start from the initial position A, and revolve in either direction, to the position OP. Let 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 = 1 tan0 = sin sinfl COS0 COS0' C0t0 = 1 tan0 For - b o cos c II. sin 2 + cos 2 = 1. For sin 2 + cos 2 = ^ + ^ = c 2 c 2 sin tan Formulae I., II., III., 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. 30 PLANE TRIGONOMETRY. 24. Use of the Preceding Formulae . I. To express all the other functions in terms of the sine, Since sin 2 + cos 2 = 1, . : cos = Vl sin 2 0. COS0 tan0 COS0 cosec = sin0 II. To express all the other functions in terms of the tan- gent. Since tan0 = ^, COS0 sec Vl+tan 2 cos0 = !-= sec (9 Vl + tan 2 sec0= Vl sin 6 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 6 lies. 25. Graphic Method of finding All the Functions in Terms of One of them. To express all the other functions in terms of the cosecant. Construct a right triangle ABC, hav- ing the side BC = 1. Then RELATIONS BETWEEN FUNCTIONS. 31 A AB AB AT3 cosec A = = - = A>. BC 1 Now .-. AC = Vcosec 2 A 1. sinA = = T) AB cosec A A AC , Vcosec 2 A 1 cos A = = - - , AB cosec A A^ Vcosec 2 A 1 and similarly the other functions may be expressed in terms of cosec A. 26. To find the Trigonometric Functions of 45. Let ABC be an isosceles right triangle in which B CA = CB. Then CAB = CBA = 45. Let AC = m = CB. Then AB 2 = AC 2 + CB 2 = m 2 + m 2 = 2 m 2 . = mV2. m mV2 V2 m 1 AB m v2 V2 tan 45 = = - = 1. cot 45 =1. AC m 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 BAG and the side BC. Therefore BAD = 30, and ABD = 60. 32 PLANE TRIGONOMETRY. Let BA = 2m. .-. BD = m. Then AD = V<4 m 2 m 2 = m V3. = ^ = - t -=1V3. .-. cosec60=-^. AB 2m y^ cos 60 =?5 = 1. .-. sec 60 = 2. Jj A. *a .-. cot60 = -. V3 BD m Also vers60= 1 - cos 60= 1 - * = 1. 2 2 sin30 o = BD = ^_ == l > . Cosec30 o =2 cos30^A5 = ^ = iV3. .-. sec30 = A. AB 2m V3 tan30=~? = ^^: = ~ .-. cotSO^VS. mV3 V3 EXAMPLES. 1. Given sin^ = -; find the other o trigonometric functions. Let BAG be the angle, and BC be perpendicular to AC. Kepresent BC by 3, AB by 5, and consequently AC A 4 by V26-9 = 4. A n A Then AB = 5 AC~4' REDUCTION OF FUNCTIONS. 33 o 35 2. Given sin0 = -: find tan and cosec 0. Ans. -> - 5 4 6 3. Given cos<9 = i; find sin and cot 0. |V2, - 3 2v2 4. Given sec0 = 4: find cot and sin0. > - Vl5 4 5. Given tan0 = V3; find sin and cos 9. |V3, - 6. Given sin = ; find cos 0. 13 lo 7. Given cosec = 5 ; find sec and tan 6. 2V6 2V6 8. Given sec = ; find sin and cot 0. ? -. 9 41 41 9. Given cot = r ; find sin and sec 0. -> - V5 32 10. Given sin = - ; find cos 0, tan 0, and cot 0. V7 3V7 V7 T' T' "3"' 11. Given gin = - ; find tan 0. c 12. Given sin0 = / 2 ; find tan 0. 28. Reduction of Trigonometric Functions to the 1st Quadrant. All mathematical tables give the trigonometric functions of angles between 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. 34 PL A NE TRIG ON OMETli Y. 29. Functions of Complemental Angles. Let AA', BB' be two diameters of a circle at right angles, and let OP and OP' be the positions of the radius for any angle AOP = A, and its comple- ment AOP'= 90- A (Art. 12). Draw PM and P'M' at right angles to OA. Angle OP'M'= BOP'= AOP = A. Also OP= OP'. Hence the triangles 0PM and OP'M' are equal in all respects. .-. P'M'=OM. .-. !M.' = ^I and f 6. If sin x = -, and cos ?/ = -, find sin (a; + ?/) and cos(a; 2/). V3 ^Ins. 1, and --- 2 FORMULA FOR TRANSFORMATION. 55 45. Formulae for the Transformation of Sums into Prod- ucts. From the four fundamental formulae of Arts. 43 and 44 we have, by addition and subtraction, the following: sin (x + 2/) + sin (x y) = 2 since cosy . . . (1) sin (x + y) sin (x y) = 2 cos x sin y ... (2) cos (x + y) + cos (x ?/) = 2 cos x cos y . . . (3) cos (x y) cos (x + y) = 2 since siny . . . (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 replace the product of a sine or a cosine into a sine or a cosine by half the sum or half the difference of two such ratios. Let x-\-y = A, and x y = B. , and y = -J(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 sin x + sin y = 2 sin %(x-\-y) cos ^ (# ?/) . . (5) sin x sin y = 2 cos ^ (x + y) sin ^- (# y) . . (6) cos x -f cos y = 2 cos i (a; -f- y) cos (a? y) . . (7) cos ?/ cos # = 2 sin (a; + y) 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 coirx mit the following enunciations to memory : Of any two angles, the Sum of the sines = 2 sin sum -cos^diff. Diff. " " " =2cosjsum.sinidiff. 56 PLANE TRIGONOMETRY. Sum of the cosines = 2 cos ^ sum cos % diff. Diff. " " = 2 sin sum- sin diff. EXAMPLES. 1. sin 5 a cos 3 # = (sin 8 # -f- sin2#). For, sin5cccos3a;= Jjsin (5x + 3x) + sin (5x 3x) \ = \ (sin 8 x + sin 2 x) . 2. Prove sin sin 30 = (cos 2(9 - cos 40). 3. " 2sin0cos< = sin (0 + <) + sin (6 <). 4. " 2sin20cos3< = sin(20-f 3<) + sin (20 3<). 5. sin 60 + sin 30 = 2 sin 45 cos 15. 6. sin 40 -sin 10 =2 cos 25 sin 15. 7. " sinl00+sin60=2sin80cos20. 8. " sin8 sin4a = 2cos6asin2a. 9. " sin 3 x + sin a; = 2 sin 2 a; cos x. 10. " sin 3 sin a; = 2 cos 2 a; sin x. 11. " sin 4 07 + sin 2 a; = 2 sin 3 x cos a;. 46. Useful Formulae. The following formulae, 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 : ^ since -f- sin?/_ 2 sin %(x -\-y) cos^(x,y) sina; smy 2cos(# + y) sin|(x y) = tan $(x -+-?/) cot %(x y) ). (Art . 24) y) The following may be proved by the student in a similar manner : 2. !H+liM cos x -f cos y THE TANGENT OF TWO ANGLES. 57 3. ""s + 'foy-cotiCs-y), cos y cos a; 4. - = tani (g-y), COSX + COS?/ 5. cos y cos a; 6. cosy cos a; 47. The Tangent of the Sum and Difference of Two Angles. Expressions for the value of tan(# + 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) = sin ( x + y) = sin a? cos y + cos g sin y cos(ie-fy) cos x cos y sin x sin y Dividing both terms of the fraction by cos x cos y, sin x cos y cos a; sin y cos a; cos y cos a? cos?/ tan (x + y) = - 2 - . - __2 cos a; cosy sin a; sin y cos a; cosy cos a; cos a; = tana + tany 1} 1 tana;tany v In the same manner may be derived tan(s-y)= ~ (2) 1-ftanxtany Also, cot( a? + y) = COta?COty "" 1 (3) cot x + cot y and eot(x-y) = (4) coty cot a; 58 PLANE TRIGONOMETRY. EXERCISES. Prove the following : 1 tana; 2. tan(s-45') = 1 + tana; 3 sin (x + y) _ tan x -f tan y sin (a; y) tan x tan ?/ 4 cos (a; y) _ tan x tan y -f 1 cos (z -f ?/) 1 tan x tan y 5. sin (x + y) sin (x y) = sin 2 a; sin 2 ?/ = cos 2 ?/ cos 2 a;. 6. cos (x + y)cos(x y) = cos 2 x sin 2 ?/ = cos 2 ?/ sin 2 a?. 7. tail x tan y =>"(***). cos a; cos?/ 8. cotxcoty = sn a; sn y sin 2 a; cos2a; 9. = sec x. sin x cos x 10. If tan a; = and tan ?/ = J, prove that tan (a; -f y) = f , and tan (a; y) = -J. 11. Prove that tan 15 = 2 - V3. 12. If tan x= f , and tan y = Jy, prove that tan (x + y)= 1. What is (x H- y) in this case ? 48. Formulae for the Sum of Three or More Angles. Let x } y, z be any three angles ; we have by Art. 43, EXAMPLES. 59 sin (x + y -f z) = sin (a; -|- y) cos 2 -f cos (a; -f y) sinz = sin ic cos y cos 2 -f- cos x sin ?/ cos z + cos x cosy sin z sin a; sin?/ sinz . . (1) In like manner, cos (x + y + z) = cos a; cos y cos z sin x sin ?/ cos z sin a; cosy sinz cos x sin y sinz . . (2) Dividing (1) by (2), and reducing by dividing both terms of the fraction by cos x cosy cos 0, we get tanfff + sH-g) = tana? + tany + tans - tana? tany tanz ^ 1 tana; tan ?/ tany tanz tanz tana; EXAMPLES. 1. Prove that sin x + sin y + sin z sin (a; -f y + z) = 4sin JO -f y) sin$(y + z) sin J(z + JB). By (6) of Art. 44 we have sin a; sin (a; -f y -f z) = 2cos J(2# + y -f z) sin J(?/ + z), and siuy -\- sinz = 2 sin ^(y + 2) cos^(?/ z). . . sin a; + sin y -f- sin z sin (a; + y -f- 2) = 2sin(#+z) cos^(2/ z) 2cos^(2a;+2/+z) sin =2sin|-(2/ + z) |cosj(y z) cos %(2x + y + z) in^(a;4- y) sin J(a? + z) J(2/ + z) sin|(z + ). Prove the following : 2. cos a; + cos y + cos z + cos (a; + y + z) -f z) cos(z + a?) cos J 60 PLANE TRIGONOMETRY. 3. sin (x + y z) = sin x cos y cos z + cos cc sin i/ cos z cos a; cosy simz -f sin a? siny sinz. 4. sin x + sin i/ sin z sin (cc -f y z) = 4 sin $(x z) sinj(;z/ 2) s 5. sin (y z) + sin (2 a?) -f- sin (a; y) 2;) sin ^(2 x) sin * (x t/) = 0. 49. Functions of Double Angles. To express the trigo- nometric functions of the angle 2 a? in terms of those of the angle x. Put y = x in (1) of Art. 42, and it becomes sin 2 x = sin x cos a; + cos x sin a;, or sin 2 ie = 2 sin x cos a; ....... (1) Put y = x in (2) of Art. 42, and it becomes cos 2 x = cos 2 x sin 2 a; ....... (2) = l-2sin 2 a; ........ (3) or =2cos 2 #-l ........ (4) Put y = x in (1) and (3) of Art. 47, and they become tan2a;= 2tana ; ........ (5) 1 tan 2 a (6) 2 cot a Transposing 1 in (4), and dividing it into (1), we have sin2 * = tan*. (7) EXAMPLES. 61 NOTE. These seven formulae are very important. The student must notice that x is any angle, and therefore these formulas will be true whatever we put for x. Thus, if we write - for x, we get sin a; = 2 sin? cos- ... ........ (8) --6in2- 2 2 or = l-28in2- = 2cos2--l . . (10) 2 2 and so on. EXAMPLES. Prove the following : 1. 2cosec2x = sec#cosec#. 2. __ =8ec 2. cosecx 2 3. 1 + tan 2 x 4. l' 5. tan#+cot# = 2cosec2x. 6. cotx tan = 2 cot 2 a;. sina; + oos a; sinz 1 coso; 9. Given sin 45=-L; find tan 22f. Ans. V2-1. V2 10. Given fean# = -; 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 = 2# in (1) of Art. 42, and it becomes 62 PLANE TRIGONOMETRY. = sin (2#-f x) = sin2a; cosse + cos2x sin a; = 2 sin a; cos 2 a; + (1 2 sin 2 a) sin X (Art. 49) = 2sinic(l sin 2 a?) -f- sinx 2 sin 3 a; = 3 sin a; 4sin 3 #. = cos 2 x cos x sin 2 x sin x = (2cos 2 # 1) cos a; 2sin 2 iccosx (Art. 49) = 4cos 3 ic tan2a;4-tana; 1 tan 2 x tan x 2 tana; 1 tan 2 a; -f tanx 1 _ 2 tan 2 a; 1 tan 2 a; 3 tan a; tan 3 a; 1 3 tan 2 a; EXAMPLES. Prove the following : 2. sma; Sin3a: - sina; cos 3 x + cos x 3. sin3x + cos * cos x sin x 4. FUNCTIONS OF THE HALF ANGLE. 63 - 1 - + _ - tan 3x tana; cot# cot3# 1 COS X 51. Functions of Half an Angle. To express the func- tions of - in terms of the functions of x. Since cosx= 1 2 sin 2 -, or . 9 X sin = 2cos'|-l . 1 cos x . . [Art. 49, (10)] and . . 0111 cos 2 * 3 1 -f- cos x (2} 2 2 Or sin x /I COS X (3\ 2 - V 2 and cos x /I -f cos a; 2 -\ 2 Bv division tan /I cos aj 1 cos a: x-v *->y 2i *" \1 -f- cos a; sinaj By formulae (3), (4), and (5) the functions of half an angle 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 cos a (nothing else being known about the angle x) gives two values each for sin- and cos-, 2 one positive and one negative. But if the value of x be 64 PLANE TRIGONOMETRY. COS- 2 IS given, we know the quadrant in which - lies, and hence we know which sign is to be taken. Thus, if x lies between and 360, - lies between and 180, and therefore sin- is positive; but if x lies between 360 and 720, ? lies between 180 and 360, and hence i sin- is negative. Also, if x lie between and 180, cos^ " 2 is positive; but if x lie between 180 and 360 negative. The case may be investigated geometrically thus : Let OM = the given cosine (radius being unity, Art. 16), = cos x. Through M draw PQ per- pendicular to 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 case 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, fche 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 Og and 0?*, but the latter are negative ; hence we obtain two, and only two, values of cos- from a given value of cosx. 2 Also, the tangent of half the angle whose cosine is given is two-valued. This follows immediately from (5) of Art. 51. FUNCTIONS OF THE HALF ANGLE. 65 53. If the Sine of an Angle be given, the Sine and the Cosine of its Half are each Four- Valued. 2sin-cos- = sina; . . . (Art. 49) w We have and By addition, [sin f + cos ? ] =14- sin a;. sin 2 - 4- cos 2 - =1 .... (Art. 23) 2 2 / x x \2 By subtraction, I sin- cos- j = 1 sin a;. .-. sin| + cos| . . (1) and and sin-- cos- = Vl-sinz . . (2) .-. 2sin^= 2 sinx Vl sina . . (3) 2cos-= Vl + since =F Vl sin a; . . (4) a Thus, if we are given the value of sin a; (nothing else being known about the angle x), it follows from (3) and (4) that sin- and cos- have each four values equal, two 2 2 by two, in absolute value, but of contrary signs. The case may be investigated geometrically thus : Let ON = the given sine (radius being unity) = sin x. Through N draw PQ parallel to OA ; and draw OP, OQ. Then all angles whose sines are equal to sin a? are terminated either by OP or OQ, and the halves of these angles are termi- nated by the dotted lines Op, Og, Or, or Os. The sines of angles ending at Op, Oq, Or, and Os are all different Q 66 PLANE TRIGONOMETRY. in value ; and so are their cosines. Hence we obtain four T* 'T* values for sin-, and four also for cos-, in terms of x. Jj 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 2 2 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 -, 2 22 and is positive. Therefore (1) is positive, and (2) is negative and hence (3) and (4) become 2 sin - = Vl 4- sin x Vl sin x, 2 2 cos - = VT 4- sin x + Vl sin x. When - lies between 45 and 135, sin -> cos-, and is 2 22 positive. Therefore (1) and (2) are both positive; and hence (3) and (4) become 2 sin- = Vl 4- sin a; 4- Vl sin#, 2 2 cos - = Vl 4- sin x Vl sin x. And so on. 54. If the Tangent of an Angle be given, the Tangent of its Half is Two-Valued. 2 tan i We have tan0 = - 1 ..... (Art. 49) FUNCTIONS OF THE HALF ANGLE. 67 Put tan- = #; thus 2 tan0 . 2 tan 6 Thus, given tan0, we find ;o unequal values 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 sin 3 x = 3 sin x 4 sin 3 x . . (Art. 50) f\ Put x = -, and we get 3 3 3' a cubic equation, which therefore has three roots. EXAMPLES. 1. Determine the limits between which A must lie to satisfy the equation 2 sin A = Vl + sin 2 A Vl sin 2 A. By (1) and (2) of Art. 53, 2 sin A can have this value only when sin A + cos A = Vl + sin 2 A, and sin A cosA = Vl sin2A; i.e., when sin A > cos A and negative. 68 PLANE TRIGONOMETRY. Therefore A lies between 225 and 315, or between the angles formed by adding or subtracting any multiple of four right angles to each of these ; i.e., A lies between 2tt7r + and 2mr + 1 , 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 2 cos A == Vl + sin 2 A Vl sin 2 A. By (1) and (2) of Art. 53, 2 cos A can have this value only when cos A + sin A = Vl -f- sin 2 A, and cos A sin A = Vl sin2A; i.e., when sin A > cos A and positive. Therefore A lies between 2W7T + - and 2tt7r + ^", 4 4 where n is any positive or negative integer. 3. State the signs of (sin 6 + cos 0) and (sin cos 0) when has the following values: (1) 22; (2) 191; (3) 290; (4) 345; (5) -22; (6) -275; (7) -470; (8) 1000. Am. (1) +, -; (2) -, +; (3) -, -; (4) + , -; (5) +, -; (6) +, +; (7) -, -; (8) -, -. 4. Prove that the formulae which give the values of sin- and of cos^ in terms of sin a; are unaltered when x 2i ' has the values (1) 92, 268, 900, 4wir + Jir, or (4n + 2) TT- |TT; (2) 88, - 88, 770, - 770, or 4n7r \ VALUES OF SPECIAL ANGLES. 5. Find the limits between which A must lie when 2sinA= Vl + sin2A Vl sin2A. 56. Find the Values of the Functions of 22|. In (3), (4), and (5) of Art. 51, put x = 45. Then cos 22i= A /I + cos 45 \/2 + V2 2 ~~ sin 45 Since 22 is an acute angle, its functions are all positive. The above results are also the cosine, sine, and cotangent respectively of 67, since the latter is the complement of (Art. 15). 57. Find the Sine and Cosine of 18. Let cc=18; then 2 a; = 36, and 3 x = 54. .-. 2x + 3z = 90. .-. sin 2 x = cos 3 x ..... (Art. 15) .. 2 sin x cos x = 4 cos 3 x 3 cos x . . (Art. 50) or 2sinjc = 4cos 2 ic 3 = 1 4 sin 2 x. Solving the quadratic, and taking the upper sign, since sin 18 must be positive, we get sin 18=^*. + 2 Also, cos 18= Vl - sin 2 18= 4 Hence we have also the sine and cosine of 72 (Art. 15). 70 PLANE TRIGONOMETRY. 56. Find the Sine and Cosine of 36. cos36=l-2sin 2 18 . . . [(3) of Art. 49] A/10 - 2 V5 .-. sin 36= Vl - cos 2 36 = 4 The above results are also the sine and cosine, respec- tively, of 54 (Art. 15). Otherwise thus: Let x = 36; then 2 x =72, and 3 x =108. (Art. 29) 2 sin & cos x = 3 sin # 4 sin 3 a?, 2 cos #== 3 4 sin 2 a; = 4 cos 2 x 1. Solving, cos = But 36 is an acute angle, and therefore its cosine is positive. ... oos 36'=^+! 4 59. If A + B + C = 180, or if A, B, C are the Angles of a Triangle, prove the Following Identities: ABC (1) sin A + sinB + sin C = 4cos cos cos 222 ABC (2) cos A -f cosB + cos C = 1 +4 sin sin sin 222 (3) tan A -f- tan B + tan C = tan A tan B tan C. ANGLES OF A TRIANGLE. 71 We have A + B + C = 180. /. sin(A+B) =sinO, and sin^I> = cos-- (Arts. 15 and 30) and .-. sin A + sinB + sin C = 2 cos - cos - 22 J AQLIL UUO - . ^.rvirb. *uy . (Art. 15) D = 2sin-cos- . . . 2 2 . (Art. 49) 9 A + B C . (Art. 15) rt O ^ 2 cos C cos A ~ B -4-2< C A+B = 2 cos ^ (2 cos cos l^j . (Art. 45) 2 y 2 2 J = 4cos|cos|cos| .... (1) Again, cos A + cos B = 2 cos -- cos A ~ B . (Art. 45) and cosC = l-2sin 2 5 .... (Art. 49) .-. cosA+ msin S in . . (2) 72 PLANE TRIGONOMETRY. Again, tan (A + B) = tanC ..... (Art. 30) = tanA + tanB (A 4 } 1 -tan A tan B .-. tanA + tanB = tanC (1 tanAtanB). .-. tanA + tanB+tanC = tanAtanB tanC .... (3) NOTE. The student will observe that (1), (2), and (3) follow directly from Examples 1 and 2, and formula (3), respectively, of Art. 48, by putting A + B + C = 180. EXAMPLES. Prove the following statements if A + B + C = 180 : 1. cos(A + B-C) = -cos2C. ABC 2. sinA + sinB sinC = 4sin sin cos-- 222 3. sin2A + sin2B + sin2C = 4sinAsinB sinC. 4. sin 2 A + sin 2 B sin 2 C = 4 sin C cos A cos B. 5. tan 7 A tan 4 A tan 3 A = tan 7 A tan 4 A tan 3 A. 6. sin A sin B -f sin C = 4 sin cos - sin 222 7. C ot - + cot ? + cot- = cot ^ cot cot?- 8. tan A cot B = sec A cosec B cos C. 60. Inverse Trigonometric Functions. The equation sin 6 x means that is the angle whose sine is x ; this may be written 6= sin" 1 ^, where sin" 1 ^ is an abbreviation for the angle (or arc) whose sine is x. So the symbols cos" 1 ^ tan" 1 ^ and sec" 1 !/, are read "the angle (or arc) whose cosine is x" " the angle (or arc) whose tangent is #," and " the angle (or arc) whose secant is y." These angles are spoken of as being the inverse sine of x, the INVERSE TEIGONOMETEIC FUNCTIONS. 73 inverse cosine of x, the inverse tangent of x, and the inverse secant ofy, respectively. Such expressions are called inverse trigon ometric functions. NOTE. The student must be careful to notice that 1 is not an exponent, sin' 1 a; is not (sin a:)' 1 , which = sin a Notice also that sin" 1 = cos" 1 - is not an identity, but is true only for the par- ticular angle 60. This notation is only analogous to the use of exponents in multiplication, where we have a- 1 tt = = l. Thus, cos" 1 (cos x) = x, and sin (sin' 1 ) =x; that is, cos" 1 is inverse to cos, and applied to 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" 1 \. We know that siii30= \. .-. 30 is an angle whose sine is J; or 30= sin- 1 . 2. Prove that tan- 1 + tan- 1 \ = 45. tan" 1 ^- is one of the angles whose tangent is J, and tan" 1 -^ is one of the angles whose tangent is J. Let a = tan- 1 ^-, and /8 = tan~ 1 ^; then tan a = J and tan (3 = ^. Now tan( + /3)= . . . (Art. 47) 1 tanatan/J -ti_ = l i-fxi But tan 45= 1, .-. a + = 45; that is, taa- 1 + tan' 1 J = 45. Therefore 45 is one value of tan -1 |- + tan- J |-. PLANE TRIGONOMETRY. i-i^tau-'-^L 3. Prove that tan- 1 a + tan- 1 ?/ " 1 ?/ = B. .. tan B = y. Now xy l-xy . Any relations which have been established among the trigonometric functions may be expressed by means of the inverse notation. Thus, we know that cos x = Vl sin 2 a;. This may be written x = cos" 1 Vl sin 2 a; . . (1) Put sino; = 0; then a; = siii~ 1 0. Thus (1) becomes sin- 1 B = cos" 1 Vl - 0*. 5. By Art. 49, cos 26 = 2 cos 2 - 1, which may be written 26 = cos" 1 (2 cos 2 1). Put cosO = x. .-. 2 cos- 1 x= cos- 1 2^ 1. 6. By Art. 49, sin 2 = 2 sin cos 0, which may be written 20 = sin" 1 (2 sin cos 0) . Put sin = x. .-. 2 sin- 1 a = sin- 1 (2 a Vl- TABLE OF USEFUL FORMULAE. 75 7. Prove sin- 1 a; = cos- 1 Vl x 2 = tan- 1 8. " tan" 1 x = sin- 1 x = cos' Vl + x 2 Vl + x 2 9. " 1 or* 10. sin (2 sin- 1 a?) = 2 a; vT^ 11. 12. cos- 1 - + 2 sin- 1 - = 120. 13. " cot- 1 3 + cosec- 1 V5 = - 4 14. 61. Table of Useful Formulae. The following is a list of important formulae proved in this chapter, and summed up for the convenience of the student : . . (Art. 43) 2. cos (a; + y) = cos x cos y sin x sin y. 3. sin(a; y) = sin x cos y cos x sin y . . (Art. 44) 4. cos (x y) = cos x cosy + sinx siny. 5. 2 sin xcosy = sin (a; -f y) + sin (a; y) . (Art. 45) 6. 2coscc siny = sin (x -f- y) sin (a; y). 7. 2 cos a; cos y = cos (# -f-2/) + cos (x y). 8. 2sino;sin2/ = cos (x y) cos (a; + y) . 9. sinas-f sin?/ = 2sin-j-(a; + y) co^J(a; y). 10. sinz siny = 76 PLANE TRIGONOMETRY. 11. coscc-j- cos?/ = 2cos-J-(# -|- y) cos^(x y). 12. cosy cos x = 2 sin -J (x + y) siu^(x y). 13. sin a; + sin?/ = tan^(gp + y) sin a; sin y tan J (a; y ) 14. tan (x + ?/) = . (Art. 47) l-tanztan?/ 15. tan(s-y) = 1 + tan x tan y 16. cot cot x -f- coty 17. COt 07 COt?/ 18. tan(a;45 ) = tana;Tl .... (Art. 47) tan x 1 19. sin (#-f ?/) sin (x y) = sin 2 a; sin 2 ?/ = cos 2 ?/ cos 2 a?. 20. cos (x+ y) cos (a; y) = cos 2 a? sin 2 ?/ = cos 2 ?/ sin 2 a?. 21. tan* tony = ?![*!. cos a? cos ?/ 22. sin a; sin?/ 23. sin2a? = 2sina;cosa;= 2tana; (Art. 49) 24. cos 2 a? = cos 2 a; sin 2 #= 1 2 sin 2 a; = 2 cos 2 a _ 1 tan 2 a; ~ 1 + tan 2 a;' 1-|- cos 2 x 2 cos 2 a? 26. tan 2 a; = 27. cot 2 a; = EXAMPLES. 77 2 tana 1 tan 2 a; COt 2 a? 1 2cot 28. sin3x = 3 sin a; -4 sin 3 a; (Art. 50) 29. cos 3 x = 4 cos 3 x 3 cos x. 3 tan a; tan 3 a; 30. tan3a = - 1 3 tan 2 a; 31. sin 2 ^ = 1 - < cosa? (Art. 51) 9 X 1 4- COS X 32. cos 2 - = ^- 2 2 33. tan" 1 a; -\- tan" 1 y = tan" 1 ^ .... (Art. 60) EXAMPLES. 1 2 1. If sin a = -, and sin /? = -, find a value for sin (+/?), 3 3 and sin (a ft). A V5 + 4V2. V5-4V2 T 2. If cos a = -, and cos fi = , find a value for sin ( a -f /?) , o 41 andcos(a + /8). . 156 133 ! ' 205' 205' o o 3. If cos = -, and cos /? = -, find a value for sin (a + )8), 4 5 and sin ( -j8). ^ ng 2 V7 + 3 V21 2V7-3V21 4. If sin = -, and sin/S = -, find a value for sin ( + /?), andcos( + /3). ^ Q 1 34. ' 25 78 PLANE TRIGONOMETRY. 5. If sin a = .6, and sin (3 = , find a value for sin (a /?), and cos ( + /). 16 33 65' 66' 6. If sin= -, and sin 5 = = , show that one value V5 Vio of a + is 45. 7. Prove cos 6 4- cos 3 =2 cos 2 cos 0. 8. " 2cosacos/8 = cos (a (3) + cos (a + /?). 9. " 2sin30cos50 =sin80-sin20. 10. " 2cos|0cos- =cos^-f cos20. 11. " sin 40 sin = (cos30 - cos 50). 12. " 2 cos 10 sin 50 = sin 60 + sin 40. 13. Simplify 2 cos 2 9 cos 6 2 sin 4 sin 0. ^Ins. 2 cos 30 cos 20. 50 00 S0 14. Simplify sin cos sin cos cos 40 sin 20. 22 22 Prove the following statements : 15. cos3 cos7a== 2 sin 5 sin 2 a. 16. sin 60+ sin 20= 2 sin 40 cos 20. 17. sin30-fsin50 =2sin40cos0. 18. sin70-sino0 =2cos60sin0. 19. cos50 + cos90 = 2cos70cos20. 20. Sin2 * + sin * =tan^- cos0 + cos20 2 21. cos (60 + A) + cos (60 - A) = cos A. EXAMPLES, 79 22. cos (45+ A) + cos (45- A) = V2cos A. 23. sin (45+ A) -sin (45- A) = V2sinA. 24. cos20 + cos40 = 2cos30cos0. 25. cos40 cos60 = 2sin50sin0. 26. cos + cos 3 6 + cos 5 + cos 7 = 4 cos cos 2 cos 4 0. 27. sma cos/3 28. ooU-tan/8 = cos <" ! + ^ sin a cos /? 29. sinA V2 30. V2 sin (A + 45) = sin A -f cos A. 31. cos (A + 45) + sin (A- 45) = 0. 32. tan (*-*)+ ten* = 1 tan (0 <) tan 33. = 34. cos (6 + <) - sin (0 - <) = 2sin - - 0}cos - - V 4 / 35. sin ?i0 cos + cos nO sin = sin (n -f 1 ) 0. 36. c l-cot0 37. tan ($-*] + cot (o + - = 0. V 38. cotf0- V 39. tan (n + 1)0- tan n^ = tap 80 PLANE TRIGONOMETRY. 40. If tanx = 1, and tany = -, prove that V O tan (a? + y) = 2 + VS. 41. If tana = , and tan/? = - , prove that m+1 2m +1 tan(4-/3) = l. 42. If tan a = m } and tan ft = w, prove that 1 ~ mn cos 43. If tan0 = (a -f- 1), and tan < = (a 1), prove that 2 cot (0 -) = a 2 . Prove the following statements : 44. cos OK 2/4- z) = cosx cosy cos z -\- cosx sin?/ sinz sin x cos y sin 2 -f sin x sin ?/ cos 2. 45. sin (x y z) = sin x + sin?/ + sins; 4- 4sin J(x y) sin J(x 2) sini(?/-f 2 46. sin (B -f- y z) + sin (# -f- z y) -f sin (y -\- z x) = sin (a; -}- y -f- z) 4- 4 sin a; sin ?/ sin z. 47. sin2# + sin2?/-f- sin 2z sin2 (x + y -f) = 4 sin (ic + y) sin (y -\- z) sin (z-\- x). 48. cos 2 a; 4- cos 2y -{- cos 2 2 4- cos 2 (x + y -\-z) = 4 cos (a; 4- ?/) cos (y 4- z) cos ( 4- a). 49. cos (x + i/ 2) 4 cos (y + 2 0,-) 4- cos (2 + a; y) 4- cos (a; 4- y + 2) = 4 cos x cos ?/ cos z. EXAMPLES. 81 50. sin 2 a; -f- sin 2 ?/ -f sin 2 z -f- sin 2 (x -f y -f z) = 2jl cos (# + ?/) cos (y+z) cos (z-f-a) \. 51. cos 2 a; -f cos 2 ?/ + cos 2 z + cos 2 (aj + y z) = 2jl -f- cos (a?-f ?/) cos (xz) cos (yz) \. 52. cosa; sin (y z) -f- cos?/ sin(z a;) + cosz sin(a; ?/) = 0. 53. sin x sin (?/ z) + sin ?/ sin (z a;) -f sin z sin (x y) = 0. 54. cos (a; -f- y) cos (# ?/)-}- sin (y + z) sin (?/ z) cos (a? -f 2) cos (a; z) = 0. O 2/1 55. sec 2 56. cos 2 0(1- tan 2 0) = cos 2 0. 57. cot20 = cot ^^l. 2cot0 58. sec20=" cot 2 0-l 59. ( sin - + cos- ) = 1 + sin ft V 2 2y 60. (sin - cos - ) = 1 sin 0. \ 2 2y 61. 2 62 cos 2 1 tan 63. l-tan| 82 PLANE TRIGONOMETRY. gj 1 + sin x -f- cos x _ x 1 + sin x cos a; 2 c ~ cos 3 x -h sin 3 x bo. cos x -f sin x 2 ~~ sin 3 x 2 + sin2a; DO. - - ; - = - cos# smas 2 67. cos 4 0-sin 4 <9 = cos 20. aQ sin 30 cos 30 bo. --- = . smO cosO 69. cosSf sinSS =2cot2ft sin ^ cos 70. sin 20 71. sin cos 72. tan (45 + a?) - tan (45 - a?) = 2 tan 2 a?. 73. tan (45 - x) + cot (45 - a;) = 2 sec 2 a. 74. tanW+x)-!^^ tan 2 (45+ a) + 1 -rf COS (iC + 45) 75. . __v Z - i = sec 2 ic tan 2 ic. cos (a? - 45) 76. tans= sin a; + sin 2 a; t 1 + cos a; + cos 2 x 77. 1 cos x -f- cos 2 a; 78. Cos3a; cos a; EXAMPLES. 83 QOIYI/Y* 01 n Q /> 79. cos 3 x + 3 cos x cot 8 a? 3 cot a; 80. cot3# = 3 cot 2 a; 1 Q1 1 cos3ir 51. 1 cos a; 82. cos x sin x go cos 2 x -f- cos 12 x cos 7 a cos 3 x 2sin4# cos 6 x -f cos 8 x cos cos 3 a? sin2x 84. sin2a; sm2y = sin 2 ( + y) sin 2 (o; y). 85. tan50H-cot50=2seclO. 86. sin 3 a = 4 sin a sin (60 + a) sin (60 - x) . 87. cot --tan- = 2. 8 8 88. 89. 90. (3sin0 - 4sin 3 0) 2 + (4cos 3 - 91 sin 2 6> cos (9 = t^ e (lH-cos2(9)(l+cos0) 2 92. If tan = -, and tan #> = , prove tan (2 + <) = i- 95. Prove that tan- and cot- are the roots of the 2 2 equation 84 PLANE TRIGONOMETRY. 94. If tan0 = -, prove that a 6 la -b 2cos0 b \ a + b Vcos 2 95. Find the values of (1) sin 9, (2) cos 9, (3) sin 81, (4) cos 189, (5) tan202i, (6) tan97|. Ans. (1) i ( V3 + V5 - V5 - V5), (2) 1(^3 + V5+- V5- V5), (3) sin 81= cos 9, (4) cos 189= -cos 9, (5) V2 - 1, (6) - 96. If A = 200, prove that (1) 2sin^ = + Vl-f-sinA + Vl-sjnA. 2 tan A 97. If A lies between 270 and 360, prove that (1) 2sin- = + Vl - sin A - Vl + sinA. 2 (2) tan - = cot A + cosec A. ft 98. If A lies between 450 and 630, prove that . A (1) 2 sin = Vl + sin A Vl sin A. (2) 2cos = Vl -fsinA + Vl sin A. EXAMPLES. 85 Prove the following statements, A, B, C being the angles of a triangle. 99. 100. sin A + sin B 2 2 sin3B sin3C _tan3A cos3C-cos3B~ 101. sin - cos - + sin 2 cos ?+ sin- cos - 222222 102. cos 2 ^ + cos 2 |- cos 2 ^ = 2 cos- cos? sin 5. 103. sin A cos A sin B cos B -f sm C cos C = 2 cos A sin B cos C. 104. cos 2 A + cos 2 B + cos2C = 1 4 cos A cos B cos C. 105. sin 2 A sin 2 B + sin 2 C *= 2 sin A cos B sin C. 106. tan ? tan - + tan - tan - + tan - tan - = 1. 2 2 2 2 2 2 Prove the following statements when we take for sin" 1 , cos" 1 , etc., their least positive value. 107. sin- 1 - = 008-^ = cot" 1 V3. 2i 2i 108. 2tan-'(cos2e) = 109. 86 PLANE TRIGONOMETRY. 111. tan- 1 V5 (2 - V3) - cof 1 V5 (2 + V3) = coir 1 VB. 112. sec- 1 V3 1 4- X 2 113. 2 cot" 1 a; = cosec" 1 115. Bin- l =: V5 116. 117. If 6 = sin- 1 -, and < = cos- 1 ? then + 4 = 90. 5 5 1 _ ^,2 118. Prove that cos (2 tan- 1 a) = '> 1 -f- x" 119 " " tan- 1 ^ ^^ -h cosec- 1 VlO = ^- 2 4 2 5 33 120. " " 2tan-V cosec- 1 - = sin- 1 O O OO 121. 122. " " sin- 1 (cos x) + cos" 1 (sin y) + cc + y = TT. 1 12 123. " tan- 1 h tan- 1 - - + tan" 1 -5 = TT. 1 _i_ x 1 a; or 124. " " tan- 1 - + tan- 1 - = nir + - 125. " " sin- 1 x sin- 1 y = cos- 1 (xy VI - w 2 - 2/ 2 4- *T NATURE AND USE OF LOGARITHMS. 87 CHAPTER IV. LOGARITHMS AND LOGARITHMIC TABLES, TRIGO- NOMETRIC 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 given base is the exponent of the power to which the base must be raised to give the number. Thus, if a x = m, x is called the " logarithm of m to the base a," and is usually written o? = log a m, the base being put as a suffix.* The relation between the base, logarithm, and number is expressed by the equation, (base) log = number. Thus, if the base of a system of logarithms is 2, then 3 is the logarithm of the number 8, because 2 3 = 8. If the base be 5, then 3 is the logarithm of 125, because 5 3 = 125. 63. Properties of Logarithms. The use of logarithms depends on the following properties which are true for all logarithms, whatever may be the base. *From the definition it follows that (1) log a a* = x, and conversely (2) a lo &a" = m. Taking the logarithms of both sides of the equation a x = m, we have log a a x = x=\og m. Conversely, taking the exponentials of both sides of x = log a m to base a, we have a x = o lo ?a m = m. a x = m and x = \og n m are thus seen to be equivalent, and to express the same relation between a number, in, and its logarithm, x, to base a. 88 PLANE TRIGONOMETRY. (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 l =i a, whatever a may be ; therefore log a a = 1. (3) The logarithm of zero in any system whose base is greater than 1 is minus infinity. For or 00 = = - - = : therefore log = cc . a 00 co (4) The logarithm of a product is equal to the sum of the logarithms of its factors. For let x = log a w, and y log a ?i. .-. m = a x , and n = a y . .. mn = a x+y . .. log a mn = x + y = log a m + log a n. Similarly, log a mnp = log a m + log a w + \gaP> and so on for any number of factors. Thus, log 60 = log (3 x 4 x 5), = log 3 + log 4 + log 5. (5) The logarithm of a quotient is equal to the logarithm of the dividend minus the logarithm of the divisor. For let # = log a ra, and y = log a w. . . m = a x , and n = a*. .-. log a = x y = log a m log a n. Thus, log = log 17 - log 5. PROPERTIES OF LOGARITHMS. 89 (6) The logarithm of any power of a number is equal to the logarithm of the number multiplied by the exponent of the poiver. For let aj = log a m. .-. m == a*. .-. m p = a px . . : log a m p = px = p log a m. (7) The logarithm of any root of a number is equal to the logarithm of the number divided by the index of the root. For let x = log a m. .-. m = a*. 1 X .-. m r = a r . 1 x 1 .-. log(m r ) = - = -log a m. It follows from these propositions that by means of logarithms, the operations of multiplication and division are changed into those of addition and subtraction; and the operations of involution and evolution are changed into those of multiplication and division. 1. Suppose, for instance, it is required to find the product of 246 and 357; we add the logarithms of the factors, and the sum is the logarithm of the product : thus, Iog 10 246 = 2.39093 Iog 10 357 = 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 w 371.49 = 2.56995 log M 52.376 = 1.71913 0.85082 which is the logarithm of 7.092752, the quotient required. 90 PLANE TRIGONOMETRY. 3. If we have to find the fourth power of 13, we proceed thus: Iog 10 13 = 1.11394 4 4.45576 which is the logarithm of 28561, the number required. 4. If we are to find the fifth root of 16807, we proceed thus : 5)4.22549 = Iog 10 16807, 0.845098 which is the logarithm of 7, the root required. 5. Given Iog 10 2 = 0.30103 ; find Iog 10 128, Iog 10 512. Ans. 2.10721, 2.70927. 6. Given Iog 10 3 = 0.47712 ; find Iog 10 81, Iog 10 2187. Ans. 1.90849, 3.33985. 7. Givenlog 10 3; find log ]0 -tys*. 0.28627. 8. Find the logarithms to the base a of o -V> 4/~ 3/ r, -5 a 3 , a 3 , y a, y a 2 , a 2 . 9. Find the logarithms to the base 2 of 8, 64, , .125, .015625, -v/64. Ans. 3, 6, -1, -3, -6, 2. 10. Find the logarithms to base 4 of 8, A/16, ^.5? AV.015625. Ans. f, f, - J, - 1. Express the following logarithms in terms of log a, log 6, and logc: 11. logVO'^c) 6 . Ans. 61og + 91og6 + 31ogc. floga + | log 6 -f !logc. 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 m = 10*, for all positive values of m, integral or fractional. Thus, if we suppose m to assume in succession every value from to oo, the corresponding values of x will form a system of logarithms, to the base 10. Such a system is formed 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. Now 10 =1, .-. logl =0; 10 T =:10, .-. log 10 =1; 10 2 =100, .-. log 100 =2; 10 3 =1000, .-. log 1000 = 3. and so on. Also, 10^= T V = .1, .-. log.l =-1; 10 " 2 =yU =- 01 > -- log.Ol =-2; 10- 3 = T VTF - -001, .-. 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 called from its inventor, Baron Napier, a Scotch mathematician. f First introduced in 1G15 by firiygs, a contemporary of Napier. 92 PLANE TRIGONOMETRY. 10 and 100 is some number between 1 and 2 ; i.e., 1 -f- a decimal ; 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 + 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 always kept positive. NOTE. It is convenient to keep the decimal part of the logarithms always posi- tive, in order that numbers consisting 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, -5, 4. NOTE. When the characteristic is negative, 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 10 , 2 12 , 16 20 , 2 100 . 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 ra be any number whose logarithm to base b is required. Let o; = log 6 m; then b* = m. log a m; or xlog a b = log a m. Hence, to transform the logarithm of a number from base a to base b, we multiply it by -- log. b 94 PLANE TRIGONOMETRY. This constant multiplier is called the modulus of log. & the system of which 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 logelO Putting a for m in (1), we have log, a = ^ = JL, by (2) of Art. 63. log a & Iog 6 .-. Iog 6 a x log a & = 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 3 )* = 5 to . .'. 3z = log 5 m. Thus, the logarithm of any number to base 5, divided by 3 (i.e., by Iog 5 125), is the logarithm of the same number to the base 125. Otherwise by the rule given in (1). Thus, Show how to transform 2. Logarithms with base 2 to logarithms with base 8. Ans. Divide each logarithm by 3. TABLES OF LOGARITHMS. 95 3. Logarithms with base 9 to logarithms with base 3. Am. Multiply each logarithm by 2. 4. Find Iog 2 8, Iog 5 l, Iog 8 2, Iog 7 l, Iog 32 128. Ans. 3, 0, |, 0, 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 mantissas have to be given in the tables. (2) The mantissas are the same for the logarithms of all numbers which have the same significant digits, in the same order, so that it is sufficient to tabulate the mantissas 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 P and N -5- 10* are numbers whose significant digits are the same as those of K Then log (N x 10 P ) = logN +ploglO = logN +p. (1) Also, log (N -s- 10') = log N - q log 10 = log N - q. (2) 96 PLANE TRIGONOMETRY. 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 log 787 =2.895975. Now log 7.87 = log ~ = log 787 - 2 100 = 0.895975. Also, log .0787 = log = log 787 - 4 = 2.895975. Also, log 78700 = log (787 x 100) =. log 787 + 2 = 4.895975. NOTE 1. We do not write log to 787 ; for so long as we are treating of logarithms to the particular base 10, we may omit the suffix. NOTE 2. Sometimes in working 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. NOTE 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~.60S582, we may write 7.603582 10. In calculations with negative characteristics we follow the rules of Algebra. EXAMPLES. 97 EXAMPLES. 1. Add together 2.2143 1.3142 5.9068 7.4353 Ans. 2. From 3.24569 take 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 characteristic has to be divided by a number which is 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 + 2.7268 Given that Iog2=.30103, Iog3=.47712, and Iog7=.84510; find the values of 5. log 6, log 42, log 16. Ans. .77815, 1.62325, 1.20412. 98 PLANE TRIGONOMETRY. 6. log 49, log 36, log 63. Am. 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. Ana. 1.77815, 2.47712, .02119, 5.63548. 9. Given Iog2=.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) 3 , (.00042) 5 , (.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 we shall use tables of seven, places of decimals. f These tables are arranged so as to give the mantissse 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-figure tables used in this country are Stanley's, Vega's, Bruhns', etc. In the appendix to the Elementary Trigonometry are given five-figured tables, which are sufficiently near for most prac- tical applications. 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 limits 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 largest 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 number are those in the left column headed N, followed by one in larger type at the top of the page. The first three figures of the mantissas 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 which we retain, that figure which gives the nearest approach 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 figures, 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 we require the logarithm of 62541. The table gives .7961648 as the mantissa, and the characteristic is 4, by the rule ; therefore log 62540. = 4.7961648. Similarly, log .006^81 = 3.7980288 . . (Art. 64) 100 PLANE TRIGONOMETRY. Suppose, however, that the given number has more than five digits. For example : Suppose we require to find log 62761.6. We find from the table log 62761 = 4. 7976899 log 62762 = 4.7976968 and diff. for 1 = 0.0000069 Thus for an increase 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 an 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 this 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 rasps fiiifticiently exact for practical purposes. Had we taken a whole number or a decimal, the process would have been the same. TABLE OF 1,01 N. O 1 2 3 4 5 6 9217 7 8 1 9 F 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 '.P. 6250 51 52 53 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 6300 795 8800 8870 8939 9009 9078 9148 9,^v 53A6 0051 0745 1440 2134 2829 3523 4217 4911 5605 9126 0120 0815 1509 2204 2898 3592 4286 4980 5674 6368 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 62.1 68 6.8 13.6 20.4 27.2 34.0 40.8 47.6 54.4 61.2 9495 796 0190 0884 1579 2273 2967 3662 4356 5050 9564 0259 0954 1648 2343 3037 3731 4425 5119 9634 0329 1023 1718 2412 3106 3800 4494 5188 9703 0398 1093 1787 2481 3176 3870 4564 5258 9773 0468 1162 1857 2551 3245 3939 4633 5327 9842 0537 1232 1926 2620 3314 4009 4703 5396 9912 0606 1301 1995 2690 3384 4078 4772 5466 9981 0676 1370 2065 2759 3453 4147 4841 5535 7965743 5813 5882 5951 6645 7339 8032 8725 9419 0112 0805 1498 2191 6021 6090 6160 6229 6298 6992 7685 8379 9072 9765 0458 1151 1844 2537 6437 7131 7824 8517 9211 9904 797 059 7 1290 1983 6506 7200 7893 8587 9280 9973 0666 1359 2052 6576 7269 7963 8656 9349 0043 0736 1428 2121 6714 7408 8101 8795 9488 0181 0874 1567 2260 6784 7477 8171 8864 9557 0250 0943 1636 2329 6853 7547 8240 8933 9627 0320 1013 1706 2398 6923 7616 8309 9003 9696 0389 1082 1775 2468 3160 3853 4545 5237 5930 6622 7314 8006 8697 9389 7061 7755 8448 9141 9835 0528 1221 1913 2606 797 2675 27452814 2883 2952 3022 3091 3229 3299 3368 4060 4753 5445 6137 6829 7521 8213 8905 797 9596 3437 4130 4822 5514 6207 6899 7590 8282 8974 9666 3507 4199 4891 5584 6276 6968 7660 8351 9043 9735 0426 1118 1809 2500 3191 3882 4573 5263 5954 3576 4268 4961 5653 6345 7037 7729 8421 9112 9804 3645 4337 5030 5722 6414 7106 7798 8490 9181 9873 3714 4407 5099 5791 6483 7175 7867 8559 9251 3784 4476 5168 5860 6553 7245 7936 8628 9320 3922 4614 5307 5999 6691 7383 8075 8766 9458 3991 4684 5376 6068 6760 7452 8144 8836 9527 9942 0011 0080 0150 6219 0910 1601 2293 2984 3675 4366 5056 5747 6437 798 0288 0979 1671 2362 3053 3744 4435 5125 5816 0357 1048 1740 2431 3122 3813 4504 5194 5885 0495 1187 1878 2569 3260 3951 4642 5333 6023 0565 1256 1947 2638 3329 4020 4711 5402 6092 0634 1325 2016 2707 3398 4089 4780 5471 6161 0703 1394 2085 2776 3467 4158 4849 5540 6230 0772 1463 2154 2846 3536 4227 4918 5609 6299 0841 1532 2224 2915 3606 4296 4987 5678 6368 798 6506 6575 6645 6714 6783 6852 6921 6990 7680 8370 9060 9750 0440 1130 1820 2509 3199 7059 7749 8439 9129 9819 0509 1199 1889 2578 3268 7128 7197 7887 8577 9267 9957 799 0647 1337 2027 2716 799 3405 7266 7956 8646 9336 0026 0716 1406 2096 2785 7335 8025 8715 9405 0095 0785 1475 2164 2854 7404 8094 8784 9474 0164 0854 1544 2233 2923 7473 8163 8853 9543 0233 0923 1613 2302 2992 7542 8232 8922 9612 0302 0992 1682 2371 3061 7611 8301 8991 9681 0371 1061 1751 2440 3130 3819 7818 8508 9198 9888 0578 1268 1958 2647 3337 3474 3543 3612 3681 3750 3888 3957 4026 N. 1 2 3 4 5 6 7 8 9 P.P. 102 PLANE TRIGONOMETRY. Thus,' suppose' we require to find log 627616 and log .627616. The mantissa is exactly the same as before (Art. 66), and the only difference to be made in the final result is to change the characteristic according to rule (Art. 64). Thus log 627616 = 5.7976942, and log .627616 = 1.7976942. 69. To find the Number corresponding to a Given Loga- rithm. If the decimal part of the logarithm is found ex- actly in 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 find 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. 64). 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 log 62726 = 4.7974476 log 62725 = 4.7974407 diff . for 1 = .0000069 Thus for a difference of 1 in the numbers there is a difference of .0000069 in the logarithms. The excess of the given mantissa 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. .0000046 46 69) 46.0 (.666 .-. number to be added = - = = .667 .0000069 69 41 4 .-. log 62725.667 = 4.7974453, ^60 and .-. log 627.25667 = 2.7974453; 414 therefore number required is 627.25667. 460 ARITHMETIC COMPLEMENT. 103 We might have saved the labor of dividing 46 by 69, by using the table of proportional parts as follows : given mantissa = .7974453 mantissa of 62725 = .7974407 diff. of mantissse = 46 proportional part for .6 = 41.4 4.6 " " " .06 = 4.14 .46 .006= .414 and so on. .-. number = 627.256666-... 69 a. 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 - b = a + (10 - b) - 10, where a and b are logarithms, and 10 6 is the arithmetic complement of b. 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 the 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 2. Given find 3. Given find 4. Given find 5. Given EXAMPLES. log 52502 =4.7201758, log 52503 =4.7201841; log 52502.5. log 3.0042 = 0.4777288, log 3.0043 = 0.4777433; log 300.425. log 7.6543 = 0.8839055, log 7.6544 = 0.8839112; log 7.65432. log 6.4371 = 0.8086903, log 6.4372 = 0.8086970; log 6437125. Ans. 4.7201799. 2.4777360. .8839066. log 12954 =4.1124039, log 12955 =4.1124374; find the number whose logarithm is 4.1124307. 6. Given log 60195 = 4.7795532, log 60196 =4.7795604; find the number whose logarithm is 2.7795561. 7. Given log 3.7040 = .5686710, log 3.7041 = .5686827; find the number whose logarithm is .5686760. 8. Given log 2.4928 = .3966874, log 2.4929 = .3967049; find the number whose logarithm is 6.3966938. 6.8086920. 12954.8. 601.95403. 3.70404. 2492837. NATURAL TRIGONOMETRIC FUNCTIONS. 105 9. Given log 32642 = 4.5137768, log 32643 =4.5137901; find log 32642.5. Arts. 4.5137835. 10. Find the logarithm of 62654326. 7.7969510. Use specimen page. 11. Find the number whose logarithm is 4.7989672. Ana. 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 = 1; that is, .5 exactly. 2i 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 io 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. 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. 72. To find the Sine of a Given Angle. Find the sine of 25 14' 20", having given from the table sin 25 15' = .4265687 sin 25 14' = .4263056 diff. for 1' == .0002631 Let d = diff. for 20" ; and assuming that an increase in the angle is proportional to an increase in the sine, we have 60 : 20 : : .0002631 : d. ^ = 20 x. 0002631 60 .-. sin 25 14' 20" = .4263056 + .0000877 = .4263933. NOTE. We assumed here that an increase in the angle is proportional to the increase in the corresponding sine, which is sufficiently 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 36' = .7120260 diff. for 1' = .0002043 observing that the cosine decreases as the angle increases from to 90. EXAMPLES. 107 Let d = decrease of cosine for 25" ; then 60 : 25 : : .0002043 : d. f)K ... d = x .0002043 = .0000851. .*. cos 44 35' 25" = .7122303 - .0000851 = 7121452. Similarly, we may find the values of the other trigono- metric functions, remembering that, in the first 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 5. Given find EXAMPLES. sin 44 35'= .7019459, sin 44 36'= .7021531; sin 44 35' 25". sin 42 15' =.6723668, sin 42 16' =.6725821; sin 42 15' 16". sin 43 23'= .6868761, sin 43 22'= .6866647; sin 43 22' 50". sin 31 6' =.5165333, sin 31 7' =.5167824; sin 31 6' 25". cos 74 45'= .4265687, cos 74 46' =.4263056; cos 74 45' 40". Ans. .7020322. .6724242. .6868408. .5166371. .4263933. 108 PLANE TRIGONOMETRY. 6. Given cos 41 13'= .7522233, cos 41 14'= .7520316; find cos 41 13' 26". Ans. .7521403. 7. Given cos 47 38'= .6738727, cos 47 39'= .6736577; find cos 47 38' 30". .6737652. 74. To find the Angle whose Sine is Given. Find the angle whose sine is .5082784, having given from the table gin 3Q o 33 , = 5082 901 sin 30 32' = .5080396 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. Find the angle whose cosine is .4043281, having given from the table cos 66 9' = .4043436 cos 66 10' = .4040775 diff. for 1' = .0002661 cos 66 9' = .4043436 given cosine = .4043281 diff. = .0000155 EXA3IP 109 Let d = diff. between 66 9' and required angle; tken .0002661: .0000155 : : 60 : d. Required angle is greater than 66 9 1 because its cosine is less than cos 66 9'. .-. required angle =66 9' 3".5. EXAMPLES 1. Given sin 44 12' = .6971651, sin 44 11'=. 6969565; find the angle whose sine is .6970886. Ans. 44 11' 38". 2. Givej sin 48 47' = . 7522233, sin 48 46' = .7520316; find the angle whose sine is .752140. 48 46' 34". 3. Given sin 24 11' = .4096577. sin 24 12' = .4099230; find the angle whose sine is .4097559. 24 11' 22".2. 4. Given cos 32 3T = .8432351, cos 32 32* = .430787; find the angle whose cosine is .8432. 32 31* 13".5. 5. Given cos 44 11' = .7171134, cos 44 12* = . 7169106; find the angle whose cosine is .7169848. 44 11' 38". 6. Given cos 70 32' = .3332584, 0870-31' = . 3335326; find the angle whose cosine is .3333333. 79 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 which 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 + logi = 10 - log 2 = 9.6989700. In calculations we have to remember and allow for this increase 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 the logarithm of the sine of 31 15' is 9.7149776 - 10 = 1.7149776. Similarly with the logarithms of other functions. TABLES OF LOGARITHMIC FUNCTIONS. Ill NOTE. English authors usually denote these tabular logarithms by the letter L Thus, Lsin A denotes the tabular logarithm of the sine of A. French authors use the logarithms 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 angle less than 45 is the cosine, cotangent, etc., of another angle greater than 45 (Art. 16). Hence the degrees at the top of the tables are generally marked from 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 there- fore in eacji 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. * Many tables are calculated for angles at intervals of 10". 112 PLANE TRIGONOMETRY. 77. To find the Logarithmic Sine of a Given Angle. Find log sin 38 52' 46". We have from page 113 log sin 38 53' = 9.7977775 log sin 38 52' = 9.7976208 cliff, for 1' = .0001567 Let d diff. for 46", and assuming that the chancre in the log sine is proportional to the change in the angle, we have 60 : 46 : : .0001567 : d. d=s 46 x. 0001567 ^ 0001201 60 .-. log sin 38 52' 46"= 9.7976208 + .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. for 1'= .0011017 Let d = decrease of log cosine for 23" ; then 60 : 23 : : .0011017 : d. 23x1017 .-. log cos 83 27' 23"= 9.0571723 - .0004223 = 9.0567500. EXAMPLES. 1. Giren log sin 6 33'= 9.0571723, log sin 6 32' =9.0560706; find log sin 6 32' 37". Ana. 9.05675. 38 Deg. TABLE OF LOGARITHMS. 113 ; Sine. Diff. Tang. Diff. Cotang. Diff. Cosine. f o I 2 3 i 4 5 9.7893420 9.7895036 9.7896652 9.7898266 9.7899880 9.7901493 1616 1616 1614 1614 1613 9.8928098 9.8930702 9-8933306 9.8935909 9.8938511 9.8941114 2604 2604 2603 2602 2603 10.1071902 10.1069298 10.1066694 10.1064091 10.1061489 10.1058886 987 988 988 989 990 QQO 9.8965321 9.8964334 9.8963346 9.8962358 9.8961369 9.8960379 60 59 58 55 6 7 8 9 10 9.7903104 9.7904715 9.7906325 9.7907933 9.7909541 1611 1610 1608 1608 1607 9.8943715 9.8946317 9.8948918 9.8951519 9.8954119 2602 2601 2601 2600 2600 10.1056285 10.1053683 10.1051082 10.1048481 10.1045881 991 992 992 992 QQO 9-8959389 9.8958398 9.8957406 9.8956414 9.8955422 54 53 52 So ii 12 13 14 15 9.7911148 9.7912754 97914359 9-7915963 9.7917566 1606 1605 1604 1603 9.8956719 9-89593 J 9 9.8961918 9.8964517 9.8967116 2600 2599 2599 2599 2CQ8 10.1043281 10.1040681 10.1038082 10.1035483 10.1032884 994 995 995 995 QQ7 9.8954429 9-8953435 9.8952440 9.8951445 9.8950450 49 48 45 16 17 18 19 20 9.7919168 9.7920769 9.7922369 9.7923968 9.7925566 1601 1600 1599 1598 9.8969714 9.8972312 9.8974910 9.8977507 9.8980104 2598 2598 2597 2597 10.1030286 10.1027688 10.1025090 10.1022493 10.1019896 996 99 8 99 8 99 8 IOOO 9.8949453 9.8948457 9.8947459 9.8946461 9.8945463 44 43 42 40 21 22 23 24 25 9.7927163 9.7928760 97930355 97931949 9.7933543 1597 1595 1594 1594 9.8982700 9.8985296 9.8987892 9.8990487 9.8993082 2596 2596 2595 2595 2CQC 10.1017300 10.1014704 10.1012108 10.1009513 10.1006918 999 IOOI IOOI IOOI 1003 9-8944463 9.8943464 9.8942463 9.8941462 9.8940461 P P 35 26 27 28 2 9 3 9.7935I35 9.7936727 9-793 8 3 I 7 9.7939907 9.7941496 1589 9.8995677 9.9003459 9.9006052 2594 2594 2594 2593 10.1004323 10.1001729 10.0999135 10.0996541 10.0993948 1002 1004 1004 1004 9.8939458 9.8938456 9.8937452 9.8936448 9.8935444 34 33 32 3 1 30 32 33 34 35 9.7943083 9.7944670 9-7946256 9.7947841 9.7949425 1587 1586 1585 1584 9.9008645 9.9011237 9.9013830 9.9016422 2592 2593 2592 2591 2CQI 10.0991355 10.0988763 10.0986170 10.0983578 10.0980987 1006 1007 1007 1007 1008 9.8934439 9.8933433 9-8932426 9.8931419 9.8930412 27 26 25 36 37 38 39 40 9.7951008 9.7952590 9.7954171 9.7955751 9-7957330 1582 9.9021604 9.9024195 9.9029376 9.9031966 2591 2591 2590 2590 10.0978396 10.0975805 10.0973214 10.0970624 10.0968034 1009 1010 1010 1010 IOII 9.8929404 9.8928395 9.8927385 9.8926375 9.8925365 24 23 22 21 2O 4i 42 43 44 45 9.7958909 9.7960486 9.7962062 9-7963638 9.7965212 1 9.9034555 9.9037144 9.9039733 9.9042321 2589 2589 2588 2589 2^87 10.0965445 10.0962856 10.0960267 10.0957679 10.0955090 1012 1013 1013 1013 IOI4 9.8924354 9.8923342 9.8922329 9.8921316 9.8920303 19 18 17 16 15 46 47 48 49 50 9.7966786 9-7968359 9.7969930 9.7971501 9.7973071 s? 1571 1570 1560 9.9047497 9.9050085 9.9052672 9.9055259 9.9057845 2588 2587 2c;86 10.0952503 10.0949915 10.0947328 10.0944741 10.0942155 IOI5 1016 1016 1016 1018 9.8919289 9.8918274 9.8917258 9.8916242 9.8915226 14 13 12 II 10 52 53 54 55 9.7974640 9.7976208 9.7977775 9.7979341 9.7980906 1568 1567 1566 '565 9.9060431 9.9063017 9.9065603 9.9068188 9.9070773 2586 2586 2585 2585 2^84 10.0939569 10.0936983 10.0934397 10.0931812 10.0929227 1017 1019 1019 IO2O 9.8914208 9.8913191 9.8912172 9.8911153 9.8910133 9 8 7 6 5 56 1 9.7982470 9.7984034 97985596 9.7987158 9.7988718 1564 1562 1562 1560 9-9073357 9.907594I 9.9078525 9.9081109 9.9083692 2584 2584 2584 2583 10.0926643 10.0924059 10.0921475 10.0918891 10.0916308 IO2I IO2I IO22 1023 9.8909113 9.8908092 9.8907071 9.8906049 9.8905026 4 3 2 I O t Cosine. Diff. Cotang, Diff. Tang. Diff. Sine, t 51 Deer. 114 PLANE TRIGONOMETRY. 2. Given log sin 55 33'= 9.9162539, log sin 55 34'= 9.9163406 ; find log sin 55 33' 54". Ans. 9.9163319. 3. Given log cos 37 28'= 9.8996604, log cos 37 29'= 9.8995636 ; find log cos 37 28' 36". 9.8996023. 4. Given log cos 44 35' 20"= 9.8525789, log cos 44 35' 30"= 9.8525582 ; find log cos 44 35' 25".7. 9.8525671. See foot-note of Art. 76. 5. Given log cos 55 11'= 9.7565999, log cos 55 12'= 9.7564182 ; find log cos 55 11' 12". 6. Given log tan 27 13'= 9.7112148, log tan 27 14'= 9.7115254 ; find log tan 27 13' 45". 9.7565636. 9.7114477. 79. To find the Angle whose Logarithmic Sine is Given. Find the angle whose log sine is 8.8785940, having given from the table log sin 4 21 '=8.8799493 log sin 4 20' = 8.8782854 diff. for 1'= .0016639 given log sine = 8.8785940 log sin 4 20'= 8.8782854 diff. = .0003086 Let d = diff. between 4 20' and required angle ; then .0016639 : .0003086 : : 60 : d. 3086 x 60 cl = = 24, nearly. 16639 .-. required angle = 4 20' 24". EXAMPLES. 115 80. To find the Angle whose Logarithmic Cosine is Given. Find the angle whose log cosine is 9.8934342. ^Ve have from page 113 log cos 38 31'= 9.8934439 log cos 38 32'= 9.8933433 diff. for 1'= .0001006 log cos 38 31'= 9.8934439 given log cosine = 9.8934342 diff. = .0000097 Let d = diff. between 38 31' and required angle ; then .0001006 : .0000097 : : 60 : d. . ,0000097 x .. .0001006 1006 .-. required angle = 38 31' 5".8. XOTE. In using both the tables of the natural sines, cosines, etc., and the tables of the logarithmic sine.-*, cosines, etc., the student will remember that, in the first quadrant, as the angle increases, the sine, tangent, and secant increase, but the cosine, cotangent, and cosecant decrease. EXAMPLES. 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; fmd 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. 116 PLANE TRIGONOMETRY. 4. Given log cos 80 53'= 9.1998793, log cos 80 52' 50"= 9.2000105 ; find the angle whose log cosine is 9.2000000. Ans. 80 52' 51". 5. Given log tan 35 4'= 9.8463018, log tan 35 5'= 9.8465705; find the angle whose log tangent is 9.8464028. 35 4' 23". 6. Given log sin 44 35' 30"= 9.8463678, log sin 44 35' 20"= 9.8463464; find the angle whose log sine is 9.8463586. 44 35' 25". 7. 7. Find the angle by page 113 whose log tangent is 10.1018542. Ans. 51 39' 28". 7. 81. Angles near the Limits of the Quadrant, It was assumed in Arts. 72-80 that, in general, the differences of the trigonometric functions, both natural and logarithmic, are approximately proportional to the differences of their corresponding angles, with certain exceptions. The excep- tional cases are as follows : (1) Natural functions. For the sine the differences are insensible for angles near 90; for the cosine they are in- sensible for angles near 0. For the tangent the differences are irregular for angles near 90 ; for the cotangent they are irregular for angles near 0. (2) Logarithmic functions. The principle of propor- tional parts fails both for angles near and angles near 90. For the log sine and the log cosecant the differences are irregular for angles near 0, and insensible for angles near 90. For the log cosine and the log secant the differences are in- sensible for angles near 0, and irregular for angles near 90. For the log tangent and the log cotangent the differences are irregular for angles near and angles near 90. EXAMPLES. 117 It follows, therefore, that angles near and angles near 90 cannot be found with exactness from their log trigono- metric functions. These difficulties may be met in three ways. (1) For an angle near use the principle that the sines and tangents of small angles are approximately proportional to the angles themselves. (See Art. 130.) (2) For an angle near 90 use the half angle (Art. 99). (3) In using the proportional parts, find two, three, or more orders of differences (Alg., Art. 197). Special tables are employed for angles near the limits of the quadrant. EXAMPLES. 1. Given log, 7 = .8450980, find Iog 10 343, Iog 10 2401, and Iog 10 16.807. Ans. 2.5352940, 3.3803920, 1.2254900. 2. Find the logarithms to the base 3 of 9, 81, , ^, .1, V Ans. 2, 4, - 1, - 3, - 2, - 4. 3. Find the value of Iog 2 8, Iog 2 .5, Iog 3 243, Iog 5 (.04), Iog 10 1000, log w .001. Ans. 3, - 1, 5, - 2, 3, - 3. 4. Find the value of log a a% Iog 6 ^&*, Iog 8 2, log^ 3, Iog lfl0 10. . |, |, , , i Given Iog 10 2 = .3010300, log, 3 = .4771213, and Iog 10 7 = .8450980, find the values of the following : 5. Iog 10 35, Iog 10 150, Iog 10 .2. Ans. 1.544068, 2.1760913, 1.30103. 6. lo glo 3.5, lo glo 7.29, log ]0 . 081. Ans. .5440680, .8627278, 2.9084852. 7. lo glo i, Io glo 3 5 , lo Ans. .3679767, 2.3856065, .0780278. 118 PLANE TRIGONOMETRY. 8. Write down the integral part of the common loga- rithms of 7963, .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 1() , 49 6 , 343 A '*, (V ) 20 , (4.9) 12 , (3.43) 10 . Ans. 9, 11, 85, 4, 9, 6. 12. Find the position of the first significant figure in the numerical value of 20 7 , (.02) 7 , (.007) 2 , (3.43)^, (.0343) 8 , (.0343) T V 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 .30103. 14. Logarithms with base 3 to common logarithms. Ans. Multiply each log by .4771213. 15. Given Iog 10 2 = .3010300, find Iog 2 10. 3.32190. 16. Given Iog 10 7 = .8450980, find Iog 7 10. 1.183. 17. Given Iog 10 2 = .3010300, find Iog 8 10. 1.10730. 18. The mantissa of the log of 85762 is 9332949; find (1) the log of ^.0085762, and (2) the number of figures in (85762) ", when it is multiplied out. Ans. (1) 1.8121177, (2) 55. EXAMPLES. 119 19. What are the characteristics of the logarithms of 3742 to the bases 3, 6, 10, and 12 respectively ? Ans. 7, 4, 3, 3. 20. Prove that 7 log j-f + 6 log f + 5 log + log f f = log 3. 21. Given log, 7, find Iog 7 490. Ans. 2 -f - logio 7 22. From 5.3429 take 3.6284. 3.7145. 23. Divide 13.2615 by 8. 2.4076. 24. Prove that 6 log f + 4 log T \ + 2 log y = 0. 25. Find log [297 VII}* to the base 3Vll. 1.8. Given log 2 = .3010300, log 3 = .4771213. 26. Find log 216, 6480, 5400, f. Ans. 2.3344539, 3.8115752, 3.7323939, 1.6478174. 27. Find log .03, 6~ Ans. 2.4771213, 1.7406162, 1.6365006. 28. Find log .18, log 2.4, log T 3 g, -4ns. 1.2552726, .3802113, 1.2730013. 29. Find log (6.25)*, log4V005. .1136971, 1.45154. 30. Given log 56321 = 4.7506704, log 56322 = 4.7506781; find log 5632147. 6.7506740. 31. Given log 53403 = 4.7275657, log 53402 = 4. 7275575; find log 5340234. 6.7275603. 32. Given log 56412 = 4.7513715, log 56413 = 4.7513792; find log 564.123. 2.7513738. 120 PLANE TRIGONOMETRY. 33. Given log 87364 = 4.9413325, log 87365 = 4.9413375; find log .0008736416. Ans. 4.9413333. 34. Given log 37245 = 4.5710680, log 37246 = 4.5710796; find log 3.72456. .5710750. 35. Given log 32025 = 4.5054891, log 32026 = 4.5055027; find log 32.025613. 1.5054974. 36. Given log 65931 = 4.8190897, log 65932 = 4.8190962; find log .000006593171. 6.8190943. 37. Given log 25819 = 4.4119394, log 25820 = 4.4119562; find log 2.581926. .4119438. 38. Given log 23454 = 4.3702169, log 23453 = 4.3701984; find log 23453487. 7.3702074. 39. Given log 45740 = 4.6602962, log 45741 = 4.6603057; find the number whose logarithm is 4.6602987. 45740.26. 40. Given log 43965 = 4.6431071, log 43966 = 4.6431170; find the number whose logarithm is 4.6431150. .000439658. 41. Given log 56891 = 4.7550436, log 56892 = 4. 7550512; find the number whose logarithm is .7550480. 5.689158. EXAMPLES. 121 42. Given log 34572 = 4.5387245, log 34573 = 4.5387371; find the number whose logarithm is 2.5387359. Ans. 345.7291. 43. Given log 10905 = 4.0376257, 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. Ans. .00000025725982. In the following six examples the student must take his logarithms from the tables. 45. Kequired the product of 3670.257 and 12.61158, by logarithms. Ans. 46287.74. 46. Required the quotient of .1234567 by 54.87645, by logarithms. Ans. .002249721. 47. Kequired the cube of .3180236, by logarithms. Ans. .03216458. 48. Required the cube root of ,3663265, by logarithms. Ans. .7155216. 49. Required the eleventh root of 63.742. 1.45894. 50. Required the fifth root of .07. .58752. 51. Given sin 42 21' = .6736577, sin 42 22' =.6738727; find sin 42 21' 30". .6737652. 52. Given "sin 67 22'= .9229865, sin 67 23' = .9230984 ; find sin 67 22' 48".5. .9230769. 122 PLANE TRIGONOMETRY. 53. Given sin 7 17' = .1267761, sin 7 18' = .1270646 ; find sin 7 17' 25". Ans. .1268963. 54. Given cos 21 27' = .9307370, cos 21 28' = .9306306; find cos 21 27' 45". .9306572. 55. Given cos 34 12' = .8270806, cos 34 13' = .8269170 ; find cos 34 12' 19".6. .8270272. 56. Given sin 41 48' = .6665325, sin 41 49' = .6667493; find the angle whose sine is .6666666. 41 48' 37". 57. Given sin 73 44' = .9599684, sin 73 45' = .9600499; find the angle whose sine is .96. 73 44' 23".2. 58. Given cos 75 32' = .2498167, cos 75 31 '=.2500984; find the angle whose cosine is .25. 75 31' 21". 59. Given cos 53 7' = .6001876, cos 53 8' = .5999549; find the angle whose cosine is .6. 53 7' 48 ".4. 60. Given log sin 45 16' = 9.8514969, log sin 45 17' = 9.8516220 ; find log sin 45 16' 30". 9.8515594. 61. Given log sin 38 24' = 9.7931949, log sin 38 25' = 9.7933543 ; find log sin 38 24' 27". 9.7932666. EXAMPLES. 123 62. Given log sin 32 28' = 9.7298197, log sin 32 29' = 9.7300182 ; find log sin 32 28' 36". Ans. 9.7299388. .63. Given log sin 17 1' = 9.4663483. log sin 17 0' = 9.4659353 ; find log sin 17 0'12". 9.4660179. 64. Given log sin 26 24' = 9.6480038. log sin 26 25' = 9.6482582; find log sin 26 24' 12". 9.6480547. 65. Given log cos 17 31' = 9.9793796, log cos 17 32' = 9.9793398 ; find log cos 17 31' 25".2. 9.9793629. 66. Given log tan 21 17' = 9.5905617, log tan 21 18' = 9.5909351 ; find log tan 21 17' 12". 9.5906364. 67. Given log tan 27 26' = 9.7152419, . log tan 27 27' = 9.7155508 ; find log tan 27 26' 42". 9.7154581. 68. Given log cot 72 15' = 9.5052891, log cot 72 16' = 9.5048538 ; find log cot 72 15' 35". 9.5050352. 69. Given log cot 36 18' = 10.1339650, log cot 36 19' = 10.1337003 ; find log cot 36 18' 20". 10.1338768. 70. Given log cot 51 17' = 9.9039733, log cot 51 18' = 9.9037144 ; find . log cot 51 17' 32". 9.9038352. 124 PLANE TRIGONOMETRY. 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. 6 53' 8". 73. Given log cos 22 28' 20"= 9.9657025, log cos 22 28' 10"= 9.9657112; find the angle whose log cosine is 9.9657056. 22 28' 16". In the following examples the tables are to be used : 74. Find log tan 55 37' 53". Ans. 10.1650011. 70. Find log sin 73 20' 15". 7. 9.9813707. 76. Find log cos 55 11' 12". 9.7565636. 77. Find log tan 16 0'27". 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' 26". 9.9164762. 82. Find log cos 37 19' 47". 9.9004540. 83. Find log sin 37 19' 47". 9.7827599. 84. Find log tan 37 19' 47". 9.8823059. 85. Find log sin 32 18' 24".6. 9.7279096. 86. Findlogcos3218'24".6. 9.9269585. 87. Find log tan 32 18' 24".6. 9.8009511. EXAMPLES. 125 Prove the following by the use of logarithms : ^512 x ^.00003075 = . 000232432 . SX80 -=- -v/.OOOOOOl 90 . W(2002)""j = a (1) m cos < = & (2) where a and b are given, and the values of m and < are required. Dividing (1) by (2), we get tan 4 = -, b which gives two values of <, differing by 180, and there- fore two values of m also from either of the equations a b m = - = sin cos < The two values of m will be equal numerically with opposite signs. In practice, m is almost always positive by the conditions of the problem. Accordingly, sin < has the sign of a, and cos < the sign of b } and hence < must be taken in the quail- rant denoted by these signs. These cases may be considered as follows : (1) Sin < 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. TRIGONOMETRIC EQUATIONS. 129 (2) Sin positive and cos < negative. This requires that $ be taken in the second quadrant, because only in this quad- rant is sin positive and cos < negative for the same angle. (3) Sin (f> and cos < both negative. This requires that be taken in the third quadrant, because only in this quad- rant are sin and cos both negative for the same angle. (4) Sin negative and cos positive. This requires that < be taken in the fourth quadrant, because only in this quadrant is sin negative and cos positive for the same angle. Ex. 1. Solve the equations msin < = 332.76, and mcos < = 290.08, for m and <. log m sin < = 2.52213 log m cos < = 2.46252 log tan < = 0.05961 .-. < = 4855'.2. log m sin = 2.52213 log sin 4 = 9.87725 log m = 2.64488 .-. m = 441.45. Ex. 2. Solve m sin < = - 72.631, and m cos $ = 38.412. Aiis. = 117 52'.3, m = - 82.164. 84. Solve the equation a sin < + & cos = c (1) a, 6, and c being given, and < required. Find in the tables the angle whose tangent is - ; let it be/?. Then - = tan /?, and (1) becomes a (sin + tan ft cos <) = c ; /sin < cos (3 + cos < sin J3\ or ci ( ] c j V cos ft ) or sin(-f ^) = -cos^ = -siny8 .... (2) a o 130 PLANE TRIGONOMETRY. There will be two solutions from the two values of < -f- ft given in (2). Find from the tables the value of cos ft. Next find from the tables the magnitude of the angle a whose sine = - cos /3, and we get sin (< + /?) = sin a, .-.$ + = wir + (-!)" . . (Art. 38) nir -l' where n is zero or any positive or negative integer. In order that the solution may be possible, it is necessary to have - cos J3 =, or < 1. a NOTE. This example might have been solved by squaring both sides of the equation; but in solving trigonometric equations, it is important, if possible, to avoid squaring both sides of the equation. Thus, solve COB 9 = k sin ........... (3) If we square both sides we get cos2 = Jfl sin2 = fc2(i - cos2 0) . (4) Now if a be the least angle whose cosine = , we get from (4) *S\ + k 2 = mra ........... (5) But (3) may be written cot 9 = k. .'. e = nn + a ........... (6) (6) is the complete solution of the given equation (3), while (5) is the solution of both cos = k sin 0, and also of cos = k sin 0. Therefore by squaring both mem- 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 b = 0.02112-* log a = 1.87314 log tan = 0.14798 - .-. ft = 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. TRIGONOMETRIC EQUATIONS. 131 log sin ft = 9.91111 log c= 1.63517 - colog& = 9.97888- log sin (< + )= 9.52516+ ... < + ft = 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 + 2 cos = V2. Ans. - -+mr + (-l) n ^- 4 6 4. . o b 5. sin^-cos^ = l. - + W7r + (-l) n ^. 6. " V3 sin - cos 6 = V2. ^-TT + WTT + ( I) n i7r. 85. /Sofae ^/ie equation sin (a + ) = msinx ...... (1) in which a and m are given. From (1) we have sin (a + #) +- sin a; _ m + 1 /2\ sin (a -f- x) sin x m 1 (Art. 46) . . (3) tan which determines x + ^-a, and therefore a?. If we introduce an auxiliary angle, the calculation of equation (3) is facilitated. 132 PLANE TRIGONOMETRY. Thus, let m = tan<; then we have by [(14) of Art. 61] m il l = tan^ + l = co ra-1 tan<-l which in (3) gives tan (x + } = cot (< - 45) tan a . (4) \ */ This, with tan < = m, gives the logarithmic solution. The logarithmic solution of the equation sin (a x) = m sin x is found in the same manner to be tan = m, and tan fx - 1^ = cot (< + 45) tan which the student may show. Example. Solve sin (106+ a;) = - 1.263 sin a (a < 180). log tan< = logm = log (- 1.263) = 0.10140 -. .-. < = 128 22'.3. 4>_45= 8322'.3; log cot (< - 45) = 9.06523 i= 53 O'.O, log tan^a = 10.12289 -] = 9.18812 V 2 / | = 8 46'.0 or 188 46'. .-. x = -44 14' or 135 46'. 86. Solve the equation tan(tt-f-#) = m tana; (1) in which a and m are given. TRIGONOMETRIC EQUATIONS. 133 From (1) we have tan ( + x) + tan x _ m + 1 tan (a + x) tan x m 1 = si "("+ 2 *) [(21) of Art. 61] sma fffll -..,_ 1 = cot(< - 45) sin a . (Art. 85) where tan = m. Example. Solve tan (23 16'+ x) = .296 tana;, log tan 4> = logm == log(.296) = 1.47129. .-. < = 16 29'.3. < - 45= - 28 30'.7 ; log cot(< - 45) = 10.26502- = 2316'.l, log sin a = 9.59661 logsin(a + 2z) = 9.86163- a + 2 x = 226 38'.9 or 313 21M. .-. x = 101 41'.5 or 1452'.6. 87. Solve the equation tan(a + a;).tanx = m (1) in which a and m are given. From (1) we have 1 + tan (a + a;) tan x _ 1 + m 1 tan (a + x) tan x 1 m = ^V . . (Ex. 4 of Art. 47) cos (a + 2 a) -| : ^ COS 1 + m = tan (45- <) cos a [(16) of Art. 61] where tan = m. 184 PLANE TRIGONOMETRY. Example. Solve tan (65 + a;) tana; = 1.5196 (a < 180). log tan < = log w = 0.18173. .-. < = 5039'9"; 45 _ < = _ 11 39' 9" ; log tan(45 -<) - 9.31434- a = 65 0' 0" ; log cos a = 9.62595 log cos (a + 2 x) = 8.94029- a + 2 = 95or 265. .-. 2 x = 30 or 200. /. x = 15 or 100. 88. Solve the equations m sin (6 + x) = a .......... (1) m sin (< -f x) = b .......... (2) for m and x, the other four quantities, 0, , a, b, being known. Expanding (1) and (2) by (Art. 44), we get m sin cos x + m cos sin x = a ..... (3) m sin (f> cos x -\- m cos < sin x = b ..... (4) Multiplying (3) by sin < and (4) by sin 0, and subtracting the latter from the former, we have m sin x (sin < cos 9 cos sin 0) = a sin < 6 sin 0. ...... (5) To find the value of mcoso;, multiply (3) and (4) by cos < and cos 0, respectively, and subtract the former from the latter. Thus m cos a? (sin < cos cos < sin 0) = b cos 6 a cos <. (6) Having obtained the values of m sin x and m cos x from (5) and (6), m and a; can be calculated by Art. 83. TRIGONOMETRIC EQUATIONS. 135 EXAMPLES. 1. Solve m cos (6 + x) = a, and m sin ( < + a; ) =6, for m sin z and cos x. ^ m gin ^ = 6_coe cos (# - 2. Solve m cos (0 + cc) = a, and mcos (< x)= b, for m sin x and m cos x. A b cos a cos < sin (0 + *) sin (0 -f 89. /SWve ^/ie equation m ....... (1) a; sin a y cos a = n ....... (2) for x and y. Multiplying (1) by cos a and (2) by since, and adding, we get x = m cos a + w sin . To find the value of y, multiply (1) by sin a and (2) by cos a, and subtract the latter from the former. Thus y = m sin a n cos a. Example. Solve x sin a + y cos a = a, x cos a y sin a = b. 90. To adapt Formulae to Logarithmic Computation. As calculations are performed principally by means of logarithms, and as we are not able by logarithms directly* to add and subtract quantities, it becomes necessary to know how to transform sums and differences into products * Addition and Subtraction Tables are published, by means of which the logarithm of the sum or difference of two numbers may be obtained. (See Tafeln der Addi- tions, und Subtractions, Logarithmen flir sieben Stellen, von J. Zech, Berlin.) 136 PLANE TRIGONOMETRY. and quotients. An expression in the form of a product or quotient is said to be adapted to logarithmic computation. 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 into a product, so as to adapt it to logarithmic computation. Put - = tan d> ; * thus a / T S a cos 6 b sin 6 = a ( cos - sin V = a (cos B tan < sin 0) a cos (0 T <) COS< 2. Similarly, a sin 6 cos = ^ sin (0 <). 3. Transform a b into a product, a + & = a/l-|--) = a(l4- tan 2 ) = a sec 2 <, if - = tan 2 . a a b = a( 1 ) = V a / if - = sin 2 4>. a * The fundamental formulae cos (a; #) and sin (x + #) (Art. 42) afford examples of one term equal to the sum or difference of two terms ; hence we may transform an expression a cos 6 sin into an equivalent product, by conforming it to the for- mulae just mentioned. Thus, comparing the identity, m cos cos 9 m sin sin B = m cos ( T 0) or m COB ), with a cos 6 sin 0, we will have a cos 0+ &sin0 = mcoB(0 t ) if we assume and 6 = m sin ; i.e. (Art. 83), if tan = - and m= a See Art. 84. cos * ADAPTATION TO LOGARITHMIC COMPUTATION. 137 .-. log (a + 6) = log a + 2 log sec <; and log (a 6) = loga-f21ogcos<. 4. Transform 1239.3 sin - 724.6 cos 6 to a product. log 6 = log (- 724.6) = 2.86010- loga = log (1239.3) = 3.09318 log tan < = 9. 76692- .-. < = -3018'.8. log a = 3.09318 log cos < = 9.93615 log _^_ = 3.15703 = 1435.6. .-. 1239.3 sin - 724.6 cos = 1435.6 sin (0 - 30 18'.8) . 91. Solve the equations r cos < cos = a ........ (1) (2) (3) for r, <, and 0. Dividing (2) by (1), we have tan 0=5, a from which we obtain 0. From (1) and (2) we have . . (4) . . COS0 S1110 from which we obtain rcos <. From (3) and (4) we obtain r and < (Art. 83). 138 PLANE TRIGONOMETRY. EXAMPLES. 1. Solve r cos cos = 53.953, r cos sin = 197.207, rsin< = - 39.062, for r, 0, 0. log 6 = 2.29493 log r cos < = 2.31060 log cos = 9.99221 logr = 2.31839 .-. r = 208.16. log a = 1.73201- log tan = 0.56292 - .-. = 10518'.0. log sin = 9.98433 log r cos = 2.31060 log r sin = 1.59175- log tan = 9.28115- 92. Trigonometric Elimination. Several simultaneous equations may be given, as in Algebra, by the combination of which certain quantities may be eliminated, and a result obtained involving the remaining quantities. Trigonometric elimination occurs chiefly in the applica- tion of Trigonometry to the higher branches of Mathematics, as, for example, in Physical Astronomy, Mechanics, Analytic Geometry, etc. As no special rules can be given, we illus- trate the process by a few examples. EXAMPLES. 1. Eliminate < from the equations x = a cos <, y = b sin <. From the given equations we have - = cos <, ^ = sin <, a b which in cos 2 < -}- sin 2 = 1, gives + y=l. TMIGONOMETEIC ELIMINATIONS. 139 2. Eliminate from the equations a cos + b sin < = c, 6 cos < -f- c sin < = a. Solving these equations for sin < and cos <, we have ,_bc a 2 U ^-^I~^ -- - ac b 2 which in cos 2 < + sin 2 < = 1, gives (be - a 2 ) 2 + (c 2 - ab) 2 = (ac - b 2 ) 2 . 3. Eliminate from the equations y cos < x sin = a cos 2 <, y sin <^> + x cos <^> = 2 a sin 2 <. Solve for x and y, then add and subtract, and we get - ?/) 2 = a 2 (l - si 4. Eliminate a and ft from the equations a = sin a cos ft sin 6 -f- cos a cos . . . (1) b = sin a cos /2 cos cos a sin . . . (2) c = sin a sin/? sin (3) Squaring (1) and (2), and adding, we get 2 i ,2 *2 2Oi 2 //l\ - = sin 2 sin 2 /? (5) sin 2 140 PLANE TRIGONOMETRY. Adding (4) and (5), we have a 2 + b 2 + --^ = 1. sin 2 5. Eliminate from the equations a sin + 6 cos = c, acos< bsiu = d. Ans. a 2 + & 2 <= c 2 + c? 2 . 6. Eliminate from the equations m = cosec 6 sin 0, n = sec cos 0. mV (m* +n%) = 1. 7. Eliminate and < from the equations sin 6 -j- sin < = a, cos + cos < = b, cos(0 - 0) = c. a 2 + V 2 - 2c = 2. 8. Eliminate x and ?/ from the equations tan x -f- tan 37 = a, cot x + cot y = 6, a + y = c. cot c = - -. a b 9. Eliminate from the equations x = cos 2 < + cos <, y = sin 2 < + sin <. EXAMPLES. Solve the following equations : 1. tan + cot = 2. Ans. 45. 2. 2 sin 2 + V2 cos = 2. 90, or 45. 3. 3tan 2 0-4sin 2 = l. 45. 4. 2sin 2 + V2sin0 = L > . 45. EXAMPLES. 141 5. cos 2 0-V3 cos + f = 0. Ans. 30. 6. sin50 = 16 sin 5 0. TITT, or mr - 6 7. sin 9 sin 6 = sin 4 0. Jn?r, or f n?r -^. To 8. 2 sin = tan 0. mr, or 2 TITT - o 9. 6cot 2 0-4cos 2 = l. 7i7r-. 3 10. tan + tan (0 -45) = 2. mr -. o 11. cos 4- V3 sin = V2. 2 WTT - 3 12. tan (6 + 45) = 1 + sin 2 0. ww - , or 7i7r. 13. (cot0-tan0) 2 (2 + V3) = 4(2 14. cosec cot = 2 V3. 2nir-- 6 15. cosec + cot (9 = V3. 16. sin- = cosec cot 0. 2?*7r. 17. sin 50 cos 30 = sin 90 cos 70. J^-WTT + (- 1)|- 18. siri 2 + cos 2 2<9 = f. mr , or mr T % TT. 19. V3sin0-COS0 = V2. TITT + ^ + (- l) n r 6 4 20. tan + cot = 4. WIT + 21. si 22. Solve m sin <#> = 1.^9743, and m cos <#> = 6.0024. 142 PLANE TRIGONOMETRY. 23. Solve m sin <= -0.3076258, and m cos 0=0.4278735. (m positive.) Ans. < = 324 17' 6".6, ?>i = 0.52698. 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 = c, and a cos + 6 sin 6 = c sin0 cos 0, show that sin 2 <9(c' 2 - a 2 - 6 2 ) = 2 a&. Solve the following equations : 27. V2 sin + V2 cos = V3. ^bis. - + mr + ( l) n -. 28. 2 sin a; + 5 cos a; = 2. Sug. [2.5 = tan 68 12']. Ans. x = -6S 12' -f 71 180 + ( - 1)"(21 48'). 29. 3 cos a 8 sin a; = 3. Sug. [2.6 = tan 69 26' 30"]. Ans. x = - 69 26' 30" + 2 n 180 (69 26' 30"). 30. 4 sin x 15 cos x = 4. Sug. [3.75 = tan 75 4']. .4HS. x = 75 4' + n 180 + (- 1) H (14 56'). 31. cos (a -)-)= sin (a -f x) -f- V2cos^8. ^l?is. ic = - + 2 n?r /?. 4 32. cos + cos 3 H- cos 5 = 0. Ans. l(2n + I)TT, or 1(3 ?i I)TT. 33. sin 5 = sin 3 + sin = 3 - 4 sin 2 0. , or |(2?i + I)TT. 3 34. Ans. 35. or cos" 1 - EXAMPLES. 143 36. Solve m sin(0 -\-x) = a cos /?, and m cos (0 #) = ei sin /?, for msina; and racosx. (Art. 67.) Ans. cos 20 a sin (5 0) m cos a; = ^- cos 2 (9 37. Solve m cos (0 + <) = 3. 79, and ra cos (0 <) = 2.06, for m and 0, when < = 3J 27'.4. (Art. 67.) 38. Solve r cos < cos = 1.271, r cos 4> sin = - 0.981, r sin 4 = 0.890, for r, <#>, 0. (Art. 70.) 39. Solve rcos = 12.124, for r, <, (9. 41. Solve cos(2aj + 3y) = J, cos(3x + 2?/) = |V3. ^TiS. i=|W7r T ^7r LTT, y = |7l7r |TT ^ 42. Solve cos 3 + cos 50 +V2 (cos + sin0) cos = 0. Ans. 4(90 = 2?i7r TT or 43. Solve c Ans. sin = 0, or tan0 = lv / 2. 44. Solve 3 sin -f cos = 2 x, sin -f 2 cos = x. Ans. = 71 34', aj = 144 PLANE TRIGONOMETRY. 45. Solve 1.268 sin < = 0.948 + m sin (25 27'.2), 1.268 cos 4 = 0.281 + m cos (25 27'.2). Ans. = 60 53'.8, m = 0.372. 46. Transform at + y* + z 4 2 y*z* 2 zW 2 a?y* into a product. ^.7?s. (#+?/+2) (y+z x) (z+x y) (x-\-y z). 47. Eliminate from the equations m sin 2 = n sin 0, p cos 2 == g cos 0. m 2 2 48. Eliminate and from the equations x = a cos m cos m , y = b cos 6 sin w <, 2 = c sin" 49. Eliminate 6 from the equations a sin + 6 cos 6 = h, a cos b sin = k. Ans. a 2 + b 2 = h 2 + fc 2 . 50. Eliminate from the equations a tan + 6 sec B = c, a' cot + 6' cosec = c'. Ans. (a'b + cb') 2 + (06' + c'&) 2 = (cc f - aa') 2 . 51. Eliminate from the equations x = 2 a cos cos 2 a cos 0, y = 2a cos sin 2 a sin 0. ^.ns. ic 2 + y 2 = a 2 . 52. Eliminate from the equations x = a cos + 6 cos 2 0, and y = a sin + 6 sin 2 0. Ans. a 2 [(x + 6) 2 + /] = [x 2 + 7/ 2 - 6 2 ] 2 . 53. Eliminate a and /3 from the equations 6 + c cos a = u cos (a 0), & + CCOS/? = MCOS (ft 0), a /? = 2<; and show that u 2 2 we cos + c 2 = 6 2 sec 2 <. EXAMPLES IN ELIMINATION. 145 54. Eliminate and from the equations x cos 6 + y sin = a, b sin (0 -f- <) = a sin <, a cos (0 + 2cf>) - i/ sin (0 + 2<) = a. 55. Eliminate from the equations #_ sec 2 cos 2 a sec 2 + cos 2 0' ^ = sec 2 + cos 2 0. y 56. Eliminate from the equations (a + 6) tan (0 - ) = (a - b) tan (0 + <), 57. Eliminate ^ from the equations 58. Eliminate ^ and <^> from the equations a 2 cos 2 b 2 cos 2 (f> = c 2 , acosQ + b cos < = r, a tan & = b tan <. 59. Eliminate <^> from the equations n sin m cos = 2 m sin , n sin 2 m cos 2 < = n. Ans. (nsin#-f- mcos#) 2 = 2m (m + n). 60. Eliminate a from the equations x tan (a /?) = y tan ( + /?), (# y) cos 2 a + ( + y) cos2fi = z. Ans. 146 PLANE TRIGONOMETRY. CHAPTER VI. EELATIONS BETWEEN THE SIDES OP A TKIANGLE AND THE FUNCTIONS OF ITS ANGLES, 93. Formulae. In this chapter we shall deduce formulae 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 triangles. 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 : ^ I 'c a = csin A = ccosB = 6 tan A = &cotB . . (1) b =csinB = ccosA= atanB = a cot A . . (2) c = 6 sec A = asecB = frcosecB = acosec A . (3) which may be expressed in the following general theorems : * The student must remember that a, 6, c, are numbers expressing the lengths of the sides in terms of some unit of length, such as a foot or a mile. The unit may be whatever we please, but must be the same for all the sides. OBLIQUE TRIANGLES. 147 I. In a right triangle each side is equal to the product of the hypotenuse into the sine of the opposite angle or the cosine of the adjacent angle. II. In a right triangle each side is equal to the product of the other side into the tangent of the angle adjacent to that other side, or the cotangent of the angle adjacent to itself. III. In a right triangle the hypotenuse is equal to the product of a side into the secant of its adjacent angle, or the cosecant of its opposite angle. EXAMPLES. In a right triangle ABC, in which C is a right angle, prove the following : 1. tan B = cot A + cos C. 2. sin2A = sin2B. 3. cos2A + cos2B = 0. 4. 7. tan2A = a b ~2b 2a 2ab & ft2 ~2 6. cos2A = -- c 2 8. RinSA- 3 ^'-" 3 , b 2 - a? c 3 OBLIQUE TRIANGLES. 95. Law of Sines. In any triangle the sides are pro- portional to the sines of the opposite angles. Let ABC be any triangle. Draw CD perpendicular to AB. We have, then, in both figures CD = a sin B = b sin A. (Art. 94) .. asinB = b sin A. a b sin A sin B Similarly, by drawing a perpen- dicular from A or B to the opposite side, we may prove that 148 PLANE TRIGONOMETRY. or sinB a sinC b and c sin C sin A sin A sinB sinC a : 6 : 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) BC 2 = AC 2 + AB 2 -2AB x AD, A' or a = But AD = b cos A. In an obtuse-angled triangle (see second figure) we have (Geom., Book III., Prop. 27) D A BC 2 = AC 2 + AB 2 + 2ABx AD, c B or AD = b cos CAD = - b cos A. But Similarly, b 2 = c 2 + a 2 - 2 ca cos B, NOTE. When one equation in the solution of triangles has been obtained, the other two may generally be obtained by advancing the letters so that a becomes b, b becomes c, and c becomes a; the order is abc, bca, cab. It is obvious that the formulae 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 their difference. By Art. 95, a : b = sin A : sin B, By composition and division, a + b __ sin A -f- sin B a b sin A sin B c-a tan(C-A) Since tan(A + B) = tan (90 - C) = co the result in (1) may be written a + b _ cot^C ,. a-b tanJ(A-B) and similar expressions for (2) and (3). 98. To show that in any triangle c = a cos B + b cos A. In an acute-angled triangle (first figure of Art. 96) we have = a cos B + 6 cos A. In an obtuse-angled triangle (second figure of Art. 96) we have C = DB-DA = a cos B b cos CAD. .*. c = acosB + 6 cos A. Similarly, 6 = c cos A + a cos C, a = b cos C -f- c cos B. 150 PLANE TRIGONOMETRY. EXAMPLES. 1. In the triangle ABC prove (1) a + b:c = cos ^ (A B) : sin 1 C, and (2) a b:c = sin \ ( A - B) : cos 1 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 J- ^cos^Acos * (B - C) DC asinB and also JL = 2sin| Asin i( 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 = b * + 2 ~ a " ' = 1 - 2 sin 2 - . . (Art. 49) Zi OC 2bc 2 be _ (a + b c) (a b + c) 2 be Let a-|-Z>-|-c = 2s; then a + b c = 2(s c), and a 6 + c = 2(s b). 2 2bc OBLIQUE TRIANGLES. 151 Similarly, sm^ = A /- v - '- (2) Ln 2 II. cosA = 2cos 2 --! (Art. 49) S 2 ~ 2 be 2 be _ (a + b -f- e) (b -f- c a) * 2 be = 2s'2(s-a) 2 be ... cos|= ^ 7 Similarly, C os? = A /^--^. . ..... (5) III. Dividing (1) by (4), we get r /77 ^TT; ^\ J_ v / \ " "~~" Cv I I O "" " " C/ I / f\\ 2 = V / _ x 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. 152 PLANE TRIGONOMETRY. 100. To express the Sine of an Angle in Terms of the Sides. sin A = 2 sin A cos ..... (Art. 49) _ 2 , ___ .-. sin A = V s (s a) (s 6) (s c). be 2 . ____ . Similarly, sin B = Vs (s a) (s b) (s c), ac o ab Cor. sin A == V2 b*c?+~2cW+ 2 a 2 6 2 - a 4 -^c" 4 , and similar expressions for sin B, sin C. EXAMPLES. In any triangle ABC prove the following statements : 1. a (6 cos C ccosB) = 6 2 c 2 . o sin A -f- 2 sin B _ sin C sin 2 A m sin 2 B sin 2 C 4. = a 2 ra& 2 c 2 5. a cos A -f&cosB ccosC = 2ccos AcosB. K cos A cosB cos C sin B sin C sin C sin A sin A sin B 7. a sin (B - C) + 6 sin (C - A) + c sin ( A - B) = 0. 8. tan A tan B = ^^. s 9. taniA-s-taniB=.(s-&)-s-(*-c). AEEA 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, = 6sinA. D Similarly, S = ^ ac sin B, S = i ab sin C. (2) Given one side and the angles. Since which is Similarly, a : b = sin A : sin B (Art. 95) sin A ' S = i ab sin C, gives q 2 sinB sinG 2 sin A 6 2 sin A sin C c 2 sin A sin B 2 sin B (3) Given the three sides. 2sinC sin A = ~ Vs(s - a) (s - b) (s - c) (Art. 100) be Substituting in S = \ be sin A, we get S = Vs(s - a) (s - b} (s - c) . 154 PLANE TRIGONOMETRY. 102. Inscribed Circle. To find the radius of the inscribed circle of a triangle. 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 S = area of ABC = A AOB + A BOC + A COA = r = rs (Art. 99) 103. Circumscribed Circle. To find the radius of the circumscribed circle of a triangle in ^^__^ terms of the sides of the triangle. Let be the centre of the circle A/ described about the triangle ABC, and R its radius. Through draw the diameter CD and join BD. Then Z BDC = Z BAG = Z A. = 2KsinA, or a = 2RsinA. = a = b = c 2 sin A 2 sin B 2 sin C But be (1) (Art. 101) (2) RADII OF THE ESCRIBED CIRCLES. 155 104. Escribed Circle. To find the radii of the escribed circles of a triangle. A circle, which touches one side of a triangle and the other two sides produced, is called an escribed circle of the triangle. Let 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 ?v We then have from the figure A ABC = A AOB + A AOC - A BOG. Q _ cr i i b r i ar i S s a + c - a) = r, (s - a) . (Art. 99) . . (1) Similarly it may be proved that if r 2 , r s are the radii of the circles touching AC and AB respectively, S S c 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, IA and 1C bisect the angles BAG and BC A ; * Often called the incentre and circumcen- tre of a triangle. 156 PLANE TRIGONOMETRY. therefore the arc BD is equal to the arc DC, and DOH bisects BC at right angles. Draw IM perpendicular to AC. Then Z DIC = A -L^ = BCD + BCI = DCI. Also, AI = IMcosec = rcosec 2 2 that is, (K + 01) (E - 01) = 2 Kr. 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 2 cot A + cot B 3. Find the area of the triangle ABC when (1) a = 4, b = 10 ft., C = 30. Ans. 10 sq. ft. (2) 6=5, c = 20 inches, A = 60. 43.3 sq. in. (3) a = 13, b = 14, c = 15 chains. 84 sq. chains. 4. Prove = + + . r r r 5. Prove r = -_ cosiA EXAMPLES. 157 6. Prove that the area of the triangle ABC is represented by each of the three expressions : 2 R 2 sin A sin B sin C, rs, and Rr(siii A + sin B + sin C). 7. If A = 60, a = V3, 6=V2, prove that the area = i(3+V3). 8. Prove R (sin A + sin B -+- sin C) = s. 9. Prove that the bisectors of the angles A, B, C, of a triangle are, respectively, equal to 2 be cos 2 ca cos - 2 ab cos - 222 , , - o-fc c-\- a a-{- 106. To find the Area of a Cyclic* Quadrilateral. A Let ABCD be the quadrilateral, and a, b, c, and d its sides. Join BD. Then, area of figure = S = ^ ad sin A -f ^ be sin C = ^(ad + 6c)sin A . . . (1) Now in A ABD, BD 2 = a 2 + d 2 - 2 ad cos A, and in A CBD, 51?= 6 2 + c 2 - 26ccos C = 6 2 + c 2 2 be cos A. . r a 2 _&2_ c 2_ h ^ ' SmA= V 1 - 2(ad + 5c) - (a 2 - 5 2 - c 2 * See Geometry, Art. 251. 158 PLANE TRIGONOMETRY. 2 (ad -f be) = 2V (s - a) (s - 6) (8 - c) (s - d) ad -f &c (where 2s = a-|-& + Substituting in (1), we have S = V (s - a) (s - b) (s - c )(s-d). The more important formulae proved in this chapter are summed up as follows : 1. _^_ = -_ = _^_ (Art. 95) sin A sin B sin C 2. a 2 =& 2 + c 2 -26ccosA ....... (Art. 96) a == (Art. 97) a- 5 tan^(A-B) 4. B iniA=:J<'- ft) <'- c >. ..... (Art. 99) \ be Is (s a) 5 . 6 . 7. sin A =-^s(s-a)(s-b)(s-c) . . (Art. 100) oc V2 6 2 c 2 + 2 c 2 a 2 -f- 2 a 2 6 2 - a 4 - 6 4 - c 4 . 8. Area of A = Vs (s - a) (s-b) (s-c). . (Art. 101) EXAMPLES. 159 9. Area of A = (a + 6 + c) = rs . . . . (Art. 102) 10. S a) (s b) (s c) s 11. K = (Art. 103) EXAMPLES. In a right triangle ABC, in which C is the right angle, prove the following : 2 - sin 2 A + sin 2 B . 9 B c a 3. cos 9 A b 4- c 4. cos 2 = - 2 2c 5. sin(A-B) + cos2A = -. . 1. A T> 6. 2 7. sin(A-B) + sin(2A 8. c 9. (sin A - sin B) 2 + (cos A + cos B) 2 = 2. ^ Q /a + & i lQ> b_ 2 sin A \a 6 \a + 6 Vcos2B In any triangle ABC, prove the following statements : 11. 160 PLANE TRIGONOMETRY. 12. (& -c) COT = a sin 2 2 a b cos B cos A 14. c 1 4 cos C 1 ~ b 4 c cos B -f- cos C lo. - - - - a 1 cos A 16. b + c 17. a-f-6+c = (6 4 c)cos A 4 (c 4 a)cos B 4- (a 4 6)cos C. 18. b + c a = (6 + c)cos A (c a)cosB -f (a 6)cos C. 19. a cos (A + B -f C) - b cos (B + A) - c cos ( A + C) = 0. orv cos A cos B cos C _ a 2 -f- 6 2 4- c 2 a b c 2abc A a sin C 21. tanA = - - b a cos C n TI 22. 2 2 T? n ^ \ 23. 2 2 b+c+a 24. tan-(&4c-a) = tan| 25. c 2 =(a + 6) 2 sin 2 2 + ( a -6) 2 cos 2 -. 2 2 26. ccosA4-cosB = 2a4-&sin 2 5. 27. 28. tan B -*- tan C = (a 2 + b 2 - c 2 ) -*- (a 2 - b 2 4 c 2 ). EXAMPLES. 161 29. a 2 -f b' 2 + c 2 = 2(a6 cos C + 6c cos A + ca cos B). 30. cos 2 -^-cos 2 ? = (s-a)--6(s-6). 2 2 r^ Ti 31. 6 sin 2 |- c sin 2 = s a. 2 Z 32. If p is the length of the perpendicular from A on BC, sinA = ^. be 33. If A = 3B, 6 ' gin B 34. If V6c sin B sinC = , then B = C. b + c B C A 35. a cos cos cosec = s. 36. If cos A = |, and cos B = |f , then cos C = if. 37. If sin 2 B -f sin 2 C = sin 2 A, then A = 90. 38. If D is the middle point of BC, prove that 39. If a = 26, and A = 3B, prove that C = 60. 40. If D, E, F, are the middle points of the sides, BC, CA, AB, prove 4(AIT + BE 2 + CF 2 ) = 3(a 2 + 6 2 + c 2 ). 41. If a, 6, c, the sides of a triangle, are in arithmetic progression, prove tan - tan 5 = !. 2 2 3 A f> T-P ta n A tan B c 6 , , /.AO 42. If - = -- prove that A = 60 . tan A 4- tan B c ' 43. If cos B = ^BA, prove that B = C. 2sinC J 162 PLANE TRIGONOMETRY. 44. If a 2 = b 2 - be + c 2 , prove that A = 60. 45. If the sides of a triangle are a, 6, and V 2 + ab -f 6 2 , prove that its greatest angle is 120. 46. Prove 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, prove (1) (b 2 - c 2 ) tan ADB = 2 be sin A, and (2) cot BAD + cot DAC = 4 cot A + cot B + cot C. 48. Find the area of the triangle ABC when a = 625, b = 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 a = 13, 6 = 14, c = 15. Ana. 4; 10.5; 12; 14. 50. Prove the area S = -| a 2 sin B sin C cosec A. 51. " " " "=Vnyvv 52. "= 2abc /cos -cos- cos 2 2 53. 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. 54. Prove that the angles of the pedal triangle are, respectively, TT 2 A, IT 2 B, TT 2 C. 55. Prove r.r^ = r 8 cot 2 - cot 2 - cot 2 -. A T? r 1 56. Prove r cos = a cos cos 2 22 57. Prove that the area of the incircle : area of the tri- T> /~1 angle : : TT : cot cot cot 222 EXAMPLES. 163 Prove the following statements : 58. If a, b, c, are in A. P., 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. be = 4 R 2 (cos A + cos B cos C) . 61. If C is a right angle, 2 r + 2 R = a + 5. 62. r 2 r 3 + ?v*i + Wz = s 2 - 63 . 1 + 1 + 1= be ca ab 2rR C 64. i 1 ! -f- r 2 = ccot A B . C 65. r cos = a sin sin 2 22 66. If pu p z , p 3 be the distances to the sides from the circumcentre, then a_. b^ , _c __abc_ Pi Pi Ps ^PiPiPs 67. The radius R of the circumcircle 1 s/ _ abc _ 2 Af sin A sin B sinC 68. S = -sin2B-f-sin2A. 4 4 69. s a s 6 s c s S 70. a6cr = 4R(s-a)(s-6)(s-c). 71. The distances between the centres of the inscribed j^ and escribed circles of the triangle ABC are 4 R sin , 4Rsin?, 4Rsin^. 2' 2 72. If A is a right angle, r 2 + r 3 = a. 164 PLANE TRIGONOMETRY. 73. In an equilateral triangle 3 K = 6 r = 2 ?*j. 74. If r, r ly r 2 , r s denote the radii of the inscribed and escribed circles of a triangle, - 2 T 2 r 3 75. The sides of a triangle are in arithmetic progression, and its area is to that of an equilateral triangle of the same perimeter as 3 is to 5. Find the ratio of the sides and the value of the largest angle. Ans. As 7, 5, 3 ; 120. 76. If an equilateral triangle be described with its angular points on the sides of a given right isosceles triangle, and one side parallel to the hypotenuse, its area will be 2 a 2 sin 2 15 sin 60, where a is a side of the given triangle. 77. If h be the difference between the sides containing the right angle of a right triangle, and S its area, the diam- eter of the circumscribing circle = V^ 2 + 4S. 78. Three circles touch one another externally : prove that the square of the area of the triangle formed by join- 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. 80. If G!, 2 , 3 are the centres of the escribed circles of a triangle, then the area of the triangle OiOsOg = area of triangle ABcfl + j-y-5 - + b + -- c - - 1- 6 -f- c a a + c b a + b c J 81. If the centres of the three escribed circles of a tri- angle are joined, then the area of the triangle thus formed is - where r is the radius of the inscribed circle of the ,2r original triangle. SOLUTION OF TRIANGLES. 165 CHAPTER VII. SOLUTION OP TKIATOLES, 107. Triangles, In 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, the other three can be calculated. The process of determining the unknown elements from the known is called the solution of triangles. NOTE. If the three angles only of a triangle are given, it is impossible to deter- mine the sides, for there is an infinite number of triangles 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. 108. There are Four Cases of Right Triangles. I. Given one side and the hypotenuse. II. Given an acute angle and the hypotenuse. III. Given one side and an acute angle. IV. Given the two sides. Let ABC be a triangle, right-angled at C, and let a, &, and c, as before, be the sides opposite the angles A, B, and C, respectively. A b The formulae for the solution of right triangles are (1), (2), (3) of Art. 94. 166 PLANE TRIGONOMETRY. 109. Case I. Given a side and the hypotenuse, as a and c; to find A, B, b. We have sin A = -. c . *. log sin A = log a log c, from which A is determined ; then B = 90 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 = 536, c = 941 ; find A, B, b. Solution by Natural Functions. We have sin A = - = = .569607. c 941 From a table of natural sines we find that A = 34 43' 22". .-. B = 55 16' 38". Lastly, b = c cos A = 941 x .821918 = 773.425. Logarithmic Solution. log sin A = log a log c. log o = 2.7291648 log c = 2.9735 log sin A* = 9.7555752 .-. A = 34 43' 22". .-. B = 55 16' 38". log b = log c + log cos A. log c = 2.9735896 log cos A = 9.9148283 log b = 2.8884179 .'. 6 = 773.424. Our two methods of calculation give results which do not quite agree. The discrepancies arise from the defects of the tables. * Ten is added so as to get the tabular logarithms (Art. 76). EIGHT TRIANGLES. 167 The process of solution by natural sines, cosines, etc., can be used to advantage only in cases in which the measures of the sides are small numbers. We might have determined b thus : b = V(c-o)(c + o); or thus : b = a tan B. NOTB. It is generally better to compute all the required parts from the given ones, so that if an error is made in determining one part, that error will not affect the computation of the other parts. To test the accuracy of the work, compute the same parts by different formulae. Ex. 2. Given a = 21, c = 29 ; find A, B, b. Ans. A = 46 23' 50", B = 43 36' 10", b = 20. NOTE. In these examples the student must find the necessary logarithms from the tables. 110. Case II. Given an acute angle and the hypotenuse, as A and c; to find B, a, b, We have B = 90 - A. Also a = c sin A, and b = c cos A. .-. log a = log c + log sin A ; and log 6 = log c H- log cos A. Thus B, a, and b are determined. Ex. 1. Given A = 54 28', c = 125 ; find B, a, b. B = 90 - A = 35 32'. Solution by Natural Functions. We have a = c sin A, and b = c cos A. Using a table of natural sines, we have a = 125 x .813778 = 101.722, and b = 125 x .581177 = 72.647. 168 PLANE TRIGONOMETRY. Logarith m ic Solut ion. log a = log c -f- log sin A. log c = 2.0969100 log sin A = 9.9105057 log a* = 2.0074157 .-. a = 101. 722. log b = log c + log cos A. log c = 2.0969100 log cos A = 9.7643080 log b * = 1.8612180 .-. 6 = 72.647. Ex. 2. Giveji A = 37 10', c = 8762 ; find a and b. Ans. 5293.4; 6982.3. 111. Case III. Given a side and .an acute angle, as A and a; to find B, 6, c. We have B = 90 - A. Also -, and c = and tan A sin A .-. log b = log a log tan A, 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' 36". log b = log a log tan A. log a = 3.7381858 log tan A = 9.8001090 log b = 3.9380768 .-. 5 = 8671.152. log c = log a log sin A. log a = 3.7381858 log sin A = 9.7273076 log c = 4.0108782 .. c= 10253.64. Ex. 2. Given A = 34 18', a = 237.6 ; find B, 6, c. Ans. B = 5542'; 6 = 348.31; c = 421.63. * Ten ia rejected because the tabular logarithmic functions are too large by ten (Art 76). EIGHT TRIANGLES. 169 112. Case IV. Given the two sides, as a and b; to find A, B, c. We have tan A = - ; then B = 90 - A. b Also c = a cosec A = - sin A .*. log tan A = log a log 6, and log c = log a log sin A. Ex. Given a = 2266.35, b = 5439.24 ; find A, B, c. Solution. log tan A = log a log b. log a = 3.3553270 log b = 3.7355382 log tan A = 9.6197888 .-. A = 22 37' 12". .-. B = 67 22' 48". log c = log a log sin A. log a = 3.3553270 log sin A = 9. 5850266 log c = 3.7703004 .-. c = 5892.51. NOTE. In this example we might have found c by means of the formula c = \/a 2 + 6 2 ; but we would have had to go through the process of squaring the values of a and b. If these values are simple numbers, it is often easier to find c in this way; but this value of c is not adapted to logarithms. A. formula which consists entirely of factors is always preferred to one which consists of terms, 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 Case 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 from one another (Art. 81). It is therefore desirable, in this case, to find B first, by either of the following formulae : fr <" 1TO PLANE TRIGONOMETRY. rJ5l? (Art. 50) 1 + cosB ^~ a (2) -v Then 6 = ccosA (3) or = V(cH-a)(c a) (4) Ex.1. Given a = 4602.21059, c = 4602.836; find B. c - a = 0.62541, log (c - a) = T.7961648 2 c = 9205.672, log 2 c = 3.9640555 2)5.8321093 log sin ? = 7.9160547 & B = 56'40".36. .-. ? = 28'20".18. NOTE. The characteristic 5 is increased numerically to 6 to make it divisible by 2 (see Note 4 of Art. 66). Ten is then added to the characteristic 3, making it 7, so as to agree with the Tables (Art. 76). There is a slight error in the abore value of B on account of the irregular differences of the log sines for angles near (Art. 81). A more accurate value may be 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 a = 60, c = 100 ; find A, B, b. Ans. A = 36 52'; B = 538'; 6 = 80. 2. Given a = 137.66, c = 240 ; find A, B, b. Ans. A = ,35 ; B = 55 ; b =, 196.59. EIGHT TRIANGLES. 171 3. Given a = 147, c = 184 ; find A, B, b. Ana. A = 531'35"; B = 36 58' 25"; 6 = 110.67. 4. Given a = 100, c = 200 ; find A, B, 6. Ans. A = 30 ; B = 60 ; b = 100 V3. 5. Given A = 40, c = 100 ; find B, a, 6. Ans. B = 50; a = 64.279; 6 = 76.604. 6. Given A = 30, c = 150 ; find B, a, b. Ans. B = 60; a = 75; b = 75 V3. 7. Given A = 32, c = 1760; find B, a, b. Ans. B = 58 ; a = 932.66 ; b = 1492.57. 8. Given A = 35 16' 25", c = 672.3412 ; find B, a, b. Ans. B = 54 43' 35"; a = 388.26; 6 = 548.9. 9. Given A = 75, a = 80; find B, 6, c. Ans. B = 15; 6 = 80(2-V3); c = 80(V6-V2). 10. Given A = 36, a = 520 ; find B, 6, c. Ans. B = 54; 6 = 715.72; c = 884.68. 11. Given A = 34 15', a = 843.2 ; find B, 6, c. Ans. B = 5545'; c = 1498.2. 12. Given A = 67 37' 15", 6 = 254.73 ; find B, a, c. Ans. B = 22 22' 45"; a =618.66; c = 669.05. 13. Given a = 75, 6 = 75 ; find A, B, c. Ans. A = 45 = B ; c^= 75 V2. 14. Given a = 21, 6 = 20 ; find A, B, c. Ans. A = 46 23' 50"; c = 29. 15. Given a = 300.43, 6 = 500 ; find A, B, c. Ans. A = 31; B = 59; c = 583.31. 16. Given a = 4845, 6 = 4742; find A, B, c. Ans. A = 45 36' 56". 172 PLANE TRIGONOMETRY. OBLIQUE TRIANGLES. 114. There are Four Cases of Oblique Triangles. I. Given a side and two angles. II. Given two sides and the angle opposite one of them. III. Given two sides and the included angle. IV. Given the three sides. The formulae for the solution of oblique triangles will be taken from Chap. VI. Special attention must be given to the following three, proved in Arts. 95, 96, 97. a b c (1) The Sine-rule, sin A sin B sin C ? + c 2 -a 2 (2) The Cosine-rule, cos A = 2 be (3) The Tangent-rule, tan^-^ = ^=^cot- 2t a -\- o 2i 115. Case I. Given a side and two angles, as a, B, C; find A, b, c. (1) A = 180 - (B + C). .-. A is found. b a , asinB sinB sin A sin A c a a sin C (2) (3) sinC sin A sin A These determine b and c. Ex. 1. Given a = 7012.6, B = 38 12' 48", C=60; find A, 6, c. Solution. A = 180 - (B + C) = 81 47' 12". log a = 3.8458729 log sin B = 9.9714038 colog* sin A = 0.0044775 log b = &8217542 b = 6633.67. log a = 3.8458729 log sin C = 9.9375306 colog sin A = 0.0044775 log c = 3.7878810 .-. c = 6135.94. * See Art. OBLIQUE TRIANGLES. 173 Ex, 2. Given a=1000, B = 45, C = 127 19' ; find A, b, c. Ans. A = 7 11'; 6 = 5288.8; c = 5948.5. 116. Case II. Given two sides and the angle opposite one of them, as a, b, A; find B, C, c. (1) sin B = ftsmA ; thus B is found. ft (2) C = 180 _ (A + B); thus C is found. /o\ a. sin C ,T_ f , (3) c = ; thus c is touna. sin A This is usually known as the ambiguous case, as shown in geometry (B. II., Prop. 31). The ambiguity is found in the equation Since the angle is determined by its sine, it admits of two values, which are supplements of each other (Art. 29). Therefore, either value of B may be taken, unless excluded by the conditions of the problem. I. If a < 6 sin A, sin B > 1, which is impossible ; and therefore there is no triangle with the given parts. II. If a = b sin A, sin B = 1, and B = 90 ; therefore there is one triangle a right triangle with the given parts. III. If a > b sin A, and < b, 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 > B, and since A is given, B must be acute ; thus there is only one triangle with the given parts. 174 PLANE TRIGONOMETRY. These 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 a<6 sin A, the circle will not meet AX, and there- fore no triangle can be formed with the given parts. II. If a = b sin A, the cir- cle touches AX in B' ; there- fore there is one triangle, right-angled at B. III. If a > b sin A, and < b } the circle cuts AX in two points B and B', on the _ same side of A; thus there B/x A are two triangles ABC and AB'C, each having the given parts, the angles ABC, AB'C being supplementary. IV. If a > &, 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 < b sin A, a *= b sin A, a > b sin A and < 6, a > b, no solution. one solution (right triangle). two solutions. one solution. These results may be obtained algebraically thus : We have a 2 == W + c 2 - 2 be cos A . . . (Art. 96) .. c = b cos A Va 2 b 2 sin 2 A, OBLIQUE TRIANGLES. 175 giving two roots, real and unequal, equal or imaginary, according as a >, =, or < b 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 are both acute, and there can be no ambiguity. In the solution of triangles there can be no ambiguity, except when an angle is determined by the sine or cosecant, and in no case whatever when the triangle has a right angle. Ex. 1. Given a = 7, 6 = 8, A = 27 47' 45"; find B, C, c. Solution. log b = 0.9030900 log sin A = 9.6686860 cologa = 9.1549020 log sin B = 9.7266780 log a = 0.8450980 log sin C = 9.9375306 colog sin A = 0.3313140 log c= 1.1139426 .-. B = 32 12' 15", or 14747'45". .-. c = 13. .% = 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 .-. c = 1.1538. Thus, there are two solutions. See Case III. Ex. 2. Given a = 31.239, b = 49.5053, A = 32 18'; find , C, c. Ans. B = 56 56' 56".3, or 123 3' 3".7 ; C = 9045' 3".7, or 2438'56".3; c = 58.456, or 24.382. 176 PLANE TRIGONOMETRY. 117. Case III. Given two sides and the included angle, as a, b, C; find A, B, c. (1) tanB = cot C . . (Art . 114) Hence =- i s known, and = 90 - - 2 22 .\ A and B are found. c = asinC r 6sinC sin A' sinB' and thus c is found and the triangle solved. In simple cases the third side c may be found directly by the formula c = Va 2 + b 2 - 2 ab cos C . . . (Art. 96) or the formula may be adapted to logarithmic calculation by the use of a subsidiary angle (Art. 90). Ex. 1. Given a = 234.7, b = 185.4, C = 84 36' ; find A, B, 2i s o (Art. 102) 2 a-c NOTE, The quantity r is the radjqe of the inscribed circle (Art. 102). 1 178 PLANE TRIGONOMETRY. Ex. 1. Given a = 13, b = 14, c = 15 ; find A, B, C. Solution. a = 13 log(s-a) = .9030900 6 = 14 log (-&) = .8450980 c = 15 log(s-c) = .7781513 2 s = 42 colog s = 8.6777807 s = 21 log?- 2 = 1.2041200 logr= .6020600. S QJ Oj 8 _ b = 7^ .-.log tan A = 9.6989700. s - c = 6 ' .-.log tan - = 9.7569620. .-. log tan - = 9.8239087. .-. A = 53 7' 48".38 ; B = 5929 f 23'M8; C = 67 22' 48".44. Without the use of logarithms, the angles may be found by the cosine formulae (Art. 96). These may 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, b = 14, c = 13. Then A is the greatest angle. A 6 2 + c 2 -a 2 14 2 + 13 2 - 15 2 ^T 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, = 64; find 6 = 71.0919. 2. Given c = 338.65, A = 53 24', B = 66 27'; find a = 313.46. . 3. Given c = 38, A = 48, B = 54; find a = 28.87, b = 31.43. 4. Given a = 7012.5, B = 38 12' 48", C = 60; find b and c. ^ ns . 5 = 4382.82 ; c = 6135.94. 5. Given a = 528, b = 252, A = 124 34'; find B and C. Ans. B = 238'33"; C = 32 17' 27". 6. Given a = 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". 7. Given a = 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". 8. Given a = 7, 6 = 8, A = 27 47' 45" ; find B, C, c. Ans. B = 32 12' 15", or 147 47' 45" ; C = 120, or 4 24' 30" ; c = 13, or 1.15385. 9. Given 6 = 55, c = 45, A = 6 ; find B and C. Ans. B = 149 20' 31"; C = 24 39' 29". 10. Given 6 = 131, c = 72, A = 40 ; find B and C. Ans. B = 108 36' 30" ; C = 31 23' 30". 11. Given a = 35, 6 = 21, C = 50 ; find A and B. Ans. A = 93 11' 49"; B = 36 48' 11". 12. Given a = 601, 6 = 289, C = 100 19' 6"; find A and B. Ans. A = 56 8' 42"; B = 23 32' 12", 180 PLANE TRIGONOMETRY. 13. Given a = 222, b = 318, c = 406 ; find A = 32 57' 8". 14. Given a = 275.35, 6=189.28, c= 301.47 ; find A, B, C. Ans. A = 63 30' 57" ; B = 37 58' 20" ; C = 78 30' 43". 15. Given a = 5238, b = 5662, c = 9384 ; find A and B. Ans. A = 29 17' 16" ; B = 31 55' 31". 16. Given a = 317, b = 533, c = 510 ; find A, B, C. Ans. A = 3518'0"; B = 76 18' 52"; C = 68 23' 8". 119. Area of a Triangle (Art. 101). EXAMPLES. Find the area : 1. Given a = 116.082, 6 = 100, C = 118 15' 41". Ans. 5112.25. 2. Gfiven a = 8, 6 = 5, C = 60. 17.3205. 3. Given 6 = 21.5, c = 30.456, A = 41 22'. 216.372. 4. Given a = 72.3, A = 52 35', B = 63 17'. 2644.94. 5. Given 6 = 100, A = 76 38' 13", C = 40 5'. 3506.815. 6. Given a = 31.325, B = 13 57' 2", A = 53 11' 18". Ans. 135.3545. 7. Given a = .582, 6 = .601, c = .427. .117655. 8. Given a = 408, 6 = 41, c = 401. 8160. 9. Given a = .9, 6 = 1.2, c = 1.5. .54. 10. Given a = 21, 6 = 20, c = 29. 210. 11. Given a = 24, 6 = 30, c = 18. 216. 12. Given a = 63.89, 6 = 138.24, c = 121.15. 3869.2. HEIGHT OF AN OBJECT. 181 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. 182 PLANE TRIGONOMETRY. 121. To find the Height of an Object standing on a Horizontal 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 the 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. r. Let CD be the object, whose base D /; is inaccessible ; and let it be required /'/ to find the height CD, and its horizon- ^'$ / \ a tal distance from A, the nearest acces- B '' a ' ] 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 = (Art. 95) sin (a - (3) HEIGHT OF AN OBJECT. 183 rD pA _ a s i n a s i n ft sin (a ft) ' a sin 8 cos a and AD = AC cos a = ?- sin (a /?) (2) TF/ien /ie Zwe BA cannot be measured directly toward the object. At A observe the vertical /. CAD = , and the horizontal Z. DAB = ft ; measure AB = a, and at B observe the Z. DBA = y. \^ Then AD= sT ^^-. \- Z 7 .-. 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 bank ; find the height of the tower. Ans. 124.26 feet. 2. At 360 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. Ans. 20 V3 ; 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 situated above a Horizontal Plane, and its 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 AE, observe the A C AE = a, and DAE = y; measure AB = a, and at B observe the Z. CBE = ft. Then CE = (Art. 122) Bin (a -ft) Also, AE = (Art. 122) sin (a ft) DE a COS a s " 1 sin (a ft) ... CD= asm fi jsiiitt sin (a- ft) 1 a sin/? sin (a y) cos y sin ( 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. 1\ 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' respectively : 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 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 ? Ana. (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 ^ P D vertical distance AD are required. Measure the angles of depression, 6BC, aAC, equal to a and (3 respectively, and denote AB by a. Then BD = CD tan , AD = CD tan/?. .-. a = CD (tan a tan 0). p-pj _ a cos a cos ft ~ sin (a - (3) ' and sin (a (3) EXAMPLES. 1. From the top of a house, and from a window 30 feet below the top, the angles of depression of an object on the ground are 15 40' and 10 : find (1) the horizontal distance of the object, and (2) the height of the house. AIM. l (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. Ans. 570.2 yards. 186 PLANE TRIGONOMETRY. 125. To find the Distant between Two Inaccessible Objects on a Horizontal Plane. Let C and D be the two inac- __- x7 cessible objects. Measure a base line AB, from whose extremities C and D are visible. At A observe the angles CAD, DAB; and at B observe the angles CBA and CBD. Then, in the triangle ABC, we know two angles and the 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 BAG, BAD = 76 30' and 44 10', respectively ; and the angles ABD, ABC = 81 12' and 46 5', respectively : find the distance between C and D. Ans. 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 = -V21. 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 height above the ground the visible horizon will be limited. THE DIP OF THE HORIZON. 187 Let be the centre of the earth, P a point above the sur- face, PD a tangent to the surface at c 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 h = OP - OA = r sec - r = r(l cos 0) cos r = hcos 6 1-COS0 rtan0 = - h sin Also 1-COS0 2 '; PD 2 = PA x PB = h(h + 2r) (Art. 48, Ex. 8) . . (Geom.) Since, in all cases which can occur in practice, h is very small compared with 2r, we have approximately Let n = the number * of miles in PD, h = the feet in PA, and r = 4000 miles nearly. Then A== PD^ = (5280 n) 2 = 2r 8000 x 5280 5280 n 2 = 5.28 8000 = 8 * It will be noticed that n is a number merely, and that the result will be in feet, since the miles 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, if ?i = 1 mile, we have h = | feet = 8 inches ; if 71 = 2 miles, h = f - 2 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. Ans. (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, ft, and the unknown angles PAC, PBC by x, y respectively ; then a and ft 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 x + y = 2ir a ft C . Also &sinx = osi M==pa sin a sin ft Assume an auxiliary angle < such that a sin a (1) tan $ = then the value of . o sin ft can be found from the tables. EXAMPLES. 189 m r Sin X Thus, = tan . smy = tan (0 - 45) sm ic + sin i/ tan < -f 1 [(14) of Art. 61]. .-. tan %(x y) = tan \ (x + y) tan (< 45) [(13) of Art. 61]. = tan (45 - <) tan \ ( + ft + C) . (2) thus from (1) and (2) a; and # are found. EXAMPLES. Solve the following right triangles : 1. Given a=51.303, c=150; find A = 20, B = 70, 6 = 140.95. 2. Given a=157.33, c=250; find A = 39, B = 51, 6 = 194.28. 3. Given a =104, c=185; find A = 3412'19".6, B = 5547'40".4, 6 = 153. 4. Given a=304, c=425; find A=4540' 2".3, B= 44 19' 57 ".7, 6 = 297. 5. Given 6=3, c=5; find A=53 7'48".4, B=3652'll".6, a=4. 6. Given 6=15, c=17; find A=28 4'20".9, B = 6155'39".l, a=8. 7. Given 6=21, c=29; find A=4336'10".l, B = 4623'49' f .9, a=20. 8. Given 6=7, c=25; find A=7344'23".3, B = 16 15' 36". 7, a=24. 190 PLANE TRIGONOMETRY. 9. Given 6=33, c=65; find A=5929'23".2, B = 3030'36".8, a=56. 10. Given c=625, A =44; find o=434.161, 6=449.587. 11. Given c=300, A=52; find a=236.403, 6=184.698. 12. Given c=13, A= 6722'48".5; find B = 2237'll".5, a = 12 6=5. 13. Given A = 7719'10".6, c=41; find B = 1240'49".4, a=40, 6 = 9. 14. Given B = 4853'16".5, c=73; find A=41 6'43".5, a=48, 6 = 55. 15. Given B = 64 0'38".8, c=89; find A=2559'21".2, a=39, 6=80. 16. Given A = 7719'10".6, a=40; find B = 1240'49".4, 6=9, c=41. 17. Given A=8712'20".3, a=840; find B= 247'39".7, 6=41, c=841. 18. Given A=3231'13".5, a=336; find B = 5728'46".5, 6=527, c=625. 19. Given A=8241'44", a=1100; find B= 718'16", 6=141, c=1109. 20. Given A=7523'18".5, 6=195; find B = 1436'41 f '.5, a=748, c=773. 21. Given B = 87 49' 10", 6 = 42536.37; find A= 210 f 50", a=1619.626, c=42567.2. EXAMPLES. 191 22. Given A= 8859' 6=2.234875; find B= 1 1', a= 125.9365, c= 125.9563. 23. Given A= 35 16' 25", a = 388.2647; find B= 5443'35," 6 = 548.9018, c=672.3412. 24. Given a= 7694.5, 6 = 8471; find A= 42 15', B = 4745', c= 11444. 25. Given a = 736, 6 = 273; find A= 6938'56".3, B = 2021'3".7, c=785. 26. Given a=200, 6 = 609; find A= 18 10' 50", B = 7149'10", c=641. 27. Given a=276, 6 = 493; find A= 2914'30".3, B = 60 45 '29. "7, c=565. 28. Given a= 396, 6=403; find A= 44 29' 53", B= 4530' 7", c=565. 29. Given a=278.3, 6=314.6; find A= 41 30', B= 4830', c=420. 30. Given a=372, 6=423.924; find A= 41 16' 2".7, B= 4843'57".3, c=564. 31. Given a=526.2, 6=414.745; find A= 5145'18".7, B= 3814'41".3, c=670. Solve the following oblique triangles : 32. Given B= 50 30', C = 122 9', a = 90; find A= 7 21', 6=542.850, c=595.638. 33. Given A= 8220', B= 4320', a=479; find C = 5420', 6=331.657, c=392.473. 34. Given A= 79 59', B= 44 41', a=795; find C= 51 20', 6=567.888, c=663.986. 192 PLANE TRIG ON OMETR Y. 35. Given B = 37 58', C= 65 2', a=999; find A = 77 0', 6 = 630.< r 71, 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 or -\ (\n 7, x 7^ . -LD , = O. O ; find C = 50 37' 44", a=4.3485, o=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, 0=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, a=221, o=222. 41. Given A= 21 14' 25", a =345, 6 = 695; find B = 46 52' 10", r\ 111 53' 25", c= 883.65. or B'=133 7' 50", C'= 25 37' 45", o'=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", o' = 20.172. 43. Given A= 21 14' 25", a=309, 6=360; find B = 24 51' 54", C = 13347' 41", c=615.67, or B'= 155 2' 6", C'= 3 43' 29", 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. 193 47. Given A= 50, = 119, 6=97; find B = 38 38' 24", C= 91 21' 36", c= 166.3. 48. Given C= 65 59', a = 25, c=24; find A= 72 4' 48", B= 41 56' 12", 6=17.56, or A'=10755'12", B'= 6 6' 48". 49. Given A = 1855'28".7, a=13, 6 = 37; find B= 6722'48".l, or B f = 11237'11".9. 50. Given C= 1511'21", a=232, 6 = 229; find A= 85 11' 68", B= 7936'40", c=61. 51. Given C=12612'14", a=5132, 6=3476; find A= 3228'19", B= 2119'27", c=7713.3. 52. Given C= 5512' 3", a = 20.71, 6 = 18.87; find A= 6728'51".5, B= 5719'5".5, c=18.41. 53. Given C= 1235' 8", a =8.54, 6 = 6.39; find A=13615'48", B= 31 9' 4", c=2.69. 54. Given C= 34 9' 16", a=3184, 6=917; find A=13351'34", B= 11 59' 10", c=2479.2. 55. Given C= 3210'63".8, a=101, 6 = 29; find A=13623'49".9, B= 1125'16".3, o=78. 56. Given. C= 9657'20".l, a=401, 6=41; find A= 7719'10".6, B= 543'29".2, c=408. 57. Given C= 3040'35", a=221,- 6=149; find A = 110 0'57".5, B= 3918'27".5, c=120. 58. Given C= 6659'25".4, a =109, 6=61; find A= 7936'40", B= 3323'54".6, c=102. 59. Given C = 13124'44", a = 229, 6 = 109; find A= 3323'54".6, B= 15ll f 21".4, c=312. 194 PLANE TRIGONOMETRY. 60. Given = 104 3' 51", a =241, 6=169; find A=4546'16".5,B = 30 9'52".5, c=332.97. 61. Given a=289, 6=601, c=712 ; find A=2332'12", B = 56 8'42", C = 10019 f 6". 62. Given a=17, 6=113, c=120; findA= 737'42", B = 6155'38", C = 11026'40". 63. Given a=15.47, 6=17.39, c = 22.88; find A=4230'44", B = 4925'49", C = 883'27". 64. Given a= 5134, 6=7268, c=9313; find A= 33 15' 39", B = 5056' 0", C = 9548'21". 65. Given a=99, 6 = 101, c=158; find A=3722'19", B = 3815'41", C = 10422'0". 66. Given a=ll, 6=13, c=16; findA = 43 2' 56", B = 5346 r 44", C = 8310'20". 67. Given a=25, 6=26, c=27; findA=5615' 4", B =59 51' 10", C = 6353'46". 68. Given a=197, 6 = 53, c=240 ; find A = 3153'26".8,B= 810'16".4, C = 13956'16".8. 69. Given a = 509, 6=221, c=480 ; find A=8432'50".5,B=2536'30".7, C=6950'38".8. 70. Given a=533, 6=317, c = 510 ; find A = 76 18' 52", B=3518' 0".9, C = 6823'7".l. 71. Given a=565, 6=445, c=606 ; find A=6251'32".9, B = 4429'53", C = 7238'34".l. 72. Given a=10, 6=12, c=14; find A=4424'55".2,B=57 7'18", C = 7827'47". 73. Given a = .8706, 6 = .0916, c=.7902; find A=14949'0".4,B= 3 1'56".2, C = 279'3".4. EXAMPLES. 195 Find the area : 74. Given o=10, 6 = 12, C = 60. Ans. 30V3. 75. " a =40, 6 = 60, C = 30. 600. 76. " 6 = 7, c=5V2, A = 135. 17. 77. a=32.5, 6 = 56.3, C=475'30". 670. 78. 6 = 149, A= 7042'30", B = 3918'28". 15540. 79. c=8.025, B = 100 5' 23", C = 31 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. a=577, 6=73, c=520. 12480. 84. a = 52.53, 6 = 48.76, c=44.98. 1016.9487. 85. a=13, 6 = 14, c=15. 84. 86. " a=242 yards, 6 = 1212 yards, c=1450 yards. Ans. 6 acres. 87. 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 : i : 1. O o 89. The area of a 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. 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 ? Ans. 138.5 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 of elevation of the top and bottom of the tower are observed to be 75 and 60 respectively : find the height of the cliff. Ans. 86.6 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 the 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 KE. ; 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. Ans. 250 V3 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. 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. Ans. 20V3, 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 6, respec- tively, prove DE = (a + 1)) 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 BCA 135, and the length of BC is 1000 yards : find the height of A over B. Ans. 500(3 + V3) yards. 109. A right triangle rests on its hypotenuse, the length of which is 100 feet; one of the angles is 36, and the inclination of the plane of the triangle to the horizon is 60: find the height of the vertex above the ground. Ans. 25 V3 cos 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. 221 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. Ans. - miles. V2 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 ( V3 + 1) . 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 23 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 671.38 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 30 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) 765.015 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 - 200 PLANE TRIGONOMETRY. 120. On the bank of a river there is a column 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. 10 V115 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. 160 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. Ans. 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 300 height of the tower is - - feet. V7 . 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. v-i 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 60(3* -3"*) feet. 131. A tower stands at the foot of an inclined plane whose inclination to the horizon is 9; aline 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 TRIGONOMETRY. 133. From a window it is observed 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. Ans. 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 J, the angle QBA is 110, and the angle PBA is 52 2Q f : 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. Ans. 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 CP. Ana. 2109.824 yards. 140. A, B, C, are three objects such that AB = 320 yards, AC = 600 yards, and BC = 435 yards. From a station P it is observed that APB = 15, and BPC = 30 : find the distances of P from A, B, and C ; the point B being near- est to P, and the angle APC being the sum of the angles APB and BPC. Ana. PA = 777, PB = 502, PC = 790. 204 PLANE TRIGONOMETRY. CHAPTER VIII. CONSTRUCTION OF LOGARITHMIC AND TRIGONOMETRIC 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 x in a series of ascending powers ofx. By the Binomial Theorem, [2 2 nx(nx \)(nx 2) 1_ ~~ ~ 12 Similarly, i-i [2 |3 (2) LOGARITHMIC SERIES. 205 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) The series in the parenthesis is usually denoted by e ; hence e *= 1 + * + ^ + ^ + ^ + ....... (3) |2 [3 [4 which is the expansion of e x in powers of x. This result is called the Exponential Theorem. If we put x = 1, we have from (3) 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 c ; then c = log e a, and a* e cx . Substi- tuting in (3), we have or a.= l + *log. + +... (4) which is the expansion of a* in powers of x. 130. Logarithmic Series. To expand log e (l + #) in a series of ascending powers of x. By the Binomial Theorem, - 1) (- 2) 206 PLANE TRIGONOMETRY. = 1 + X [ a _l _ i ( a ._ 1)2+ i . (a _ 1)3 _ (_l)4+ ...] + terms involving x 2 , x", etc. Comparing this value of a x with that given in (4) of Art. 129, and equating the coefficients of x, we have log e a = a-l-^(a-l) 2 + i(a-l) 3 -i(^-l) 4 +- Put a = 1 + # ; then log,(l+*) = *-f 2 + f-| 4 + ......... (3) 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 calculation. If x > 1, the series is altogether unsuitable. We shall therefore deduce some more convenient formulae. Changing x into x, (1) becomes log,(l-) = -*-f-f-f- ...... (2) Subtracting (2) from (1), we have Put 1 x n and (3) becomes i !Li = <>r _ i i i ge n "|_ 2w + 1 3 ( 2 " + 1 ) 5 ( 2n + 1 ) or log e (n -f 1) 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 common 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 prime 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 n = 1, 2, 4, 6, etc., suc- cessively, in (4) of Art. 130, we obtain the following Napierian Logarithms : log, 3 = log.2 + 2[J + J- + J-+^ + ..-] = 1.09861228. log, 4 = 21og e 2 =1.38629436. log. 5=lo log, 6 = log, 2 + log, 3 =1.79175946. log, 7 = 1 o g ,6 + 2[i + J^ + ,-i J+ ...] =1.94590996. log e 8 = 31og e 2 =2.07944154. log e 9 = 21og e 3 =2.19722456. log e 10 = log e 5 + log e 2 = 2.30258509. And so on. 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 of 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 logm = Napierian logm x .43429448. The number .43429448 is called the modulus of the common system. It is usually denoted by /x. 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 /x, we obtain a series by which common logarithms may be computed ; thus, lo glo (n + l) = (1) Common Logarithms. Iog 10 2 = /ilog.2 = .43429448 x .69314718 = .3010300. Iog 10 3 = /xlog e 3 = .43429448 x 1.09861228 = .4771213. Iog 10 4 = 2 Iog 10 2 = .6020600. log, 5 = /xlog 6 5 = .43429448 x 1.60943790 = .6989700. And so on. 132. If 6 be the Circular Measure of an Acute Angle, sin 0, 0, and tan are in Ascending Order of Magnitude. With centre 0, 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' = 0. Then BB' < arc BAB' < BT + B'T (Geom., Art. 246) .-. BC < arc BA < BT. LIMITING VALUES OF SIN 0. 209 EC BA BT '* OB OB OB' .-. sin < < tan 0. 133. The Limit of ^, when is Indefinitely Diminished, is Unity. We have sin < < tan ...... (Art. 132) .-. 1<-A_< sec0. sin0 Now as is diminished indefinitely, sec 6 approaches the limit unity ; then when 9 = 0, we have sec = 1. A .-. the limit of - , which lies between sec0 and unity, is unity. 8in .*. also - - approaches the limit unity. As _ = x ge(} ^ the limifc Q tan_0 when $ . g 60 6 definitely diminished, is also unity. This is often stated briefly thus : =!, and =l, when = 0. NOTE. From this it follows that the sines and the tangents of very small angles are proportional to the angles themselves. 134. If 9 is the Circular Measure of an Acute Angle, sin s B~ lies between 9 and 9 -- ; and cos 9 lies between 1 -- and (1) We have tan->- ..... (Art. 132) 2i 2i 06 ., sm- >- cos-. 210 PLANE TRIGONOMETRY. B B .\ 2 sin- cos r >0cos 2 -- 22 2i .-. sin<9><9(l-sin 2 - a -"-} . . (Art. 132) V A/ .-. sinO<0 and > - - (2) cos(9 = l-2sin 2 -. Also, . s in->--V-Yby (1). 22 4V2/ ... cose 1 - T and < 1 - - + t . 2 2 16 NOTE. It may be proved that sin 6> , as follows : 6 We have 3 sin - - sin 9 = 4 sins ? ( Art. 50) (1) 3 3 .-. 3sin - - sin -=4B\n* - (by putting - for 0) (2) 3^ 3 3^ 3 3 gin .1 - sin =4 ei^ (n) 3 n 3 n-l 3 n Multiply (1), (2), ... (n) by 1, 3, ... 3"- 1 , respectively, and add them, 3n Bin 1 _ 8 in = 4(in3 ? + 3 sins i + ... S' 1 " 1 sins -^-V 8" \ 3 3 2 3' 1 / SINE AND COSINE OF 10" AND OF 2'. 211 ^ ~ ! / 1 + S+-^ri) (Art. 182) If n = oo , then j = 1 (Art. 133) 4 / 1 1 ^_ 4 _^ _1 ? 33 A 1 + 3^ + '" g^2j ~ "sT ' 1 _ ^ ~ 7T 3 2 0-sin0<-, and /. sin0>0--. 6 6 This makes the limits for ein closer than in (1) of this Art. 135. To calculate the Sine and Cosine of 10" and of V. (1) Let be the circular measure of 10". Then g = 1Q * = 3.141592653589793 - 180 x 60 x 60 64800 or = .000048481368110 -, correct to 15 decimal places. . -. - = .000000000000032 ., 4 ... 0-~ = . 000048481368078-., " 4 Hence the two quantities and agree to 12 deci- 4 03 mal places ; and since sin 6 < and > (Art. 134), 4 .-. sin 10" = .000048481368, to 12 decimal places. We have cos 10" = VI -sin 2 10" = 1 - i sin 2 10" = .9999999988248 , to 13 decimal places. Or we may use the results established in (2) of Art. 134, and obtain the same value. 212 PLANE TRIGONOMETRY. (2) Let be the circular measure of 1'. Then = ^ = .000290888208665, to 15 decimal places. loU X oU .-.^=.000000000006 to 12 " " 4 .-. 6 - - = .00029088820 to 11 " 4 /j3 Hence and differ only in the twelfth decimal. 4 .-. sin 1' = .00029088820 to 11 decimal places. cos 1' = Vl - sin 2 1 ' = . 999999957692025 to 15 decimal places, Otherwise thus : 1 _ 6 1 =.999999957692025029 to 18 decimal places. and = .00000000000000044 to 17 decimal places. 16 Butcosl'>l-^and 1 ; then - is not greater than .0001. ... j# is not >^(.0001) 2 , i.e., not > .0000000025 ; and & is much less than this. 3n 3 .-. log (n -f- d) log n //,-, correct at least as far as seven decimal places. RULE OF PROPORTIONAL PARTS. 219 Hence if the number be changed from n to n -f d, the corresponding change in the logarithm is approximately f 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 (0 -\- h) sin 6 sin h cos sin 6 (1 cos h) = sin h cos 6 (l - tan tan -\ . (Art. 51) \ 4/ If h is the circular measure of a very small angle, sin h = h nearly, and tan ^ = ^ nearly. 2 2 .-. sin (0 + h) - sin = 7i cos fl - tan tan - \ 7,2 = 7i cos e ~ sin 0. 2i If 7i is the circular measure of an angle not > 1', then 7i is not > .0003 (Art. 135). .-. - is not > .00000005; and sin 6 is not > 1. .-. sin(0-t-7i) sin 6=hcos 0, as far as seven decimal places, which proves the proposition. Similarly, sin (0 h) sin 6 = h cos 0, approximately. 146. To prove the Rule for a Table of Natural Cosines. cos (0 h) cos = sin h sin cos 0(1 cos h) = sin h sin 0(1 cot tan - ) V 2, If h is the circular measure of a very small angle, sin h = h nearly, and tan - = - nearly. .-. cos (0 h) cos = h sin 0/1 cot tan - -COS0. 220 PLANE TRIGONOMETRY. We may prove, as in Art. 145, that cos is not > .00000005. .*. cos (0 h) cos = h sin 0, as far as seven decimal places, which proves the proposition. Similarly, cos (6 + h) cos = h sin 0, approximately. 147. To prove the Rule for a Table of Natural Tangents. sin sin * cos (6 + h) cos cos (0 + 7i) cos _ tan h ~cos 2 0(l-tan0tan7i)' If h is the circular measure of a very small angle, tan 7i = h nearly. .-. tan (0 + 7i) - tan = -I 1 sec2 l-h tan .-. tan (6 -f h) tan 6 = h sec 2 0, approximately, unless sin sec 3 6 is large, which proves the proposition. Similarly, cot (0 h) cot = 7i cosec 2 0, approximately. /#c7i. 1. If 7i is the circular measure of an angle not > l r , then h is not > .0003. Hence the greatest value of h 2 sin sec 3 6 is not > .00000009 sin sec 3 0. Therefore, when > |, 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 and 45. RULE OF PROPORTIONAL PARTS. 221 148. To prove the Rule for a Table of Logarithmic Sines. sin (0 + h) smO = h cos - - sin . . (Art. . sin (9 .-. log sin (0 + h) log sin = Jh cot (9 - h * --(h cot (9 - 7 |Y+ -.."I (Art. 130) L 2 2\ 2J J = ^ cot - (1 + cot 2 0) + ... If ft is the circular measure of an angle not > 10", then h is not > .00005, and therefore, unless cot 6 is small or cosec 2 large, we have log sin (0 + h) log sin = pli cot 0, as far as seven decimal places, which proves the rule to be generally true. JSch. 1. When is small, cosec 9 is large. If the leg sines are calculated to every 10", then h is not >. 00005, and /A is not > .5. cosec 2 is not In order that this error may not affect the seventh decimal place, 6 cosec 2 must not be > 10 3 , that is, must not be less than about 5. When 6 is small, cot is large. Hence, when the angles 222 PLANE TRIGONOMETRY. are small, the differences of consecutive log sines are irregu- lc$, and they are not insensible. Therefore the rule does not apply to the log sine when the angle is less than 5. . 2. When is nearly a right angle, cot# is small, and cosec 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 (6 - h) - cos = h sin - cos . (Art. 146) cos .-. log cos (0 h) log cos = /* [ h tan - - 1 fh tan - In this case the differences will be irregular and large when is nearly a right angle, and irregular and insensible when is nearly zero. This is also clear because the sine of an angle is the cosine of its complement. 150. To prove the Rule for a Table of Logarithmic Tan- gents. tan (6 + h) tan = h sec 2 + 7t 2 sin sec 3 . (Art. 147) tan & sin $ cos RULE OF PROPORTIONAL PARTS. 223 .-. log tan (0 + h) log tan 6 .-. log tan (0 + h) log tan ph _2 * 2 cos 20 ~ sin cos ** sin 2 20 151. 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 which are neither very small nor nearly equal to a right angle. The following exceptional cases arise : (1) The difference sin (0 -f- h) sin 6 is insensible when is nearly 90, for in that case hcosO is very small; it is then also irregular, for Jft 2 sin0 may become comparable with ft cos 0. (2) The difference cos (0 -f h) cos 6 is both insensible and irregular when 6 is small. (3) The difference tan (0 + h) tan is irregular when is nearly 90, for ft 2 sin sec 3 may then become compar- able with h sec 2 ; it is never insensible, since sec is not <1. (4) The difference log sin (0 -f- ft) log sin is irregular when 6 is small, and both irregular and insensible when is nearfy 90. (5) The difference log cos (6 + ft) log cos 6 is insensible and irregular when is small, and irregular when is 90. (6) The difference log tan (0 + 7i) log tan is irregular when is either small or nearly 90. 224 PLANE TRIGONOMETRY. 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 functions 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 from the value log cos 0, for small angles, or from the value log sin &, for angles nearly 90. When 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 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 Methods to replace the Rule of Proportional Parts. (1) The simplest plan is to have tables of log sines and log tangents, for each second, for the first 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 or 90. (2) Delambre's Method. In this method a table is con- structed which gives the value of log (- log sin I" for 9 every second for the first few degrees of the quadrant. * This article has been taken substantially from Hobson's Trigonometry. METHODS TO REPLACE THE RULE. 225 Let 6 be the circular measure of n seconds. Then, when is small, we have 6 = n sin 1", approximately. T sin0 T sinn" , ^ .-. log = log : = log sin n" log n log sm 1". n sin 1 .-. log sin n" = log n + ( log h log sin 1" Y V 0. / Hence, if the angle is known, the table gives the value of the expression in parenthesis, and log ft can be found from the ordinary table of the logs of numbers ; thus log sin n" can be found. If log sin n" is given, we can find approximately the value of n, and then from the table we have the value of the expression in parenthesis; thus we can find logw, and then n from an ordinary table of logs of numbers. Rem. When is small (less than 5), = 1 , approximately . . (Art. 134, Note) (7 6 Hence a small error in will not produce a sensible error in the result, since log - - will vary much less rapidly than 6. (3) Maskelyne' s Method. The principle of this method is the same as that of Delambre's. If is a small angle, we have /\5 sin = , approximately, 6 and cos 0=1--, " . . (Art. 134) sinfl " .-. log sin 6= log + | log cos 0, approximately. 226 PLANE TRIGONOMETRY. When 6 is a small angle, the differences of log cos 6 are insensible (Art. 149) ; hence, if be given, we can find log accurately from the table of natural logarithms, and also an approximate value of log cos ; the formula then gives log sin at once. If log sin be given, we must first find an approximate value of from the table, and use that for finding log cos 0, approximately ; 6, is then obtained from the formula. EXAMPLES. 1. Prove l 2. + i+... = 2e. q 1 / o q . Prove log- = --[ - - + - - b 2 2 1- 3 ' 23 2 - 5 5. Prove tan + \ tan 3 -f | tan 5 H 6. Prove log e 11 = 2.39789527 .> by (4) of Art. 130. 7. Prove log e 13 = 2.56494935"., " 8. Prove log e 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, 52.9381, and .085623. Ana. 7.18923, 3.96913, - 2.4578. EXAMPLES. 227 /Q 11. Prove that the limit of m sin is 0, when ra = oo. m 12. " " " ratan is 0, m 13. " " " sin is Trr 2 , " n = oc. 2 n 14. " " " Trr 2 tan _ is Trr 2 , " n = oo. n 15. " " " - is , " = 0. vers 60 o 2 16. " " " fcos^Yisl, " n = oo. / ff\r 17. " a " [sin-]isl, " n = oo. V y 18. " " " /^cos-Yisl, " n = oo. V / n /sin _N 19. " I 5]isl, " ?i = oo. / /}\ ^i^ ^^ 20. " " (cos-j ise~^, " n = oo. 21. " " " (cos-) is zero, when n = oo. 22. If is the circular measure of an acute angle, prove (1) cos0 = 2 tan : find the limit- ing values of n that these equations may coexist. Ans. n must lie between 1 and 2, or between 1 and 2. 26. Find the limit of (cos ax) cosec26x , when x = 0. 27. From a table of natural tangents of seven decimal places, show that when an angle is near 60 it may be determined within about - of a second. 28. When an angle is very near 64 36', show that the angle can be determined from its log sine within about y 1 ^ of a second ; having given (log e 10) tan 64 36' = 4.8492, and the tables reading to seven decimal places. DE MOIVRE'S THEOREM. 229 CHAPTER IX. DE MOIVKE'S THEOREM,* APPLICATIONS, 153. De Moivre's Theorem. For any value of n, positive or negative, integral or fractional. . (1) I. When n is a positive integer. We have the product (cos a + V 1 sin a) (cos ft + V 1 sin J3) = (cos a cos (3 sin a sin ft) -f V 1 (cos a sin ft + sin a cos /?) = cos (a -f ) + V^T sin (a + /?) . Similarly, the product [cos ( + ) + V 11 ! sin (a + 0) ] [cos y + V^~l sin 7 ] = cos (a 4- /? + y) + V^^ sin (a + ft -f y) - Proceeding in this way we find that the product of any number n of factors, each of the form cosa-{- V 1 sin a = cos (a + /? + y H ---- w terms) -f V^T sin (a + /? + y -| ---- n terms). Suppose now that a = /? = y = etc. = 0, then we have (cos + V^T sin 0) H = cos n6 + V^l sin w0, which proves the theorem when n is a positive integer. * From the name of the French geometer who discovered it. 230 PLANE TRIGONOMETRY. II. Wlien n is a negative integer. Let n = m; then m is a positive integer. Then (cos + V^ sin 6) * = (cos + V^T sin 0) ~ m 1 1 -(by I.) (cos0-f V 1 sili 0) m cosra0-f- V 1 sinm0 1 cosmfl V 1 sinmfl cosm^ -|- V 1 sin mO cosmO V 1 sinw0 cosm^ V 1 sinm0 / ^ f\ = cosm0 V Ismm^ cos j mv + sm j mB = cos ( mO) + V 1 sin (mO). . . (cos 6 4- V^^ sin 6) n = cos nB + V^ sin n0, which proves the theorem when n is a negative integer. III. Wlien n is a fraction, positive or negative. Let n = -, where p and q are integers. Then (cos0+V ::: Ism0) ? '=cosp0+ V^sinp^by I. and II.). / P P \ q But f cos -0+ V 1 sin-0 j cospO + V .-. (cos + V^l sin 0) p = cos -0 + V^I sin - 6> - that is, one of the values of (cos + V 1 sin 6) q pO - . p$ is cos -- hv Ism In like manner, (cos V 1 sin 0) n = cos nO V 1 sin nO. Thus, De Moivre's Theorem is completely established. 'It shows that to raise the binomial cos# + V 1 sin^ to DE MOIVRE'S THEOREM. 231 any power, we have only to multiply the arc 6 by the exponent of the power. This theorem is a fundamental one in Analytic Mathematics. 154. To find All the Values of (cos0 + V^ sin 0) When n is an integer, the expression (cos + V 1 sin0) n P can have only one value. But if n is a fraction = -, the expression becomes _ P , - - (cos 6 + V 1 sin 0)* =\ (cos + V 1 sin 0) p , which has q different values, from the principle of Algebra (Art. 235). In III. of Art. 153, we found one of the values _ P of (cos 6 -\- V 1 sin0) 7 ; we shall now find an expression p which will give all the q values of (cos 9 -f V 1 sin 0)*. Now both cos and sin remain unchanged when is increased by any multiple of 27r; that is, the expression cos0-h V Isin0 is unaltered if for we put (0-f 2r?r), where r is an integer (Art. 36) . = [cos (0 + 2rv) + V- 1 sin (0 + 2r7r)] ? ^^ (Art. 153) (1) (0 + 2nr) When ?' = g, cos - = cos fpO \ pO = COS f + 2p7TJ= COS y, the same value as when r = 0. When r = q + 1, c< the same as when r = 1, etc., from which it appears that there are q and only q different values of cos^ , since the same values afterwards recur in the same order. Similarly for sin Therefore the expression cos gives all the q values of (cos + V 1 sin 0) * and no more. And this agrees with the Theory of Equations that there must be q values of a?, and no more, which satisfy the equa- tion x q = c, where c is either real or of the form a-f-6 V 1. APPLICATIONS OF DE MOIVRE'S THEOREM. 233 APPLICATIONS OF DE MOIVRE'S THEOREM. 155. To develop cos nO and sin nO in Powers of sin and COS0. We shall generally in this chapter write i for V 1 in accordance with the usual notation. * By De Moivre's Theorem (Art. 153) we have cos ?i0H- i sinrj0 = (cos0-Msin0) n . . . . . (1) Let n be a positive integer. Expand the second member of (1) by the binomial theorem, remembering that i 2 = 1, p = i } and that i 4 = -fl (Algebra, Art. 219). Equate the real and imaginary parts of the two members. Thus, cos nO = cos n B - n ( n ~ 1 ) cos"- 2 sin 2 L? + n ^~ II n ( n - *) ( n ~ 2 ) sin n0 = n cos"- T sin - - ~ cos*- 3 <9 sin 8 + n(n-l) (n-2) (n-3) (n - 4) COB .^ sin 5 g _ etc> (3) [^ The last terms in the series for cos nO and for sin nO will be different according as n is even or odd. The last term in the expansion of (cos + isin0) n is ?'"sin n 0; and the last term but one is ni n '^cos sin"" 1 ^. Therefore : When n is even, the last term of cosw0 is t*sin*0 or n ( I)" 1 sin w 0, and the last term of sin nO is w n ~- cos 6 sin"- 1 or n( 1) 2 cos sin"" 1 6. When n is odd, the last term of cos nO is ni n ~ l cos sin"" 1 9 n\ or n( 1) 2 cos sin"" 1 0, and the last term of sinnO is nl n-l 2 n 234 PLANE TRIGONOMETRY. EXAMPLES. Prove the following statements : 1. sin 40 = 4 cos 3 sin 4 cos sin 3 0. 2. cos% = cos 4 0-6 cos 2 sin 2 + sin 4 0. 156. To develop sin and cos in Series of Powers of 0. Put nO = a in (2) and (3) of Art. 155; and let n be in- creased without limit while a remains unchanged. Then since = -, must diminish without limit. Therefore the n above formulas may be written na ( 0) n 9 /, /sin 0V = cos"0 -- ^ ^-cos n ~ 2 0( ) [2 \ J <*(<*- 0)0* -20)0* -30) cog 4 , /sin L 0V_ \ e ) and Bina^ (-fl)(-2fl) 30/sinfly . , 2 v [3 V 6 y If n = oo, then = 0, and the limit of cos and its powers is 1; also the limit of ( S ^ 11 ) and its powers is 1. Hence (1) and (2) become > cos=l-| 2 + | 4 -|'+- (3) 3 , a 5 sm = + i. In the series for since and cos , just found, a is the circular measure of the angle considered. CONVERGENCE OF THE SERIES. 235 Cor. 1. If a be an angle so small that a 2 and higher powers of a may be neglected when compared with unity, (3) becomes cos a = 1, and (4), sin a = a. If a 2 , 3 be retained, but higher powers of a be neglected, (3) and (4) give 3 2 sin = a ; cos a = 1 (Compare Art. 134) Cor. 2. By dividing (3) by (4), we obtain O , ,, O O O O O 4 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 are etc. 2-3 4-5 6-7 8-9 Hence the ratio of the (n-fl)th term to the nth term is a 2 ; and whatever be the value of a. we can take ' n so large that for such value of n and all greater values, this fraction can be made less than any assignable quantity ; hence the series is convergent. Similarly, it may be shown that the series for cos a is always convergent. 158. Expansion of cos n in Terms of Cosines of Multi- ples of 0, when n is a Positive Integer. Let x = cos + i sin ; then - = - = cos i sin 0. x cos -f i sin and x - = 2isiuO ..... (1) X 236 PLANE TRIGONOMETRY. Also x n = (cos 6+i sin 0)" = cos nO+i sin nO (Art. 153) (2) and = (cos0 isin0) M = cos?i0 i'sinntf ... (3) M? ^. 2 cos nO = x n + -i, and 2 i sin n0 = z n - i . (4) 3J 3J M Hence (2cos0) = (# + or 1 )", by (1), = x n -f nof- 2 + n ( n ~ JJ. of-4 _j_ etc. + war< M - 2) + or n I? IV nor^ + ^-\ + ^=^l x ^ + 1 V etc. a ! 2 ~ 2 cos (i - 2)6 + ~ 2cos(yi - 4)0 + etc. l .-. 2"- 1 cos n ^ = cos nO + ?i cos (w 2) + 7 ll!^ll) C os O - 4) + etc. (5) NOTE. In the expansion of (x + x~ l ) n there are w + 1 terms; thus when n is even there is a middle term, the (- + l)th, which is independent of 0, and which is Hence when n is even the last term in the expansion of 2 n ~ l cos n 6 is When n is odd the last term in the expansion of 2 n1 cos n 0"is 159. Expansion of sin" 9 in Terms of Cosines of Multiples of 0, when n is an Even Positive Integer. (2 % sin BY = (x- *Yby (1) of Art. 158 V X J iT~ .^(n-1 [2 EXPANSION OF SIN n 0. 237 I? = cos n<9 - w cos (n 2)0 + n ( n ~ 1 ) cos (w - 4) ) 4- 160. Expansion of sin rt in Terms of Sines of Multiples of 0, when n is an Odd Positive Integer. (2 f sin ^) w = fx - -Yby (1) of Art. 158 V X J [2 [ (2esin0) n = 2 1 sin w^ n2t sin (n 2)0 l liGLzi of Art 158] 238 PLANE TRIGONOMETRY. Whence dividing by 2i, we have H-l 2 s in"0 = sin nB - n sin (n -2)0-}- nn ~ s i n ( w _ 4)$ 14 (- 1)^(. -l)...i(. + 8) EXAMPLES. Prove that 1. 128 cos 8 0=cos 8 0+8 cos 60 + 28 cos 4 0+56 cos 20+35. 2. 64 cos 7 = cos 7 + 7 cos 50 + 21 cos 30 + 35 cos 0. 161. Exponential Values of Sine and Cosine. Since e = 1 + x + + + + .- . . . (Art. 129) = cos + i sin . . ..... (Art. 156) /J2 /)4 and e - = 1 _ + _et, = cos i sin 0. .-. 2 cos = e ie + e-* , and 2 i sin = e* 9 e~ i9 . . (1) e _l_ e -tfl . e _ e -f .-. cos0 = , and sin = - : ... (2) 2i l 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, (3) * Called also Euler's equations, after Euler, their discoverer. GREGORY'S SERIES. 239 Sch. These results may be applied to prove any general formula in elementary Trigonometry, and are of 'great im- portance in the Higher Mathematics. EXAMPLES. 1. Prove Sm26> = tail ft 1 + cos 20 Prove the following, by the exponential values of the sine and cosine. 2. cos 2 a = cos 2 a sin 2 a. 3. sin0 = -sin(-0). 4. cos30 = 4cos 3 3 cos ft Rem. If we omit the i from the exponential values of the sine, cosine, and tan- gent of 9, the results are called respectively the hyperbolic sine, cosine, and tangent of 0, and are written sinh 0, cosh 0, and tanh 9, respectively. Thus we have sinh = i sin iff, cosh = cos iff, tanh 6 = i tan iff. 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 such works as Casey's Trigonometry, Hobson's Trigonometry, Lock's Higher Trigonom- etry, etc. 162. Gregory's Series. To expand in powers of tan lere lies between - and + - By (3) of Art. 161, we have 1-f t'tanfl 9 \ i9 240 PLANE TRIGONOMETRY. .-. log e= log(l -h i tan 0) - log(l - i tan (9). tan 5 0-etc.) (Art. 130) n 5 0-etc ..... (1) which is Gregory's Series. This series is convergent if tan 6 = or < 1, i.e., if 6 lies between ^ and -, or between |TT and JTT. This series may also be obtained by reverting (5) in Cor. 2, Art. 156. Cor. 1. If tan 6 = x, we have from (1) fam-ia?=*4-~-etc ...... (2) o o Cor. 2. If = -, we have from (1) 4 (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. Euler's Series. tan- 1 1 + tan- 1 1 = . . . (by Ex. 2, Art. 60) Put = tan- 1 i. A tan = 1, which in (1) of Art. 162 gives Put e = tan- 1 - .-. tan = \, and (1) becomes 3 ' 3 t au - 1 i= 1 -- 1 -+-i ___ L-4-etc ..... (2) 3 3 3.3 3 ^5-3 5 7-3 7 MACHINES SERIES. 241 Adding (1) and (2) we have =./i'lJL J_ ..W3L_ii; J_ A.-/** 3 5 3 5 4 2 3-2 3 5-2 5 / \3 3.3 3 5.3 a series which converges much more rapidly than (3) of Art. 162. 164. Machin's Series. Since 2 tan- 1 - = tan- 1 -i (by Ex. 3, Art. 60)= tan" 1 4> 5 1 12 ...^tan-'i-tan-'^. 5~3~^P~5~.!?~ V239 3-(239) 3 ' 5 (239) 5 In this way it is found that TT = 3.141592653589793 . - - Cor. Since tan- + tan- = tan NOTE. The series for tan" 1 and tan" 1 are much more convenient for pur- 70 99 poses of numerical calculation than the series for tan" 1 -- 239 Example. Find the numerical value of TT to 6 figures by Machin's series. 242 PLANE TRIGONOMETRY. 165. Given sin = x sin (0 + ); expand in a Series of Ascending Powers of x. We have e ie e~ i9 = x[e i9 + ia - e~ i9 - ia -~\ . . (Art. 161) ... e 2iO - 1 = x G 2iO . e ia _ e -a- .-. 2 10 = log(l - xe~ ia ) - log(l - ae ia ) = #(e ia - 6~ ia ) 4- ^( e 2ia_ e -2ia) _|_ ^ (Art. 130) .-. ^ = a;sma + -sin2 + -sin34-"- (Art. 161) (1) 2 3 Examine. H a = ir 20 > then JB = 1. .-. (1) becomes 166. Given tan x = n tan ^ ; expand x in Powers of n. pix _ p ix aiO _ p - id ^ = n e e - .... (Art. 161) -ix iO __ -i9 = n- / 1-|A [ where m = - V 1 4- me 2 * 2io; = 2^4- log(l + me-** 9 ) log(l 4- me 2 fl ) . . . . (Art. 161) RESOLUTION INTO FACTORS. 243 RESOLUTION OF EXPRESSIONS INTO FACTORS. 167. Resolve x n 1 into Factors. Since cos 2 TIT V 1 sin 2 rir = 1, where r is any integer, and x n = 1, .-. of = cos 2r7r V 1 sin 2r7r. i .-. a; = (cos 2?-7r V- 1 sin 2r)" = cos V^l sin E. . . (Art. 153) (1) n n (1) When n is even. If r = 0, we obtain from (1) a real root 1 ; if r = -, we obtain a real root 1, and the two cor- responding factors are x 1 and x -f 1. If we put ,- = l,2,3...5-l, in succession in (1), we obtain n 2 additional roots, since each value of r gives two roots. The product of the two factors, which are n n J and ' - 2 - n n n n which is a real quadratic factor. (2) -2a; cos (3) * This expression gives the n nth roots of unity. 244 PLANE TRIGONOMETRY. (2) When n is odd. The only real root is 1, found by putting r = in (1) ; the other n 1 roots are found by putting r = 1, 2, 3, in (1) or (2) in succession. ( a; 2 2#cos- Tr+l)(x 2 2xcos- rr+l.. (4) 168. Resolve x n + l into Factors. Since cos (2r + !)* V^T sin (2r -f I)TT = - 1, where r is any integer, and x n = 1, .-. af = cos(2r + l)7r V^l sin (2r + I)TT. .-. a; =[cos(2r-f l)ir V :r I sin (2r + !)*]" / T . 2r+l = cos - 7TV Ism TT . . . (1) n n which is a root of the equation x n = 1 ; ie., 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 successively, in (1). The product of the two factors, and = x*-2xcos^-*7r + l . (2) n which is a real quadratic factor. RESOLUTION INTO FACTORS. 245 STT n (3) (2) When n is odd. The only real root is 1 ; the other n 1 roots are found by putting r = 0, 1, 2, in (1), in succession. .-. x n + 1= (x + 1/ar 2 - 2acos - + lYo 2 - 2 cos + 1 " (4) EXAMPLES. 1. Find the roots of the equation x 5 1 = 0. Ans. 1, cos (2nr)-M'sin|(2rir), where r=l, 2, 3, 4. 2. Find the quadratic factors of as 8 1. .4ns. (a 2 - 1 ) (y? - V2 a; + 1) (a? + 1) (aj 2 + V2 a; + 1) . 3. Find the roots of the equation 4 + 1 = 0, and write down the quadratic factors of x* + 1- Ana. -L V^l A:; (a? - W2 + 1) (x- 2 + W2 + 1). V2 V2 169. Resolve iB 271 2 # M cos ^ + 1 into Factors. Let a^ M -2a; n cos^ + l=:0. ... z? n 2x n cos + cos 2 = sin 2 6. .-. x n cos = V 1 sin = i sin 0. /. a; = (cos B i sin 0)" = cos 2r7r + i sin 2r7r + B (1) n n since cos is unaltered if for we put + 2 r?r. If we put r = 0, 1, 2, -"n 1, successively in (1), we find 2w differ- ent roots, since each value of r gives two roots. 246 PLANE TKlGONOMETItY. The product of the two factors in (1) / 2nr + . . = x cos ~ i sin n J sin ? v "*" n i GO n .-. x 2n 2x n cosO + l = for 2 - 2 a cos - + 1 Yor 2 - 2 z cos 2?r + ^ + A . . V w A w ) / (271-4)^ + \ ...f a; 2 2 a; cos ^ hi ) .. (3) Cor. Change x into - in (3) and clear of fractions, and we get x 2n - 2 a n x n cos + a 2 " = (a? - 2 ax cos- + a 2 V- ( x 2 2 ax cos v [- a 2 }( x 2 2 ax cos ^ |- a 2 ^ n J\ n to n factors (4) EXAMPLES. Find the quadratic factors of the following : Ans. (x*-2x cos 15 + 1) (x 2 - 2 x cos 105 + 1) X (x 2 - 2 x cos 195 + 1) O 2 - 2 x cos 285 -f 1) = 0. 2. x 10 - 2^00810 + 1=0. Ans. (x*-2x cos 2 + 1) (or 2 - 2x cos 74 + 1) X (x 2 - 2 x cos 146 + 1) (x 2 - 2 a; cos 218 -f 1) (ar' - 2 x cos 290 -f 1) = 0. DE MOIVRWS PROPERTY OF THE CIRCLE. 247 170. De Moivre's Property of the Circle. Let be the centre of a circle, P any point in its plane. Divide the circumference into D n equal parts EC, CD, DE, , begin- ning at any point B; and join and P with the points of division B, G, D, .... Let POB = 6; then will OB 2M -2 OB" OP" cos ?i0+OP 2n = PB 2 PC 2 PD 2 . to n terms. For, put OB = a, OP = a;, and = -; then PB 2 = OP 2 + OB 2 - 2 OP . OB cos = x- + a 2 2 ax cos - n (1) PC 2 = OP 2 + OO 2 - 2 OP OC cos = x 2 -f a 2 2 ax cos ; and so on n Multiplying (1), (2), (3)," together, we have PB 2 PC 2 PD 2 to n terms = [or 2 ax cos - + n ax cos - 2 ax cos + a 2 = a 2n - 2 a"^ cos + a 2 " . [by (4) of Art. 169] B 2n ... (3) which proves the proposition. 248 PLANE TRIGONOMETRY. 171. Cote's Properties of the Circle. These are particu- lar cases of De Moivre's property of c C the circle. (1) Let OP, produced if necessary, meet the circle at A, and let n then nO is a multiple of 2?r. Hence we have from (3) of Art. 170, after taking the square root of both members, OB" _ OF = PB . PC PD ... to n factors ... I. (2) Let the arcs AB, BC, ---be bisected in the points a, &,; then we have, by (1), OB 2M _ OP 2 ' 1 = Pa PB P6 . PC PC . to 2 n factors. Hence, by division, OB n + OP n = Pa P6 PC ... to n factors ... II. Cor. If the arcs AB, BC, ... be trisected in the points (*!, a 2 , 61, 6 2 , * * > then we have OB 2n + OB M .OP n +OP 2n =Pa 1 .Pa 2 .P6 1 .P6 2 ... to 2n factors. 172. Resolve sin & into Factors. (1) Put x = 1 ; then we get from (3) of Art. 169 (1) \ Put = 2n in (1), and let 2na = 1 cos = 1 cos 2n = 2 sin 2 w ; then extracting the square root, we have i = 2 n ~ 1 sin< sin (< + 2) sin (< + 4 a) x x sin (< + 2na 2 a) (2) RESOLUTION INTO FACTORS. 249 But sin ( + 2 na 2 a) = sin (< + TT 2 a) = sin (2 <) , sin (< -f- 2 n 4 ) = sin (4 <), 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 sin n = 2 n ~ l sin sin (2 a + <) sin (2 a ) sin (4 a -f ) sin (4 a ) x sin [(w 1) + <] sin [(?? 1) <^]. But sin (2 + ) sin (2 a ) = sin 3 2 a sin 2 <, and so on. .-. sinn<=2 n ~ 1 sin<(sin 2 2 sin 2 (/>) (sin 2 4 sin 2 <^>) x ... X [sin 2 (?i 1) sin 2 <^>] .... (3) Divide both members of (3) by sin <, and then diminish < indefinitely. Since the limit of sin n is n, we get w = 2 w - 1 sin 2 2asin 2 4asin 2 6x - xsin 2 (w 1) (4) Divide (3) by (4) ; thus x... (5) Put 7i^> = 0, and let n be increased while < is diminished without limit, remaining unchanged ; then since 2 na = TT, the limit of = -* (Art. sin - - sm - n \n n /i and the limit of n sin < = that of n sin - = 0; and so on. n Hence (5) becomes NOTE. The same result will be obtained if we suppose n even. 250 PLANE TRIGONOMETRY. Hem. When 0>0 and < TT, sin 9 is +, and every factor in the second member of (6) is positive; when 0>n- and <2n, sin 6 is , and only the second factor is negative; when 0>2 TT and <3n-, both members are positive, since only the second and third factors are negative ; and so on. Hence the + sign was taken in extract- ing the square root of (1). Cor. Let 0=-, then sin - = 1, and - = 1. Hence (6) be- comes 7T 2 2 2 2 2 4 2 2 1-3 3-5 5-7 7-9 which is Wall-is? s expression for TT. 173. Resolve cos<9 into Factors. In (2) of Art. 172, change < into -\- a, then n becomes n + nit, i.e., w< + - Hence (2) becomes But sin(< -f 2n a) = sin(< -f- TT a) = sin (a <), sin(^) + 2w 3)== sin(3 <^>), and so on. Hence when n is even we have from (1) cos n<=2"~ 1 sin( + ^>) sin( <^>) sin(3 + ^>) sin(3 <) = 2 n - ] (sin 2 a sin 2 <) (sin 2 3 sin 2 ^) X x[sin 2 (w-l)-sin 2 <] (2) Therefore, putting n = 0, as in Art. 172, we obtain -$\ (3) NOTB. For an alternative proof of the propositions of Arts. 172 and 173, see Lock's Higher Trigonometry, pp. 92-95. SUMMATION OF SERIES. 251 EXAMPLES. 1. If = , prove that 4 n sin a sin 5 a sin 9 a sin (4 n 3) a = 2~ n+ i 2. Show that = cos 5 9. SUMMATION OF TRIGONOMETRIC SERIES. 174. Sum the Series We have 2 sin a sin J/? = cos a ^ ) cos ( a -\- ^ ) (Art. 45) y r 2 / \ v 2 sin ( + /?) sin i)8 = cos (a + %)- cos ( + f 0), \ V 2 sin (a + 2^8) siniyg = cos ( + f ft) - cos ( + etc. = etc. 2 sin [ + (w I)/?] sin $fi = cos |- + 2^V] - eos[ + ?^l Therefore, if S n denote the sum of n terms, we have, by addition, 2S n sin %f3 = cos ( - 0) - cosf + 27l ~ 1 + 5 ^/glsiniwjg . . (Art. 45) 252 PLANE TRIGONOMETRY. 175. Sum the Series cos + cos ( + /?) + cos ( + 2/?)H ----- hcos[a+(-n I)/?]. We have 2 cos a sin J/3 = sin (a + ^-/J) sin (a %(3), 2 cos ( + /?) sin i0 = sin ( + f /?) - sin ( + /3), etc. = etc. 2 cos [ + (71 1)] sin /? Denoting the sum of n terms by S n , and adding, we get 2 S B sin = sin L + ^=11 -sin - cos fa + ^^^ pl sin | w . -_L___J 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 student is advised to commit them to memory. Cor. If we put /? = , then sin $nfi sin TT = 0. Hence we have from Arts. 174 and 175 n NOTE. These two results are very important, and the student should carefully notice them. 176. Sum the Series 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. 253 Thus, if m is even, we have from Art. 159 2 m ~ 1 sin m = ( l) 2 [cosm m cos (m 2)H ---- ] . (1) 2 m-l gin m ( a + ) m = (-If [cos m (a+p) -m cos (m-2) (+) + ...] (2) and so on ; and the required sum may be obtained from the known sum of the series [cos ma + cos m (a + (3) 4-cosra (-f2/3) H ---- ] and Jcos(ra 2) <* + cos[(m 2) (<* + /?)] + cos [(m _ 2) (a + 2 j8)] + ,,-j, etc. We may find the sum of the series cos m + cos m ( + p) + cos m ( + 2 /?) + etc. to n terms in a similar manner by the aid of Art. 158. EXAMPLES. 1. Sum to n terms the series sin 2 a + sin 2 ( + ) + sin 2 (a + 2 /3) + We have 2 sin 2 a = - (cos 2 a - 1) by (1), 2 sin 2 ( + P) = - [cos 2 (a + /?) - 1] by (2), 2 sin 2 (a + 2 ) = - [cos 2 (a + 2 /3) 1], and so on. Hence 2S n =tt-[cos2a+cos2(+/?)+cos2(+2/?) + ..-] = n - cos[2 + (n = n cos [2 + (n 1)^8] sin yi^ ~2 254 PLANE TRIGONOMETRY. 2. Sum to n terms the series cos 3 a + cos 3 2 a + cos 3 3 a + 2 cos [3 a + k ( n 1) 3 ] sin 8 sin a 6 cos a -j sin wet 8 sin i a 177. Sum the Series sin a sin ( + /?) + sin (-f- 2 /?) to n terms . (1) Change /? into (3 + TT, and (1) becomes sina + sin( + 7T + j8) + sin( + 27r-l-2^)H . . (2) Therefore we have from Art. 174 Sill inf.. + ("!.!)( Similarly, cos co ---- to n terms 8n 178. Sum the Series cosec 6 + cosec 2 -f cosec 4 -f- to n terms. f\ We have cosec = cot -- cot 0, 2 cosec 2 = cot cot 26, etc. = etc. cosec 2"- 1 = cot 2 n ~ 2 - cot 2 2n ~ 1 0. Therefore, by addition, as in Art. 174, SUMMATION OF SERIES. 255 NOTE. The artifice employed in this Art., of resolving each term into the dif- ference of two others, is extensively used in the summation of series. Practice alone will give the student readiness in effecting such transformations. If he cannot discover the mode of resolution in any example, he will often easily recognize it when he sees the result of summation. The student, however, is advised to resort to this method of solution only as a last resource. 179. Sum the Series n /a tan H- ^ tan r + \ tan- -f to n terms. 2 4 We have tan0 = cot0- 2 cot 20, |tan- = icot^ etc = etc. _ tan = cot ___ cot -i "- 1 n ~ l "- 1 n ~ 2 n ~ - i 180. Sum the Series Denote the sum by S M , and substitute for the sines their exponential values (Art. 161). Thus, x n-l j-gt( a + n/3-/3) _g-t(a + n/3-^) J _^_ x n+l |^ e t(a+n)3-j3) _ g-i(a+n/3-p) J 256 PLANE TRIGONOMETRY. sing a?sin( ft) a n sin(ft+n/3)+af +1 sin[tt+(?i I)/?] Cor. If a; < 1, and ?i be indefinitely increased, g _ sin a x sin (a ft) 1 2XCOSP + X 2 Sch. Similarly, cos a -f x cos ( + /3) + x 2 cos (a + 2 ft) -f- to w terms = cos a x eos( ct ft) x n cos( a + nft) + & n+1 cos [ + (^ / o \ We may obtain (3) from (1) by changing a to a -f- ^ 2 Also Soo = cosa-a;cos(tt-/3) (4) 1 2xcos^ + a; 2 181. Sum the Infinite Series a; sin ( + ) + ^sin ( + 2/3) + ^sin ( + 3j8) + ., I? Le and Let S denote the former series, and C the latter. Then C + S = xe 6 *(e** - 1) . . .[by (3) of Art. 129] _l-) . . . (Art 161) gX cos /3g i(a + iC sin /3) _ gta (cos a -f- * sin a) .... (Art. 161) EXAMPLES. 257 Equating real and imaginary parts, we have C = e xco *P cos (a -f x sin /?) cos a, S = e xcos P sin (a -f- x sin /3) sin a. EXAMPLES. Prove the following statements : 1. The two values of (cos 4 + V^T sin 4 0)* are (cos 20 + V^I sin 2 0) ... (Art. 154) 2. The three values of (cos + V 1 sin 0) i are 6 , / - T 27T + , / - ? 27T + COS-+V 1 sm r , cos r :!: hv Ism - ^ , 3 o o 3 47T + ^ . / - ? 47T + cos - :!: h V 1 sin - JE 3. The three values of (1)^ are l+f*. -1, f* ..... (Art. 154) 4. The six values of 1^ are contained in cos Vl sin , where r = 0, 1, or 2. 6 6 5. The three values of (1 -f V I)* are contained in ^ : lsinfl where = j, JTT, or-V^. 6 6. The three values of (3 + 4V !)* are contained in + V^l sin ^1 where r = 0, 1, or 2. 7. cos 6 = cos 6 0-15 cos 4 sin 2 + 15 cos 2 sin 4 - sin 6 0. 8. sin 90=9 cos 8 sin 0-84 cos 6 sin 3 0+126 cos 4 sin 5 -36cos 2 0sin 7 + sin 9 0. 258 PLANE TRIGONOMETRY 9. tan nO [2 10. Given - = - : show that is nearly the circular B 2166 measure of 3. Prove the following : 11. - 64 sin 7 = sin 7 1 sin 5 6 + 21 sin 3 35 sin 0. 12. -2 9 sin 10 0=cos 10 0-10cos 8 + 45 cos 6(9-120 cos 40 + 210 cos 2 - 126. 13. 2 6 (cos 8 + sin 8 0) = cos 8 + 28 cos 4 <9 -f 35. 14. cos 6 <9 + sm fi = -|(5 + 3 cos 40). 15. Expand (sin0) 4 ' l+2 in terms of cosines of multiples of 0. 16. Expand (sin 0) 4n+1 in terms of sines of multiples of 0. 17. Expand (cos 0) 2n in terms of cosines of multiples of 0. Use the exponential values of the sine and cosine to prove the following : 18. 1 - cos 2 19. If log (x 4- y V^T) = a + (3 V 1, prove that cc 2 -f i/ 2 = e 2a , and y = x tan /?. 20. If sin( + V l) = a; + y V 1, prove that x 2 cosec 2 a y 2 sec 2 = 1. 21. EXAMPLES. 259 7T 22. V^i-^e^ 2 . 23. e'(cos (9 + V^I sin 0) = eV cos . V^T s i n - \ 4 4/ 24. The coefficients of x n in the expansion, (1) of e* cos &#, and (2) of e ax sin &#, in powers of #, are 25. The coefficient of af* in the expansion of e* cos a; in n 02 ^ powers of x is cos - [n 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 a and b be the sides of a triangle, A and B the opposite angles, then will log 6 log a =cos 2 A cos 2B + ^(cos 4 A cos 4B) cos 6B) H ---- . 28. If A -f iB = log(m + m), then tan B = -, and 2 A = log (n 2 + m 2 ) . 29. cos (0 4- to) = cos 30. sin (0 + i\M = V* > / i * > J / \ J 31. 2 cos (a -f- '/?) = cos (e- 9 + e~^) i sin (e^ e-0). r a e ~^ r [cos (^8 log r + a?*) -f- i sin (/? log r -f ar) ], where a + ^ = ?*(cos r -\-i sin r) . 33. log (a + ib) = 4 log (a 2 -f 6 2 ) 4- i tan- 1 ^ . a 260 PLANE TRIGONOMETRY. 34. = sin'^afsinfa nO) + e iia sin n0]. 35. | = _i- + -i- + _L- + -'-. 36. Write down the quadratic factors of a; 13 1. Ans. (x 1) [cc 2 2 # cos Jg (2 r?r) -f 1], six factors, putting r = 1, 2, 3, 4, 5, 6. 37. Solve the equation x & 1 = 0. 38. Give the general quadratic factor of x 20 a 20 . -4?is. # 2 2 ax cos y 1 ^ (rw) + a 2 . 39. Find all the values of A/1. ^4ns. cos i(r7r)4- 1 sini(?-7r), r having each integral value from to 11. 40. Write down the quadratic factors of # 6 -j- 1. Ana. x 2 -V3o; 41. Write down the general quadratic factor of x -f 1. Ans. x 2 2 x cos (1 + 2 r) 9 + 1. 42. Find the factors of a 13 + 1 = 0. Ans. (x + 1) [# 2 2 x cos T ^ (TT + 2 PTT) + 1], seven fac- tors in all. 43. Find a general expression for all the values of -\/ 1. W -h 2 T'TT 7T -4- 2 ?*7T ^bis. cos |- i sin -I - , where r may have n n any integral value. 44. Solve x 12 - 2 0^0081^ + 1 = 0. Ans. x 2 2 x cos j- (3 r-n- + TT) + 1 = 0, six quadratics. EXAMPLES. 261 45. Solve z 10 + V3 x> + 1 = 0. Ans. x 2 + 2x cos (r x 72 + 6) + 1 = 0, five quadratics. 46. Write down the quadratic factors of yan _ 2 x n y n cos a + y~ n . Ans. y?-2xy cos " + y\ u factors. Prove the following : 47. tan^tan^ + ^tan-h tan + (1)2, where n is even. [Use (2) of Art. 172.] 48. sin 5 - cos 5 = 16 cos (0 - 27) cos (0 + 9) x X sin (6 + 27) sin (0 - 9) (cos - sin 0). 50. --.-. ^1 -4- _|_ L _L 4- . . . . ' 12 + 2 2 + 3 2 + 4 2 ~6 52 . 1 + 1+1 + 1 + ... = !?. I 2 3 2 5 2 7 2 8 53 =3 36 144 324 576 ' *" " ' ' 35 ' 143 ' 323 ' 575 ' 54 ir = 2 1.3.3.5.5.7.7.9- 55 /o^. " 3.35. 99.195-323 8. 80- 224. 440... 262 PLANE TRIGONOMETRY. 57. cos x + tan sin x 58. cos x cot sin x = 59. By aid of the formula cos = S1] and Art. 172, 2sin0 deduce the value for cos 6 obtained in Art. 173. 60. By expanding both sides of Ex. 57 in powers of x and equating the coefficients of x, prove that L TT y 7T+2/ oir y o7r-\-y OTT y 61. Prove in like manner from Ex. 58 that 29999 & i . . 6 2 y 2-rr-y 2>rr 63. Prove = 1-- + -+ 64. Prove that -J- = _ y 7r-y 2-rr-y tr+y 2-r+y STT y ir-y 3-r+y Sum the following series to n terms : 65. sinaH-sin2+sin3aH ----- \-s'mna = sn EXAMPLES. 263 66. cos a + cos 2 a + cos 3 a -\ ----- h cos na = nrr 67. si 68. sn sin 2 ?i sin sm " 2 sin a nn .o . 90 . 20 wsin sinncos(?i 69. sm 2 tt + sm 2 2+sm 2 3H = 2sm 70. 2 -sin a 71. sin 3 -f sin 3 ( + ft) + sin 3 ( + 2/3) + ... 72. 73. sin a sin 2 a -f- sin 2 a sin 3 + sin 3 a sin 4 4- _ n sin a cos sin na cos (n + 2) 2 sin a 74. tan-f-2tan2tt + 2 2 tan2 2 H ---- = cot 2 n cot 2 n . 75. (tan + cot a) + (tan 2 a + cot 2 ) + (tan 2 2 -f cot 2 2 ) H ---- = 2 cot a 2 cot 2 W a. 76. sec a sec 2 a 4- sec 2 a sec 3 a + = cosec a [tan (n + 1) a tan ] . 264 PLANE TRIGONOMETRY. 77. cosec a cosec 2 a + cosec 2 cosec 3 a -\ ---- = cosec a [cot a cot (?i + 1) ]. 78 sin20 sin40 ... ,.._. sec (2 71 + 1)0- sec cos cos 3 6 cos 3 cos 5 2 sin 79. cos 4 a + cos 4 ( + /?) + cos 4 (<* + 2/3) + ... = f n + cos[2 + (n l)/3]sinyiff cos [4 + (K 1)2/8] sin 2n 2 sin 8 sin 2 /? 80 t n nO _ s i n ^ + s i n 3 + sin 5 H ---- to n terms cos + cos 30 + cos 5 -f- to n terms 81. cos $ cos(0 + a) + cos(0 + )cos(0 -f 2 a) + cos(0 + 2 a)cos((9 + 3 a) + = ^ cos a + 2 2 sin sin sin 20+sin 30 ---- to 91 terms cos cos20+cos30 ---- to n terms =tan 83. sm(p + 1)0 cos + sin(p + 2)0 cos 2 + ... sin ( p + 1 + ?i ) sin ?i0 2sin0 84. sin 3 sin + sin 6 sin 2 + sin 12 sin 4 + ... = 1 (cos 20 -cos 2^0), 85. sin (sin- ) +2 sin- (sin ) +4 sin (sin = 2 n ~ 2 sin -~ - J sin 2 0. A A A A A A 86. tan sec0 + tan- sec - + tan-sec- + ."=tan tan. 2 4284 2 n 87. cot cosec 0+2 cot 2 cosec 2 0+2 2 cot 2 2 cosec 2 2 0+ EXAMPLES. 265 oo -L i -* I -*- j sin 20 siu~20 sin 3 sin 3 sin 4 sin = -- -(cot 30 -cot 40). sm0 v 89. sin cos 2 cos 2 sin 30 sin 3 cos 4 = cose \ 2, L . 1 (0 + |Y tan (n + 1) f$ + |V tai/0 + |YI 90. tan- 1 91. 1 _l_ ^ _l_ ^ 2 l -t-o 4- o -Et n- 1 i- ~4 aU n + = tan" 1 nx. 92. sin a sin 3 a, + sin - sin - + sin ^ sin " 93. -H -- - -- 1 -- _ + cos + cos 30 cos + cos 5 cos + cos 7 = cosec [tan ( w + 1 ) tan 0] . >sec20sec2 2 0+... = sin0(cot0 cot2"0). 94. -sec0 + isec0sec20-hisec0sec20sec2 2 0+ 22 2 95. = log 2 sin 2 - log 2 sin 2 n+1 0. Sum the following series to infinity : 96. [2 [3 = e co * 2e cos (0 + sin cos 0). 266 PLANE TRIGONOMETRY. If 98. l- C 99. 2 cos + 1 cos 2 4- 1 cos 3 + cos 4 l9 -f cos 6 = - -Iog(l-cos0). 1 cos 100. s [2 [S "T" ' [2 ' L- i i i /Q n 102. si |2 == 6 sin ^p 4~ sin (/) 103. cos0-|cos204-icos30 = log (% cos -Y 104. cos 20 + ^ cos 60 + I cos 100 4 ... = ilog cot-. 105. 106. x cos - - cos 2 4- ~ cos 3 - - cos 4 + . . . 234 = log(l 4- 2x cos0 4- x'J. 107. sin sin ^ _ sin 2 ^5_? 4. s in 3 -^ = cot"^! 4- cot 2 4- cot0). 108. - + i + 1 + 1+...=^!. I 4 2 4 3 4 4 4 90 109. 1 a. i _|- 1 4. 1 _i =?*-. I 4 3 4 5 4 7 4 96 PART II. SPHERICAL TRIGONOMETRY. CHAPTER X. FORMULA EELATIVE TO SPHEKIOAL TRIANGLES, 182. Spherical Trigonometry has for its object the solu- tion of spherical triangles. 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, BOG, COA its face-angles. We may construct a sphere with its centre at 0, 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, BOG, 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 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 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 - b', 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 between the sides and angles of a spherical triangle, it holds of course for the polar triangle also. Hence, by means of the above formulas, any theorem of a spherical triangle may be at once transformed into another theorem by substituting 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- rectangular triangle ; 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. 270 SPHERICAL TRIG ON OMETR Y. XOTE. It is shown in geometry that a spherical triangle may, 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 be com- puted from the above data. EXAMPLES. 1. In the spherical triangle whose angles are A, B, C, prove B + C-A<7r (1) C + A - B < TT (2) A + B-C<7r (3) 2. If C is a right angle, prove A + B < ITT (1), and A - B < (2). 3. The angles of a triangle are A, 45, and 120 ; find the maximum value of A. Ans. A < 105. 4. The angles of a triangle are A, 30, and 150 ; find the maximum value of A. Ans. A < 60. 5. The angles of a triangle are A, 20, and 110; find the maximum value of A. Ans. A < 90. 6. Any side of a triangle is greater than the difference between the other two. RIGHT SPHERICAL TRIANGLES. 185. Formulae for Right Triangles. Let ABC be a spherical triangle in which C is a right angle, and let be the centre of the sphere ; then will OA, OB, OC be radii: let a, 6, c denote the sides of the triangle O- opposite the angles A, B, C, re- spectively ; then a, &, and c are the measures of the angles BOG, COA, and AOB. RIGHT SPHERICAL TRIANGLES. 271 . From any point D in OA draw DE _L to )&, and from E draw EF _L to erf/Snd join DF. Then BE is JL to EF (Geom. Art. 537). Hence (Geom. Art. 507), DF DF is JL to O^rf .-. Z DFE = Z ( Art - 183 ) OTP OTT OT? v ^ Now - ; that is, cos fc = cos a cos to S . (1) OD OE OD ^ = il'^ ; thatis ' sin6 ^ sillBsinc - (2) Interchanging a's and &'s, sin a = sin A sin c . . (3) | = || ^f; that is, tan o= cos B tan c. . (4) Interchanging a's and 6's, tan b = cos A tan c . . (5) 5E = 55.5Z ; that is, tan 6 = tan B sin a. . (6) Interchanging a's and 6's, tan a = tan A sin ?> . . (7) Multiply (6) and (7) together, and we get tan A tan B = - - = -J , by (1) cos a cos b cos c .-. cos c = cot A cot B ($) Multiply crosswise (3) and (4), and we get sin a cos B tan c = tan a sin A sin c. -r, sin A cos c A * 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, A'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, AC, 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 also 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 formulae 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 b sin C (2) EXAMPLES. 275 sin A = sin a sin C (3) cos b = tan A cot C (4) cos a = tan B cot C (5) sin A = tan B cot b (6) sin B = tan A cot a (7) cos C = cot a cot b (8) c6s 6 = cos B sin a (9) cos a = cos A sin b (10) EXAMPLES. In the right triangle ABC in which the angle C is the right angle, prove the following relations : 1. sin 2 a -f sin 2 b sin 2 c = sin 2 a sin 2 b. 2. cos 2 A sin 2 c = sin 2 c sin 2 a. 3. sin 2 A cos 2 c = sin 2 A sin 2 a. 4. sin 2 A cos 2 b sin 2 c = sin 2 c sin 2 &. 5. 2 cos c = cos (a + &) + cos (a b). 6. tan i (c + a) tan 1 (c a) = tan 2 1 6. 7. sin 2 -= sin 2 -cos 2 -4- cos 2 -sin 2 -- 2 2222 8. sin (c- 6) = tan 2 sin(c + 6). 2i 9. If b = c = ^, prove cos a = cos A. 10. If a = b = c } prove sec A = 1 -f sec a. 11. If c < 90, show that a and 6 are of the same species. 12. If c > 90, a and b are of different species. 13. A side and the hypotenuse are of the same or oppo- site species, according as the included angle < ; or >- 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, 6, c denote the sides of the triangle opposite the angles A, B, C, respectively. Then a, 6, and c are the measures of the angles BOC, CO A, and AOB. From any point D in OA draw DG _L to the plane BOC, and from G draw GE, GF J_ to OB, OC. Join DE, DF, and GO. Then DG is J_ to GE, GF, and GO (Geom. Art. 487). Hence, DE is _L to OB, and DF _L to OC (Geom. Art. 507). .-. Z DEG =F 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 b _ sin c sin A sin B sin C NOTE. The common value of these three ratios is called the modulus of the spherical triangle. JSch. In the figure, B, C, 6, c are each less than a right angle ; but it will be found 011 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 will 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, &, c the sides of the triangle opposite the angles D^-^1\ A, B, C, respectively. Then ^^n IV a = ZBOC, 0<^ /% / ) c b = Z COA, ^"^l^X c = Z AOB. B From any point D in OA draw, in the planes AOB, AOC, respectively, the lines DE, DF 1_ to A. Then Z EDF = Z A ....... (Art. 183) Join EF; then in the plane triangles EOF, EDF, we have OE 2 + OF 2 -20E-OFcosEOF . . (1) DE 2 + DF 2 -2DE.DFcosEDF . . (2) also in the right triangles EOD, FOD, we have OE 2 - DE 2 = OD 2 , and OF 2 -DF 2 = OD 2 . (3) Subtracting (2) from (1), and reducing by (3), and transposing, we get 2 OE . OF cos EOF = 2 OD 2 + 2 DE DF cos EDF. or cos a = cos b cose + sin b sine cos A (4) 278 SPHERICAL TRIGONOMETRY. By treating the other edges in order in the same way, or by advancing letters (see Note, Art. 9G) we get cos b = cos c cos a 4 sin c sin a cos B . . (5) cos c = cos a cos b 4 sin a sin & cos C . . (6) Sch. Formula (4) has been proved only for the case in which the sides b 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 than 90. Produce B A, BC to meet m B', and put AB'=c', CB'=a'. Then, from the triangle AB'C, we have by (4) cos a' = cos b cos c f 4 sin b sin c f cos B'AC, or COS(TT a) = cos b cos(?r c) 4 sin b sin(?r C)COS(TT A). . . cos a = cos b cos c -f- sin b sin c cos A. (2) Suppose both b 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 b' cos c' 4- sin b' sin c' cos A r ; but b 1 = TT - 6, c' = TT - c, A' = A. .-. cos a = cos b cos c 4 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 4 sin B sin C cos a. 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, b' = TT - B, c' = TT - C, etc. . (Art. 184) Hence, substituting, we get cos A = cos B cos C + sin B sin C cos a . . . . (1) Similarly, cos B = cos C cos A -j- sin C sin A cos & .... (2) cos C = cos A cos B -f- sin A sin B cos c . . . . (3) 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 sides, and sides of angles. 193. To show that in a spherical triangle ABC, cot a sin b = 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 2 b + sin a sin b cos b cos C -f- sin b sin c cos A. Transpose cos a cos 2 b, and divide by sin a sin b ; thus, cot a sin b = cosb cos C + Sm C S A 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 formulae are as follows : cot a sin b = cot A sin C + cos C cos b . . . . (1) cot a sin c = cot A sin B + cos B cos c . . . . (2) cot b sin a = cot B sin C -f- cos C cos a . . . . (3) cot b sin c = cot B sin A -f- cos A cos c .... (4) cot c sin a = cot C sin B + cosB cos a .... (5) cot c sin b = cot C sin A + cos A cos b . . . . (6) 280 SPHERICAL TRIGONOMETRY. EXAMPLES. 1. If a, b, c be the sides of a spherical triangle, a', b', c' the sides of its polar triangle, prove sin a : sin b : sin c = sin a' : sin 6' : sin c'. 2. If the bisector AD of the angle A of a spherical triangle divide the side BC into the segments CD = b', BD = c', prove sin b : 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 BAG. cot ABC sin DC + cot ACB sin BD = cot ADB sin BC. 4. If a, /?, y be the perpendiculars of a triangle, prove that sin a sin a = sin b sin (3 sin c sin y. 5. In Ex. 4 prove that sin a cos a = Vcos 2 b + cos 2 c 2 cos a cos b cos c. 194. Useful Formulae. 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 cose cos A . . (1) sin a cos C = sin b cos c cos b sin c cos A . . (2) sin b cos A = cos a sin c sin a cose cos B . . (3) sin b cos C = sin a cos c cos a sin c cos B . . (4) sin c cos A = cos a sin b sin a cos b cos C . . (5) sin c cos B = sin a cos b cos a sin b cos C . . (6) FORMULA FOR THE HALF ANGLES. 281 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 . . (7) sin A cos c = sin B cos C + cos B sin C cos a . . (8) sin B cos a = cos A sin C + sin A cos C cos b . . (9) sin B cose = sin A cos C + cos A sin C cos b . . (10) sin C cos a = cos A sin B -f- sin A cos B cos c . . (11) sin C cos b = sin A cos B -f- cos A sin B cos c . . (12) 195. Formulae 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 a- cos fr cose = _ 2 sin , A (Art 49) sin b sin c 2 o 2-A.__-i cos a cos 6 cos c 2 sin b sin c _ cos (b c) cos a sin b sin c / Art 45) Let 2s = a + 6 + 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 b) sin (s c) .'. sin = -- * - ! ; - sin b sin c 2 sin b sin c A /sin (s b} sin (s c) /^ ^ .-. 8in-=-J ' . . (l) 2 \ sin b sine 282 SPHERICAL TRIGONOMETRY. Advancing letters, *>^<-<- sn c sn a sin C /!i!L(l^)si"( 2 \ sin a sin 6 II. 2cos 2 = l + cosA ..... (Art. 49) 2 -, . cos a cos b cos c s= i-j -- ; - : - sin o sin c _ cos a cos (b + c) sin 6 sin c 2 A_ sin4-(a 4- b + c) sin|(6 4- c a) .*. COS --- - - ; - : - 2 sin b sin c _ sins sin(s a) sin b sine sin 6 sin c Advancing letters, C = Isins 2 \ si sin a sin b III. By division, we obtain 2 \ sin s sin (s a) -& . . . (8) 2 \ sin s sin (N tan = / sin ( g ~ q ;) i"(-'* 2 if sins sin (s c) FORMULA FOR THE HALF ANGLES. 283 Sch. The positive sign must be given to the radicals in each case in this article, because JA, J-B, ^C are each less than 90. Cor. 1. tan tan s in (s c) nch tcUl Ld,lJ sins in(s a) . ^-LV; ni\ tan 2 tan - j-qi-j till sins C12^ belli belli sins {-*-*) Cor. 2. Since sin A = 2 sin -cos A , in A _ 2Vsins sin(s a) sin(s b) sin(.s- c) ,|ox sin 6 sine = ^~ (14) sin o sin c where n 2 = sins sin(s a) sin(s b) sin(s c). 1. Prove sin 2 A = EXAMPLES. 1 cos 2 a cos 2 6 cos 2 e -f 2 cos a cos b cos c sin" b siir'c sin 2 6 sin 2 c' where 4 ir = 1 cos 2 a cos 2 6 cos 2 c -f 2 cos a cos b cos c. 2. Prove cose = cos (a + b) sin 2 f- cos (a b) cos 2 - . A . B . C sin(s a) sin (s-^fr)sin(s e) 3. Prove sm sin sin = . 222 sin a sin b sine 4. p r ove cos A + cos B . = sin (a + b) 1 cosC sine 5. Prove S C sA + COsB sin (a -6)sinc = 0. 1 cos C 6. Prove CQS A - cos B = sin (a~b\ 1 + cosC sine 284 SPHERICAL TRIGONOMETRY. 196. Formulae 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 cos a = + cos B cos C = _ 2 gin2 a A sin B sin C 2 2 sin 2a = cos A + cos ( B + C ) . 2 sin B sin C (Art 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 : . a_ / cos S cos (8 A) ,^ m 2~ \ sin B sin C cos S cos (S B) ,\ 2 \ sin C sin A fcW sm ^,_cosScos(S-C) sin A sin 5 cos- a /cos (S - B) cos (S - C) m 2 \ sinBsinC -v cos 6 _ . /cos(S-C)cos(S-A) 2 \ sin C sin A og c_ cos (S - A) cos (S - B) 2 Af sin A sin B t L /3 cosS cos (S A) m 2 " \ ~ cos (S - B) cos (S - C) FORMULAE FOE THE HALF SIDES. 285 t 5= /_ cos S cos (S - B) 2 \ cosS-CcosS cos(S-C)cos(S-A) tan-= / cos S cos (S - C) 2 \ cosfS-AUosrS- 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 1 to the above radicals, because ^, -, ^, are each less than 90. a t 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). .'. 7T A<7T B + 7T 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 -, _2V cosScos(S A)cos(S B)cos(S C) . *. Sill Ct ~ ~ sin B sin C 2N ;) sinB sinC where N= V- cos S cos (S - A) cos (S - B) cos (S - C). 286 SPHEEICAL TRIGONOMETRY. EXAMPLES. 1. Prove cosC=-cos(A + B)cos 2 --cos(A-B)sin 2 -. -r, . a . b c N cos S 2. Prove sin - sin - sin - = , 222 sin A sin B sin C where N = V cos S cos (S A) cos (S B) cos (S C). 197. Napier's Analogies. T i sin A sin B , A , -4<\r\\ /- \ Let m = = - (Art. 190) (1) sin a sin 6 sin A + sin B , . , , = -r- -(Algebra) (2) sin a 4- sin o or = sin A- sin B (3) sin a ~ sin 6 cos A 4- 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 b . . (5) . . (cos A + cos B) (1 4- cos C) = m sin C sin (a 4- b) , (6) from (4) and (5) Dividing (2) by (6), sin A 4- sin B _ sin a 4- sin b 1 4- cos C cos A 4- cos B sin (a. 4- b) sin C ... tani(A + B) = cos ti tt -;icc4 (7) cos 4 (Oi 4- b) 2, (Arts. 45, 46, and 49) Similarly, tan^ (A - B) = s | n ^ a ~ ^ cotg . . (8) sin \ (a 4- o) * DEL AMBERS ANALOGIES. 287 Writing TT A for a, etc., by Art. 184, we obtain from (7) and (8) Sch. The formulae (7), (8), (9), (10) are known as Napier 1 s Analogies, after their discoverer. The last two may be proved without the polar triangle by starting with the formulae of Art. 191. Cor. In any spherical triangle whose 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 b) and cot - are necessarily positive, therefore by (7) tan (A -f- B) and cos J (a + b) are both positive or both negative. .-. ( A + B) and (a -f b) are both > or both < or both = 90. 198. Delambre's (or Gauss's ) Analogies. = sin cos j- cos sin a ~& a ' Vsin (s b) sin (s c) /sin s sin (s 6) sin b sin c Ai sin c sin a _i_ /sins sin (.s a) /sin (s c) sin (s ^) A/ sin b sin c Al sin c sin a _ sin (s 6) -j- sin (s a) /sin .9 sin (s c) sin c \ sin a sin b a b) C . K , ., nKN cos ..... (Arts. 45 and 195) cos- 288 SPHERICAL TRIGONOMETRY. .-. sin (A + B) cos | = cos | (a- 6) cos- . . . (1) +j Similarly, we obtain the following three equations : sini(A-B)sin- = sini(a-&)cos- ... (2) 2 2 cos i (A + B) cos = cos \ (a + 6) sin - . . . (3) 2 2 cosf (A B) sin- = sin j- (a + 6) sin. - ... . (4) ^ Sch. 1. When the sides and angles are all less than 180, both members of these equations are positive. Sch. 2. Napier's analogies may be obtained from De- lambre's by division. NOTE. Delambre's analogies were discovered by him in 1807, and published in the Connaissance des Temps for 1809, p. 443. They were subsequently discovered independently by Gauss, and published by him, and are sometimes improperly called Gauss's equations. 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 London Mathe- matical Society, Vol. III. [Casey's 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 2 acos 2 & = sin (c -f- 6) sin (c &). 2. tan 2 a : tan 2 b = sin 2 c sin 2 b : sin 2 c sin 2 tt. 3. cos 2 acos 2 B = sin 2 A sin 2 a. 4. cos 2 A + cos 2 c = cos 2 A cos 2 c + cos 2 a. 5. sin 2 A cos 2 B = sin 2 a sin 2 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 7. If the angle A of a right triangle be acute, show that the difference of the sides which contain it is less than 90. B sin (s a) 8. Prove tan - = '-> 2 sins 9. Prove (1) 2 w = sin a sin 6; (2) 2N = sinasinB. 10. Prove sin 2 a sin 2 b = sin 2 a -f sin 2 b sin 2 c. 11. Prove tan2 2 sin (c + 6) 12. Prove 2sin 2 ^= sin 2 (a + b) + sin 2 (a- b). 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 2 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'S = ^ + sin26 4cos 2 - 17. Prove tan S = cot - cot -. 2 2 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 SPHERICAL TRIGONOMETRY. 20. Given the sum of the sides a and b, and the sum of the base angles : solve the triangle. 21. Show that sin^=^ sin c + sin + Vsin c -sii[^ 2 2Vsinc 2 cos ft sin c 23. cosA 2 cos b sin c 24. sin (a -f &) tan| (A + B) - sin (a - b) cot (A - B). 26. sin(A B) = 1 -f- cos a cos 6 cos b cos a 1 cos a cos 6 07 / A , T \ sin a sin b 27. cos(A + T>) = 1 + cos a cos b 28. cos(A-B) = 1 cos a cos 6 or o C -o(X <> b , oCL o b 29. sin^ - = sm j - cos" 1 h cos- - snr -. 2 2222 30. sin (c - b) = sin (c + b) tan 2 -. a 31. sin (a 6) = sin a tan sin 6 tan 2 2 T> 32. sin (c a) = cos a sin b tan 2 33. If ABC is a spherical triangle, right-angled at C, and cos A = cos 2 a, show that if A be not a right angle, b + c = $7r or |TT, according as b and c are both < or both > - EXAMPLES. 291 34. If , (3 be the arcs drawn from the right angle respectively perpendicular to and bisecting the hypotenuse c, show that sin 2 - (1 -f sin 2 ) = sin 2 /?. 2t 35. In a triangle, if C be a right angle and D the middle point of AB, show that 4 cos 2 - sin 2 CD = sin 2 a + sin 2 b. 2i 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 2 p = cot 2 a + cot 2 6. 37. cos^p = cos 2 A + cos 2 B. 38. tan 2 a = =F tan a' tan c. 39. tan 2 b = tan 6' tan c. 40. tan 2 a : tan 2 b = tan a' : tan b'. 41. sin 2 /) = sin a' sin b'. 42. sinp sin c = sin a sin b. 43. tan a tan b = tan c sin p. 44. tan 2 a + tan 2 b = tan 2 c cos 2 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. 292 SPHERICAL TRIGONOMETRY. 49. If in a triangle the side a = 90, show that cos A + cos B cos C = 0. 50. If and 0' are the angles which the internal and ex- ternal bisectors of the vertical angle of a triangle make with the base, show that - cos A ~ cos B cos A 4- cos B cos = -^ , and cos 0' = 2 cos- 2sin- 2 2 51. Given the base c and cos = cos C : find the locus of the vertex. 52. Prove 4N 2 = 1 - cos 2 A - cos 2 B - cos 2 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 ship = sin b sin q = sin c sin r = 2n. (2) sinAsinp = sin B sin q = sinCsinr= 2N. 54. Prove 8?i 3 = sin 2 a sin 2 6 sin 2 c sinA sinB sinC. 55 Prove s * n2 ^ + sin 2 B -f sin 2 C _ 1 + cos A cos B cos G sin 2 a -f- sin 2 b -\- sin 2 c 1 cos a cos b 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. cos a + cos b Ans. cos I = 2cos- 2 57. If CD, CD' are the internal and external bisectors of the angle C of a triangle, prove that A. /-.T^ cot a 4- cot b , cot a ~ cot b cot CD = ^- , and cot CD' = 2 cos- EXAMPLES. 293 58. Show that the angles and 0', made by the bisectors of the angle C in Ex. 55 with the opposite side c, are thus given : 2sin cot e , = sin CD'. 2co S | 59. Show that the arc intercepted on the base by the bisectors in Ex. 55 is thus given : sin 2 A sin 2 B cot DD' = 2 sin A sinB sinC 60. Prove that cos 2 b cos 2 c cos 2 c cos 2 a cos b cot B cos c cot C cos c cot C cos a cot A cos 2 a cos 2 b cos a cot A cos b 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 by the bisector of the vertical angle, show that s s' , in >n' a b tan ^- tan - = tan'-^ 62. Prove sin b sin c 4- cos b 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 b of a triangle is thus given : 1 4- cos a 4- cos b 4- cos c cos I = 1-! * a b 4 cos - cos - 9 9 294 SPHERICAL TRIGONOMETRY. 64. If the side c of a triangle be 90, and 8 the arc drawn at right angles to it from the opposite vertex, show that cot 2 8 = cot 2 A + cot 2 B. 65. Prove that the angle < between the perpendicular from the vertex on the base and the bisector of the vertical angle is thus given : 66. In an isosceles triangle, if each of the base angles be double the vertical angle, prove that cos a cos ^ = cos f c + - ) a 2 A 67. If a side c of a triangle be 90, show that (1) cot a cot b + 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) _ Q 1 cos c sin C 69. t = tan % (a + b) : tan |(a b). 70. tan J ( A + a) : tan ( A - a) = tan J (B + b) : tan (B - b). 71. If the bisector of the exterior angle, formed by pro- ducing BA through A, meet the base BC in D f , and if BD = c", CD' = 6", prove sin b : sin c = sin I" : sin c". 72. If D be any point in the side BC of a triangle, prove sinBD _ sin BAD sin C sin CD ~~ sin CAD sinB EXAMPLES. 295 73. If A = a, show that B and b are either equal or supplemental, as also C and c. 74. If A = B -f 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 + cos b = 2 cos - cos CD. 77. In an equilateral triangle show that (1) 2cos-sin- = l. (2) tan 2 ~ -{- 2 cos A = 1. l 78. If b + c = TT, show that sin 2 B + sin 2 C = 0. 79. Show that sin b sin c -f- cos b cos c cos A = sin B sin C cos B cos C cos a. 80. If D be any point in the side BC of a triangle, show that cos AD sin a = cos c sin DC + cos b sin BD. 81. Prove cos 2C = cos 2 ^(a+6)sin 2< p + cos 2 |(a-6)cos 2 ^. 82. " 83. " sin s sin (s a) sin (s b) sin (s c) = J (1 cos 2 a cos 2 b cos 2 c -f 2 cos a cos b cos c). 84. If AD be the bisector of the angle A, prove that (1) cos B -f- cos C = 2 sin sin ADB cos AD. (2) cos C - cos B = 2 cos ^ cos ADB. 296 SPHERICAL TRIGONOMETRY. 85. Prove cos a sin b = sin a cos b cos C + cos A sin c. 86. " sin C cos a = cos A sin B -f- sin A cos B cos C. 87. In a triangle if A = -, B = -, C = -, show that 5 3 /& A\ 1 H- cos a cos 6 cose 88. Prove sin (S A) = 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 /8, prove that tan - tan " = tan a "*~ tan ^^ 90. If 6 = c, show that sin - cos 2 2 sin 6 = j-, and sin B = 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. PRELIMINARY OBSERVATIONS. 297 CHAPTER XI. SOLUTION OF SPHERICAL TRIANGLES, 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. 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 Rules (Art. 298 SPHERICAL TRIGONOMETRY. 180). 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, 6, c denote the sides opposite the angles A, B, C, respectively. We shall assume that the parts are all positive and less than 180 (Art. 182). 201. Case I. Given the hypotenuse c and an angle A ; to find a, b, B. B y (3)> (5), and (8) of Art. 185, or by Napier's Kules, we have sin a = sin c sin A, tan b = tan c cos A, cotB = 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, (!)] B and b are determined imme- diately without ambiguity. If a be very near 90, we 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 against numer- ical errors, it is often expedient to check the logarithmic work, which may be done in every case without the neces- sity of new logarithms. To check the work, we make up a formula between the three required parts, and see whether CASE ii. 299 it is satisfied by the results. In the present case, when the three parts a, b, B have been found, the check formula is sin a = tan b cot B . . . . [(6) of Art. 185] Ex. 1. Given c = 81 29' 32", A = 32 18' 17" ; find a, 6, B. Solution. log sine =9.9951945 log sin A = 9.7278843 log sin a = 9". 7230788 .-. a =31 54' 25". log cose =9.1700960 log tan A =9.8009157 log cot B = 8.9710117 .-. B = 84 39' 21". 33. log tan c = 10.8250982 log cos A = 9.9269687 log tan b =10.7520669 .-. 6 = 79 51' 48".65. Check. log tan b =10.7520669 log cot B = 8.9710117 log sin a = 9.7230786' Ex. 2. Given c = 110 46' 20", A = 80 10' 30" ; find a, 6, B. Ans. a = 67 5' 52". 7, b = 155 46' 42". 7, B = 153 58' 24".5. 202. Case II. Given the hypotenuse c and a side a ; to find 6, A, B. By (1), (3), (4) of Art. 185, or by Napier's Rules, we have , cos c A sin a -p, tan a cos b = , sm A = , cos B = cos a sin c tan c The check formula involves 6, A, B ; therefore, from (9) of Art. 185 we have cos B = sin A cos b. In this case there is an apparent ambiguity in the value of A, but this is removed by considering that A and a are always of the same species (Art. 187). 300 SPHERICAL TRIGONOMETRY. Ex. 1. Given c = 140, a = 20 ; find 6, A, B. Solution. log cose = 9.8842540- colog cos a = 0.0270142 log cos b = 9.9112682- .-. b = 144 36' 28".4. log tan a =9.5610659 colog tan c = 0.0761865- logcosB = 9.6372524- .-. B = 115 42' 23".8. log sin a = 9.5340517 colog sine = 0.1919325 log sin A =9.7259842 .-. A =328'48".l. Check. log sin A =9.7259842 log cos b = 9.9112682 log cos B =9.6372524 Ex. 2. Given c = 72 30', a = 45 15' ; find b, A, B. Ans. b = 64 42' 52", A = 48 7' 44".5, B = 71 27' 15". 203. Case III. Given a side a and the adjacent angle B ; to find A, 6, c. By (10), (6), (4) of Art. 185, we have cos A = cos a sin B, tan b = sin a tan B, cot c = cot a cos B. Check formula, cos A = tan b cot c. In this case there is evidently no ambiguity. Ex. 1. Given a =31 20' 45", B = 5530'30"; find A, 6, c. Solution. log cos a =9.9314797 log sin B =9.9160371 log cos A = 9.8475168 .-. A = 45 15' 30". 6. log cot a =0.2153073 log cos B = 9.7530361 log cote =9.9683434 .-. c =47 5' 11". log sin a =9.7161724 log tan B =0.1630010 log tan 6 =9.8791734 .-. b =37 7' 50". Check. log tan b = 9.8791734 log cote =9.9683434 log cos A = 9.8475168 CASE IV. 301 Ex. 2. Given 0=112 0'0",B = 152 23' 1".3; find A, 6, c. Ans. A = 100, b = 154 7' 26".5, c = 70 18' 10".2. 204. Case IV. Given a side a and the opposite angle A ; to find b, c, B. B J ( 7 )> ( 3 )> ( 10 ) of Art. 185, we have v sin 6 = tan a cot A, sinc = ^A sin B = sin A cos a (7/iecfc formula) sin 6 = sin c sin B. In this case there is an ambiguity, as the parts are deter- mined by their sines, and two values for each are in general admissible. But for each value of b there will, in general, be only one value for c, since c and b are connected by the relation cos c = cos a cos b (Art. 185); and at the same time only one admissible value for B, since cos c = cot A cot B. Hence there will be, in general, only two triangles having the given parts, except when the side a is a quadrant and the angle A is also 90, in which case the solution becomes indeterminate. It is also easily seen from a figure that the ambiguity must occur in general (Art. 188). When a A, the formulae, and also the figure, show that b, c, and B are each 90. Ex. 1. Given a = 46 45', A = 59 12' ; find b, c, B. Solution. log tan a = 0.0265461 log cot A =9.7753334 log sin b =9.8018795 .-. b =3919'23".5, or 14040'36".5. log sin a =9.8623526 log sin A =9.9339729 log sine =9.9283797 .-. c =57 59' 29", or 122 0'31". 302 SPHERICAL TRIGONOMETRY. log cos A =9.7093063 log cos a =9.8358066 log sin B =9.8734997 .-. B =48 21' 28", Check. log sine =9.9283797 log sin B = 9.S734997 log sin b =9.8018794 or 131 38' 32". Ex. 2. Given a= 1.12, A = 100: find &, c, B. ^t/is. & = 154 7'26".5, c = 7018'10".2, B = 15223' 1".3, or 2552'33".5, or 10941 r 49".8, or 2736'58".7. 205. Case V. Given the two sides a and b ; to find A, B,c. B y ( 7 )> ( 6 )> (!) of Art 185 > we have cot A = cot a sin b, cot B = cot b sin a, cos c = cos a cos b. Check formula, cos c = cot A cot B. In this case there is no ambiguity. Ex. 1. Given a = 54 16', & = 3312'; find A, B, c. Ans. A = 68 29' 53", B = 38 52' 26", c = 6044'46". .Ex. 2. Given a = 56 34', b = 27 18'; find A, B, c. . A = 73 9' 13", B = 31 44' 9", c = 60 41' 9". 206. Case VI. Given the two angles A and B ; to find a, b, and c. By (10), (9), (8) cos a = ^A cos b = ^-5, cos c = cot A cot B. sin B sin A Check formula, cos c = cos a cos b. There is no ambiguity in this case. Ex. 1. Given A = 74 15', B = 32 10' ; find a, 6, c. Ans. a = 59 20' 44", b = 28 24' 54", c = 63 21' 24".5. Ex. 2. Given A = 91 11', B = 111 11' ; find a, b, c. Ans. a = 91 16' 8", b = 111 11' 16", c = 89 32' 29". EXAMPLES. 303 207. Quadrantal 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 formulae 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 + b = TT, or A + B = TT, can be made to depend on the solution of right triangles. Thus (see the second figure of Art. 191) the triangle B'AC has the two equal sides, a' and b, given, or the two equal angles, A and B', given, according as a -f- b = TT or A + B = IT in the triangle ABC. EXAMPLES. Solve the following right triangles : 1. 4. Given c=3234 f , find a=2215'43", Given c= 69 25' 11", find a = 50 0' 0", Given c=55 9' 32", find 6=5153', Given c=12712', find 6=39 6' 25", Given a = 118 54', find A =95 65' 2", a = 4444'; 6 = 24 24' 19", A = 54 54' 42"; b = 56 50' 49", a=2215' 7"; A=2728'37'.5, A=1285'54", = 10 49' 17", B = 508'21". = 6325'4". = 5221'49". c= 118 20' 20". * Quadrantal triangles are generally avoided in practice, but when unavoidable, they are readily solved by either of these methods. 304 6. 7. SPHERICAL TRIGONOMETRY. 1.0. 11. 12. Givena=2946' 8", findA=54 1'16", Given a= 77 21' 50", find 6=2814'31".l, or 6'=15145'28".9, Given a =68, find 6=2552'33".5, or 6' = 1547'26".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", find a= 67 33' 27", B = 13724'21"; 6 = 155 27'54" ? c= 142 9' 13". A= 83 56' 40"; c=7853'20", B=2849'57".4, c' = 1016'40", B'=15110'2".G. c=7018'10".2, B=2736'58".7, c'=10941'49".8, B' = 15223'l".3. B = 47 13' 43", c= 133 32' 26". 6 = 4332'31"; B = 5759'19".2, c=5420'. B = 135 33' 39"; 6=143 5' 12", c= 120 55' 34". B = 99 57' 35"; 6 = 100 45', c=945'. 13. Solve the quadrantal triangle in which c = 90, A = 42l', B = 12120'. Ans. C = 6716'22", 6 = 11210'20", 14. Solve the quadrantal triangle in which a = 17412'49".l, b = 94 8' 20", c = ( Ans. A = 17557'10", B = 13542'55", = 4631'30". = 13534'8". SOLUTION OF OBLIQUE SPHERICAL TRIANGLES. 208. The Solution of Oblique Spherical Triangles presents Six Cases ; as follows : I. Given two sides and the included angle, a, b, C. II. Given two angles and the included side, A, B, c. III. Given two sides and an angle opposite one of them, a, b, A. CASE I. 305 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. Given 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) = cos a- 6 C cos i (a 4- &) 2 . sin \ (a 4- 6) 2 These determine ^(A + B) and -J-(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 (a 4- &) sin^- 2 306 SPHERICAL TRIGONOMETRY. Check, tan (a + 6) cos (A + B) = cos $ (A - B) tan |. There is no ambiguity in this case. Ex. 1. Given a = 43 18', b = 19 24', C = 74 22' ; find A, B, c. Solution. (a + 6) = 31 21', (a - 6) = 11 57', C = 37 11'. log cos (a 6) = 9.9904848 logseci(a + 6) = 0.0685395 log cot 5 =10.1199969 2 logtan4(A+B) =10.1790212 0".5 .-. A = 84 10' 17".5 B = 28 48' 16".5. c = 4135'48".5. log sin 4- (a -6)= 9.3160921 logcosec(a+&) = 0.2837757 log cot 5 = io. 1199969 logtan|(A-B) = 9.7198647 ... i(A-B)=2741'0".5. log cos | (a + 6) = 9.9314605 logseci(A + B) = 0.2579737 log sin - = 9.7813010 log cos -= 9.9707352 .-. - = 20 47' 54".25. 2 Otherwise thus: Let fall the perpendicular BD, dividing the triangle ABC into two right tri- angles, BDA, BDC. Denote AD by ra, the angle ABD by <, and A BD by p. Then by Napier's Rules, we have cos C = tan (b m) cot a ; sin (b m) = cot C tanp ; sin m = cot A .-. tan(6 m) = tan a cos C (1) and tan A sin m = tan C sin (b m) (2) CASE II. 307 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 Kules cos a *= cos (b m) cosp ; cos c = cos m cos p. .'. cos c cos (6 m) = cos m cos a, 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- viously determining A and B. This formula may be adapted to logarithms by the use of a subsidiary angle (Art. 90). Ex.2. Given 6 = 120 30' 30", c=7020'20", A=50iO r 10"; find B, C, a. Ans. B = 135 5' 28".8, C = 50 30' 8% a = 69 34' 56". 210. Case II. Given two angles, A, B, and the included side c ; to find a, 6, C. By Napier's Analogies (9) and (10) of Art. 197, cosi(A4-B) 2 sm(A + B) 2 from which a and b are found. The remaining part C may be found by (2) of Art. 198. sin %(a b) cos ~ = sin | (A B) sin - Check, cos J (a - b) cot 5 = cos (a + b) tan ( A + B) . There is no ambiguity in this case. 308 SPHERICAL TRIGONOMETRY. Ex. 1. Given A =68 40', B = 5620 f , c=8430'; find a, b, C. Solution. (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 log tan ^(a + b) =10.2913206 .'-. i(a + 6) = 6255' 9" |(a-&)= 6 16' 39" a = 69 11' 48" b = 56 38' 30". = 97 19' 3".5. logsini(A-B) =9.0310890 logcoseci(A+B) =0.0520711 log tan -=9.9582465 2 log tan i(a - 6) =9.0414066 ... -i- (a -b) =6 16' 39". log sin (A-B) =9.0310890 log cosec i(a - b) =0.9612050 log sin -=9.8276063 log cos -=9.8199003 . 5 _ 48 39' 31f ". a Otherwise thus : Let fall the perpendicular BD (see last figure). Denote, as before, AD by 972, the angle ABD by <, and BD by p. Then by Napier's Eules, we have cos c = cot < cot A ; cos < = cote tan jo; cos(B <) = cot a tan^>. .-. cot < = tan A cos c (1) tan a cos (B <)=cos< tanc (2) From (1) is determined, and from (2) a is found. Similarly b may be found. Also, from the same triangles, we have cos C sin $ = cos A sin(B <), from which C is found. CASE 111. 309 Ex. 2. Given A = 135 5' 28".C, 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". 211. 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, sin B=- sin A ..... (Art. 190) (1) sin a Then C and c are found from Napier's Analogies, ^,, , sin A _ sinB _ sinC sin a sin b sin c Since B is found from its sine in (1), it will have two values, if sin A sin b < sin a, 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 tan and tan -, or which is the same thing, that A B and a b have the same 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 two complete solutions. If only one of the values of B satisfies this condi- tion, there is only one triangle that satisfies the problem, since 310 SPHERICAL TRIGONOMETRY. 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', 6 = 47 15', A = 56 30'; findB,C,c. Solution. log sin b = 9.8658868 cologsiria =0.1682577 log sin A ='9.9211066 log sin B =9.9552511 .-. B = 6426'4", B' = 115 33' 56". + B) = |(A-B) = 45 0' 0". - 2 15' 0". 60 28' 2". - 3 58' 2". 86 1'68". -2931'5S". 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) When B = 64 26' 4". logsinJ(o-6) = 8.5939483- colog sin i (a + 5) =0.1505150 log cot |(A - B) = 1.1589413- log tan -=9.9034046 ... = ,38 40' 48". .-. = 77 21' 36". logsini(A + B) =9.9395560 cologsini(A-B) = 1.1599832- log tan i(a - b) =8.5942832- log tan -=9.6938224 .-. =26 17' 40". .-. c= 53 35' 20". CASE III. 311 (2) logsini(a-6) =8.5939483- cologsin(a+6) =0.1505150 log cot i(A-B') = 0.2467784- log tan =8.9912417 ...5-'= 535'50J". .-. C'=ll H'40i". B') =9.9989581 cologsm-J(A-B') = 0.3072223- log tan|(a - !>) =8.5942832 - log tan -=8.1 o 9004636 .-. - = 432'47i". .-. c'=9 5'34". ?. B = 64 26' 4", C =77 21' 36", c =53 35' 20"; B' = 115 35' 56", C' = llll'40", c'= 9 5' Otherwise thus: Let fall the perpendicular CD ; denote AD by m, the angle ACD by <, and CD by p. Then we have cos A = tan m cot b ; .-. tan m = cos A tan b (1) cos b = cot A cot 0. . . cot = cos b tan A Again, cos a = cos (c m) cos p ; cos b = cos m cos p. .-. cos (c m) = cos a cos m-^-b . . Also, cos (C <) = cot a tan p ; cos < = cot b tanp. .-. cos (C (j>) = cot a tan 6 cos < . . Lastly, (2) (4) sin a The required parts are given by (1), (2), (3), (4), (5). Ex. 2. Given a = 7349'38", 6=120 53' 35", A=8852'42": find B, C, c. Ans. B = 116 44' 48", C = 116 44' 48", c = 120 55' 35". 812 SPHERICAL TRIGONOMETRY. 212. Case IV. Given two angles, A, B, and the side a opposite one of them ; to find b, c, C. This case reduces, by aid of the polar triangle, to the preceding case, and gives rise to the same ambiguities. Hence the same remarks made in Art. 211 apply in this case also, and the direct solution may be obtained in the same way as in Case III. Thus, The side b is found from the formula sin b = sinB sin A sin a (1) Then c and C are found from Napier's Analogies. ta c cosHA + B) _ 2 cosf (A B) Check, 2 sin A cos % (a + b) sin B sin C (2) (3) sin a sin b sin c Ex. 1. Given A=667'20", B=5250'20", a=5928'27"; find 6, c, C. Solution. By (1) we find b = 48 39' 16", or 131 20' 44". i(A + B)=5928'50". i(A-B) = 6 38' 30". log cos |(A + B) = 9.7057190 cologcosi(A-B)= 0.0029244 log tan J (a + b) = 10.1397643 log tan - = 9.8484077 - = 35 II 1 %(a-b)= 5 24' 351". log cos % (a - b) =9.9980612 cologcos^(a + &) =0.2314530 log cot J(A + B)= 9.7704854 log tan - =9.9999996 = 45'. The second value of b is inadmissible (see Eule of Art. 211), and therefore there is only one solution. Ans. b = 48 39' 16", c = 70 23' 4H", C = 90. CASE V. 313 Ex. 2. Given A = 110 10', B = 133 18', a = 147 5' 32" ; find 6, c, C. -4ns. b = 155 5' 18", c = 33 1' 45", C = 70 20' 50". 213. Case V. Given the three sides, a, &, c ; to find the angles. The angles may be computed by any of the formulae of Art. 195 ; but since an angle near 90 cannot be accurately determined by its sine, nor one near by its cosine (Art. 151), neither of the first six formulae can be used with advan- tage in all cases. The formulae for the tangents however are accurate in all parts of the quadrant, and are therefore to be preferred for the solution of a triangle in which all three sides or all three angles are given. By (7) of Art. 195 we have tan - / sin ~ 6) sin - c) 2 \ sin s sin (s a) 1 /sin (s a) sin (s b) sin (s c) siu(s a)\ sins Since the part under the radical is a symmetric function of the sides, it is in the formulae for determining all three angles A, B, C, and when once calculated, it may be utilized in the calculation of each angle. For convenience in com- putation, denote this term by tan r. Then tanr= /sin (s - a) sin (s - b) sin (s - c) . \ sins and (7), (8), (9) of Art. 195 become (1) sin (s - a) tanr (2) 2 sin (s - b) g *!L^ ....... (3) 2 sin (s c) 314 Check, SPHERICAL TRIGONOMETRY. sin A sin B sin C sma sinB sin 6 sine Ex.1. Given a = 46 24', 6 = 67 14', c = 8112'; ftnd A, B, C. Solution. a = 46 24' b= 67 14' c = 81 12' 2s = 194 50' s= 97 25', -a= 51 1', *_& = 30 11', s-c = 16 13'. log sin (s - a) = 9.8906049 log sin ( 8 -b) = 9.7013681 log sin (s - c) = 9.4460251 colog sin s = 0.0036487 log tan 2 r = 9.0416468 log tan r= 9.5208234. tan r = 9.5208234 sin(s-a) = 9.8906049 tan -=9.6302185 2 .-. -=23 6' 45". A =46 13' 30". tan r = 9.5208234 sin(-6) = 9.7013681 tan -=9.8194553 2 .-. ?= 33 25' 10". B= 66 50' 20". tanr= 9.5208234 sin(s-c)= 9.4460251 tan -=10.0747983 2 X ~ =49 54' 35". A C= 99 49' 10". Ex. 2. Given a = 100, b = 37 18', c = 62 46' ; find A, B, C. Ans. A = 17615'46".56, B = 217'55".08, C = 322'25".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 B = we have, from (7), (8), (9) of Art. 196, tan ^ = tan R cos (S A), CASE VI. 315 tan - = tan R cos (S - B), 2 tan = tan K cos (S - C), Zi by which the three sides may be found, sin A sin B sin C Check, sin a sin b sine Ex.1. Given A = 68 30', B = 7420', C = 8310' ; find a, fe, c. Solution. 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 tan 2 R = 9.9078418 log tan E = 9.9539209.* 8 = 113 0' S-A = 44 30' S-B= 38 40' S - C = 29 50' log tan - = 9.8071630 a log tan | = 9.8464574 log tan - = 9.8921785 a = 65 2 & = 70 9' 9i c = 7555' 9". Check, sin a sin b sine sin A sin B sin C Ex.2. Given A=5955'10", B = 8536'50 , C = 5955'10"; find a, b, c. Ans. a = 129 11' 40", b = 63 15' 12", c = 129 11' 40". * The necessary additions may be conveniently performed by writing log tan R on a slip of paper, and holding it successively over log cos (S A), log cos (S B), and logcos(S-C). 316 SPHERICAL TRIGONOMETRY. EXAMPLES. Solve the following right triangles : Given c= 84 20', find a= 35 13' 4", Given c= 67 54', find a= 3935'51", 3. Given c= 2218'30", find a= 16 17' 41", 4. Given c = 145, find a= 13 12' 12", 5. Given c= 98 6' 43", find a=137 6', 6. Given c= 46 40' 12", find a= 2627'23".8, 7. Given c= 76 42', find b= 70 10' 13", 8. Given c= 91 18', find 6= 94 18' 53 ".8, 9. Given c= 8651', find a= 86 41 '14", 10. Given c= 2349'51", find 6= 19 17', 11. Given c= 97 13' 4", find b= 7913'38".2, 12. Given c= 3740'20", find 6= 026'37".2, A= 35 25'; 6= 83 3' 29", B= 85 59' 1". A= 4328'; b= 60 46' 25J", B= 7022'21". A= 47 39' 36" 5 b= 1526'53", B= 4433'53".4. A= 23 28'; 6=14717'15", B = 109 34' 33". A = 138 27' 18"; b= 7751', B= 8055'27". A= 37 46' 9"; b= 3957'41".4, B= 62 0' 4". a= 4718'; A= 49 2'24".5, B== 75 9' 24 ".75. a= 72 27'; A= 72 29' 48", B= 94 6'53".3. 6= 18 1'50"; A= 88 58' 25", B= 18 3' 32". a= 1416'35"; A= 3736'49".3, B= 5449'23".3. a= 132 14' 12"; A = 131 43' 50", B= 8158'53".3. a= 37 40' 12"; A-: 8925'37", B= 043'33". EXAMPLES. 317 Given a= 82 6', B= 43 28'; find A= 84 34' 28", b= 43 11' 38", c= 84 14' 57". Given a= 4230'30", B= 53 10' 30"; find A= 53 50' 12", b= 42 3' 47", c= 56 49' 8". Given a= 2020'20", B= 38 10' 10"; find A= 6435'16".7, b= 1516'50".4, c= 25 14' 38 ".2. Given a= 92 47' 32"; B= 50 2' 1"; find A= 92 8' 23", b= 50, c= 9147'40". Given b= 5430', A= 35 30'; find B= 7017'35", a= 30 8'39".2, c= 59 51' 20". Given b= 155 46' 42". 7, A- 8010'30"; find B = 15368'24".5, a= 67 6'52".6, c=11046'20". Given a= 35 44', A= 37 28'; find 6= 69 50' 24", c= 73 45' 15", B= 77 54', or &'=110 9' 36", c'= 106 14' 45", B = 102 6'. Given a=12933', A = 104 59'; find b= 1854'38", c=127 2' 27", B= 2357'19", or 6'=161 5' 22", c'= 52 57' 33", B' = 156 2' 41". Given a= 21 39', A= 42 10' 10"; find 6= 2559'27".8, c= 3320'13".4 ? B= 52 23' 2".8, or 6'=154 0'32".2, c'=14639'46".6, B'=12736'57".2. Given a= 42 18' 45", A= 4615'25"; find 6= 60 36' 10", c= 68 42' 59", B= 6913'47", or &'=11923'50", c'=lll17' 1", B' = 11046'13". Given & = 160, B = 150; find a= 39 4'50".7, c=13650 f 23".3, A= 67 9'42".7, or a'=14055 f 9".3, c'= 43 9'36".7, A'=11250'17".3. 318 SPHERICAL TRIGONOMETRY. 24. Given a= 25 18' 45", A= 15 58' 15". Ans. Impossible ; why? 25. Given a= 32 9' 17", b= 32 41'; find A= 49 20' 17", B= 50 19' 16", c= 44 33' 17". 26. Given a= 55 18', b= 3927'; find A= 66 15' 6", B = 45 1'31", c= 6355'21". 27. Given a= 56 20', b= 78 40'; find A= 56 51' 7", B= 8031'48", c= 8344'44i". 28. Given a= 86 40', 6= 32 40'; find A= 8811'57".8, B= 3242'37".8, c= 8711'39".8. 29. Given a= 37 48' 12", b= 5944'16"; find A= 41 55' 45", B= 70 19' 15", c= 66 32' 6". 30. Given a=116, b= 16; find A= 9739'24".4, B= 1741'39".9, c=11455'20".4. 31. Given A = 52 26', B= 49 15'; find o= 3624'34".5, b= 3433'40", c= 48 29' 20". 32. Given A = 64 15', B= 48 24'; find a= 54 28' 53", b= 4230'47", c= 64 38' 38". 33. Given A = 54 1'15", B = 137 24' 21"; find a= 29 46' 8", b = 155 27' 55", c=142 9' 12". 34. Given A= 46 59' 42", B= 57 59' 17"; find a= 36 27', 6= 4332'37", c= 54 20' 3". 35. Given A = 55 32' 45", B = 101 47' 56"; find a= 54 41' 35", 6=10421'28", c= 98 14' 24". 36. GivenA= 6027'24".3, B= 5716'20".2; find o= 5432'32".l, 6= 5143'36".l, c= 68 56' 28". 9. EXAMPLES. 319 Solve the following quadrantal triangles : 37. Given B= 74 45', find 6= 8517'15".5, a= 18 12', A= 17 34' 2", c= 90; = 104 31' 13". 38. Given A = 110 47' 50", find a =104 53' 0".S, B = 13535'34''.5, 6=133 39' 47".7, c= 90; C = 10441'37".2. Solve the following oblique triangles : 39. Given a= 73 58', find A=116 8' 28", 6= 38 45', B= 35 46' 39", 0= 46 33' 39"; c= 51 I'll". 40. Given a= 96 24' 30", find A= 9753'Qi", 6= 6827'26", B= 6759'39i", 0= 84 46' 40"; c= 8731'37". 41. Given a= 76 24' 40", find A= 63 C 48'35V, 6= 58 18' 36", B= 5146'12i", = 116 30' 28"; c=10413'27". 42. Given a= 8618'40", find A= 6448'53|", 6= 4536'20", B= 4023'15|", = 120 46' 30"; c= 108 39' 111". 43. Given a= 88 24', find A= 65 13' 3%", 6= 56 48', B= 49 27' 51", = 128 16'; c= 120 10' 52". 44. Given a= 68 20' 25", find A= 56 16' 15", 6= 52 18' 15", B= 45 4' 41", = 117 12'20"; c= 9620'44". 45. Given a= 88 12' 20", find A= 63 15' 12", 6=124 7'17", B = 132 17' 59", 0= 50 2' 1"; c= 59 4'25". 46. Given a= 3223'57", find A= 60 53' 2", 6= 3223'57", B= 60 53' 2", 0= 66 49' 17"; c= 34 19' 11". 47. Given 6= 99 40' 48", find B= 95 38' 4", c=10049'30", 0= 9726'29".l, A= 6533'10"; a= 6423'15".l. 48. GivenA= 3134'26", find a= 40 1' 5J", B= 30 28' 12", 6= 3831' 3J", c= 70 2' 3"; = 130 3' 50". 320 SPHERICAL TRIGONOMETRY. GivenA = 1305'22".4, B= 3226'6".41, c= 51 6'11".6; find a = 84 14' 29", 6= 44 13' 45", C^ 36 45' 26". Given A = 96 46' 30", B= 84 30' 20", c=12646'; find a=10221'42", 6= 78 17' 2", C -125 28' 13 J". Given A = 84 30' 20", B= 76 20' 40", c=13046'; find a= 94 34' 52 J", b= 76 40' 48i", C = 130 51' 33|". Given A= 107 47' 7", B= 3858'27", c= 51 41' 14"; find a= 70 20' 50", b= 38 27' 59", C= 52 29' 45". GivenA = 12841'49", B = 107 33' 20", c=12412'31"; find a=12544'44", b= 8247'35", C = 12722' 7". GivenA=12958'30", B= 34 29' 30", c= 50 6'20"; find a= 85 59', b= 4729'20", C= 36 6' 50". Given A= 95 38' 4", C= 9726'29", 6= 64 23' 15"; find a= 99 40' 48", c= 100 49' 30", B= 65 33' 10". Given A = 70, B = 13118', c116 ; find a= 57 56' 53", b = 137 20' 33", C= 9448'12". Given a= 62 15' 24", & = 10318'47", A= 53 42' 38"; find B= 6224'24".8, C = 15543'll".3, c=153 9' 35J", or B'=11735'35".2, C f = 59 6'10".6, c'= 7025'26". Given a= 52 45' 20", b= 7112'40", A= 46 22' 10"; find B= 59 24' 15 1", C = 115 39' 55J", c= 9733'18".8, or B'=12035'44i", C'= 2659'55".2, c'= 2957'10".5. Given a= 4845'40", b= 6712'20", A= 42 20' 30"; find B= 55 39' 57", = 116 34' 18", c= 93 8' 9".6, or B'=12420' 3", C'= 24 32' 15", c'= 27 37' 20". Given a= 46 20' 45", b= 65 18' 15", A= 40 10' 30"; find B= 54 6' 19", C = 116 55' 26", c= 90 31' 46", or B' = 12553'41", ; C'= 2412'53".3, c'= 27 23' 14". EXAMPLES. 321 61. Given a=15057' 5", 6 = 134 15' 54", A = 14422'42"; find B = 12047'44", 0= 97 42' 55", c= 55 42' 8", or B'= 59 12' 16", C'= 29 9' 9", c'= 2357'29". 62. Given a= 5045'20", b= 6912'40", A= 44 22' 10"; find B= 5734'51".4, C = 11557'50".6, c= 9518'i^".4, or B'=12225' 8".6, C'= 2544'31".6, c'= 28 45' 5".2. 63. Given a= 40 5'25".6, 6==11822' 7".3, A= 2942'33".8; find B= 4237'17".5, C = 160 1'24".4, c=15338'42".4, or B'=13722'42".5, C'= 5018'55".2, c'= 90 5'41".0. 64. Given a= 9940'48", b= 6423'15", A= 95 38' 4"; find B= 65 33' 10", 0= 9726'29", c= 100 49' 30". (No ambiguity ; why?) 65. Given A = 79 30' 45", B= 46 15' 15", a= 53 18' 20"; find b= 36 5'34f", c= 5024'57", C= 70 55' 35". (No ambiguity ; why?) 66. Given A = 73 11' 18", B = 61 18' 12", a= 46 45' 30"; find b= 4152'34f", c= 41 35' 4", C= 6042'46".5. (Only one solution ; why?) 57. GivenA= 4630'40", B= 36 20' 20", o= 4215'20"; find b= 3318'47", c= 60 32' 6", = 110 3'14".6. (Only one solution ; why?) 68. Given A = 61 29' 30", B= 24 30' 30", a= 34 30"; find b= 1530'30".5, c= 39 33' 52", 0= 98"48'58".5. (Only one solution ; why?) 69. Given A = 36 20' 20", B= 4630'40", a= 42 15' 20"; find b= 55 25' 2f , c= 8127'26J", C = 11922'27i", or &'=12434'57i", c'=16234'27", C'=16441'55". 322 SPHERICAL TRIGONOMETRY. 70. Given A = 52 50' 20", B = 66 7' 20", a= 59 28' 27"; find b= 81 15' 15", c=11010'50|", C = 119 43' 48", or b'= 98 44' 45", c'=13845'26", C'=14224'59". 71. Given A = 115 36' 45", B= 80 19' 12", b= 8421'56"; * find a = 114 26' 50", c= 82 33' 31", C= 79 10' 30". 72. Given A= 61 37' 52". 7, B = 13954'34".4, 6 = 15017'26".2; find a = 4237'17".5, c=12941' 4".8, C= 8954'19".0, or a'=13722'42".5, c f = 1958'35".6, C'= 2621'17".6. 73. Given A = 70, B = 120, b= 80. Ans. Impossible ; why ? 74. Given a =108 14', b= 75 29', c= 56 37'; find A=12353'47", B= 57 46' 56", C= 4651'51".5. 75. Given a= 57 17', b= 2039', c= 76 22'; find A = 21 1' 2", B= 8 38' 46", C = 15531'36".5. 76. Given a= 68 45', b= 53 15', c= 46 30'; find A= 94 52' 40", B= 58 5' 10", C= 5050'52i". 77. Given a = 63 54', b= 47 18', c= 53 26'; find A = 86 30' 40", B= 54 46' 14", C= 6312'55i". 78. Given a= 70 14' 20", b= 49 24' 10", c= 38 46' 10"; find A= 110 51 '16", B= 48 56' 4", C= 38 26' 48". 79. Given a=12412'31", 6= 54 18' 16", c= 9712'25"; find A = 127 22' 7", B= 5118'11", 0= 72 26' 40". 80. Given a = 50 12' 4", 6=116 44'48", c=129ll'42"; find A= 59 4' 25", B= 94 23' 10", = 120 4' 50". 81. Given a =100, b= 50, c= 60; find A = 13815'45".4, B= 3111'14".0, C= 3549'58".2. 82. Given A = 86 20', B= 76 30', C= 94 40'; find a= 87 20' 28", 6= 76 44' 2V, c= 93 55' 31". EXAMPLES. 323 83. 84. 85. 86. 87. 89. Given A = 96 45', find a= 88 27' 49", Given A = 78 30', find a= 7457'46", Given A = 57 50', find a= 58 8' 19", Given A = 129 5' 28", find a = 135 49' 20", Given A = 138 15' 50", find a=100 0' 8".4, GivenA=10214'12", find a=10425' 8", Given A= 20 9' 56", find a= 20 16' 38", B = 10830', b = 107 19' 52", B = 11840', 6=120 8'49", B= 98 20', b= 83 5' 36", B = 142 12' 42", 6=14437'15", b= 49 59' 56". 4, B= 54 32' 24", b= 53 49' 25", B= 55 52' 32", c=11528'13j". C= 93 20'; c=10018'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"; b= 5619'41", i c= 6620'43". 90. If a, 6, c are each < ^, show that the greater angle may u exceed - 91. If a alone > ^TT, show that A must exceed - 2 92. If a and b are each >^-TT, and c TT ; (2) B ma?/ be > TT; (3) C may or may not be < TT. 93. If cos a, cos 6, cos c are all negative, prove that cos A, cos B, cos C are all necessarily negative. 94. In a spherical triangle, of the five products, cos a cos A, cos b cos B, cos c cos C, cos a cos b cos c, cos A cos B cos C, show that one is negative, the other four being positive. 324 SPHERICAL TRIGONOMETRY. CHAPTER XII. THE IN-OIEOLES AND EX-OIEOLES, AEEAS, 215. The Ill-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 AO, BO; from O 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. Now tan OF = tan OAF sin AF . (Art. 186) or, denoting the radius OF by ?, we have j^ tan r = tan sin (s a) (1) or tan r = y/ sin ( * ~ a ^ sin *.* ~ 6) si " ^ ~ ^ Af sins n sins (Art. 195) (2) THE ESCRIBED CIRCLES. 325 Also, sin(s a) c) cos^a cos (& + c) sinja (Art. 198) sin- _ sin a sin^B sin-|-C sin^-A which in (1) gives B . C sin sin o o tan r = - = ? sin a ......... (3) cos^A . . (Art. 196) (4) 2 cos ^ A cos B cos ^ C an equation which is equivalent to the following : cotr = ^-[cosS+cos(S-A)+cos(S-B)+cos(S-C)](5) 216. The Ex-Circles. To find the angular 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 ex-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, TT 6, TT c, A, TT B, TT C, the problem becomes identical with that A< of Art. 215 ; and we obtain the value for the in-radins of the colunar triangle A'BC, by substituting for 6, c, B, C, their supplements in the five equations of that article. 326 SPHERICAL TRIGONOMETRY. Hence, denoting the radius by r a , we get tan r a = tan i A sin s (1) (2) sin(s a) == cos j _ >S]L_ gina . . . . cos^- A 2cos-i-AsiniB sin-|C cotr a =^-[ cosS cos(S A)-f-cos(S B) + cos(S 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. tSch. 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, 6, A, B. The values for the radii r b and r c are therefore found in the same way as the above values for r a \ or they may be obtained from the values of r a by advancing the letters. Thus. tan r b =* tan 4- B sin s = , etc., sm(s b) and tan r c = tan i C sin s = . n -, etc. sm(s c) 217. The Circumcircle. To find the angular radius of the circumcircle of a triangle. The small circle passing through the vertices of a spheri- cal triangle is called the circumscribing circle, or circumcircle, of the triangle. THE CIECUMCIECLE. 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 circimi- 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 = OB A, OBC = OCB, OCA = OAC Let OC = R ; then, in the triangle ODC, we have cos OCD = tan CD cot CO = tan 1 a cot R . (Art. 186) tan i a tanR or tan R = cos(S- A) cos S . . . . (1) (Art. 196) (2) Also cos(S - A)= cos i[(B + C) - A] = cosi(B + C) cos| A + sin|(B + C) si: -c)] (Art. 198) sin A cos ^6 cos | c, cosset which in (1) gives tanR = sin-^q sin A cos ^b cos ! c _ 2 sin a sin | b sin j . . . . (3) (Art. 195) (4) 328 SPHERICAL TRIGONOMETRY. which may be reduced to the following : tan R = [sin(s a) + sin(s b) + sin(s c) sins] (5) ft 218. Circumcircles of Colunar Triangles. To find the angular radii of the circumcircles of the three colunar triangles. Let KU R 2 , R 3 be the angular radii of the circumcircles of the colunar triangles on the sides a, b, c, respectively. Then, since Rj is the circumradius of the triangle A'BC whose parts are a, TT 6, TT c, A, TT B, TT C, we have, from Art. 217, (2) ,r, -p sin A a /Q \ tanK 1= =- - f ........ (3) sin A sin b sin c (4) tan Rj = [sins sin(s a)+sin(s 6)+sin(s c)] (5) ^ n Similarly, -P tan-i-6 cos(S B) tan K 2 = --- ^ = - i '- = etc., cosS N and tanR 3 = -^ n ^ = cos ( S - C )=etc. cosS N EXAMPLES. Prove the following : 1. cos s + cos (s a) -f cos (s b) -f- cos (s c) = 4 cos \ a cos \ b cos J c. 2. cos (s 6) -f cos (s c) cos (s a} cos s = 4 cos J a sin 6 sin -J- c. PROBLEM. 329 3. cosJB 2 cos ^-B sin ^C sin J A cos 4- A cos 4- B . 4. tan r c = % sin c = cos-J-0 2cosCsin^-Asin^-B 5. cot r : cot ^ : cot r 2 : cot r 3 = sins : sin(s a) : sin(s 6) : sin(s c). 6. tan r tan i\ tan r 2 tan r 3 = n 2 . 1. cot r tan r tan ?' 2 tan r 3 = sin 2 s. 8. q , T> _ 2 cos a cos J b sin J c 71 10. tanE 1 :tanE 2 :tanK 3 =cos(S-A):cos(S-B):cos(S-C). 11. cot E cot E! cot E 2 cot E 3 = N 2 . 12. tan E cot Ej cot E 2 cot E 3 = cos 2 S. AREAS OF TRIANGLES. 219. Problem. To find the area of a spherical triangle, having given the three angles. Let r = the radius of the sphere. E = the spherical excess = A + B + C - 180. K = area of triangle ABC. It 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. 330 SPHERICAL TRIGONOMETRY. Cor. The areas of the colunar triangles are 180 180 180 220. Problem. To find the area of a triangle, having given the three sides. Here the object is to express E in terms of the sides. I. CagnolCs Theorem. sin |E = sin ( A + B + C - TT) = sin J(A -f B) sin $ C cos (A + B) cos JC = sin fr C cos |C p cQs ^, _ ^ _ GQg ^ q + ^ j (Art. 198) cos^c sin 1 a sin 4 6 sin C sin a sin -J 6 2w /A , iar\si\ - - -- -- : - : - l Art. iyo ) (^i ) cos^c cos^-c sin a sin o (2) 2 cos a cos J 6 cos c II. Lhuilier's Theorem. sin j(A + B) - sin ^(TT - C) cos i(A + B) + cos |(TT - C) sin \ ( A + B) cos \ G a 6) cos-^c ^ cos^-C + cos^c sin-JC cos Js cos ^(s c) ( Art = Vtan|stanJ(s-a)tan^(s-6)tani-(s-c) (Art. 195) (3) AREAS OF TRIANGLES. 331 221. Problem. To find the area of a triangle, having given two sides and the included angle. cos^E^ cos [ (A + B) - (iTr - iC)] B) cos^C os 2 |-C (Art. 198) ]sec|-c . . (1) Dividing (1) of Art. 220 by this equation, and reducing, we have t iF tan % a tan ^ b cos EXAMPLES. (2) 1. Given a = 1132'56".64, b = 82 39' 28". 4, c = 7454 f 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 b= 82 39' 28 ".40 c= 74 54' 31 ".06 2s = 27036'56".10 s = 13518'28".05 s-a= 22 15' 31 ".41, s-b= 52 38' 59". 65, s- c = 6023'56".99. is = 6739'14".025 -a) = ll 7'45".705 -6) = 2619'29".825 _c = 30ll'58".495 log tan |s = 0.3860840 log tan |0 - a) = 9.2938583 log tan l(s -b) = 9.6944058 log tan i( - c) = 9.7649261 log tan 2 iE = 9.1392742 log tan IE = 9.5696371. 1E = 2021'58".25. E = 81 27' 53" = 293273". 7T?- 2 . . . . [(1) of Art. 219] 332 SPHERICAL TRIGONOMETRY. 2. Given A = 84 20' 19", B = 27 22' 40", C = 7533'; find E = 715'59". 3. Given a = 46 24', b = 67 14', c = 8112'; find K = ii 4. Given a = 108 14', b = 75 29', c = 56 37' ; find E = 48 32' 34".5. 5. Prove cosjE = 1 + cosa + cos6 + cosc 4cosiacos|6cosic _ cos 2 ig -f cos 2 i& -f cos 2 ic cos^c 6. " sin * E= / si \ 7. u cos 1 E= /CQ \ sinC _ cot|-6 cot^c + cos A sin A cot ^ c cot ^ a -\- cos B sinB EXAMPLES. Prove the following : 1. sin (s a) -f sin (s b) + sin (s c) sin s = 4 sin la sin J6 sin^c. 2. sin s -f sin (s 6) + sin (s c) sin (.9 a) = 4 sin i a cos ^ 6 cos ^ c. EXAMPLES. 333 3. sin (s b) sin (s c) 4- sin (s c) sin (s a) 4- sin (s a) sin (s b) 4- sin s sin (s a) 4- sins sin(s b) 4- sins sin(s c) = sin b sin c 4- sin c sin a 4- sin a sin b. 4. sin (s b) sin (s c) 4- sin (s c) sin (s a) sin(s a)sin(.s b) 4- sins sin(s a) -f sin s sin (s b) sin s sin (s c) = sin 6 sin c 4- sin c sin a sin a sin b. 5. sin 2 s + sin 2 (s a) 4- sirr(s b) + sin 2 (s c) = 2(1 cos a cos 6 cos c) . 6. sin 2 s 4- sin 2 (s a) sin 2 (s 6) sin 2 (s c) = 2 cos a sin 6 sin c. 7. cos 2 s 4- cos 2 (s a) 4- cos 2 (s 5) 4- cos 2 (s c) = 2 (1 4- cos a- cos b cos c) . 8. cos 2 s 4- cos 2 (s a) cos 2 (* b) cos 2 (s c) = 2 cos a sin b sin c. 9. tan r cot ^ tan r 2 tan r 3 = sin 2 (s a) . 10. tanr tanr x cotr 2 tanr 3 = sin 2 (s 6). 11. tan r tan r tan r 2 cot ?* 3 = sin 2 (s c) . 12. cotr sins = cot^-Acot^-B cot^C. 13. tan T-J -h tan r 2 4- tan r 3 tan r = 14. cot 7'j 4- cot r 2 4- cot ?: 3 cot r = 15. tan r, sin A sin B sin C 4 sin ^ a sin ^ b sin c sin b sin c 1 + cos A 1 + cos B 1 + cos C -i - tan 7% 4- tan r 2 4- tan r tan r 16. - =|(l4-cosa4-cos&4-cosc). cot TI 4- cot r 2 4- cot r B cot r 334 SPHERICAL TRIGONOMETRY. 17. cot'r, + cot 2 ,-, + cot'r, + cotV = gjl-eosacogftcogc). n 2 18 1 _i_ 1 1 1 _ 2 cos a sin b sin c sin 2 r sin 2 ?^ sin 2 r 2 sin 2 7* 3 n 2 19. cot r 2 cot r 3 4- cot r s cot 7\ + cot r cot 7* 2 + cot 7* (cot ?*! 4- cot r 2 + cot r 3 ) _ sin 6 sin c -f- sin c sin a 4- sin a sin 6 7r 20. tan ?* 2 tan ?* 3 + tan 7* 3 tan ^ + tan ^ tan r 2 4- tan r(tan 7^ 4- tan r 4- tan r s ) = sin b sin c + sin c sin a 4- sin a sin &. 21. cot R tan E-! cot B 2 cot K 3 = cos 2 (S A) . 22. cot E cot R! tan E 2 cot R 3 = cos 2 (S - B) . 23. cot E cot E! cot E 2 tan E 3 = cos 2 ( S G ) . 24. tan E x 4- tan E 2 = cot r + cot r 3 . 25. tan E x 4- tan E 2 4- tan E 3 tan E = 2 cot r. 26. tan E tan Ej + tan E 2 4- tan E 3 = 2 cot i\. 27. tan E 4- tan Ej tan E 2 4- tan E 3 = 2 cot 7' 2 . 28. tan E 4- tan Ej 4- tan E 2 tan E 3 = 2 cot r s . 29. cot 7*! 4- c t ^2 + c t ^*3 cot r = 2 tan E. 30. cot r cot 7*j 4- cot r 2 4- cot r 3 = 2 tan E x . 31. cot r 4- cot 9'j cot r 2 4- cot ? 3 = 2 tan E 2 . 32. cot r 4- cot r x -f cot r 2 cot ?* 8 = 2 tan E 3 . 33. tan E 4- cot r = tan E, 4- cot r, = etc., = (cot r 4- cot TI 4 cot r 2 + cot r 3 ) . 34. tan 2 E + tan 2 E 1 4-tan 2 E 2 4-tan 2 E 3 _ 2(1 + cos A cos B cos C) ~~ EXAMPLES. 335 35 tan 2 B 4- tan 2 B^ + tan 2 E 2 + tan 2 E 3 = ^ cot 2 r 4- cot 2 TI 4- cot 2 r 2 +' cot 2 r 3 36. tan 2 E + tan 2 Ej - tan 2 E 2 - tan 2 E 3 2 (cos A sin B sin C) N 2 O rr ,o 2 cos a sin 6 sin c 37. eot 2 r 4- cot 2 1\ cot 2 r 2 cot 2 r 3 = n 2 38 tan 2 E 4- tan 2 E! - tan 2 E 2 - tan 2 E 8 = ._ cosA < cot 2 r + cot 2 r x cot 2 r 2 cot 2 r 3 cos a 39. tan E cot Ej = tan & tan ^ c. 40. (cot r + tan B)'+ 1 = ( gin ffi + si " & + sin C Y. \ 2n y -r, \ o , /sin 6 + sin c sin a\ 2 41. (cot rj tan E) 2 + 1 = ( - ) \ 2n J N 42. tan ^ A sin (s a) = 2 cos J A cos ^B cos ^ C jo tanr _cos(S A)cos(S B) cos(S C) tanE 2 cos JAcos^B cos^C 44. cot (s b) cot (s c) -f- cot (s c) cot (s a) 4- cot(s a) cot(s 6) = cosec 2 r. 45. eot (s b) cot (s c) cot s cot (s b) cot s cot (s c) = cosec 2 7y 46. cot (s c) cot (s a) cot s cot (s c) cot s cot (s a) = cosec 2 r 2 . 47. cot(s a)cot(s b) cot s cot (s a) cot s cot (s 6) = cosec 2 r 3 . ^o cot(g a) , cot(s 6) , cot(s c) , 2 cots ^"* : ^ 1 :; * r . i . sjirrj sm 2 r 2 sm 2 r 3 sm 2 r = 3cot(s a) cot( 6)cot(s c). 336 SPHERICAL TRIGONOMETRY. 49. cosec 2 1\ -f cosec 2 r 2 -f- cosec 2 r 3 cosec 2 r = 2 cot s[cot(s a) + cot(s b) + cot(s c)]. 50. -1- + -J--+ * +./ sm 2 r 2 sin 2 r 3 -22 tan (s - a) tan (s - b) tan s tan (s a) tan (s b) tan (s c) 2 N cos s 51. cot B. cot Bj cot B 2 cot B 3 = n 52. sin(A-E) = 2 cos -j- a sin -j- 6 sin ^ c 53. sin(B-*E) = - n ^ . . J sin ^ a cos f o sin ^ c 54. sin(C-iE) = 2 sin a, sin J 6 cos -J c t-t- /A 1 TT<\ sin 2 4- 6 -f- sin 2 4- c sin 2 4-a 55. cos(A ^E)= 2 2 cos -J- a sin -J- 6 sin J c 56. cos(B - JE) = 57. 58. sinC _ tan ^ 6 tan ^c + cos A _ tan |q cot ^6 cosB sin A sin B 59. 60. 61. 62. If S, Sj, S 2 , S 3 denote the sums of the angles of a triangle and its three colunars, prove that S -j- Sj + ^2 + S 3 = 3 IT. 63. In an equilateral triangle, tan E = 2 tan?*. EXAMPLES. 337 64. If EU E 2 , E 3 denote the spherical excesses of the cohmars on a, b, c, respectively, show that E -f E x + E 2 + E 3 = 2ir ; and therefore the sum of the areas of any triangle and its colunars is half the area of the sphere. 65. Given a = 108 14', b = 75 29', c = 5637'; find E = 48 32' 34".5. 66. Given a = 63 54', b = 47 18', c = 5326'; find E = 24 29' 49*-". 67. Given a = 69 15' 6", b = 120 42' 47", c = 159 18' 33"; find E = 216 40' 23". 68. Given a = 33l'45", b = 155 5' 18", = 110 10'; find E = 133 48' 55". 69. Given a = b 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=~ t prove sinE = J, and cosE = J. 3 2 72. If G = ~ 9 prove sin ^ E = sin a sin \ b sec c, and & cos^-E = cosiacos& sec^c. 73. If a = b, and C = -, prove tan E = \ tan a sec a. 74. If A-hB-f-C = 27r, prove cos 2 Ja+cos 2 ^& + cos z -Jc=l. 75. If a + 6 = TT, prove that E = C ; and if E' denote the spherical excess of the polar triangle, prove that = sin a cos 76. Prove Bin'jE = Vsi " * E si " * E ' f 1 " cot a cot % b 338 SPHERICAL TRIGONOMETRY. CHAPTER XIII. APPLICATIONS OF SPHERICAL TRIGONOMETRY, SPHERICAL ASTRONOMY. 222. Astronomical Definitions, The celestial sphere is the imaginary concave surface of the visible heavens 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. Because 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 underneath 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 west 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 declination, 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. 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, 340 SPHERICAL TRIGONOMETRY. passing through the star; it is generally reckoned from the south point of the horizon round by the west, from 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. The 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 to 360, or from O h to 24 h . 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 to 90. It corresponds to terrestrial latitude. The polar 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 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. SPHERICAL COORDINATES. 341 224. Graphic 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 zenith is Z, QZ is the ter- restrial latitude ; and since QZ == PR, .-. 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 : VK = right ascension of the body = a, KS = decimation " " " = 8, ZPS or QK = hour angle " " " = *, PS = north polar distance " " " =p, HT = azimuth " " " = a, TS = altitude " " =h, ZS = zenith distance " " " = z, QZ = PR = latitude of the observer = . The triangle ZPS is called the astronomical triangle; ZP = 90 - 4 = co-latitude of the observer, PS = 90 - 8, SZ = 90 - h. 342 SPHERICAL TRIGONOMETRY. Let the small circle MM', passing through S, and parallel to the equinoctial, represent the apparent diurnal motion 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 hemisphere 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 character as follows : (1) Given the latitude 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 APE,, right angled at E, we have cos EPA = - cos ZPA = tan EP cot AP. .*. cos t = tan < tan 8 (1) from which the hour angle is found. 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. Eequired the apparent time of sunrise at a place whose latitude is 40 36' 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 4 = 9.9331352 loff tan 8 = 9.6250362 log cos t = 9.5581714- .-. t =111 11' 44" = 7h 24 m 47 s , nearly, PROBLEMS OF SPHERICAL ASTRONOMY. 343 which taken from 12 h , the time of apparent noon, gives 4 h 35 m 13 s , the time of apparent sunrise.* (2) Given the latitude of a place and the declination of a star; to find its azimuth from the north at rising. Let A = the azimuth = AR. Then in the triangle APR we have sin AP = cos AR cos PR, or sin 8 = cos A cos <. .*. cos A= sin sec < (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 h 12 m 46 8 .3 ; K 63 15' 11" E. (3) Given 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-<, PS = 90 8, and ZPS = t. Let A = the azimuth from the north = RT, p = the angle ZSP, and z = ZS. Then by Delambre's Analogies (Art. 198), siu.%(p + A) cos % z = cos ^(8 ) cos^J, sin ^(p A) sin z = sin ^-(8 <)cos^, + A) cos $z = sin i(S+ <)sin, A) sin z = cos |(8 + <) sin^-Z. 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 examples no corrections are applied for refraction, serai-diameter of the sun, change in declination from noon, etc. 344 SPHERICAL TRIGONOMETRY. (4) Given the right ascensions and declinations oftivo stars; to find the distance between them. Let P be the pole, S and S' the two stars. p a _ a / Let a and a' 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, s , PS = 90-S=p, PS' = 90-S'=p', 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 = m. Then cosP = tanPDcotPS. .-. tan m = cos P tan p. Also cos SS' = cos PS cos S'D sec PD . (Art. 209) .. cos d = cos p cos S'D seem. Ex. Required the distance between Sirius and Aldebaran, the right ascensions being 6 h 38 m 37 8 .6 and 4 h 27 25 8 .9, and the declinations 16 31' 2" S. and 16 12' 27" K, respectively. Here P = 2 h ll m 11 8 .7 log cos P = 9.9245789 = 32 47' 55" log tan p = 0.5279161- p = 106 31' 2" log tan m = 0.4524950- m= 109 25' 55" m = 109 25' 55". '= 7347 ; 33" S'D= 35 38' 22" log cos S'D = 9.9099302 log cos^> = 9.4537823- m = 109 25' 55", co i g cos m = 0.4779643- = d= 46 0'44". log cos cZ = 9.8416768 = 106'31' 2", PROBLEMS OF SPHERICAL ASTRONOMY. 345 (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, SD, SL perpendicular to VD, VL. Then VD = right ascension = , SD = dec- lination = 8, VL = longitude = A, SL = latitude = /. Denote the ob- liquity of the ecliptic DVL by ) = 9.9634401 log tan a = 10.6255266 cologcos<9= 0.1632816 log tan A =10.7522483 s log sin (6 - ) = 9.5950996 log sin 8 = 9.8558743 cologsin<9= 0.1384575 log sin/ = 9.5894314 346 SPHERICAL TRIGONOMETRY. 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. Ans. 4 h 56 m . 2. Given the latitude of a place = 40 36' 23".9, the hour angle of a star=4640'4".5, and its declination = 23 4' 24".3 ; to find its azimuth and altitude. Ans. 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' 1ST., when the hour angle of the star is 3 h 15 m 20 s W., and the declination is 12 42' K Ans. 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' K, respectively, and their longi- tudes 740'24"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 3956 miles. NOTE. This is evidently a case of (4) where two sides and the included angle are given, to find the third side. Ans. 3305 miles. 5. The latitudes of Paris and Pekin are 48 50' 14" N. and 39 54' 13" K, and their difference of longitude is 114 7' 30"; find the distance between them in degrees. 'Ans. 73 56' 40". GEODESY. 226. The Chordal Triangle. Given tivo sides and the included angle of a spherical triangle; to find the correspond- ing angle of the chordal triangle. GEODESY. 347 The chordal triangle is the triangle formed by the chords of the sides of a spherical triangle. Let ABC be a spherical triangle, the centre of the sphere, A'BC the colunar triangle, and M, N the middle points of the arcs A'B, A'C. Then the chord AB is parallel to the radius OM, since they are both perpendicular to the chord A'B. Simi- larly, AC is parallel to ON". In the spherical triangle A'MN", we have cos MN = cos A'N cos A'M + sin A'X sin A'M cos A' (Art. 191) Denote the angle BAG of the chordal triangle by Aj. Then arc WN or angle MON = A 1? A f N = !(-&), A'M = J(ir-c), and A' = A. .-. cos A! = sin -J- b sin c-f- cos % b cos -J-c cos A with similar values for cos B! and cos C^ (1) Cor. 1. If the sides b and c are small compared with the radius of the sphere, A l will not differ much from A. Let Aj = A 0; then cos AJ = cos A -f- sin A, nearly. But sin-|-& sin-J-c = sin 2 J(6 + c) sin 2 1(6 c), and cos ^6 cos c = cos 2 \(b + c) sin 2 J Substituting in (1) and reducing, we get 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 the circular measure by the circular measure of one second, or, approxi- mately, by the sine of one second. 348 SPHERICAL TRIGONOMETRY. Cor. 2. The angles of the ch or dal triangle are, respectively, equal to the arcs joining the middle points of the sides of the colunar triangles. 227. Legendre's Theorem. If the sides of a spherical triangle be small compared ivith the radius 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 r neglecting powers of above the fourth, a be cos cos - cos - cosA = r T c (Art. 191) sin -sin - r r te / ^ J*r/ !\ ( Art - 156 > 7 2 bc\ 2r* 24 r 4 A Or 2 q __ c _ c 26c 246CT 8 6? ' 2 ca 2 be * The term ? is the circular measure of the angle which the arc a subtends at the centre of the sphere; and similarly for _ and . GEODESY. 349 Now if A', B' ? C' denote the angles of the plane triangle whose sides are a, b, c, respectively, we have 2 be (Art. 96) A , ----- , and sin 2 A' = - (Art. 100) 4crc~ Therefore (1) becomes co S A = co S A'- 6c ^ A ' .... (2) Let A = A' + 0, where is a very small quantity ; then cos A cos A' sin A', nearly. where A denotes the area of the plane triangle whose sides are a, 6, c. ^ We have therefore A = A' -\ -- - Similarly B = B' + A C = C' + A 3r 2 Sr 2 ... A + B + C - A' - B' - C' = -; r 2 or A + B + C TT = ~ = spherical excess (Art. 219) .-. A-A'-B-B'=C-C'-:^ 2 = i spherical excess. 3 1 Cor. 1. If the sides of a spherical triangle be very small compared with the radius of the sphere, the area of the spherical triangle is approximately equal to 350 SPHERICAL TRIGONOMETRY. For, tanJE= A /tan tan^Uan tan (Art. 220) \ 2r 2r 2r 2r and far, S __ S I -1 _i I . for, " *-" _ 1~~"'| 1 i V"~~" V V I PrP lic *' 11 ~ ~ I -L ~T~ . I tctll - _ I J- T t) U CUO. (Art. 156) (Arts. 101 and 156) A/t -L r6- + & 2 + c 2 V_ A _L ~ ~ ( 12 r 2 y 4r 2i 24?- That is : the area of the spherical triangle exceeds the area the plane triangle by - "*" /" C parf of the latter. Cor. 2. If we omit terms of the second degree in -, we have Hence, if the sides of a spherical triangle be very small compared with 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 area of a spherical triangle on the Earth's surface being known, to establish a formula for com- puting the spherical excess in seconds. Let A be the area of the triangle in square feet, and n the number of seconds in the spherical excess. Then we have 180 x 60 x 60 nr 2 ROY'S RULE. 351 . . . (Art. 219) _ (1} 180 x 60 x 60 206265 " Now, the length of a degree on the Earth's surface is found by actual measurement to be 365155 feet. ... -^=365155. .. r== 180x365155. 180 . 7T 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 plane 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", 701'48", 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 apparent spherical excess is A + B + C - 180 = - 1". The area of the triangle is calculated from the expression a2sinBsinC (Art. 101) 2 sin A and by Roy's Rule the computed spherical excess is found to be .23". Now since the computed spherical excess may be supposed to be the real spherical excess, the sum of the observed angles ought to have been 180 -f .23". Hence it appears that the sum of the errors of the obser- vations is .23" (-!") = 1".23, which the observer must 352 SPHERICAL TRIGONOMETRY. add to the three 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. 229. Reduction of an Angle to the Horizon. Given the angles of elevation or depression of two objects, which are at a small angular distance from the horizon, and the angle which the objects subtend, to fyid the horizontal angle between them. Let a, b be the two objects, the angular distance between which is measured by an observer at ^ O ; let OZ be the direction at right angles to the observer's horizon. De- scribe a sphere round O 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, AB = 6 -f x, Aa = h, B6 = It. Then in the triangle aZb we have cos ab cos aZ cos bZ cos AB = cos aZb = or cos (6 H- x) = sin aZ sin 6Z cos sin h sin k cos h cos k This gives the exact value of AB ; by approximation we obtain, where x is essentially small, . n cos 6 hk .-. x sin = hk (h 2 + A; 2 ) cos 0, nearly. 2 hk - (h 2 + k 2 ) /"cos 2 1 - sin 2 "' X= 2sin0 = i CO + &) 8 tan i - (ft - &) 2 cot i 0]. SMALL VARIATIONS IN PARTS OF TRIANGLES. 353 EXAMPLES. 1. Prove that the angles subtended by the sides of a spherical triangle at the pole of its circumcircle are respec- tively double the corresponding angles of its chordal tri- angle. 2. If AU B 1? Ci ; A 2 , B 2 , C 2 ; A 3 , B 3 , C 3 ; be the angles of the chordal triangles of the colunars, prove that cos A^cos^asinS, cos B^ sin &sin(S C), cos C^ sin 2 csin(S B), cosA 2 =:sin2 asin(S C), cos B 2 = cos -I &sinS, cosC 2 = sin2Csin(S A), j^sin^ asin(S B), cosB 3 =sin.l 6sin(S A), cosC 3 =c 3. Prove Legendre's Theorem from either of the formulae for sin |- A, cos \ A, tan J A, respectively, in terms of the sides. 4. If C = A + B, prove cos C = tan a tan -J- b. 230. Small Variations in the Parts 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 tivo parts. Suppose C and c to remain constant. (1) Eequired 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 c?A; then we require the ratio of da and dA when both are extremely small. Thus H^ sin (A + dA)sinc = sinC sin (a 354 SPHERICAL TRIGONOMETRY. or (sin A cos dA -f cos A sin dA) sin c = sin C (sin a cos da + cos a sin da) (2) Because the arcs dA and da are extremely small, their sines are equal to the arcs themselves and their cosines equal 1 : therefore (2) may be written sin A sin c + cos A sin cdA = sin C sin a -f sin C cos ada (3) Subtracting (1) from (3), we have cos A sin cdA = sin C cos ada. da _ cos A sin c _ tan a dA sin C cos a tan A (2) Required the relation between the small variations of the other sides (a, b). We have cos c = cs a cos b -f sin a sin b cos C (1) .. cosc = cos(a-j-da)cos(6-f- db) -|-siii(a-|-da)sin(&-4-d6)cosC, or = (cos a sin ada) (cos b sin bdb) 4- (sin a + cos ada) (sin 6 -f cos bdb) cos C . (2) Subtracting (2) from (1) and neglecting the product da db, we have = (sin a cos b cos a sin 6 cos C) da + (cos a sin fr sin a cos 6 cos C) a7>, .v _(cot6sina cosacosC) , (cot a sin 6 cos 6 cos C) sin a sin 6 ^ _ cot B sin C da cot A sin C db (\t 1931 sin a sin 6 da _ cos A db cos B (3) Required the relation between the small variations of the other angles (A, B). SMALL VARIATIONS IN PARTS OF TRIANGLES. 355 By means of the polar triangle, we may deduce from the result just found, that dA _ cos a dB cos b (4) Required the relation between the small variations of a side and the adjacent angle (b, A). We have cot c sin b = cot C sin A + cos b cos A ... (Art. 193) Giving to b and A very small increments, and subtracting, as before, we get cot c cos bdb = cot C cos AdA sin b cos Adb cos b sin Ad A. (cote cos b + sin 6 cos A)db = (cot C cos A cos b sin A) dA. db = _ ^^ da . . (Arts. 191 and 192) sin c sin C db _ _ cos B sin b _ sin b cot B dA cos a sin B cos a EXAMPLES. 1. If A and c are constant, prove the following relations between the small variations of any two parts of the other elements : da _ tan a t db _ sin a dC ~ ~ tanC ' dB ~~ sinC* db _ _ tan a m dC _ _ dC ~ ~ sin C ' dB~ 2. If B and C remain constant, prove the following : = ten6 * = sin B sinC. dc tan c da * sina dc dc sin c cos 6 156 SPHERICAL TRIG ONOMETR Y. POLYEDRONS. 231. To find the Inclination of Two Adjacent Faces of a Regular Polyedron. 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 AB. .-. 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 DE, it is perpendicular to the plane 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 of sides in each face, and n the number of plane angles in each solid angle. Then - 2m m and Zcae = |Z of the planes OAC and OAD. In the right triangle cae we have cos cae cos ce sin ace. But cos ce = cos COE = cos ( ]= sin--- 2 2 cos- = sin -sin n m sin - = cos - cosec n m VOLUME OF A PARALLELOPIPED. 357 Cor. 1. If r be 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 = CE tan CEO = AE cot ACE tan CEO = 2 COt m tan 2' Cor. 2. J/*R be the circumradius of the polyedron, then R = - tan- tan-- 2 n 2 For, r = A cos aoc = R cot eca cot eac = R cot cot - m n .-. R = -tan^tan-. 2 n 2 Cor. 3. The surface of a regular polyedron, F being the number of faces, =t cot 4 m For, the area of one face = m cot .. etc. 4 m Cor. 4. TVie volume of a regular polyedron 12 m For, the volume of the pyramid which has one face of the polyedron for base and for vertex r ma 2 , TT cot - .-. etc. 34 m 232. Volume of a Parallelepiped. To find the volume of a parallelepiped in terms of its edges and their inclinations to one another. 358 SPHERICAL TRIGONOMETRY. Let the edges be OA = a, OB = b, OC = c, and let the inclinations be BOC = a, COA = ft AOB = y. Draw CH perpendicular to the face AOBE. Describe a sphere o< round as centre, meeting OA, OB, OC, OE, in a, b, c, e, respectively. The volume of the parallelo- E piped is equal to the area of the base OAEB multiplied by the altitude CH ; that is, volume = ab sin y CH = dbc sin y sin ce where ce is the perpendicular arc from c on ab. .'. volume = dbc sin y sin ac sin bac . . . (Art. 186) 9m = abc sin y sin ft (Art. 195) H sin ft sin y = abc Vl cos 2 cos 2 (3 cos 2 y -f 2 cos a cos ft cos y. (7or. 1. The surface of a parallelepiped = 2 (6c sin -f- ca sin ft -j- aft sin y). Cor. 2. T/ie volume of a tetraedron = i abc Vl cos 2 a cos 2 /? cos 2 y + 2 cos a cos /? cos y. For, a tetraedron is one-sixth of a parallelepiped which has the same altitude and its base double that of the tetraedron. 233. Diagonal of a Parallelepiped. To find the diagonal of a parallelepiped in terms of its edges, and their mutual inclinations. Let OD (figure of Art. 232) be a parallelepiped, whose edges OA = a, OB = 6, OC = c, and their inclinations BOC = , COA = ft, AOB = y ; let OD be the diagonal required, TABLE OF FORMULA. 359 and OE the diagonal of the face OAB. Then the triangle OED gives OD 2 = OE 2 + ED 2 + 2 OE - ED cos COE = a 2 + tf + 2 ab cos y + c 2 + 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 must be 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 +a cos /?, which in (1) gives OD 2 = a 2 + b- + c 2 -f 2bc cos + 2ca cos p + 2ab cos y . (2) 234. Table of Formulae in Spherical Trigonometry. For the convenience of the student, many of the preceding formulae are summed up in the following table : 1. cos c = cos a cos b ........ (Art. 185) 2. sin b = sin B sin c. 3. sin a = sin A sine. 4. cos C = cos A cos B ....... (Art. 189) 5. sin B = sin 6 sin C. 6. sin A = sin a sin C. r _ 1 in^ = sh L 6 =i inc (Art. 190) sin A sin B sin C 8. cos a = os b cose -f sin b sine cos A . . (Art. 191) 9. cos b = cos c cos a + sin c sin a cos B. 10. cos c = cos a cos b + sin a sin 6 cos C. 11. cos A = cos B cos C + sin B sin C cos a (Art. 192) 12. cos B = cos C cos A -f sin C sin A cos b. * From the nature of projections (Plane and Solid Geom., Art. 326). 360 SPHERICAL TRIGONOMETRY. 13. cos C = cos A cos B -f sin A sin B cos c. 14. cot a sin b = cot A sin C -f cos C cos b . . (Art. 193) 15. cot a sin c = cot A sin B -f cos B cos c. 16. cot b sin a =cot B sin C -f cos C cos a. 17. cot b sin c = cot B sin A -f cos A cos c. 18. cot c sin a = cot C sin B -f- cos B cos a. 19. cot c sin b = cot C sin A -f cos A cos b. 20. sin a cos B = cos b sin c sin b cos c cos A (Art. 194) 21. sin a cos C = sin b cos c cos 6 sin c cos A. 22. sin b cos A = cos a sin c sin a cos c cos B. 23. sin b cos C = sin a cos c cos a sin c cos B. 24. sin c cos A = cos a sin b sin a cos b cos C. 25. sin c cos B = sin a cos 6 cos a sin b cos C. 26. sin^A=J si "( s - 6 ) sin ( s - c ) . . . (Art. 195) \ sin b sin c sin b sin c 27. sin b sin c 28. tan 1 A sin s sin (s a) 29 sin A ~ c ) sin 6 sin c sin 6 sin c where n = Vsin s sin (s a) sin (s b) sin (s c). 30. sin4a=J- cosScos ( S - A J (Art. 196) \ sin B sin O 31. cos i a = Jco Li 8_BlcoB{S_CJ \ sin B sin O TABLE OF FORMULAE. 361 00 cos (S A) 32. - -C) QQ . _2V cosScos(S A) cos(S B)cos(S C) oo. Sill Cf ; ~ sin B sin C 34. tan KA + B) = cos K ft - &) cot ^Q . . (Art. 197) cos |(a + 6) 35. ta sin 37 - tan *( a - 6 ) = 38. sin(A + B)cos!c = cosi(a- & ) cos iC (Art. 198) 39. sin J(A - B) sin Jc = sin ^(a - 6) cos JC. 40. cos |( A + B) cos J c = cos (a + 6) sin C. 41. cos J(A - B) sin c = sin J(a + &) sin JC. = v p \ 42. tanr ~ a sin g ~ 6 sn sns sins ...... (Art. 215) 43. tanK = [sin(s a) +sin(s b) -fsin(s c) sins] (Art. 217) 44. K = areaof A = -^l-Trr 2 ...... (Art, 219) 180 45. sin | E = - ... (Art. 220) 2 cos % a cos f o cos c 46. tan \ E = Vtanstan(s a) tan (s b) tan J(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. Ans. 5 h 9 m 13V 2. Find the time of sunset at Cincinnati, lat. 39 6' 1ST., when the sun's declination is 15 56' S. Ans. 5 h 6 in . 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' N. Ans. 4 h 32 m 16 8 .4. 4. Find the time of sunrise at Boston, lat. 42 21' N., when the sun's declination is 8 47' S. Ans. 6 h 14 m . 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. 15 h 5 m 50 1 . 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' S., when it sets to an observer at Philadelphia, lat. 39 57' N. Ans. 4 h 23 m 5 9 .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 h 5 m 36 s east of the meridian. Ans. Azimuth = S. 27 18' 40" E. ; Altitude = 11 41' 37". EXAMPLES. 363 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 5 h 17 m 40 s east, and the declination is 62 33' N. Ans. Azimuth = N. 35 54' E. ; Altitude = 39 24'. 12. Find the hour angle (t) and declination (8) -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. t = 46 40' 4".53 ; 8 = 23 4' 24".33. 13. Find the distance between Regulus and Antares, the right ascensions being 10 h O m 29M1 and 16 h 20 m 20 8 .35, and the polar distances 77 18' 41".4 and 116 5' 55".5. Ans. 9955'44".9. 14. Find the distance between the sun and inoon when the right ascensions are 12 h 39 m 3 8 .22 and 6 h 55 m 32 s . 73, and the declinations 9 23' 16".7 S. and 22 50' 21".9 K Ans. 8952'55".5. 15. Find the shortest distance on the earth's surface, in miles, from New York, lat. 40 42' 44" K, 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 h l m 9 S .34, and its declination 12 37' 36".8 K ; to find its latitude and longi- tude, the obliquity of the ecliptic being 23 27' 19".45. Ans. Latitude = ; Longitude = 18. Given the obliquity of the ecliptic w, and the sun's longitude A ; to find his right ascension a and declination 8. Ans. tan a cos sin A. 364 SPHERICAL TRIGONOMETRY. 19. Given the obliquity of the ecliptic 23 27' 18".5, and the sun's longitude 59 40 f 1".6; to find his right ascension (a), and declination (8). Ans. a = 3 h 49 m 52 8 .62 ; 8 = 20 5' 33".9 K 20. Given the sun's declination 16 0' 56". 4 K, and the obliquity of the ecliptic 2327'18".2; to find his right ascension (), and longitude (A). Ans. a = 9 h 14 m 19 8 .2 ; A = 136 7' 6".5. 21. Given the sun's right ascension 14 h 8 m 19 S .06, and the obliquity of the ecliptic 23 27' 17".8 ; to find his longitude (A), and declination (8). Ans. X = 214 20' 34". 7 ; 8 = 12 58' 34".4 S. 22. Given the sun's longitude 280 23' 52". 3, and his declination 23 2' 52".2 S. ; to find his right ascension (a). Ans. a = 18 h 45 m 14 s . 7. 23. In latitude 45 N., prove that the shadow at noon of a vertical object is three times as long when the sun's dec- lination is 15 S. as when it is 15 N. 24. Given the azimuth of the sun at setting, and also at 6 o'clock ; find the sun's declination, and the latitude. 25. If the sun's declination be 15 N., and length of day four hours, prove tan < = 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 2 8 = sin h sin h'. EXAMPLES. 365 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 \ 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 Trr 2 . 39. Two circles of angular radii, and /?, intersect orthogonally on a sphere of radius ?*; find in any manner the area common to the two. 40. If E be the spherical excess of a triangle, prove that 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 TEIGONOMETEY. right angles, the sides of the original triangle being regarded as the bases of the colimars. 42. Prove that cos 2 Ja sin 2 S + sin 2 1 6 sin 2 (S - C) + sin 2 -Jo sin 2 '(S - B) -f- 2 cos J a sinifr sin Jc sinS 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 \ b sin \ c sin ( S A) -f cos 1 b cos J c sin S = cos -Jo. 45. If A + B + C = 2 TT, prove that cos 2 -J a + cos 2 1 b + cos 2 J c = 1, and cos C = cot -| a cot J 6. 46. Solve the equations, sin b cos c sin Z + sin c cos 6 sin Y = sin a, sin c cos a sin X -f sin a cos c sin Z '= sin 6, sin a cos 6 sin Y + sin b cos a sin X = sin c, for sin X, sin Y, and sin Z. 47. If b 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 : tfB tanB da dC tanC' = sinBsinc; dA. da dB dA. sin A dB dA sinB cosC sin A dC dC cosB sinC 48. If A and c remain constant, prove the following : ^ = cos C. dB tanC db EXAMPLES. 367 49. If B and C remain constant, prove the following : ** db db sin b cos c 50. If A and a remain constant, prove the following : db _ tan b t dc _ tan c . dB ~ tan B ' dC ~ tan C ' db cosB d& 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 is described, and a great circle is described having its pole on the small circle ; show that the length o _ of their common chord is - V cos 2 0. 53. Given the base c of a triangle, and that tan |-a tan \ b = tan 2 ^- B, B being the bisector of the base, find a b in terms of c. 54. If C = A + 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 of its polar triangle, show that cos A cos A' = cos A + cos A f . 56. Show that cos a cos B cos b cos A _ cos C + cos c sin a sin b sin c 57. Prove cos A = cos a sin b ~ sin a cos b cos C , sine and cosA + co S B = ^!lL(i*l sine 368 SPHERICAL TRIGONOMETRY. 58. Prove Legendre's Theorem by means of the relations sin A _ sin B _ sin C sin a sin b sin c 59. Two places are situated on the same parallel of lati- tude ; find 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 2 A. Ans. 2 r [A cos sin" 1 (cos sin A) ] . 60. If the sides of a triangle be each 60, show that the circles described, each having a vertex for pole, 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 icosa'edron, the edge being one inch. Ans. (1) 117.85 cu. in., 70 31'43".4; (2) .4714 cu. 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 the 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. 14 DAY USE RETURN TO DESK FROM WHICH BORROWED LOAN DEPT. This book is due on the last date stamped below, or on the date to which renewed. Renewals only: _,. Tel. No. 642-3405 1 Renewals may be made 4 days prior to date due. Renewed jbo^ks^are^ subject to immediate recall. i j^ 1 -^ & & ran 7 7 J6 W3TDLD .-, KG'/ .. 160 Y* LD General Library 911383T THE UNIVERSITY OF CALIFORNIA LIBRARY