^1 %^ IN MEMORIAM FLORIAN CAJORl J^'atc (^^>^i^r^ PLANE TRIGONOMETRY BY JAMES M. TAYLOR, A.M., LL.D. Professor of Mathematics, Colgate University BOSTON, U.S.A. GINN & COMPANY, PUBLISHERS C^e 3ttl)cnaciim prefiSfi 1904 Copyright, 1904 By JAMES M. TAYLOR ALL BIGHTS RESERVED V T-5 PREFACE This book is designed to meet the needs of beginners who wish to master the fundamental principles of Trigonometry. The author's aim has been to prepare a text-book which shall be clear and practical, yet thoroughly scientific. The proofs of formulas are simple but rigorous. The use of directed lines is consistent; the directions of such lines in the figures are usually indicated by arrowheads, and these lines are always read from origin to end. Both trigonometric ratios and trigonometric lines are employed, but at first the ratios are used exclusively until they have become fixed in mind and familiar. It is proved that the ratios are the measures of the lines, and that, therefore, any relation which holds true for the one holds true for the other also. The distinction between identities and equations is empha- sized in definition, treatment, and notation. The solution of trigonometric equations is scientific and general. The trigo- nometric ratios are defined in pairs as reciprocals of each other both to aid the memory and to emphasize one of the most important of their fundamental relations. In the reduc- tion of the functions of any angle to those of an acute angle, the theorems concerning the functions of — yl and 90° -f A are made fundamental. The addition formulas are proved for positive or negative angles of any quadrant and from them are deduced the other formulas concerning the functions of two or more angles. When two or more figures are used in a proof, the same phraseology always applies to each figure. M«n«i>4A iv • PREFACE In the first chapter will be found a table of natural functions at intervals of 5°, which the student is to verify to two places by construction and measurement, and many interesting prob- lems have been introduced in order that the subject may be made more attractive and profitable to the beginner. The angles and numbers in these problems are so chosen that this table contains all that is needed for their solution. In Chapter VIII complex number is expressed as an arith- metic multiple of a quality unit in its trigonometric type form, and the fundamental properties of such number are demonstrated. The proof of De Moivre's theorem is simple and general, and its meaning and use are fully illustrated. Before beginning the study of complex number one should gain a clear idea of the sum, or resultant, of two directed forces, and by repeated experiment make this idea familiar. It is believed that the order of the text is the best for begin- ners ; but, with the exception of a few articles, Chapter I may be omitted by those who are prepared to take up at once the general treatment in Chapter II. Too much stress cannot be laid on careful and accurate construction and measurement in the first chapter. Chapters VII and VIII and the latter parts of Chapter VI may be omitted by those who wish a shorter course. In preparing this book the author has consulted the best authorities, both American and European. Many of the exam- ples have been taken from these sources. The author takes this opportunity to express to many teachers and other friends his appreciation of their valuable suggestions and their en- couragement in the preparation of this work. James M. Taylor. Colgate University, December, 1903. CONTENTS CHAPTER I TRIGONOMETRIC RATIOS OF ACUTE ANGLES Section Page 1-3. Trigonometric ratios defined, one-valued 1, 2 4. Construction of angles having given ratios 3, 4 5, 6. Approximate values, and changes from 0° to 90° . . . 5-7 7-9. Trigonometric ratios of angles of triangle and of co-angles 8, 9 10,11. Trigonometric ratios of 45°, 30°, 60° 10,11 12,13. Solving right triangles . 11-13 14, 15. Definitions and problems 14-21 CHAPTER II TRIGONOMETRIC RATIOS OF POSITIVE AND NEGATIVE ANGLES OF ANY SIZE 16. Positive and negative angles of any size 22, 23 17-19. Coterminal angles. Quadrants 24,25 20,21. Trigonometric ratios of any angle 26,27 22,23. Laws of quality. Sin-^c, cos-^c, •• • 28-30 24, 25. Fundamental identities. Proof of identities 31-35 26. Changes in ratios as angle changes from 0° to 360° . . . 36-38 27. Trigonometric ratios of 0°, 90°, 180°, 270° ...... 39 28, 29. Trigonometric ratios of - ^ and 90° + ^ 40-43 30, 31. Trigonometric ratios of 90° - A, 180° - ^, n 90° ± ^ . . 44, 45 32, 33. Trigonometric lines representing the ratios 46-51 V vi CONTENTS CHAPTER III TRIGONOMETRIC RATIOS OF TWO ANGLES Section Page 34-37. Trigonometric ratios of the sum and difference of two angles 52-56 38, 39. Trigonometric ratios of twice and half an angle .... 57-59 40. Sums and differences of trigonometric ratios 60-63 CHAPTER IV SOLUTION OF RIGHT TRIANGLES WITH LOGARITHMS 41-48. Properties and computation of logarithms 64-68 49. Right-angled triangles ". . 69-71 50,51. Isosceles triangles and regular polygons 72,73 CHAPTER V SOLUTION OF TRIANGLES IN GENERAL 52. Law of sines and law of cosines 74, 75 53-55. The four cases solved without logarithms 76, 77 56. Law of tangents 78, 79 57-59. First three cases solved with logarithms 80-87 60. Trigonometric ratios of half angles and Case (iv) solved with logarithms 88-91 61. Areas of triangles 92 62-64. Circumscribed, inscribed, escribed circles 93, 94 CONTENTS vii CHAPTER VI RADIAN MEASURE, GENERAL VALUES, TRIGONOMETRIC EQUATIONS, INVERSE FUNCTIONS Section Paok 65-67. Radian measure of angles 95-97 68. Principal values 98 69-71. Angles having the same trigonometric ratio .... 99, 100 72, 73. Solution of trigonometric equations 101-104 74. Trigonometric functions 105 75, 76. Inverse trigonometric functions 106-109 CHAPTER VII PERIODS, GRAPHS, IMPORTANT LIMITS, COMPUTATION OF TABLE, HYPERBOLIC FUNCTIONS 77. Periods of the trigonometric functions 110 78-81. Graphs of the trigonometric functions 111-114 82. Limit of the ratio of sin d or tan ^ to 115 83, 84. Computation of trigonometric functions 116, 117 85,86. Hyperbolic functions 118,119 CHAPTER VIII COMPLEX NUMBERS. DE MOIVRE'S THEOREM 87. Quality units ±1, ± V^ 120 88, 89. Directed lines and forces 121, 122 90. Complex numbers 123,124 91,92. General quality unit. Products of quality. units . . 125,126 93. De Moivre's theorem. Quotients of quality units . . 127, 128 viii CONTENTS Section Page 94, 95. Products and quotients of complex numbers .... 128, 129 96, 97. The gth roots of cos + i sin and (cos + i sin 0) r . 130, 131 98. Exponential form for cos + i sin 132, 133 CHAPTER IX ' MISCELLANEOUS EXAMPLES Trigonometric identities 134-136 Trigonometric equations and systems 137-141 Problems involving triangles 142-146 Problems involving areas and regular polygons 146-147 Formulas 148-151 ANSWERS 153-171 PLA^E TRIGOI^OMETRY CHAPTER I TRIGONOMETRIC RATIOS OF ACUTE ANGLES 1. Let A denote the number of degrees in the acute angle XOB] then A is the numerical measure, or measure, of this angle, and we can write Z XOB = A. From any point in either side of the ' angle XOB, as P, draw PM perpendicular to the other side. Observe that is the vertex of the angle, and M is the foot of the perpendicular drawn from P. This lettering should be fixed in '' no i mind so that in the following defi- nitions the lines MP, OM, and OP shall always mean the same lines as in fig. 1. The six simple ratios (three ratios and their reciprocals) which can be formed with the three lines MP, OM, OP are called the trigonometric ratios of the angle XOB, or A. These ratios are named as follows : The ratio MP /OP is the sine of A ; and its recijirocal OP /MP is the cosecant of A. The ratio OM/OP is the cosine of A ; and its reciprocal OP/OM is the secant of A. The ratio MP/OM is the tangent of A ; and its reciprocal OHL/IS.'P is the cotangent of A, I PLANE TRIGONOMETRY For brevity the sine of A is written sin A ; the cosine of Aj cos A; the tangent of A, tan A; the cotangent of A, cot A; the secant of A, sec A ; and the cosecant of A, esc A. Observe that sin ^ is a compound symbol which, taken as a whole, denotes a number. The same is true of cos A, tan A, etc. Ex. 1. What four trigonometric ratios of the angle A involve the line MP? the line OM? the line OP? Ex. 2. What trigonometric ratios are reciprocals of each other ? Ex. 3. Which is the greater, tan A or sec A? cot A or esc A ? Why ? Ex. 4. Can sin A or cos A exceed 1 ? Why ? 2. An]/ trigonometric ratio of a given angle has 07ily one value. Let XOB be any acute angle. Draw PM_L OX, P'M' ± OX, P"M" ± OB ; then, by § 1, ^ . MP M'P' or M"P"/OP". (1) From the similarity of the A [^//'^ OMP, OM'P', OM"P" it follows that the three ratios in (1) (or their reciprocals) are all equal; hence sin XOB (or esc XOB) has but one value. Also, from the similarity of these A, each of the other trigo- nometric ratios of Z XOB has only one value. 3. Two acute angles are equal if any trigonometric ratio of the one is equal to the same ratio of the other. Take O^P^ in fig. 3 equal to OP in fig. 2, and draw Pi Ml ± OjXi- We are to prove that if sin ZiOiPi = sin XOP, i.e. if M,P^/OiPi = MP /OP, (1) then ZXiOiPi = ZXOP, (2) TRIGONOMETRIC RATIOS OF ACUTE ANGLES By construction, O^Px = OP. Hence, from (1), M^P^ = MP. Therefore, by Geometry, the right tri- angles OiPiMi and 0PM are equal in all their parts. O^ Hence Z X^O^P^ = Z XOP. In like manner the student should prove the equality of two acute angles when any other trigonometric ratio of the one is equal to the same ratio of the other. 4. Having given the value of any trigonometric ratio of an oAiute angle ^ to construct the angle and obtain the values of its other trigonometric ratios. This problem will be illustrated by particular examples. Ex. 1. If ^ is an acute angle and cos^ =3/5, construct A and find the values of its other trigonometric ratios. Here cos A = DM/ OP = S/b. Hence, if OP = 5 units, DM = 3 units. Let 0, in fig. 4, be the vertex of the angle A, and OX one of its sides. On OX, to some scale, lay off OM equal to 3 units, and at M draw MS ± OX. With O as a center and with a radius equal to 5 units, draw an arc cutting MS in some point as P. Draw OPB. Then ZXOB = A. For cos XOB = 3 / 5 = cos ^. Hence, by § 3, Z XOB = A. Again, MP = V52 - 32 units = 4 units. Hence sin ^=4/5, esc ^ = 5/4 cos^ = 3/5, sec ^=5/3; tan^ = 4/3, cot ^ = 3/4. Observe that 6 is the numerical measure of OP, 4 of MP, and 3 of OM. Fig. 4 51 PLANE TRIGONOMETRY Ex. 2. If A is an acute angle and sin ^ = 2/3, construct A and find the values of its other trigonometric ratios. Here sin A = MP/ OP = 2/3. Hence, if OP = 3 units, MP = 2 units. At M, in fig. 5, draw MS ± OX and lay off MP equal to 2 units. With P as a center and 3 units as a radius, strike an arc cutting OX in some point, as 0. Draw OPB. Then ZXOB = A. For sinXO^ = 2/3 = sin^. Hence, by § 3, Z XOB = A. M X Fig. 5 Again, Hence §3 OM = V 32 - 22 units = V5 units, sin J. = 2/3, CSC ^=3/2; cos^=V5/3, sec^ = 3/V5; tan^=2/V5, cot^=V5/2. If the vertex of the angle A were required to be at some fixed point on OX, as 0, we would draw OK ± OX, lay off OQ equal to 2 units, through Q draw QN parallel to OX, and with as a center and 3 units as a radius, strike an arc cutting QN at P ; then draw OPB as before. By using a protractor, i. e. a graduated semicircle, we find that Z XOB, or A, is an angle of about 42°. Note. The student should form the habit of gaining a clear idea of the size of an angle from the value of any one of its trigonometric ratios. EXERCISE I Construct the acute angle A, obtain the values of all its trigonometric ratios, and find its size in degrees, when : 1. sin^=2/5. 6. tan^ = 4/3. 2. sin^ =4/5. 7. cot^ =5/2. 3. cos^=3/4. 8. cot^=l/3. 4. cos^=l/3. 9. sec^ =5/3. 5. tan^=l/4. 10. sec^=4/3. 11. csc^ =5/2. 12. csc^ =3/2. 13. tan^ = 4, or 4/1. 14. cot J. = 7, or 7/1. 15. tan A = 9, TRIGONOMETRIC RATIOS OF ACUTE ANGLES 16. Express each of the trigonometric ratios of an acute angle A in terms of its sine, writing (sin^)^ in the form sin^^. Infig. 1, let OP=l. Then Whence and Hence MP / OP = MP /I, i.e. sin A is the measure of MP. MP = sin A, OM = ^OP^ - MP^ = Vl - sin2^. cos A = OM/OP .-, sec A = tan A = .-. cot A = Vl — sin2^/sin A. CSC A = OP /MP = 1/sin A. Vl - sin2^ ; 1/Vl -sin2^. MP/ OM = sin A/Vl - sin'M §1 17. Express each of the trigonometric ratios of an acute angle in terms of its cosine. In fig. 1, let 0P=1. Then OM/ OP = OM/l, i.e. cos A is the measure of OM. Whence OM = cos A, and MP = V^ - OM^ = Vl - cos^^, etc. 18. Express each of the trigonometric ratios of an acute angle in terms of its tangent. Infig. 1, let 0M= 1. Then MP/ OM = MP /I, i.e. tan A is the measure of MP. Whence MP = tan A, and OP = Vo^^ + I^ = Vl + tan^^, etc. 5. To find approximately by measurement the values of the trigonometric ratios of any given angle. Ex. 1. Find by construction and measurement the values of the six trigo- nometric ratios of 40°. With a protractor lay off Z XOB = 40°. Take OP any convenient length, say 10 units (the longer the better), and draw PM ± OX. By careful measurement we find that MP = 6.4 units, OM = 1.1 units, approximately. Fig. 6 6 PLANE trigonomj:try Hence, as approximate values, we have sin 40° = 6.4/10 = 0.64, esc 40° = 10/6.4 = 1.56; cos40° = 7.7/10 = 0.77, sec 40° = 10/7.7 = 1.3; tan40° = 6.4/7.7 = 0.83, cot40° = 7.7/6.4 = 1.2. Observe that instead of taking OP equal to 10 units, we could take OM equal to 10 units. Ex. 2. By construction and measurement find the approximate values of the sine, cosine, and tangent of 5°, 10°, 15°, 20°, 25°, 35°, and compare the results obtained with their values given in the table below. On a scale of 20 to an inch take OM equal to 100 units and draw MP ± OM. With O as their common vertex and OM as their common lower side, draw the angles 5°, 10°, etc. Observe that the cosecant, the secant, and the cotangent of any angle can be obtained by taking the reciprocal of the sine, the cosine, and the tangent respectively of the same angle. Angle sin CSC cos sec tan cot 1° 0.0175 57.2987 0.9998 1.0002 0.0175 57.2900 5° 0.0872 11.4737 0.9962 1.0038 0.0875 11.4301 10° 0.1736 5.7588 0.9848 1.0154 0.1763 5.6713 15° 0.2588 3.8637 0.9659 1.0353 0.2679 3.7321 20° 0.3420 2.9238 0.9397 1.0642 0.3640 2.7475 25° 0.4226 2.3662 0.9063 1.1034 0.4663 2.1445 30° 0.5000 2.0000 0.8660 1.1547 0.5774 1.7321 35° 0.5736 1.7434 0.8192 1.2208 0.7002 1.4281 40° 0.6428 1.5557 0.7660 1.3054 0.8391 1.1918 45° 0.7071 1.4142 0.7071 1.4142 1.0000 1.0000 50° 0.7660 1.3054 0.6428 1.5557 1.1918 0.8391 55° 0.8192 1.2208 0.5736 1.7434 1.4281 0.7002 60° 0.8660 1.1547 0.5000 2.0000 1.7321 0.5774 65° 0.9063 1.1034 0.4226 2.3662 2.1445 0.4663 70° 0.9397 1.0642 0.3420 2.9238 2.7475 0.3640 75° 0.9659 1.0353 0.2588 3.8637 3.7321 0.2679 80° 0.9848 1.0154 0.1736 5.7588 5.6713 0.1763 85° 0.9962 1.0038 0.0872 11.4737 11.4301 0.0876 89° 0.9998 1.0002 0.0175 57.2987 57.2900 0.0175 In the above table observe how each trigonometric ratio of A changes as A increases from 1° to 89°. TRIGONOMETRIC RATIOS OF ACUTE ANGLES 6. Changes of the trigonometric ratios of the angle A as A increases from 0° to 90°. In fig. 7 conceive the line OP to revolve from OX to OB, i.e. suppose Z XOP, or A^ to increase from 0° to 90°. When OP coincides with OX, MP = 0, and OM = OP. When OP coincides with 05, MP = OP, and OM = 0. O M Hence sin 0° = 0, sin 90° = 1, cos 0° = 1, and cos 90° = 0. Therefore when A increases from 0° to 90° sin A increases from to 1 and cos A decreases from 1 to 0. Example. What is the vahie of tan 0° ? sec 0° ? cot 90° ? esc 90° ? When OP is very near to OB, OM is very small ; hence MP / OM is very large, and the nearer OP is to OB, the larger is MP / OM, or tan .1. So that as A approaches 90°, tan ^ increases rapidly and can exceed any constant number how- ever great ; that is, tan A becomes an infinite (denoted by oo). Hence as A increases from 0° to 90°, tan A increases from to CO ; likewise sec A increases from 1 to oo. Similarly, when OP approaches nearer and nearer to OX, OM / MP, or cot A, and OP / MP, or esc A, become infinites. Hence as the angle A increases from 0° to 90°, sin A increases from to 1, cos A decreases from 1 to 0, tan A increases from to oo, cot A decreases from oo to 0, sec A increases from 1 to oo, CSC A decreases from oo to 1. PLANE TRIGONOMETRY 7. Trigonometric ratios of the acute angles of a right triangle. Let A and B be the measures of the acute angles of the right tri- angle ABC, a the measure of the ct side opposite the angle A, b that of the side opposite B, and c that of G the hypotenuse ; then, by § 1, we have Fig. 8 sin ^ = - = -r ^ , c hypotenuse CSC A = -^ a b side adiacent cos ^ = - = — 7^* J c hypotenuse sec ^ =75 b a side opposite side adjacent b cot A =-' a Similarly we have sin B = b/c, esc 5 = c/b', cos B = a/c, sec 5 = c/a-, t2in.B = b/a, cot B = a/b. 8. Complementary angles. Two angles are said to be com plementary when their sum is 90°. E.g., the complement of 35° is 90° — 35°, or 55° ; the complement of 70° is 90° - 70°, or 20° ; the complement of A is 90° - ^ ; and in any right triangle, as ABC fig. 8, the acute angles A and B are complementary angles. 9. Trigonometric ratios of complementary angles, li Z.CAB = A, then Z CBA = 90° - A. Hence, by § 7, we have sin (90° — A)=b/c = cos A, cos (90° — A) = a/c= sin A, tan (90° -A) = b/a = cot A, cot (90° - A) = a/b = tan A, sec (90° — A) = G/a = esc A, G8C{90° - A)=c/b = sec A. TRIGONOMETRIC RATIOS OF ACUTE ANGLES 9 If we call the cosine the co-ratio of the sine, the sine the co-ratio of the cosine, the cotangent the co-ratio of the tangent, the tangent the co-ratio of the co- tangent, etc., then the six identities above can be summed up as follows : Any trigonometric ratio of an acute angle is equal to the co-ratio of its complement. E.g., since 60° and 30° are comple- mentary angles, we have sin 60° = cos 30°, tan 60° = cot 30°, esc 60° = sec 30° Again, since 45° is the complement of itself, we have sin 45° = cos 45°, tan 45° = cot 45°, esc 45° = sec 45°. Ex. 1. From the upper half of the table in § 5 obtain the lower half. Since 50° -f 40° = 90°, sin 50° = cos 40° = 0. 7660 ; since 55° + 35° = 90°, sin 55° = cos 35° = 0.8192 ; since 60° -f- 30° = 90°, sin 60° = cos 30° = 0.8660 ; etc. Ex. 2. The angle A being acute, find the value of A in the equation sin^ = cos2^. (1) If in equation (1) we substitute for cos 2 A its * identical expression sin (90° — 2 A), by Algebra we obtain the equivalent equation (2). * Algebraic definitions. Two numeral expressions which denote the same number, or any two mathematical expressions which denote equal numbers for all values of their letters, are called identical expressions. E.g., the numeral expressions 4x3 and 8 + 4 are identical, so also are the literal expressions (a -\-b)(a - b) and a"^ - 62, or cos A and sin (90° - A). An equality is the statement that two mathematical expressions denote the same number. An equality whose members are identical expressions is called an identity. An identity is to he proved. An equality whose members are not identical expressions is called an equation. An equation is to be solved. The sign of identity, =, read "is identical with," is often used instead of the sign of equality = in writing an identity whose members involve one or more letters. E.g., to indicate that the equality sin^ = cos (90° — A) is an identity and not an equation we write sin A = cos (90° — A) . Since we know that any equality which involves only numerals must be an identity, the sign of identity is used only in writing literal identities. 10 PLANE TRIGONOMETRY From (2), by § 3, sin A A sin (90° - 2 A). 90°- 2 A, ..A (2) Equation (1) is a trigonometric equation, and the only value of the acute angle A which will satisfy it is 30°. EXERCISE n 1. By § 9, cos 30° equals what ? sin 60°? cot 35°? tan 15°? sec 85°? CSC 76° ? sin 73° 14' ? cos 65° 43' ? A being an acute angle, find its value in each of the following equations: 2. sec ^= CSC ^. 6. sec (75° + ^) = esc 2^. 3. tan ^ = cot 2 A. 7. cot {A + 50°) = tan 7 A. 4. sin 2 ^ = cos 3 A. 8. sin nA = cos niA. 5. tan ^ /2 = cot 2 A. 9. tan cA = cot (30° - A). 10. Trigonometric ratios of 45°. Let ABC be an isosceles right triangle in which AB = 2 units, A=B = 45°. AC = BC = V2 units, sin 45° = V2/2 = cos 45°, tan 45° = 1 = cot 45°, sec 45° = V2 = CSC 45°. Fig. 10 11. Trigonometric ratios of 30° and 60°. Let ABD be an equilateral tri- angle in which AB = 2 units. Draw BC±AD. Then AC = 1, BC = ^S, A = 60°, Z ABC = 30°. TRIGONOMETRIC RATIOS OF ACUTE ANGLES 11 By§l, sin 30° = 1/2, CSC 30° = 2 ; cos30° = V3/2, sec 30° = 2/ V3; tan30° = l/V3, cot 30° = V3. Also, by § 1, sin 60° = V3/2, CSC 60° = 2/ V3; cos 60° = 1/2, sec 60° = 2 ; tan 60° = V3, cot60° = l/V3, or V3/3. To aid the memory observe that sin 30°, sin 45°, and sin 60° are respectively equal to Vl* V2, and V3, divided by 2. It is easy to read off the trigonometric ratios of 30°, 45°, and 60°, vsrhen we keep in mind the figures 10 and 11. Example. By § 9 obtain the values of the trigonometric ratios of 60° from those of 30°, and those of 30° from those of 60°. 12. Approximate measurements and computations. The student should remember that in all actual measurements the results are only approximate. It is impossible to measure any quantity with absolute accuracy. The degree of accuracy sought will depend upon the importance of the results. The degree of accuracy secured will depend upon the instruments, methods, and care which are used. Likewise, in practical computations, a sum, difference, product, or quotient of two approximate values will not have a greater degree of accuracy than that of the least accurate of the two values. E.g., if one numerical measure is accurate to two figures and another to three figures, their sum, product, or quotient will not in general be accurate to more than two figures. If each of two numerical measures has three-figure accuracy, or if one has four-figure accuracy and the other only three-figure accuracy, their product or quotient will not in general have more than three-figure accuracy. The values tabulated in § 5 have only four-figure accuracy. 12 PLANE TRIGONOMETRY 13. Solving right triangles. Of the six parts (three sides and three angles) of a right triangle, one part (the right angle) is always known. If, of any right triangle, two other parts are given (one at least being a side), Geometry proves that the triangle is entirely determined, and shows how to construct it. Trigonometry shows how to compute the numerical values of the unknown parts of a triangle when the known parts are sufficient to determine it. This process is called solving the triangle. Hence, in solving right triangles, we must consider the two following cases : (i) Given one side and one acute angle. (ii) Given any two sides. Case (1). Ex. 1. In the right triangle ^^C, ^ = 35° and BC = 20 feet ; find the numerical values of the other parts. Construction. To some scale (as 30 ft. to an inch) construct as accurately as possible a right a triangle having the given parts A = 35° and Solution. B = 90° - A=90° -Sb°= 65°. Now, the ratio of either of the unknown sides Fig. 12 , ' . , • X • X • .• J^ to the known side a is a trigonometric ratio of 35°, and any trigonometric ratio of 35° can be obtained from the table in § 5. Thus, b/a = cot ^ = cot 35°. § 1 .-. 6/20 = cot 35° = 1.4281. by table .-. &=: 1.4281 X 20=1:28.562. Again, c/20 = esc 35° = 1.7434. § 1, table .•.c = 1.7434 X 20 = 34.868. If we regard 20 as exact, or at least accurate to four figures, the values of 6 and c are accurate to only four figures ; for cot 35° and esc 35° are accurate to only four figures (§12). SOLUTION OF TRIANGLES 13 Check. Measure AC and AB, and multiply the number of inches in each by the number of feet which an inch represents. We thus obtain h = 28.6, c = 34.9. As a numerical check we could use a2 = c2 - 62^ or cfi = (c + 6) (c - h). But for simplicity and to emphasize the importance of accurate construction, we shall, in this chapter^ use only the check by construc- tion and measurement. Fig. 13 Case (ii). Ex. 2. In the right triangle ABC, BC = 83.91 ft. and AC = 100 ft. ; find the other parts. r^ = 4o°, Given |" ~ ^'^•^^' to find \ B = 50°, ''='''■' [C=1S0.64. Construct the triangle ABC having the given parts. rtan^ = a/6. (1) Formulas J ^ = 90° - ^. (2) [ c/6 = sec^. (3) Computation. From (1), tan A = 83.91 / 100 = 0.8391 = tan 40°. by table Hence, by § 3, A= 40°. From (2), B = 90° - 40° = 50°. From (3), c = 100 • sec 40° = 100 X 1.3054 = 130.54. by table Check. By measurement A - 40°, B = 50°, AB = ISl ft. nearly. The solution above illustrates the five steps which, in the first solutions at least, should be kept separate and distinct. (i) Statement of the problem, (ii) Construction of the triangle, (iii) Writing the needed formulas. (iv) Making the computations. (v) Applying some check or test to answers, 14 PLANE TRIGONOMETRY EXERCISE in Solve the right triangle ABC^ when : 1. A = 25°, a = 30. 9. 6 = 93.97, c = 100. 2. B = 55°, & = 10. 10. a =17.1, c = 50. 3. A = 65°, c = 70. 11. B = 75°, c = 40. 4. B = 15°, 6 = 20. 12. A = 10°, 6 = 30. 5. B = 35°, a = 50. 13. A = 20°, c = 80. 6. B = 55°, c = 60. 14. B = 25°, a = 30. 7. a = 36.4, b = 100. 15. a = 30.21, c = 33^. 8. a = 23.315, 6 = 50. 16. a = 13.4, 6 = 50. 14. A vertical line at any point is the line determined by the plumb line at that point. A horizontal line (or plane) at any point is the line (or plane) which is perpendicular to the vertical line at that point. A horizontal angle is an angle whose sides are perpendicular to the vertical line at its vertex. A vertical angle is an angle whose plane contains the vertical line at its vertex. A vertical angle of which one side is horizontal is called an angle of elevation or an angle of depression, according as the second side is above or' below the horizontal side. Note. All vertical lines converge towards the center of the earth. But in the next two definitions any two vertical lines are regarded as parallel. This is approximately true for short distances and is always assumed as true for such distances unless very great accuracy is required. The horizontal distance between two points is the distance from one of the two points to a vertical line through the other. The vertical distance between two points is the distance from one of the two points to the horizontal plane through the other. PROBLEMS 15 Fig. 14 E.g.^ let MP be the vertical line at P and let the horizontal plane at A cut this vertical line in M ; then AM is called the horizontal distance, and MP is called the vertical distance, between the points A and P. Moreover Z MAP is the angle of elevation of P, as seen from A. Also, if PN is horizontal at P and in the plane AMP, Z NPA is the angle of depression of A, as seen from P. Now if we assume that the vertical lines at A and P are parallel, the lines AM and NP will be parallel also and the angles MAP and NPA will be equal. 15. Solving problems. The practical problems which follow will illustrate the utility of the trigonometric ratios of angles in computing heights, distances, angles, areas, etc. In solving problems it will be helpful to observe the fol- lowing general method of procedure. First step. Construct accurately to some convenient scale a drawing which will show the relations of the given angles and lines to those which are required. Second step. Draw any auxiliary lines which may be help- ful in the trigonometric solution. By examining the drawing, fix upon the simplest steps which are necessary to solve the problem. Third step. Write the needed for- mulas. Make the computations, and check the answers. Ex. 1. A man, standing on the bank of a river at P, wishes to find how far he is from a tree at T on the opposite bank. He locates a staff at S so that PS±PT. By measure- ment he finds that the horizontal distance PS = 250 ft., and that the horizontal angle P-Sr= 40<^. Find the distance PT. Fig. 15 16 PLANE TRIGONOMETRY By § 1, FT / PS = tan 40° = 0.8391. by table .: PT= 250 ft. X 0.8391 = 209.77 ft. Check. By measurement PT = 210 ft. nearly, when PS = 250 ft. Ex. 2. A vertical flagstaff stands on a horizontal plane. At a point 200 ft. from the foot of the staff the angle of elevation of its top is found to be 20°. Find the height of the flagstaff. Let MP (fig. 14) represent the flagstaff, and A the point from which the angle of elevation is taken. Then ^Jlf=200ft., and ZMAP = 20°. By § 1, MP /AM = tan 20° = 0.364. by table .-. MP = 200 ft. X 0.364 = 72.8 ft. Check. By measurement MP = 73 ft. nearly, when AM = 200 ft. Ex. 3. A man wishes to find the height of a tower DB which stands on a horizontal plane. From a point A on this plane he finds the angle ^ of elevation of the top of the tower to be 35°. From aj point C, which is in the horizontal plane at A and 100 ft. nearer the tower, he finds the angle of elevation to be 65°. Find the. height of the tower. Solution 1. Let DB = y ft. Then AD / y = cot 35° = 1 . 4281 , Fig. 16 and CD/y = cot 65° = 0.4663. .-. 100 = AD- CD = (1.4281 - 0.4663) y. .: y = 103.97. Solution 2. From C draw CE ±AB, thus forming the right triangles ^CJ^and CEB. Now Z CBE = Z ABD - Z CBD = 55° - 25° = 30°. Let CE = z ft. and CB = w ft. Then ;2 / 100 = sin 35° = 0. 5736, (1) w/z = cscS0° = 2, (2) and y/w = sin 65° = 0. 9063. (3) PROBLEMS 17 Multiplying together (1), (2), and (3), member by member, we obtain y/100 = 1.1472 X 0.9063, or y = 103.97. Check. By measurement DB = 104 ft. nearly, when ^C = 100 ft. Ex. 4. From the top of a hill 300 ft. higher than the foot of a tree, the angles of depression of the top and the foot of the tree are found to be 20° and 25° respectively. Find ^ p the height of the tree. Let P be the top of the hill, B the foot of the tree, and C its top. In the plane PBC draw PA hori- zontal at P and prolong it until it intersects the vertical line BC pro- ducedm A. Fig. 17 Then Z APC = 20° and Z APB = 25°. Let BC = x ft. Then AC= (300 -X) ft. Hence ^P/300 = cot 25° = 2.1445, (1) and (300 -x)/AP = tan 20° = 0.3640. (2) Multiplying together (1) and (2), member by member, we obtain (300 - a;)/300 = 2.1445 x 0.3640. .-. X = 65.82 nearly. ' Check. By measurement BC = 66 ft. nearly, when BA = 300 ft. Note. All the problems in the following exercise need not be solved before beginning Chapter II. The solution of a few problems at a time, while the student is pursuing the more abstract and theoretic portions of the science, will serve to keep before him its practical utility and maintain his interest. EXERCISE IV 1. The length of a kite string is 250 yds. and the angle of elevation of the kite is 40°. Find the height of the kite, supposing the line of the kite string to be straight. Ans. 160.7 yds. 2. A stick 10 ft. in length stands vertically in a horizontal area, and the length of its shadow is 8.391 ft. Find the angle of elevation of the sun. Ans. 50°. 18 PLANE TRIGONOMETRY 3. A tree is broken by the wind so that its two parts form with the ground a right-angled triangle. The upper part makes an angle of 35° with the ground, and the distance on the ground from the trunk to the top of the tree is 60 ft. Find the length of the tree. Ans. 96.06 ft. 4. The distance between two towers on a horizontal plane is 60 ft. , and the angle of depression of the top of the first as seen from the top of the second, which is 150 ft, high, is 25°. Find the height of the first tower. 5. At a point 200 ft. from the base of an unfinished tower, the angle of elevation of its top is 20° ; when completed, the angle of elevation of its top at this point will be 30°. How much higher is the tower to be built? 6. The angle of elevation of the sun is 65° and the length of a tree's shadow on a level plane is 50 ft. Find the height of the tree. 7. A chimney stands on a horizontal plane. At one point in this plane the angle of elevation of tlie top of the chimney is 30°, at another point 100 feet nearer the base of the chimney the angle of elevation of the top is 45°. Find the height of the chimney. 8. A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 50° ; walking 40 ft. from the bank he finds the angle to be 30°. Find the height of the tree and the breadth of the river, if the two points of observation are in the same horizontal line at the base of the tree. 9. The shadow of a tower standing on a horizontal plane is found to be 60 ft. longer when the sun's altitude is 30° than when it is 45°. Find the height of the tower. 10. At a point midway between two towers on a horizontal plane the angles of elevations of their tops are 30° and 60° respectively. Show that one tower is three times as high as the other. 11. Two observers on the same horizontal line and in the same vertical plane with a balloon, on opposite sides of it and 2500 ft. apart, find its angles of elevation to be 35° and 55° respectively. Find the height of the balloon. 12. A man in a balloon observes that the bases of two towers, which are a mile apart on a horizontal plane, subtend an angle of 70°. If he is exactly above the middle point between the towers, find the height of the balloon. PROBLEMS 19 13. From the foot of a tower the elevation of the top of a church spire is 55°, and from the top of the tower, which is 50 ft. high, the elevation is 35°. Find the height of the spire and the distance of the church from the tower, if both stand on the same horizontal plane. 14. From the top of a tower whose height is 108 ft. the angles of depression of the top and bottom of a vertical column standing on a level with the base of the tower are found to be 25° and 35° respectively. Find the height of the column and its distance from the tower. 15. Two pillars of equal height stand on opposite sides of a horizontal roadway which is 100 ft. wide. At a point in the roadway between the pillars the angles of elevation of their tops are 50° and 25° respectively. Find the height of the pillars and the position of the point of observation. 16. A house 50 ft. high and a tower stand on the same horizontal plane. The angle of elevation of the top of the tower at the top of the house is 25°, on the ground it is 55°. Find the height of the tower and its distance from the house. 17. On the top of a bluff is a tower 75 ft. high ; from a boat on the bay the angles of elevation of the top and base of the tower are observed to be 25° and 15° respectively. Find the horizontal distance of the boat from the tower, also the distance of the boat from the top of the tower. 18. One of the equal sides of an isosceles triangle is 50 ft. and one of its equal angles is 40°. Find the base, the altitude, and the area of the triangle. 19. The base of an isosceles triangle is 68.4 ft. and each of its equal sides is 100 ft. Find the angles, the height, and the area. 20. The base of an isosceles triangle is 100 ft. and its height is 35.01 ft. Find its equal sides and the angles. 21. The base of an isosceles triangle is 88 ft. and its vertical angle is 70°. Find the height, the equal sides, and area. 22. The base of an isosceles triangle is 100 ft. and the equal angles are each 65°. Find the equal sides, the height, and the area. 23. The height of an isosceles triangle is 60 ft. and its vertical angle is 30°. Find the sides and the area. 20 PLANE TRIGONOMETRY 24. A man's eye is on a level with and 100 ft. distant from the foot of a flag pole 36.4 ft. high. When he is looking at the top of the pole, what angle does his line of sight make with a line from his eye to the foot of the pole ? 25. A circular pond has a pole standing vertically at its center and its top is 100 ft. above the surface. At a point in the circumference the angle subtended by the pole is 20°. Find the radius and the area of the pond. 26. A ladder 33^ ft. long leans against a house and reaches to a point 30.21 ft. from the ground. Find the angle between the ladder and the house and the distance the foot of the ladder is from the house. 27. From the summit of a hill there are observed two consecutive milestones on a straight horizontal road running from the base of the hill. The angles of depression are found to be 10° and 5° respectively. Find the height of the hill. 28. At the foot of a hill the angle of elevation of its summit is observed to be 30°; after ascending the hill 500 ft., up a slope of 20° inclination, the angle of elevation of its summit is found to be 40°. Find the height of the hill if the two points of observation and the summit are in the same vertical plane. One method of solution is similar to that of the second solution of example 3 in § 15. 29. At the foot of a mountain the angle of elevation of its summit is 35°; after ascending an opposite mountain 3000 ft,, up a slope of 15° inclination, the angle of elevation of the summit is 15°. Find the height of the first mountain if the points of observation and the summit are in the same vertical plane. 30. From the extremities of a ship 500 ft. long the angles which the direction of a buoy makes with that of the ship are 60° and 75°. Find the distance of the buoy from the ship, having given that cot 76° = 2 — y/3. Ans. 125(V3 + 3)ft. 31. There are two posts which are 240 and 80 ft. high respectively. From the foot of the second the elevation of the top of the first is found to be 60°. Find the elevation of the second from the foot of the first. Am. 30°. PROBLEMS 21 32. A boy standing c feet behind and opposite the middle of a football goal sees that the angle, of elevation of the nearer crossbar is A, and the angle of elevation of the farther one is B. Show that the length of the field is c (tan AcotB — 1). 33. A valley is crossed by a horizontal bridge whose length is I. The sides of the valley make angles A and B with the horizon. Show that the height of the bridge above the bottom of the valley is I /{cot A + cot B). 34. Two forces of a and b lbs. respectively act in the same direction. Find their resultant. Illustrate the problem geometrically. 35. Two forces of a and b lbs. respectively act in opposite directions. Find their resultant when a > 6, when a = b, and when a Ex. 1. State identities [!]••• [6] in words. Ex. 2. sec J. = - 4 ; find the values of the other ratios of A. Since sec ^ is — , ^ is in the second or the third qnndrant. §22 sec ^=-4; .-. cos J. =- 1/4. Dy[i] sin ^ = ± Vl - cos2 J. by [4] = ±Vl-l/16 = ±V15/4. .-. CSC ^ = ± 4/V15 = ± 4 V15/15. by[i] tan A = sin A /cos A =^ V15. by [2] cot^=Tl/V15= T V15/15. ty[i] Check. Construct A from sec A = — 4c, and then find the other trigo- nometric ratios of J., as in § 23. Ex. 3. Express the other trigonometric ratios of A in terms of sin^. CSC A = 1 /sin A. by [1] cos ^ = ± Vl -sin2^. by [4] .-. sec J. = ± 1/ Vl - sin2 ^. by [1] tan ^ = sin yl /cos A by [2] = ±smA/-Vl- sin^A. .-. cotA=± Vl - sin2 A /sin A. by [1] When sin A is positive A is in the first or the second quadrant ; when A is in the first quadrant all the trigonometric ratios of A are + ; when A is in the second quadrant only sin A and esc A are + . The signs as written above are for sin A positive. Check. Find these relations as in example 13 of Exercise VI. ^ _ li*^ \, . v^tv- ' \ PROOFS OF IDENTITIES 33 EXERCISE vn By § 24, compute the other trigonometric ratios of A, having given : 1. smA=-2/S. 5. tan4=-4/3. 9. csc^=-V3. 2. cos^ = 1/3. 6. cot ^ = - 2. 10. sec A = i. 3. sin A = 0.2. 7. cot A = 3/2. 11. tan ^ = - V7. 4. cos^=-3/4. 8. tan ^ = 2.5. 12. cos^=7n/c. Express each of the trigonometric ratios of A in terms of : 13. cos^. 14. tan^. 15. cot -4. 16. sec ^. 17. esc -4. 25. Proofs of identities. Of the different ways of proving an identity, the three following are the more common and important. (i) Derive the required identity from one or more known Ex. 1. Prove that ± Vsec2 A + csc^ A = tan ^ + cot A. (1) " Adding identities [5] and [6] in § 24, we obtain sec2 A + csc2 A = tan2 ^ + 2 + cot2 A = (tan A + cot ^)2. by Algebra, [1] Extracting the square root of both members of the last identity, we obtain identity (1). (ii) Reduce one member of the required identity to the form of the other member, iising any known identities. ^ Ex. 2. Prove that a. /- = esc A — cot A. cos A A .A 1 COS -4 , ^,^ ^„, csc^-cot^ = ^ — - - -; — - l>y [1], [3] sm J. sm J. 1 - cos ^ t, A1 1, TA^ by Algebra, [4] Vl -cos2^ + cos -4 V(l - cos AY /l - cos ^ ^ , , ^ \ ITT — \/ T-, 7 • ^y Algebra 1 - cos2 A \ 1 + cos 4 34 PLANE TRIGONOMETRY (iii) Reduce one member to its simplest form,, and then reduce the other member to the same form. When an identity contains any other trigonometric ratios than the sine and cosine, it is usually best in this method to replace these other ratios by their values in terms of the sine and cosine. Ex. 3. Prove that siii2 A tan A + cos^ ^ cot ^ + 2 sin ^ cos ^ = tan A cot^. (1) , . „ ^ sm A „ . cos A ^ . . . First member = sm^ A 1- cos^ A h 2 sm ^ cos A cos A sin A _ sin* A + cos* A + 2 sin^ A cos^ A ~ sin A cos A ^ (sin2 A + cos2 A)^ ~" sin A cos A = l/(sin A cos^). Similarly show by [3], [4], and Algebra, that second member = 1 / (sin A cos A). From the last two identities we obtain identity (1). by Algebra by Algebra by [4] EXERCISE Vra Prove each of the following identities : 1. cos A tan A = sin A. 2. sin ^ sec .4 = tan A. 3. cos J. CSC -4 = cot ^. 4. sin ^ cot J. = cos ^. 5. cos'-^^ — sin^A = 1 — 2 sin^^. 6. GOS^A - sin2 J. = 2 cos^ A - 1, sin A 1 — cos A 1 4- cos A sin A 1 + sin J. cos A Obtain from [2] From [2] by [1] From [3] by [1] From [4] by Algebra From [4] by Algebra cos -4 sin A IDENTITIES 36 sec ^ + 1 tan -4 From [6] by Algebra 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. From [6] by [1] From [2] by [1], [5] tan A sec A — \ sec ^ + tan A = \/ (sec A - tan A). (1 + tan2^)cos2^ = l. (1 + cot2^)sin2^ = l. sin2 A + sin2 ^ tan2 A = tan2 A . {csc^A — l)csc2J. =cos2^. cos* J. — sin*^ + 1=2 cos2^. cos*^ - sin*^ -f 1 =(cos2^ + sin2^)(cos2^ - sin2^) + 1 = cos2^ + (1 - sin2^) = 2 cos2^. tan2^/(l + tan2^) = sin2yl. cos J. Vl - sin2^ ^ sin A ~ Vl - cos2^ C0t2^ — C0S2^ =C0t2^ C0S2^. sec2^ + csc2^ =8ec2^ csc2^. tan ^ 4- cot ^ = sec A esc A. cot A cos A _ cot A — cos A cot A + cos A " cot A cos A sec2 J. + csc2^ tan A + cot A sec A CSC A 1/Vsec2^ - 1 = Vcsc2^ - 1. 1 + tan2^ .sin2^ esc A 1 + cot2 A cos2^ cot ^ + tan J. = cos A. cos^ + sin A sin A + cos J^. 1 — tan A 1 — cot ^ sin^A cos A + cos^^ sin ^ = sin 2I cos A. sin2 J. cos2^ + cos*^ = 1 - sin2^. 4 I — sin ^ 1 + sin 4 sec^ — tan^. 36 PLANE TRIGONOMETRY 29. ^^^A_ + l±^^^^2csc^. 1 + cos A sin A 30. l/(cot -4 + taii-4) = sin^ cos J.. 31. l/(sec^ — tan^) = sec J. + tan-4. 1 — tan A _ cot A — \ 32. rzr • 1 + tan A cot ^ + 1 33. ^— -^^^^=cos2^-sin2^. 1 + tan2 J. 34. CSC A / (cot A + tan A) = cos A. 35. csc^tI (1 - cos*^) - 2 cot2^ = 1. 26. Changes of the trigonometric ratios of A as A increases from 0° to 360°. To simplify this discussion, let OP have the same length in all of its positions. Changes of sin A. Let A = XOP, and let OP revolve coun- ter-clockwise about from the position OX. Y p x^: CHANGES OF TRIGONOMETRIC RATIOS 37 While A increases from 0° to 90°, MP increases from to OP ; hence MP / OP, or sin .1, increases from to + 1. While A increases from 90° to 180°, MP decreases from OP to ; hence MP /OP, or sin A, decreases from 4- 1 to 0. While A increases from 180° to 270°, MP decreases from to — OP ; hence MP /OP, or sin A, decreases from to — 1. While A increases from 270° to 360°, MP increases from — OP to ; hence MP / OP, or sin A, increases from — 1 to 0. Changes of cos A . While A increases from 0° to 90°, OM decreases from OP to ; hence OM / OP, or cos A, decreases from -|- 1 to 0. While A increases from 90° to 180°, OM decreases from to — OP ; hence OM / OP, or cos A, decreases from to — 1. Similarly the pupil should obtain the other changes of cos A given in the table below. Changes of tan A. Let XOP approach 90° or 270° (either from a less or a greater value) so that OM decreases in size to 1/2 its value the first second of time, to 1/2 its remaining value the next second, to 1/2 its second remaining value the third second, and so on indefinitely. Then, since MP increases slightly, MP / OM, or tan A, more than doubles its value the first second of time, more than doubles its nev?" value the next second, more than double's its last value the third second, and so on indefinitely. Thus tan A will exceed in arithtnetiG (or absolute) value any assignable constant num- ber however great ; that is, when A approaches very near 90° or 270°, tan ^ = -h 00 or — 00. 38 PLANE TRIGONOMETRY Also, when A = 0°, 1S0°, or 360°, MP = 0, and therefore tan ^ = 0. Hence we have the changes of tan A found in the table below. Changes of cot A. Let XOP approach 0°, 180°, or 360° (either from a less or greater value) so that MP decreases in size to 1/2 its value the first second of time, to 1/2 its new value the next second, and so on ; then OM/MP, or cot A , becomes -{- cc ot — oo ; that is, when A approaches near 0°, 180°, or 360°, cot ^ = -f oo or - oo. Also, when A = 90° or 270°, OM = 0, and therefore cot ^ = 0. Hence we have the changes of cot A given in the table below. Similarly the changes of sec A and esc A, which are tabu- lated below, should be proved by the student. By remembering that two reciprocal numbers are like in quality, that when the one increases the other decreases, and that their corresponding values are reciprocals of each other, the changes of esc A, sec A, and cot A are known from the changes of sin A, cos A, and tan A respectively. A increases from 0° to 90° 90° to 180° 180° to 270° 270° to 360° sin A varies from to +1 + 1 toO Oto -1 -ItoO csc^ " " + 00 to +1 + 1 to + 00 — 00 to — 1 - 1 to - X cos A " + 1 toO Oto -1 -ItoO Oto + 1 sec ^ " " + 1 to +00 — 00 to — 1 — 1 to — oo + 00 to + 1 tan A increases from to +00 -00 to to + 00 - 00 to cot ^ decreases " + 00 to 1 to -00 + 00 toO Oto -00 From what precedes it follows that : The tangent or the cotangent can have any real value. The sine or the cosine can have any value from — 1 ^o -|- 1 inclusive. The secant or the cosecant can have any value from — cc to — 1 or from -\- 1 to -{- cc iiiclusive. RATIOS OF 0°, 90°, 180°, 270^ 39 Observe that neither the sine nor the cosine can have a value greater than + 1 or less than — 1 ; and that neither the secant nor the cosecant can have any value between — 1 and + 1. E.g., + 3"/ 4 is the sine of some angle, the cosine of some angle, the tangent of some angle, or the cotangent of some angle ; but it can be neither tlie secant nor the cosecant of any angle. Again, — 3/2 can be neither the sine nor the cosine of any angle. 27. Trigonometric ratios of 0°, 90°, 180°, 270°. When A = 90° or 270°, MP = -^ OP or — OP and OM = ; hence tan A or sec A assumes the form ± OP/0. Now the division of OP by zero is impossible ; hence, strictly speaking, 90° or 270° has no tangent or secant. But when A approaches very near to 90° or 270°, by § 26 tan A or sec ^ is + x or — oo ; hence it is customary to say that the tangent or secant of 90° or 270° is oo, meaning thereby that however near A approaches to 90° or 270°, tan A or sec ^ is + oo or — oo. Again, when .1 = 0° or 180°, MP = and OM = + OP or — OP; hence cot ^ or esc J assumes the form ± OP/0. Therefore, strictly speaking, 0° or 180° has no cotangent or cosecant. But when A approaches very near to 0° or 180°, by § 26 cot A or esc .1 is -h oo or — oo ; hence it is customary to say that the cotangent or cosecant of 0° or 180° is oo. The trigonometric ratios of 0°, 90°, 180°, and 270° are tabu- lated below. To aid the memory, the reciprocal ratios are grouped together. Angle 0° I 90° 180° 270*= sme . . cosecant cosine . secant . tangent cotangent oo + 1 + 1 oo + 1 + 1 oo QO QO - 1 -1 00 -1 -1 QO 00 40 PLANE TRIGONOMETRY Note, Putting OP = a, tan 90° assumes the form a/0, where a ji^O. The form a/0 could be used as the tangent of 90°; then, whether we regarded a/0 as a symbol without numerical meaning, as a symbol of impossibility, or as a symbol of absolute infinity, when a/0 'appeared as the tangent of A, the value of A would be known as definitely as when the tangent of A is any finite number. 28. The trigonometric ratios of — A in terms of the ratios of A. ^X Fig. 24 In each figure let A denote any angle, positive or negative, which is coterminal with XOP ; then — A will be coterminal with XOP'. Angle A is in the first quadrant in fig. a, in the second quadrant in fig. b ; and so on. Take OP = OP', and draw PM ± OX and P'M' ± OX. Then in each figure the acute angles MOP and Af'OP' will be equal in size. Hence any two corresponding sides of the A OMP and OM'P' will be equal in length. Therefore, as directed lines, M'P'/ OP' = -MP I OP, i.e. sin ( - A) = - sin A, (1) and OM'/ OP' = OM/ OP, i.e. cos ( - A) = cos A. (2) TRIGONOMETRIC RATIOS OF - ^ 41 Dividing (1) by (2), tan(- A) = -tSinA. Dividing (2) by (1), cot (- J) = - cot A. From (2) by [1], sec (—A)= sec A. From (1) by [1], esc (— A) = — esc A. Identity (1) states that sin(— ^) and sin A are arithmetic- ally equal but opposite in quality; that is, when sin(— ^) is — , sin ^ is + ; and when sin (— .4) is -f, sin A is — . Identity (2) states that cos (—A) and cos A are arithmetically equal and like in quality. The six identities just proved can be summed up as follows : Any trigonometric ratio of — A is equal arithmetically to the same ratio of A ; but only the cosines (or the secants) of ^ A and A are like in quality. E.g. , sin ( - 35°) = - sin 35°, cos ( - 98°) = cos 98°, tan (- 212°) = - tan 212°, esc (- 317°) = - esc 317° Ex. 1. Express each trigonometric ratio of — 22° in terms of a ratio of 22°. Ex. 2. Express each trigonometric ratio of 320° in terms of a ratio of a positive angle less than 45°. An angle of 320° is coterminal with one of — 40° ; hence any trigono- metric ratio of 320° is equal to the same ratio of — 40° (§ 21). Whence sin 320° = sin (- 40°) = - sin 40°, cos 320° = cos (- 40°) = cos 40°, tan 320° = tan ( - 40°) = - tan 40°, etc. Similarly the trigonometric ratios of any angle in the fourth quadrant can be found in terms of those of some positive acute angle. Ex. 3. Express each trigonometric ratio of — 325° in terms of a ratio of a positive angle less than 45°. An angle of — 325° is coterminal with one of 35° ; hence sin ( - 325°) = sin 35°, cos ( - 325°) = cos 35°, etc. Similarly the trigonometric ratios of anij angle in the first quadrant can be found in terms of those of some positive acute angle. 29. The trigonometric ratios of 90° + A in terms of the ratios of A. In each figure let A denote any angle, positive or 42 PLANE TRIGONOMETRY Degative, which is coterminal with XOP, and let POP' = 90' then A + 90°, or 90° + A, is coterminal with XOP'. Fig. 25 Take OP = OP', and draw PM _L OX and P'J/' _L OX. Then the acute angles MOP and M'P'O will be equal in size. Hence any two corresponding sides of the A MOP and M'OP' will be equal in length. Therefore, as directed lines, M'P'/OP' = OM/ OP, i.e. sin (90° + A) = cos A ; (1) and OM'/OP' = - MP /OP, i.e. cos (90° + A) = - sin A. (2) .-. tan (90° + ^4) = - cot J, cot (90° -f-^) = - tan A, sec (90° + ^) = -CSC .4, esc (90° + ^4) = sec ^. Since the angle A is 90° less than the angle 90° 4- A, the six identities just proved can be summed up as follows : Ani/ trigonometric 7^atio of an angle is equal arithmetically to the co-ratio of this angle less 90°, but only the sine (or the cosecant) of the first angle has the same quality as the co-ratio of the seco7id angle. TRIGONOMETRIC RATIOS OF 90° -{-A 43 E.g., since 130° - 90° = 40°, we have sin 130° = cos 40°, tan 130° = - cot 40° ; cos 130° = - sin 40°, cot 130° = - tan 40°. Ex. 1. Express in terms of a trigonometric ratio of some positive angle less than 45° each trigonometric ratio of 126° ; 492° ; - 220°. sin 126° = cos 36°, cos 126° = - sin 36°, etc. § 29 An angle of 492° is coterminal with one of 132° ; hence sin 492° = sin 132° = cos 42° ; §§ 21, 29 cos 402° = cos 132° = - sin 42° ; etc. An angle of — 220° is coterminal with one of 140° ; hence sin (- 220°) = sin 140° = cos 50° = sin 40 ; §§ 21, 29, 9 cos (- 220°) = cos 140° = - sin 50° = - cos 40° ; etc. Similarly the trigonometric ratios of any angle in the second quadrant can be found in terms of those of some positive acute angle less than 45°. Ex. 2. Express in terms of a trigonometric ratio of some positive angle less than 45° each trigonometric ratio of — 130° ; 230°. sin (- 130°) = - sin 130° = - cos 40° ; §§ 28, 29 cos ( - 130°) = cos 130° = - sin 40° ; tan ( - 130°) = - tan 130° = cot 40°. Applying § 29 twice in succession and then § 9 once, we have sin 230° = cos 140° = - sin 50° = - cos 40° ; cos 230° = - sin 140° = - cos 50° = - sin 40° ; tan 230° = - cot 140° = tan 50° = cot 40° ; etc. Similarly we can find, in terms of the trigonometric ratios of a positive angle less than 45°, the ratios of any positive or negative angle in the third quadrant. The principles in §§ 9, 28, 29 have an important bearing on the construction and use of trigonometric tables and on the solution of triangles. By them, as is seen above, the trigonometric ratios of any angle can be expressed in terms of the trigonometric ratios of some positive angle less than 45°. Hence, from a table which contains the trigonometric ratios of all angles between 0° and 45°, we can obtain the trigonometric ratios of any angle whatever. 44 PLANE TRIGONOMETRY 30. Trigonometric ratios of complementary and supplementary angles. Applying § 29 twice and then § 28 once, we obtain sin (180° -A)= cos (90° - .4) = - sin (-A)= sin A; (1) cos (180° -A)=- sin (90° -A) = - cos (- ^1) = _ cos A ; (2) tan (180°-^)= -cot (90°-^)= tan(-^)= - tan^. (3) Comparing the last members of (1), (2), (3) with their first members we have (i) Any trigonometric ratio of an angle is equal arithmetic- ally to the same ratio of its supplement; hut only the sines (or the cosecants) of two supplementary angles have the same quality. E.g., sin 150° = sin 30°, tan 165° = - tan 15° ; cos 135° = - cos 45°, cot 155° = - cot 25° Ex. 1. Express in terms of a trigonometric ratio of its supplement each trigonometric ratio of 125° ; 143° ; 157°. Comparing the last members of (1), (2), (3) with their second members we have § 9 generalized ; that is, (ii) Any trigonometric ratio of an angle is equal to the co-ratio of its complement. Ex. 2. Applying § 29 three times in succession and § 28 once, we have sin (270°±^)= cos(180°±^)=-sin (90°±^) =-cos(±^)=-cos^ ; cos (270°±^) =-sin {1S0°±A) =-cos (90°±^) = sin (±^j =±sin A ; tan(270°±^)=-cot(180°±^)= tan(90°±^)=-cot (±^)==Fcot ^. Ex. 3. Prove (ii) by putting - ^ for ^ in (1) and (2) of § 29. 31. Trigonometric ratios of n . 90 ± A, in terms of the ratios of A. To obtain in terms of a trigonometric ratio of A any trigonometric ratio of n • 90° -f- A (where n is sl positive inte- ger), we apply § 29 t^ times in succession ; and to obtain in terms of a ratio of A any ratio of ^ -90° — ^4, we first apply § 29 w times and then § 28 once. In each case we change from ratio to co-ratio n times ; hence TRIGONOMETRIC RATIOS OF n • 90° ± ^ 45 (i) When n is even, any trigonometric ratio of n . 90° ± A is equal arithmetically to the same ratio of A. (ii) When n is odd, any trigonometric ratio of n • 90° ^ A is equal arithmetically to the co-ratio of A. When ^ is a positive acute angle, any trigonometric ratio of A is positive ; hence (iii) The two trigonometric ratios in (i) or (ii) will be oppo- site in quality when^ and only when, the ratio o/ n . 90 ± A is negative for A positive and acute. Any positive angle can be written in the form n • 90° ± A where A has some positive value less than 45°. E.g., 580° = 6 • 90° -f- 40" ; here n is even, and the angle is in the third quadrant. Hence, by (i) and § 22, we have sin 580° = sin (6 • 90° + 40°) = - sin 40° ; cos 580° = cos (6 • 90° + 40°) = - cos 40° ; tan 580° = tan (6 • 90° + 40°) = tan 40° ; etc. Again, 270° + ^ ^3 • 90° + J. ; here n is odd, and 270° + ^ is in the fourth quadrant when A <90°. Hence, by (ii) and (iii), we have sin (270° -\-A) = - cos A, tan (270° + ^) = - cot ^ ; cos (270° ^- A) = sin A, cot (270° + ^) = - tan ^ ; etc. Example. Express in terms of a trigonometric ratio of A each trigo- nometric ratio of 180° + A ; 180° - A ; 270° - A ; 360° ± A. EXERCISE IX Express each of the following trigonometric ratios in terms of the ratio of some positive acute angle less than 45°. 1. sin 168°. 6. cos (-84°). 11. cot 1054°. 2. tan 137°. 7. tan (-246°). 12. sec 1327°. 3. cos 287°. 8. cos (-428°). 13. esc 756°. 4. sin 834°. 9. cos 1410°. 14. tan (- 196° 54'). 5. sin (-65°). 10. tan 1145°. 15. cot (- 236° 21'). 46 PLANE TRIGONOMETRY 16. Prove sin 420° • cos 390° + cos ( - 300°) • sin ( - 330°) = 1. 17. Prove cos 570° • sin 510° - sin 330° • cos 390° = 0. 32. Trigonometric lines representing the trigonometric ratios. Any trigonometric ratio is a positive or a negative number, but it can always be represented by a directed line, as below. Let A denote any angle cotenninal with Z. XOP in each of the four figures. Take OP as a positive unit line, and draw PMl. OX. Then sin A = MP J OP = the numerical measure of MP ; (1) hence sin A is represented hy the directed line MP. Also, cos A = OM I OP = the numerical measure of OM; (2) hence cos A is represented by the directed line OM. T b M ; X' 1 y ^ T M ^ ^ X /- \ y^ N ,y / /• \^ c c__^ Fig. 26 TRIGONOMETRIC LINES 47 To obtain directed lines which shall represent the four other trigonometric ratios, draw OY 1. OX at 0, and take OX = OY = OP = a positive unit line. At X draw XT-JL OX, at Y draw YC _L OF, and prolong each until it meets the final side OP (produced through P or O) in some point as T or C. According to the laws assumed in § 21 for the quality of MP, OM, and OP, the directed line AT is positive or negative according as it extends upward or downward from its origin A; YC is positive or negative according as it extends to the right or to the left from its origin F; and OT or OC is posi- tive or negative according as it extends from the origin in the direction of the final side OP or in the opposite direction. E.g., XT or YC is + in fig. a or c, and — in fig. h or d. or is + in fig. a or tZ, and — in fig. 6 or c. OC is + in fig. a or 6, and — in fig. c or d. In each figure the triangles OMP, OXT, and OYC, being mutually equiangular, are similar. In each of the four figures we find that tan^ = MPjOM— XT /OX = the numerical measure of XT ; (3) hence tan A is represented by the directed line XT. cot A = OM/MP = YC/OY=the numerical measure ofYC; (4) hence cot A is represented by the directed line YC. sec A = OP / OM = OT/OX = the numerical measure of OT ; (5) hence sec A is represented by the directed line OT. CSC A = OP I MP =z OC / OY = the numerical measure of OC ; (6) hence esc A is represented by the directed line OC. The directed lines which represent the trigonometric ratios of an angle are called the trigonometric lines of that angle. The relations in (1) to (6) can be written briefly sin A = MP, cos A = OM, tan A = XT, cot A = YC, sec A = OT, esc A = OC, 48 PLANE TRIGONOMETRY Since the trigonometric lines represent graphically the trigo- nometric ratios, or, in other words, the trigonometric ratios are the numbers which measure the trigonometric lines, it follows that if we prove any relation between the trigonometric lines, we know that the same relation exists between the correspond- ing trigonometric ratios, and vice versa. 33. Use of trigonometric lines in proofs and discussions. To fix in the pupil's mind the trigonometric lines which represent the trigonometric ratios, to help familiarize him with the use of directed lines to represent positive and negative real num- bers, and to show him how the use of the trigonometric lines sometimes simplifies trigonometric proofs and discussions, we give below illustrative examples, which can be taken or omitted at the option of the teacher. Ex. 1. Using trigonometric lines, prove the relations in § 24. In each of the four figures in § 32 we have JWP VOJW"^ = 0P^ .-. sin2^ + cos2^= 1; ■qx^ + ^T^ = ~OT^, .-. 1 + tan2^ = sec2^; '6Y^+YC'^ = ~0C'^, .-. 1 4-COt2 ^=CSC2^; XT/ OX = MP/ OM, .'. tan A=8mA/cosA; YC/ OY = OM/MP, .-. cot ^ = cos ^ /sin A ; OT/OX='OP/OM, .-. sec J. = l/cos^; OC/OY= OP/ON, .'. CSC A = l/siuA. Ex. 2. Using the trigonometric lines, determine the quality of each trigonometric ratio in each quadrant. In the figures of § 32, the quality of MP, or sin A, is easily deter- mined. So also is the quality of OM, or cos -4. XT, or tan A, is positive when A is in the first or the third quadrant, and negative when A is in the second or the fourth quadrant. or, or sec A, extends in the direction of OP, and is therefore positive when A is in the first or the fourth quadrant ; and OT, or sec A, extends in the direction opposite to that of OP, and is therefore negative when A is in the second or the third quadrant. In like manner determine the quality of YC, or cot A, and of OC, or esc A. TRIGONOMETRIC LINES 49 Ex. 3. Using trigonometric lines, trace the changes of the trigono- metric ratios of A while A increases from 0° to 360°. In the figures of §32, the changes of JfP, or sin -4, and of OM^ or cos^, are easily followed. While A increases from 0° to 90° (fig. a), XT beginning at zero increases without limit as A approaches 90°; i.e. tan A increases from to + oo. While A increases from 90° to 180° (fig. 6), XT is at first of infinite length and negative, and becomes when A = 180° ; i.e. tan A increases from — GO to 0. While A increases from 180° to 270° (fig. c), XT beginning at zero increases without limit as A approaches 270° ; i.e. tan A increases from to + 00. While A increases from 270° to 360° (fig. d), XT is at first of infinite length and negative, and becomes when A = 360° ; i.e. tan A increases from — GO to 0. In like manner the student should trace the changes of cot^, sec J., and CSC A. Ex. 4. Using the trigonometric lines, find the trigonometric ratios of 180° - A and 180° -f ^ in terms of those of A, when A is in the first C^ Y C quadrant. Let the angles XOP, M'OP\ and M"OP" be equal in size ; then if A is coterminal with XOP^ 180° — A will be coterminal with XOP", and 180° + A will be coter- minal with XOP". Draw the trigonometric lines of A, 180° - A, and 180° + A. Fig. 27 Then Again, M'P' = MP, OM' = - OM, XT' = - XT, YC = - YC, 0T=- OT, OC = OC, M"P" = - MP, OW = - OM, XT = XT, YC = YC, .-. sin (180° - ^) = sin ^ ; .-. cos (180° - ^) = - cos ^ .-. tan (180° - ^) = - tan ^ .-. cot (180° - ^) = - cot ^ .-. sec (180° - ^) = - sec ^ .-. CSC (180°- ^) = csc^. .-. sin (180° + ^) = - sin ^ .-. cos (180° + ^) = - cos 4 .-. tan (180° + ^) = tan 4 ; .-. cot (180° + A) = cot A. 50 PLANE TRIGONOMETRY OT and OC are both negative when the angle is XOP". Hence sec (180° ■\- A)= - sec A, esc (180° -\- A) = - esc A. A similar proof could he given when A is in any one of the other three quadrants. Ex. 5. Using the trigonometric lines, find the trigonometric ratios of — A in terms of those of A. In each figure of § 28, let OP = OP' = a positive unit line. Then MP = sin ^, OM = cos A, M'P' = sin (- A), OM' = cos (- A). But in each figure we have M'P' = - MP, OM' = OM. Hence sin (— ^) = — sin^, cos (— ^) = cos J.. (1) From the identities (1) we can obtain the other relations as in § 28. The student should draw the other trigonometric lines of A and — A in each of the four figures in § 28, and prove the last four identities by the use of these lines. Ex. 6. Using the trigonometric lines, find the trigonometric ratios of 90° -f J. in terms of those of A. In each figure of § 29, let OP = OP' = a positive unit line. Then MP = sin ^ , OM = cos ^, M'P' = sin (90° + A), OM' = cos (90° + A). But in each figure we have M'P' ~ OM, OM' = - MP. Hence sin (90° + ^)= cos ^, cos(90° + ^)= - sin ^. (1) From the identities (1) we can obtain the other relations as in § 29. But the student should draw the other trigonometric lines of A and 90° + ^ in each of the four figures in § 29, and prove the last four identi- ties by the use of these lines. Ex. 7. Using the trigonometric lines, find the trigonometric ratios of 90° — ^ in terms of those of A, when A is in the first quadrant. TRIGONOMETRIC LINES 61 In the figures of § 32, let X be the origin of arcs, and let the arc be positive or negative according as its generating point moves counter-clockwise or clockwise. Then the arc in each figure will have the same numerical value in degrees as the angle which it subtends at the center, and the trigonometric lines of the angle in each figure can be regarded as trigono- metric lines of the arc, and the ratios which these lines repre- sent can be regarded as trigonometric ratios of these arcs. Hence, if the number of degrees in a unit arc is equal to the number of degrees in an angle, the arc and the angle have the same trigonometric lines or ratios. CHAPTER III TRIGONOMETRIC RATIOS OF TWO ANGLES 34. Sine and cosine of the sum of two angles. Let XOR and ROC \)Q any two positive acute angles. Then Z XOR + Z ROC = Z XOC. Let A denote any angle, positive or negative, coterminal with Z XOR, and B any angle coterminal with Z ROC. Then the sum A -\- B will be coterminal with Z. XOC. >X Fig. 28 The sum A -\- B may be in the first quadrant, as in fig. a, or in the second quadrant, as in fig. h. In each figure, from any point on OC, as P, draw PN ± OR and PMl. OX ; also draw NQ ± OX, and ND _L MP. Then the triangles DPN and QON will be similar. Now MP=QN-\- DP. By § 21, QN=BmAON. Again, DP / NP = OQ/ ON = cos A; whence i>P = cos ^ • NP. 62 ADDITION FORMULAS 63 Hence MP = sin ^ • ON + cos ^ • NP. .'. MP I OP = sin ^ • ON I OP + cos ^ • NP/OP. (1) Substituting for the ratios in (1) their names, we have sin (A + B) = sin A cos B + cos A sin B. [7] Again, OM=OQ- DN = cos A • ON — sin A • NP. .'. OM/ OP = cos ^ . ON/ OP - sin ^ • NP/OP. .'. cos (A + B) = cos A cos B — sin A sin B. [8] Observe that thus far [7] and [8] are proved only when the angles A and B are both in the first quadrant. In § 35 it will be shown that these relations hold true in whatever quadrant J. or 1? is. 35. General proof of [7] and [8]. sin(.4 -f 90° + B) = sin (90° -\- A -\- B) = cos(A-\-B) §29 = cos A cos B -\-(— sin A) sin B by [8] = sin (A + 90°) cos B + cos (A + 90°) sin B. (1) Again, cos (A + 90° -\- B) = cos (90° -\- A -^ B) = - sin (A i-B) § 29 = (— sin A) cos B — cos A sin B by [7] = cos (A + 90°) cos B - sin (A -f 90°) sin B. (2) Now in whatever quadrant ^4 is, ^ + 90° is in the next quadrant. Hence, from (1) and (2), it follows that if [7] and [8] are true when A is in any one quadrant, they are true also when A is in the next quadrant. But, by § 34, [7] and [8] are true when A is in the first quadrant; hence they are true when A is in the second quadrant; and so on. Hence [7] and [8] hold true in whatever quadrant A is. The same reasoning applies to B. Hence [7] and [8] hold true for all values of A and B, positive or negative. 64 PLANE TRIGONOMETRY Formulas [7] and [8], often called the addition formulas, are very important and should be memorized. So many theorems can be deduced from the formulas [7] and [8] that they are often called the fundamental formulas of trigonometry. EXERCISE X 1. State in words identities [7] and [8], as generalized in § 35. The sine of the sum') _ f sin first ■ cos second of any two angles j ~~ t_ + cos first • sin second. 2. sin 75° = sin (30° + 45°) = sin 30° cos 45° + cos 30° sin 45° by [7] _ 1 V2 V3 V2 _ V2 + Ve 3. Putting 75° = 30° + 45° and using [8], find cos 75°. 4. Putting 15° = 45° + (- 30°), find sin 15° and cos 15°. 5. Putting 15° = 60° + (- 45°), find sin 15° and cos 15°. 6. Putting 90° = 60° + 30°, find sin 90° and cos 90°. 7. Putting 0° = 45° + (- 45°), find sin 0° and cos 0°. A and B being positive acute angles, find the values of sin {A + B) and cos {A + B), having given 8. sin^ = 2/5, cosJB=:l/3. 9. sin ^ = 2/3, cos B = 1/4. 10. Putting 90° + ^ for ^ in [7], deduce [8]. 11. Putting 90° + A ioT A in [8], deduce [7]. 12. Prove [7] and [8], using trigonometric lines, A and B being in the first quadrant. Take OP = + 1. Then MP = sin {A + B), ON = cos B, NP = sin B. .-. DP = NP cos DPN = sin BcosA. QN = ON sin A = cos 5 sin ^. .-. sin {A -\- B)= QN + DP = sin AcosB -h cos A sin B. SUBTRACTION FORMULAS 55 36. Sine and cosine of the difference of two angles. Substitut- ing — B for B in [7], we have sin (.4 — B) = sin A cos (— B) + cos A sin (— B). ,'. sin (A — B) = sin A cos B — cos A sin B. [9] Substituting — B foT B in [8], we have cos (A — B) = cos A cos (— B) — sin A sin (— B). .'. cos (A — B) = cos A cos B + sin A sin B. [10] Formulas [9] and [10] are often called the subtraction formulas. EXERCISE XI 1. State in words identities [9] and [10]. The sine of the difference \_( sin first • cos second of any two angles J ~ I ~~ (^os first • sin second. 2. Putting 15° = 45° - 30°, find sin 15° by [9] and cos 16° by [10]. 3. Putting 15° = G0° - 45°, find sin 15° and cos 15°. A and B being positive acute angles, find the values of sin {A — B) and cos {A — B), having given 4. sin ^ = 1/4, sin 5= 1/3. 5. cos ^ = 2/3, cos 5 = 3/4. Prove each of the following identities : 6. sin {A + B) sin {A - B) = sm^ A cos^ B - cos2 Asin^B + (sin2 ^ sin2 B - sin2 A sin2 B) = sin2 A (cos2 B + sin2 B) - sin2 5(cos2 A + sin^ A) = sin2^ -sin^B. Observe that sin 2 J. sin^ B — sin2 ^ sin 2 B is added above as one form of zero. 7. cos (A + B) cos {A - B) = cos^A - sin2 B. 8. sin {A + B) cos B - cos {A + B) sin B = sin A. 9. sin {A -{- B) + cos {A - B) = (sin A + cos A) (sin B + cos B). 10. sin A cos {B - C) - sin Bco8{A + C) = sin {A - B) cos C. 66 PLANE TRIGONOMETRY 11. tan A + t&nB = sin {A + ^)/(cos A cos B). 12. cot JB - cot ^ = sin {A - B)/{sin A sin B). 13. Prove [9] and [10] geometrically, using trigonometric lines, when A, B, and A — B are in the first quadrant. Let XOB and BOChe any two acute angles, Z BOC being negative and Z XOC being positive. Then Z XOC = Z XOB + Z BOC. Let A denote any angle coterminal with XOB, and — B any angle coterminal with BOC. 1>X Then A + {- B), or A - B, will be coterminal with XOC. Take OP equal to + 1. Draw PM ± OX, PN ± OB, NQ ± OX, and PD _L QN. Now NP = sin {- B) = - sin 5, ON = cos (- J5) = cos B, and sin {A - B) = MP = QN - DN. Also, QN= ON' sin JTOE = cos ^ sin ^, and DN = ( - NP) cos DiVP = sin BcosA. .: sin (^ - J5) = QiV — I)N = sin J. cos -B - cos A sin B. Again, OQ = OJV cos XOE = cos BcosA, and DP = {- NP) sin DJVP = sin jB sin ^. .-. cos {A - B)=: 0M= 0Q + DP = cos ^ cos ^ + sin A sin B. 37. Tangent of the sum and difference of two angles. Divide the members of [7] by those of [8] ; then by [2] we have . , ^- sin A cos B + cos A sin B tan (A + B) = : — - — -. — -• (1) ^ ^ cos A cos B — sm A sm B ^ ^ To express tan (A + B) in terms of tan A and tan B we divide the numerator and denominator of the fraction in (1) by cos A cos B ; then by formula [2] we obtain .«x tan A 4- tan B ^^ ^ _, tan (A + B) = —^ -• fill ^ ^ ^ 1 - tan A tan B *- -■ Substituting — B for B in [11], we obtain , . „. tan A — tan B tan (A - B) = ; -• [12] ^ ^ 1 + tan A tan B •- -■ TRIGONOMETRIC RATIOS OF 2 A 67 EXERCISE Xn 1. State in words identities [11] and [12]. The tangent of the sum \ _ f the sum of their tangents of any two angles j ~ \ i _ product of their tangents 2. Putting 75° = 45° + 30°, find tan 75° by [11]. 3. Putting 15° = 60° - 45°, find tan 15° by [12]. 4. If tan ^ = - 1/2 and tan B = 3, find tan {A + E) and tan {A - B). 5. If tan ^ = - 2 and tan B = - 3, find tan {A + B) and tan {A - B). Prove each of the following identities : a ^o^MRo, jx_l+tan^ , _. _cotAcotB - 1 6. tan (45° + ^) = • 8. cot (A + B) = 1 - tan ^ cot B + cot A r, 4. /AKo A\ 1 ~ t^° ^ o ^ / . «v cot A cot B 4- 1 7. tan (45°-^) = - -• 9, cot(A-B) = — — 1 + tan ^ cot B - cot A 10. Prove identity [12] by dividing [9] by [10]. 11. Prove the identities in examples 8 and 9 by taking the reciprocals of the members of [11] and [12] respectively. 12. Find tan {A + B) and tan {A - B) in terms of cot A and cot B. 13. Find cot {A + B) and cot {A - B) in terms of tan A and tan B. 38. Trigonometric ratios of twice an angle in terms of the ratios of the angle. Substituting A for B in [7], we have sin (A -{- A) = sin A cos A -{- cos ^ sin ^ ; tliat is sin 2 A = 2 sin A cos A. [13] Substituting ^ for 5 in [8], we obtain cos 2 A = cos* A — sin* A (i) ^ = 1 -2 sin* A (ii) ► [14] = 2cos*A-l. (iii) To derive Qi) or (iii) from (i), we use identity [4]. Substituting A for B in [11], we obtain ^ ^ ^ 2 tan A _ . ^_ tan2A = — -• [15] 1 - tan* A •- -■ 58 PLANE TRIGONOMETRY EXERCISE XIII 1. State in words identities [13], [14], and [15]. sin twice an angle = 2 sin angle • cos angle. cos twice an angle = (cos angle)^ — (sin angle)^. 2. From the trigonometric ratios of 30°, find sin 60°, cos 60°, tan 60°. 3. From the trigonometric ratios of 60°, find sin 120°, cos 120°, tan 120°. 4. Express sin 6^, cos 6^, tan 6 J. in terms of the trigonometric ratios of 3 -4. 5. Express sin 3^, cos 3^, tan 3^ in terms of the trigonometric ratios of S A/2. Prove each of the following identities : _ ^ _ . cot2 A -1 _ .,^ 1-COS2J. 6. cot 2 ud = 8. sm2 A = 2 cot J. 2 7. csc2^ = (sec J.csc^)/2. 9. cos^^ =(1 + cos2 ^)/2. sec2 A 1 + tan2 A 10. sec 2^ = 2-sec2^ l-tan2^ 11. cos4^=2cos2 2^- 1 = 2(1 -2sin2^)2_i = 8sin4^ -8sin2^ + 1. 12. sin 4 J. = 4 sin ^ cos ^ — 8 sin^ ^ cos ^. 39. Trigonometric ratios of half an angle in terms of the cosine of the angle. Solving (ii) and (iii) of [14] for sin^^l and cos^ A respectively and putting A/2 for A, we obtain .A /I — cos A _. -_ , A /I + cos A ^.^_ and COS - = ^— L- , [17] Divide [16] by [17], tan | ^ V^^^- [18] TKIGONOMETRIC RATIOS OF A/2 69 EXERCISE XIV 1. State in words identities [16], [17], and [18]. • 17^ 7 .. r 1 — cos angle _, ^_ sin Jialf an angle = square root of ^— . [16] 2. Find sin 22i°, cos 22|°, tan 22 1°, from cos 46°. , cos 45° _ /I -V2/2 _ V2-V2 in22i° = J- 2 \ 2 2 * 3. Find sin 16°, cos 15°, tan 15°, from cos 30°. 4. cos A = 1 /S ; find the sine, cosine, and tangent of -4/2. 5. cos A = a; find the sine, cosine, and tangent of ^/2. 6. Express sin A, cos A, and tan A in terms of cos 2 A. 7. Express sin 2 A, cos 2 A, and tan 2 ^ in terms of cos 4 J.. 8. Express sin SA, cos 3 A, and tan 3 ^ in terms of cos 6 A. Prove each of the following identities : ^A _1 + cosA_/ sin A \^ _ /I +cosA\^ 2 "1 — cos J.~ \1 — cosJ./ ~\ sin J. / •i/x X o-^ / sin^ \2 /l-cos^\2 10. tan2 — = ( ) = ( ) = (esc A - cot A)^. ,1 ^A 2 sec J. ^ _ ^A 2 sec ^ 11. sec2— = : 12. csc2 — = 2 sec -4 + 1 2 sec A — \ 13. Express cos* A in terms of cos 2 A and cos 4 ^. (cos2^)2 = (i+i cos 2^)2 = i+^cos2J. + J cos2 2 ^ = ^ + |cos2^ + ^(1 + icos4^). .-. cos* -4 = f + i cos 2 -4 + i cos 4 .4. 14. Prove sin* -4 = f — | cos 2 -4 + |- cos 4 J.. 15. Prove sin2 A cos2 .4 = i — i cos 4 .4. Suggestion, sin^ .4 cos2 ^ = (sin A cos .4)2 = i sin2 2 A. 16. Prove sin2 .4 cos* A^^^ + I sin2 2 ^ • cos 2 ^ — y^^ cos 4 .4. Suggestion. sin2 ^ cos* -4 = (sin -.4 cos -4)2 ■ cos2 A. 60 PLANE TRIGONOMETRY 40. Sum and difference of sines and cosines. Adding and subtracting [7] and [9], and [8] and [10], we obtain sin (A + B)-{- sin (A - B) ~ 2 sin ^ cos ^ ; (1) sin (A + B)- sin (^1 - B) = 2 cos A sin B ; (2) cos (A -\- B)-\- cos (A — B) = 2 cos A cos B ; (3) cos (A -{- B)- cos (A — B) = —2smA sin B. (4) Let A -^B= C and A - B = D. ') Then A = (C+D)/2, B = (C -D)/2.j ^^'^ Substituting in (1) • • • (4) the values in (5), we obtain C 4- D C — D sinC + sinD= 2 sin — ■ — cos 2 2 [19] C + D C — D sin C — sin D = 2 cos — - — sin — 2 2 [20] C+DC-D cos C + cos D = 2 cos — - — cos — [21] C + D C — D cos C — cos D = — 2 sin — ! — sin 9 9 [22] By formulas [19] • • • [22], a sum or a difference of the sines or the cosines of two angles is transformed into a product. Hence these formulas, often called product formulas, are useful in adapting other formulas to the use of logarithms. E.g., sin7 J. + sin5^ = 2sini(7^ + 5^)cosi(7^ - 5A) = 2 sin 6 A cos A . .-. log (sin 7 J. + sin 5 J.) = log 2 + log sin 6 ^ + log cos A. Again, cos 8 ^ - cos 2 J. = - 2 sin |(8 ^ + 2 A) sin |(8 ^ - 2 A) = — 2 sin 5 -4 sin 3 A. By the converses of formulas (1) • • • (4), a product involving sines or cosines or both is transformed into a sum or a differ- ence of sines or cosines. IDENTITIES 61 EXERCISE XV 1. State in words identities [19] • • • [22]. The mm of the sines ^ « . ^ ,^ . ,^ ,.«. , , , !► = 2 sin half sum • cos fialf difference. of any two angles J Prove each of the following identities : 2. sin 60° + sin 30° = 2 sin 45° cos 15°. by [19] 3. sin 50° + sin 10° = 2 sin 30° cos 20°. 4. cos 75° + cos 15° = 2 cos 45° cos 30°. by [21] 5. cos 80° - cos 20° = - 2 sin 50° sin 30°. 6. sin 7 J. — sin 3^ = 2 cos 5^ sin 2^. „ sin 7 ^ — sin 5 ^ 2 cos 6 ^ sin ^ 7. = — = tanJ.. cos 7 .4 4- cos o A 2 cos 6 A cos A 8. 'J^AjLEllA^t^r>2A. COS A + cos 3 ^ . ^ sin C + sin D ^ C + D ^ C - D tan tt (C + D) 9. = tan . cot = — '-. sin C — sin D 2 2 tan i (C — D) , - sin C 4- sin D C + I) 10. = tan cos C + cos D 2 ,, sinC + sinD C-D D-C 11. = — cot = cot cos C - COS D 2 2 ,„ sin C - sin D ^ C-D 12. = tan cos C + cos D 2 ,„ sin C- sin D ^C + D 13. = — cot cos C — cos D 2 , ^ cos (7 + cos D , O + D C-D 14. = — cot cot cos C - cos D 2 2 = coti(C + D)coti(i)- C). 15. Given sin J. = 1 /2, sin B = 1 /3, to find sin {A + B), sin {A - B), cos (^4- 5), cos(^ — J5), sin 2^, sin 2 5, cos 2^, cos2jB: (1) when A and B are both in the first quadrant ; (2) when A is in the first and B is in the second quadrant. 62 PLANE TRIGONOMETRY 16. From the answers to example 15, find in the simplest way tan {A + B), tan {A - B), cot {A + B), cot {A - B), sec {A + B), CSC {A + B), tan 2 ^, cot 2 J., sec 2 5, esc 2 B, in cases (1) and (2). EXERCISE XVI Examples for Review Prove each of the following identities : 1. sin (x + y) _ tan X + tan y 5. „ . 2 - sec2^ cos 2 ^ = sin (x — y) tan X — tan y sec^^ cos (x + y) _ 1 - tan X tan y 6. sec 2 ^ = cos (X - ?/) 1 + tan X tan y csc2 A -2 cos {x + y)_ . „ ^ 2 tan J. i cot X - tan y. v. sin2 J. = - -. sm X cos y 1 + tan2^ cos (x - y) _ sin 3 J. — sin A ■. tan X + cot y. 8. = tan A cos X sm y cos 3 J. + cos A 9. Express sin (3 x/ 4), cos (3 x/ 4), and tan (3 x/ 4) in terms of cos(3x/2). 10. Express sin (3 x/ 4), cos (3 x/ 4), and tan (3 x/ 4) in terms of the trigonometric ratios of 3x/8. Prove each of the following identities : 11. (sin4 + cos^)2 = 1 +sin2 J.. 12. (sin A — cos A)'^ = 1 — sin 2 J.. 13. tan ^ + cot ^ = 2 CSC 2 A. 14. cot J. - tan ^ = 2 cot 2 ^. tan ^ + tan 5 15. = tan A tan B. cot A + cot B 16. Given Sin ^ =2/3, COS B= 1/2, to find (1) sin (^ + B), sin(^-B), cos{A-\-B), cos{A-B), sin2^, cos2^, sin2B, cos2B; (2)tan(^4--B), cot (A + B), tan {A - B), cot {A - B), tan 2 ^, cot 2 A, tan 2 J5, cot 2 B. Prove each of the following identities : ,„ cot^ + cotB ,„ ., ,„ , .. 17. = cos IB - A) sec (B + A). cot ^ - cot 5 ^ ' ^ ' IDENTITIES 6; cos^^ COS =2 B 19. — = tan (A + B) tan (A — B). l-tan2^tan2^ v / v y 20. V2 sin (A ± 45°) = sin A ± cos A. 21. 2 sin (45° - A) cos (45° + B) = cos (^ - ^) - sin (^ + i?). 22. 2 sin (45° + A) cos (45° + 5) = cos {A + B) + sin (^ — 5). 23. 2 sin (45° + A) cos (45° - B) = cos {A - B) -}- sin (^ + 5). cot ^ - 1 11 - sin 2 ^ 1 — sin 2 ^ 24. cot(^+45°)^ :::-:_^^ -^^ p.nt y4 25. cot (^ - 45°) = cot J. + 1 \ 1 + sin 2 ^ cos 2 -4 cot ^ + 1 tan ^ + 1 1 — cot A tan A — I 26. tan {A ± 45°) + cot {A =F 45°) =0. 27. sin9ic — sin 7x = 2cos8xsinx. 28. cos 7 X + cos 6 X = 2 cos 6 X cos x. - _ sin 3 X - sin X . _ 29. = cot 2 X. cos X — cos 3 X _ - sin 5 X — sin 2 X ^ 7 x 30. = cot cos 2 X — cos 5 X 2 31. sin A -{■ sin B cos ^ + cos B cos A — cos B sin B — sin A 32. tan (x/ 2 + 45°) = tan x + sec x. CHAPTER IV SOLUTION OF RIGHT TRIANGLES WITH LOGARITHMS 41. In Chapter I, right triangles were solved without loga- rithms. In general, however, arithmetic computations are much abbreviated by using logarithms. It is assumed that the student is already familiar with the theory of logarithms from the study of Algebra ; but to bring to mind those proper- ties of logarithms which adapt them to shortening arithmetic computations, a brief review is given below. 42. Logarithms. If a = N, (1) then X, the exponent of a, is called the logarithm of N to the base a, which is written in symbols x = \og,N. (2) Equations (1) and (2) are equivalent ; (2) is the logarithmic form of writing the relation between a, x, and N given in (1). ^.gr., since 3^^ = 9, 2 is the logarithm of 9 to the base 3 ; i.e. + 2 = logs 9. Since 2~^ = 1/8, -3 = log2(l/8). Since 4^^^ = 8, +3/2 = log4 8. - Ex. 1. Express in the logarithmic form each of the following relations: 3* = 81, 4^ = 64, 6^ = 216, n<^ = 6, 5"^ = 1/125, 3"^ = 1/243. Ex. 2. Express in the exponential form each of the following relations : logs 125 = 3, log2 32 = 5, log4 64 = 3, \ogcM=h, log2(l/16) = -4. Ex. 3. When the base is 10, what is the logarithm of 1 ? 10 ? 100 ? 1000 ? 10000 ? 100000 ? 0.1 ? 0.01 ? 0.001 ? 0.0001 ? 0.00001 ? Ex. 4. What is the number when the base is 10 and the logarithm is0?l?2?3? -1? -2? -3? -4? 64 PROPERTIES OF LOGARITHMS 65 43. Properties of logarithms. Since logarithms are exponents, from the general laws of exponents we obtain the following general properties of logarithms to any base. (i) The logarithm of the product of two or more arithmetic numbers is equal to the sum of the logarithms of the factors. Let JW = a^, N = av. Then Mx N=a^+v. Hence loga(-M' + N) = x -{■ y = loga-Sf + logai^. (ii) The logarithm of the quotient of two arithmetic numbers is equal to the logarithm of the dividend minus the logarithm of the divisor. Let Jf = a^, N = aM. Then M ^ N = a^-v. Hence loga {M ^ N) = x - y = loga M — loga JV^. (iii) The logarithm of any power of an arithmetic number is equal to the logarithm of the number multiplied by the 'exponent of the power. Let M = a^. Then, for all real values of p, we have M^ = aP'^. Hence loga (^^) = px = p logaM. (1) If p = 1/r, from (1) it follows that (iv) The logarithm of any root of an arithmetic number is equal to the logarithyn of the number divided by the index of the root. An expression is said to be adapted to logarithmic computa- tion when it involves only products, quotients, powers, or roots. E.g., x'^y^'^ /z^ is adapted to logarithmic computation; for we have loga (x^y' /'■ /z^) = c \ogaX + (1 /r) loga y - 8 log„ z. (1) Observe that only the arithmetic value of a product, quotient, power, or root is obtained by logarithms ; the quality must be determined by the laws of quality. 66 PLANE TRIGONOMETRY Logarithms do not aid in the operation of addition or of subtraction. But when, as in formulas [19] • • • [22], a sum or a difference is identical with a product, the sum or difference can be obtained by computing the product. E.g., loga (x2 - y2) = iog„ [(x + y) {x - y)']= loga (a; + 2/) + loga {x-y). 44. Common logarithms. The logarithms used for abridg- ing arithmetic computations are those to the base 10 ; for this reason logarithms to the base 10 are called common logarithms. Thus the common logarithm of a number answers the qyiQS- tion, What power of 10 is the number? IMost numbers are incommensurable powers of 10; hence most common logarithms are incommensurable numbers, whose approximate values we usually express decimally. E.g., the common logarithm of any number between 10 and 100 lies between +1 and +2 ; of any number between 1 and 10 lies between and +1 ; of any number between 0. 1 and 1 lies between —1 and ; of any number between 0.01 and 0.1 lies between -2 and -1 ; etc. Hence the common logarithm of any number between 10 and 100 is +1 + a positive decimal ; of any number between 1 and 10 is + a positive decimal ; of any number between 0. 1 and 1 is -1 + a positive decimal ; of any number between 0.01 and 0.1 is -2 + a positive decimal. 45. Characteristic and mantissa. A logarithm is said to be in the type form when it is expressed as the sum of an integer, positive or negative, and a positive decimal fraction; in this form the integer is called the characteristic, and the fraction the mantissa. In the following pages, when no base is written the base 10 is understood. A negative characteristic, as ~1, is usually written in the form 1 or 9 — 10; ~2 in the form 2 or 8 — 10; etc. CHARACTERISTICS 67 The second form, which is usually the more convenient for negative chai-acteristics, is sometimes used even when the chai-acteristic is positive. E.g., log 434.1 = 2.63759 ; +2 is the characteristic and +.63759 is the mantissa ; log 0.0769 = 2.88593, or 8.88593 - 10 ; -2, or 8 - 10, is the characteristic, and +.88593 is the mantissa. In the first form of writing a negative characteristic, the sign — is written above the characteristic to show that this sign affects the char- acteristic only. One practical advantage of the second form is that we can make the positive part of any logarithm as large as we please, or the negative part any multiple of 10 we please. E.g., log 0.0769 = 2.88593 = 8.88593 - 10 = 18.88593 - 20 = • • -. Also, log 434.1 = 2.63759 = 12.63759 - 10 = 22.63759 - 20 = • • -. 46. The characteristic of the common logarithm of a number is found by the following simple rule : Calling units' place the zeroth place, if the first significant figure in any number M is in the nth place, then the character- istic of log M. is + n or — n, according as this first figure is to the left or to the right of imits' place. E.g., when the first significant figure of a number, as 5348, is in the third place to the left of units' place, then the number lies between lO^ and 10* ; hence its common logarithm is +3 + a mantissa. Again, when the first significant figure of a number, as 0.00071, is in the fourth place to the right of units' place, then the number lies between 10—* and 10— 3 ; hence its common logarithm is — 4 + a mantissa. Let the first significant figure in the number M be in the wth place to the left of units' place ; then M lies between lO" and 10" + 1; that is, IT __ 1 Q« + a positive decimal. .*. log 3/ = +71 -f- a mantissa. Again, let the first significant figure in the number M be in the nth place to the right of units' place ; then M lies between 10-«and lO-e*-^); that is, T^J ^ i Q— n + a positive decimal. .*. log M = -71 4- a mantissa. 68 PLANE TRIGONOMETRY 47. If the expressions of tivo numbers differ only in the posi- tion of the decimal point, the two numbers have the same mantissa. When, in the expression of a number, a change is made in the position of the decimal point, the number is multiplied or divided by some entire power of 10 ; that is, an integer is added to or subtracted from its logarithm ; therefore its man- tissa is not changed. E.g., 34.271 x lO^ = 34271. .-. log 34.271 + 3 = log 34271. § 43, (i) Hence the mantissa for 34.271 equals the mantissa for 34271. 48. A convenient formula for computing the common logarithms of whole numbers is log(. + l)=log. + 2m (2^ + 3(2/+ 1)'+-) (=^) where m = 0.434294, and z is any whole number. For the proof of identity (a) see § 97 in Taylor's Calculus or § 322 in Taylor's College Algebra. To compute log 2, put z = 1 in (a) ; to compute log 3, put ^ = 2 ; to compute log 4, we have log 4 = 2 log 2 ; to compute log 5, put ^ = 4 ; and so on. The series in (a) converges more and more rapidly as z increases. Note. Before proceeding farther in this chapter, the student should familiarize himself with the use of logarithmic tables, both of natural ^ numbers and of the trigonometric ratios of angles. An explanation of the tables will be found in the introduction to them. 49. Right-angled triangles. Review §13. Case (i). Ex. 1. In the right triangle ABC, A = 48° 17', and AB = 324 ft.; solve Fig. 30 the triangle. RIGHT-ANGLED TRIANGLES 69 C B = 4\° 43', Given V ~ o^/' ' to find ^ a = 241.85, i = 324; t-fi-d|- 215.6. Construct the triangle ABC, having the given parts. r J5 = 90' - ^ = 41° 43', I Formulas -i a = c sin A, I b = c cos A. ., . , , rioga = logc + logsin^, Logarithmic formulas [^^^^^ ^^g , + j^g ^os ^. log c = 2. 51055 log c = 2. 51055 log sin ^ = 9.87300- 10 log cos ^ = 9.82311 -10 .-. log a = 2.38355 .-. log b = 2.33366 .-. a = 241.85. .-. b = 215.6. In Chapter I we checked, or verified, the calculated values by construc- tion and measurement. But these values are more usually checked, or tested, by using some known relation between the sides and angles which has not been employed in solving the triangle. Thus, in the example above, we might use either the relation a2 = c^ - 62 or tan ^ = a/ 6 as a check ; but the former is the better. Check. a2 = (c + b) (c - b). ' log (c + 6) = 2.73207 Here c + 6 = 539.6, log (c - 6) = 2.03503 c- 6 = 108.4. .-.log a = 4.76710/2 = 2.38355. As this value of log a is the same as that obtained in the solution above, the answers are probably correct to four figures. Before using the tables the student should make a complete outline of the computation (such as he would have by erasing the second members of the equations following the logarithmic formulas). Note 1. The direction above enables the student to save time by writing at once all the logarithms that are found at one place in the table. Thus we find log sin A and log cos A at the same time ; then having both log a and log b, we next find a and b. Note 2. When the student has become familiar with logarithmic computations, he need not write the logarithmic formulas. By a glance 70 PLANE TRIGONOMETRY at the trigonometric formulas he will know how to combine the logarithms in the computation and can arrange his work accordingly. Note 3. As a check formula, we use a^ = (c -\- b) (c (c + a) (c — a), according asc — 6orc — ais the greater. b) or 62 Case (ii). Ex. 2. In the right triangle ABC, AB = 18.7 ft. and CB = 1G.98 ft.; solve the triangle. Given rc= 18.7, \a = 16.98; to find A = 65° 14', B = 24° 46', b = 7.8339. Fig. 31 Logarithmic formulas Construct the triangle ABC, having the given parts. sin^ = a/c, Formulas -I ^ = 90°-^, b = a cot^. f log sin J. = log a — log c, (^ log & = log a + log cot A. loga = 11.22994 log c = 1.27184 10 loga = 1.22994 log cot A = 9.66404 - 10 log sin A = 9.95810 - 10 " .-. log b = 0.89398 .-. A = 65° 14', .-. B = 24° 46'. .-. b = 7.834. Check. a2 = (c + b) (c - 6). log {c + b) = 1.42380 Here c + 6 = 26.534, log (c - &) = 1-03607 c-6 = 10.866. .-.loga^ 2.45987/2 = 1.22994. As this value of log a is the same as that obtained from the table, the answers are probably correct to four places. Ex. 3. Given to find {:: 194.5, 233.5; A = 39° 47' 36' B = 50° 12' 24' c = 303.9. Construct triangle ABC, having the given parts. Formulas Fig. 32 tan^ = a/6, J5 = 90° - A, c = b sec A = 6/cos A. ISOSCELES TRIANGLES 71 log a = 12.28892 - 10 log b = 12.36829 - 10 log b = 2.36829 log cos A = 9.88557 - 10 .-. log tan A = 9.92063 - 10 .-. log c = 2.48272 .-. A = 39° 47' 36". .-. c = 303.89. .-. B = 50° 12' 24". Observe that the subtraction above is simplified by writing the char- acteristic 2 of log a and log b in the form 12 — 10, and the characteristic — 1 of log cos A in the form 9 — 10. Check. b^= {c + a) {c - a). log {c + a) = 2.69758 Here c + a = 498.4, log {c - a) = 2.03902 c-a = 109.4. •• logft = 4.73660/2 = 2.36830. As this computed value of log b differs by only .00001 from that found in the table, the computed parts are probably correct to four places. EXERCISE XVn Solve the triangle ABC, having given : 1. B = 67°, a = 5. 9. a = 3.414, b = 2.875. 2. A= 38°, a = 8.09. 10. A = 46° 23', c = 5278.6. 3. ^ = 15°, c = 7. 11. a = 529.3, c = 902.7. 4. 5 = 50°, b = 20. 12. B = 23° 9', b = 75.48. 5. a = 0.35, C = 0.62. 13. B = 18° 38', c = 2.5432. 6. a = 273, b = 418. 14. A = 31° 45', a = 48.04. 7. b = 58.6, c = 76.3. 15. b = 617.57, c = 729.59. 8. ^ = 9°, 6 = 937. 16. B = 82° 6' 18", a = 89.32. 50. Isosceles triangles. In an isosceles triangle the perpen- dicular from the vertex to the base divides the isosceles triangle into two equal right triangles. Hence any two parts which determine one of these right triangles determine also the isosceles triangle. 72 PLANE TRIGONOMETRY In this and the next article we shall use the following nota- c. \ tion in isosceles triaDgles : r = one of the equal sides, c = base, V h \ h = altitude, ^ = one of the equal angles, \ C = angle at the vertex. / % \ Q = area of the triangle. \ 7-> . B ^ Ex. 1. Given r and c; to find A, C, A, Fig. 33 and Q. A=. cos-i(c/2r), = 180° -2^, h = Vr2 - (c/2)2 = V(r + c/2) (r - c/2). Also, h = r sin ^ or 7i = (c/2) tan ^. Q = ch/2. 51. Regular polygons. Lines drawn from the center of a regular polygon of n sides to the vertices divide the polygon into n equal isosceles triangles ; and the perpendiculars from the center to the sides of the polygon divide these n equal isosceles triangles into 2n equal right triangles. Hence any two parts which determine one of these equal right triangles determine also the regular polygon. Using the notation given in fig. 34, we have C/2 = 3607(2 n) = ISOV^i. li p = the perimeter of the poly- gon and F = the area, we have p = nc, F = ph/2. CA, or r, is the radius of the cir- cumscribed circle and CD, or h, is the radius of the inscribed circle. REGULAR POLYGONS 73 EXERCISE XVm In an isosceles triangle, having given : 1. c and A ; find C, r, h. 2. h and C ; find A, r, c. 3. c and ^; find A, C, r. 4. c = 2.352, C = 09° 49' ; find r, A, ^, Q. 5. h = 7.4847, ^ = 76° 14'; find r, c, C, Q. 6. A barn is 40 x 80 ft. , the pitch of the roof is 45° ; find the length of the rafters and the area of both sides of the roof, the horizontal projection of the cornice being 1 ft. 7. One side of a regular decagon is 1 ; find r, h^ F. 8. The perimeter of a regular dodecagon is 70 ; find r, h, F. 9. In a regular octagon h=l; find r, c, F. 10. The area of a regular heptagon is 7 ; find r, /i, p. 11. The side of a regular octagon is 24 ft. ; find h and r ; also find the difference between the areas of the octagon and the inscribed circle, and the difference between the areas of the octagon and the circumscribed circle. 12. The side of a regular heptagon is 14 ft. ; find the magnitudes as in example 11. 13. Each side of a regular polygon of n sides is c ; show that the radius of the circumscribed circle is equal to (c/2) esc (180°/n,), and the radius of the inscribed circle is equal to (c/2) cot (180°/n). 14. The radius of a circle is k; show that each side of a regular inscribed polygon of n sides is 2A:sin (180°/n), and that each sido of a regular circumscribed polygon is 2 fc tan (180°/ n). 15. The area of a regular polygon of sixteen sides inscribed in a circle is 100 sq. in. ; find the area of a regular polygon of fifteen sides inscribed in the same circle. 16. The radius of a circle is 10 ; find the area between the perimeters of two regular polygons of thirty-six sides each, one circumscribing the circle and the other inscribed in it. CHAPTER V SOLUTION OF TRIANGLES IN GENERAL 52. The two following relations which the sides of any triangle bear to the sines and cosines of its angles are funda- mental in the study and solution of triangles. They are called the law of sines and the law of cosines respectively. Law of sines. The sides of a triangle are proportional to the sines of their opposite angles. Let A^ B, C denote the numerical measures of the angles of the triangle ABC, and a, b, c the numerical measures of its sides. From C draw CD _L BA, or BA produced. In fig. X, A is acute; in fig. y, A is obtuse. In each figure we have by §7, ^mB=p/a, (1) by § 21, sin A = DC /AC =p/b. (2) Dividing the members of (2) by those of (1), we obtain sin A /sinB = a/b, or = a sin B / sin A, I c = asin C/sinA. flogb = log a + log sin B — log sin ^, \ log c = log a + log sin G — log sin A. Fig. 38 loga= 2.25527 log sin B = 9.98636 - 10 12.24163 - 10 log sin A = 9.78934 - 10 .-. log b = 2.45229 .-. b = 283.33 Check. b + a _t3in^{B + A) b — a tan 1{B — A) b + a = 463.33. b-a = 103.33. loga = log sin C = 12.21695 log sin A = 9.78934 .-. logc= 2.42761 .-. c = 267.68 2.25527 9.96168 - 10 10 10 log(6 + a) = 2.66589 log(6-a) = 2.01423 log quotient = .65166 (5 + ^)/2 = 56°51'30'^ (5-^)/2 = 18°5r30^ log tan 1{B + A) = 10. 18514 - 10 logtan ^{B-A)= 9.53347 - 10 log quotient = .65167 (1) As the logarithms of the two members of (1) differ by only 1 in the fifth place, the value of b is correct to four places. Similarly we can check the value of c. OBLIQUE TRIANGLES 81 EXERCISE XTX Solve each of the following triangles : 1. Given B = 60° 15', C = 54° 30', a = 100. 2. Given ^ = 45° 4r, C = 62° 5', 6 = 100. 3. Given B = 70° 30', C = 78° 10', a = 102. 4. Given A = 65°, B = 65°, c = 270. 5. Given a = 123, B = 29° 17', C = 135° 6. Given 6=1006.62, A = 44°, C = 70°. 7. A ship S can be seen from each of two points A and B on the shore. By measurement, AB = 800 ft., Z SAB = 67° 43', and Z SBA = 74° 21' 16". Find the distance of the ship from A. 8. A flag pole A is observed from two points B and C, 1863 ft. apart. Given Z BCA = 36° 43' and Z CBA = 57° 21', find the distance of the flag pole from the nearer point. 9. To determine the distance of a hostile fort A from a place B, a line BC and the angles ABC and BCA were measured and found to be 322.55 yd., 60° 34', and 56° 10' respectively. Find the distance AB. 10. A balloon is directly over a straight level road, and between two points on the road from which it is observed. The points are 15847 ft. apart, and the angles of elevation are found to be 49° 12' and 53° 29' respectively. Find the distance of the balloon from each point of observation. 11. To find the distance from a point ^ to a point B on the opposite side of a river, a line AC and the angles CAB and ACB were measured and found to be 316.32 ft., 68° 43', and 57° 13' respectively. Find the distance AB. 12. From points A and B, at the bow and stern of a ship respectively, the foremast, C, of another ship is observed. The points A and B are 300 ft. apart; the angles ABC and BAC are found to be 65° 31' and 110° 46' respectively. What is the distance between the points A and C of the two ships ? 82 PLANE TRIGONOMETRY 58. Case (ii). Given two sides and an angle opposite one of them. Let a, b, A be the given parts. Then, to find B, C, c, we have - ^ 7 ■ . , sinB = osinA/a, (1) C = 180° -(A -\- B), c = a sinC/sin .4. Since two supplementary angles have the same sine (§ 30), the relation in (1) gives in general two values for B, both of which are to be taken unless one is excluded by the conditions of the problem. We have to consider the three following cases : I, when a> h\ II, when a = h\ III, when a < b. I. When a > b and A is acute or obtuse, then A > B ; hence B is acute, and there is but 07ie triangle having the given parts. II. When a = b and A is acute, then A = B ; hence B is acute and the triangle is isosceles. In this case the triangle can be solved by the method in § 50 or by the law of sines and the relation A + B -\- C = 180°. If a = 6 and A = or > 90°, the triangle is impossible. Why ? Note. Example 1 below and the first three examples in Exercise XX may be solved before Case III is considered. Bi J) By ^ A y Q V a A rn B ., Fig. 39 III. When a < b and A is acute, there are two triangles, one, or no triangle, having the given parts, according as &">,==, or < b sin A. Geometric proof. In each figure, let Z XA C = A and AC = b. OBLIQUE TRIANGLES 83 Draw CD A. AX] then in each figure CD = b sin A. With C as a center and a as a radius, describe the arc mn. li a > b sin A (i.e. if a > CD), the arc mn will cut AX (fig. x) in two points, Bi and B2, on the side of A toward D, and there will be two unequal triangles having the parts a, b, A, viz., the triangles ACB^ and ACB^. Hence B has the two values Z. AB]^C and Z AB^C, which are supplementary. If a, = ^ sin .4, the arc mn will touch AX at D (fig. y); hence B = 90° and only the right-angled triangle A CD has the given parts. If a < b sin ^, the arc mn will not meet AX (fig. z) ; hence no triangle can be constructed with the given parts. E.g., if a = 5, 6 = 7, and A = 30°, then a 6 sin J. ; hence there will be two triangles having these parts. If a < 6 and ^ = or > 90°, the triangle is impossible. Why ? Trigonometric proof. From the law of sines, sin B = b sin A /a, or (b /a) sin A. (1) Also ^ < 5 and ^ -f ^ < 180°. (2) If a > ^ sin A, b sin A /a < 1, whence from (1), sin 5 < 1 ; hence B has two unequal values, which are supplementary. Since al, whence from (1), sin B > sin A ; hence each of the two supplementary values of B will satisfy both the conditions in (2). Therefore B has two values and there are two different triangles having the given parts (fig. x). If a = b sin A, sin J5 = 1 in (1), whence B = 90° ; hence the required triangle is right angled at B (fig. y). If a 1, which is impossible; hence the triangle is impossible (fig. z). From the trigonometric proof, it follows that if ab and ^ < 90° ; hence B < 90° and there £ is only one triangle having the given parts. Fig. «) Construct the triangle ABC, having the given parts. sin B = bsmA/a, Formulas •{ C = 180° - {A -\- B), c = asm C /sin A. log 6= 2.38021 log sin J. = 9.97837 - 12.35858 - loga= 2.39794 log sin B= 9.96064- 10 10 ■10 loga= 2.39794 log sin C= 9.82517-10 12.22311 - 10 log sin ^= 9.97837 -10 log c = 2.24474 .-. B = 65° 58' 24" C = 180° - (72^ .-. c = 175.69. > 4' + 65° 58' 24") = 41° 57' 36". Check. 6 + c tan i (5 + 0) b-c tan 1(5-0) ( Here b + c = 415.69, b-c = 64.31, i(B+ C) = 53°58'; i(B-C) = 12°0'24". log (6 + c) = 2.61877 log {b-c) = 1.80828 log tan i (B + C) = 10. 13821 - 10 log tan 1(5 - 0) = 9.32772 - 10 (1) log quotients .81049 log quotients .81049 As the logarithms of the two members of (1) are equal, the values obtained above are correct. Ex. 2. How many triangles are there which have the following parts ? (i) a -I 70, 6 = 90, A = 30°. (ii) a = 40, 6 = 80, A = 30°. (iii) a = 20, 6 = 50, A = 30° (iv) a = 70, 6 = 75, A = 60°. OBLIQUE TRUNGLES 85 Ex. 3. Given a = 732, b = 1015, to find A = 40°: B = 63° 2' 20" or 116° 57' 40" C = 16° 57' 40" or 23° 2' 20" c = 1109.4 or 445.66. Here ab sin A; hence by III there are two solutions. Construct the two triangles A CBi and A CB2, having the given parts. Formula for B. sin B = bsinA/a. log 6= 3.00647 log sin A = 9.80807 - 10 log a 12.81454 2.86451 10 .-.log sin ^= 9.95003-10 .CBi = 63° 2' 20", B2 = 116° 57' 40".-^ e-- jL To find the unknown parts of At^i, we have ZACBi = 180° -{A + Bi) = 76° 67' 40" ABi = a sin ^ CBi/sin A. .0 Fig. 41 To find the unknown parts of A CB2, we have ZACB2 = 180° -{A + B2) = 23° 2' 20". AB2 = a sin ^ C-S2/sin A. \oga= 2.86451 log sin A CBi = 9.98866 - 10 Check. 12.85317 - 10 log sin A = 9.80807 - 10 .'AogABi = 3.04510 .-. ABi = 1109.4. c + b tani(C + J5) c-b tan 1(0 -B) loga= 2.86451 log sin A CB2 = 9.59257 - 10 12.45708 - 10 log sin A = 9.80807 - 10 .-.log ^^2= 2.64901 .-. ABi = 445.66. (1) 86 PLANE TRIGONOMETRY Here c + 6 = 2124.4, c-b= 94.4, log (c + 6) = 3.32723 log {c-b) = 1.97497 log quotient = 1.35226 1 log tan 1 (C + 2^) = 10.43893 - 10 log tan U<^- ^) = 9.08670 - 10 log quotients 1.35223 The equality of these logarithms to five figures verifies the answers to four figures. EXERCISE XX Solve the following triangles : 1. Given a = 145, 6 = 178, 2. Given h = 573, c = 394, 3. Given a = 5.98, 6 = 3.59, 4. Given a = 140.5, 6 = 170.6, 5. Given 6 = 74.1, c = 64.2, r 6. Given a = 27.89, ^ 6 = 22.71, ^.— 7. Given 6 = 45.21, c = 50.3, 8. Given a = 34, 6 = 22, 9. Given a = 55.55, 6 = 66.66, YUO. Given a = 309, 6 = 360, B= 41° 10'. B = 112° 4'. A = 63° 50'. A= 40°. C= 27° 18'.' 4? : ^.39-^6 11. Given 6 = 19, c = 18, 5= 65° 38'. B= 40° 32' 7". B= 30° 20'. 5= 77° 44' 40". A= 21° 14' 25". C= 15° 49'. 59. Case (iii). Given two sides and their included angle. Let a, b, C be the given parts. Then, to find A, B, c, we have tan 7 cot —y a + b 2 (1) (^+jB)/2 = 90°- C/2, (2) c = a sin C/sin ^. (3) From (1) we obtain {A — B)/% Having given {A—B)/2 and {A + 5)/2, we readily obtain A and B. Then c can be found by (3). (a Example. Given = 540, 6 = 420, C=52°6'; to find r^ = 78°17'40", 49° 36' 20", 435.15. 1" OBLIQUE TRIANGLES 87 Formulas tan B a Here 6 C cot — , a + b 2 (A H- 5)/2 = 90° - C/2 = 63° 57', c = asin C/sinA. a + 6 = 960, a-b= 120, C/2 = 26° 3'. log (a - b) log cot (C/2) 2.07918 0.31086 12.39004 2.98227 10 log (a + 6) ,-. log tan ^ (^ - ^) = 9.40777 - 10 .-. {A -5)/2 = 14°20'40". Also (^ + J5)/2 = 63°67'. loga= 2.73239 log sin C = 9.89712 - 10 12.62961 - 10 log sin A = 9.99087 - 10 .-. log c = 2.63864 .•.c = 436.15. .-. A = 78° 17' 40", and B = 49° 36' 20". Check. ^ + B + C = 52° 6' + 78° 17' 40" + 49° To check c we could use c = 6 sin C/sin B. 20" = 180°. EXERCISE XXI Solve each of the following triangles : 1. Given a = 266, Given b = 91.7, Given a = 960, Given a = 886, Given b = 41.02, b = 362, c = 31.2, b = 720, b = 747, c = 45.49, C = 73°. d t 1(^-3'*'^^ ^ = 33°7'9".i^-''^2'-'^''*'^ C = 25° 40'. C = 71° 54'. A^ ^ *^ i*^ ? 62<'9'38".^-/^-;;;i,^ The dis- 3. 4. 6. Two trees A and B are on opposite sides of a pond tance of A from a point C is 297.6 ft., the distance of B from C is 864.4 ft., the angle AC Bis 87° 43' 12". Find the distance AB. 7. Two mountains A and B are respectively 9 and 13 miles from a town C, and the angle ACB is 71° 36' 37". Find the distance between the mountains. 8. Two points A and B are visible from a third point C, but not from each other. The distances AC, BC, and the angle ACB were measured and found to be 1321 ft., 1287 ft., and 61° 22' -respectively. Find the distance AB. 9. From a point 3 mi. from one end of an island and 7 mi. from the other end, the island subtends an angle of 33° 55' 15". Find the length of the island. 88 PLANE TRIGONOMETRY 10. Two stations A and B on opposite sides of a mountain are both visible from a third station C. The distance AC, BC, and the angle ACB were measured and found to be 11.5 mi., 9.4 mi., and 59° 31' respectively. Find the distance from A to B. 11. Two trains leave the same station at the same time on straight tracks that form an angle of 21° 12'. Their average speeds are 40 mi. and 50 mi. an hour respectively. How far apart will they be at the end of the first thirty minutes ? 60. Case (iv). Given the three sides. To solve this case we first obtain formulas for the sine, cosine, and tangent of A /2, B /2, and C/2 in terms of the three sides a, h, c. Let 2s = a-{-b + c. Then 2{s — a) = h + c — a, 2 (s — b) = a -\- c — b, 2 (s — c) = a -\- b — c. By [16] and [17], we have sin2(^/2) = (l -cos^)/2, cos^(A/2) =(1 + cos A)/2. Substituting for cos A its value given in [25], we obtain ^^^2=2^^ 2b^) ^"^ 2==2V^ + ^2^ 4: be 4: be _ (a — b-\-c)(a-\-b-c) _ (b -hc + a)(b-^c-a) ~ Abe 4 be (1) 4:(s — b) (s — c) _4:S (s — a) Abe Abe A (s-b)(s-c) A s(s-a) .•.sin- = \^ P -. .•.cos- = A- ^ 2 \ be 2 \ -b)(8- s(s — a) be ^. .-,. A /(s _ b)(s - c) Dmdmg, ^''°2=V s(s - a) ' by(l) K28] THREE SIDES GIVEN 89 Observe that s is the half -perimeter, a the side opposite, and b and c the sides including the angle A, Ex. 1. State in words the relations in [28]. Ex. 2. Write the values of the sine, cosine, and tangent of 5/2; C/2. Since the trigonometric ratios of A/ 2, B/2, C/2 must all be positive (?), only the positive roots in [28] are taken. Formulas [28] enable us to obtain any angle of a triangle from the sides, either through its sine, its cosine, or its tangent. Ex. 3. If a = 13, 6 = 14, c = 15, find A, B, C, having given 1/2 = tan 26° 34', 4/7 = tan 29° 45', 2/3 = tan 33° 41'. Here s = (13 + 14 + 15)/2 = 21, s - a = 21 - 13 = 8, s - 6 = 21 - 14 = 7, s - c = 6. Hence tan | = yj^^ = 1 = tan 26° 34'. .-. A/2 = 26° 34', or ^ = 53° 8'. Again, ~~~^~^^tan — = -\ / = - = tan 29° 45'. ^ ' ^^^2 \21-7 7 .-. B/2 = 29° 45', or B = 59° 30'. Similarly ^ C/2 = 33° 41', or C = 67° 22'. Check. ^ + 5 + C = 53° 8' + 59° 30' + 67° 22' = 180°. Ex. 4. If a = 35, 6 = 84, c = 91, find A, B, C, having given tan 11° 19' = 1/5, and tan 33° 41' = 2/3. Ex. 5. If a = 13, 6 = 14, c = 15, find the sines of ^/ 2, 5/2, 0/2. If we multiply the value of tan(.l /2) in [28] by 1 in the form Vs — a / Vs — a, the expression for tan (.1/2) can be put in the more symmetrical form tan:^ = -l- / (s-a)(8-b)(s-c) 2 s — a \ 8 Letting r = ^(s - a) (s - b) (s - c)/s, [29] we have tan(A/2) = r/(s — a).^ Similarly tan (B / 2) = r / (s - b), I [30] tan (C/2) = r/(s-c).J 90 PLANE TRIGONOMETRY When all the angles are required, the tangent formulas [30] are the best, as they involve the fewest logarithms. In § 63 it will be proved that r in [29] is the radius of the circle inscribed in the triangle ABC. Ex. 6. Given Formulas a = 130, 6 = 123, to find c =77; A = 77° 19' 9", B = 67° 22' 49'', C = 36° 18' 2". r = V(.s — a) (s — b) (s — c)/s, tan (A /2) z= r / (s — a), tan (5/2) =r/{s- 6), tan(C/2) = r/{s- c). Here s = 165, s \og{s-a)= 1.54407 log(s -b) = 1.62325 log {s- c) = 1.94448 5.11180 log s = 2.21748 .-. log r = 2.89432/2 a = 35, s - 6 = 42, s - c = 88. .-. log tan {A/2) = 9.90309 - 10, whence ^/2 = 38° 39' 34.6" ; log tan (B/2) = 9.82.391 - 10, whence B/2 = 33° 41' 24.4"; log tan (C/2) = 9.50268 - 10, whence C/2 = 17° 39' 1". = 11.44716 - 10. Check. A + B+C = 11°W 9" - 67° 22' 49" + 35° 18' 2" = 180^ EXERCISE XXII Solve each of the following triangles : 1. Given a = 56, b = 43, c:^49. 2. Given a = 8.5, 6 = 9.2, c = 7.8. 3. Given a = 61.3, b = 84.7, c = 47.6. 4. Given a = 705, b = 562, c = 639. 5. Given a = .0291, b = .0184, c = .0.358. 6. Given a = 85, b = 127, A = 26° 26'. 7. Given a = 5.953, b = 9.639, C = 134°. 8. Given a = 3019, b = 6731, c = 4,228. 9. Given a = 60.935, c = 76.097, A = 133° 41' THREE SIDES GIVEN 91 10. Given h = 74.8067, c = 98.738, C = 81° 47'. 11. Given b = 129.21, c = 28.63, A = 27° 13'. 12. Given a = 2.51, 6 = 2.79, c = 2.33. 13. Given a = 32.163, c = 27.083, C = 52° 24' 16". • 14. Given a = 74.8, c = 124.09, B = 83° 26' 52". 15. Given a = 86.0619, c = 63.5761, A = 19° 12' 43". 16. Given a = 93.272, b = 81.512, C = 58°. 17. The sides of a triangular field are 534 ft., 679.47 ft., and 474.5 ft. Find the angles. 18. A pole 13 ft. long is placed 6 ft. from the base of an embankment, and reaches 8 ft. up its face. Find the slope of the embankment. 19. A point P is 13581 in. from one end of a wall 12342 in. long, and 10025 in. from the other end. What angle does the wall subtend at the point P ? 20. The distances between three cities A, B, and C are as follows : ^J5 = 165 mi., .4C = 72 mi., and BC = 185 mi. B is due east from A. In what direction is C from A? 21. Under what visual angle is an object 7 ft. long seen when the eye of the observer is 5 ft. from one end of the object and 8 ft. from the other end ? 22. Prove sin ^ = 2 Vs (s - a) {s - b) (s - c) / (6c). 23. Prove cos^ = s (s - a) /(6c) - (s - 6) (s - c) /(6c). 24. If a = 18, 6 = 24, c = 30, show that sin ^ = 3/5. 25. The sides of a triangle can be substituted for the sines of their opposite angles, and vice versa, when they are involved homogeneously in the numerator and denominator of a fraction, or in both members of an equation. For this substitution is the multiplication of each term by the equal numbers, a /sin A, 6 /sin jB, c/sin C, or their reciprocals. E.g., sin A = sin (180° - A) = s\n{B-\- C). .'. sin A =smB cos C + sin O cos B. by [7] Multiplying the first term hy a/sinA, the second by 6 /sin B, and the third by c/sin C, we obtain 92 PLANE TRIGONOMETRY a = b cos C + c cos B. ^ Similarly b = a cos C + c cos A, ^ (1) and c = acosB -{- bcos A. J 2c2 + a2 2sin2C + sin2^ 26. Prove 3abc 3 6 sin A sin C 27. Multiplying the equations in (1) of example 25 by — a, 6, and c respectively, and adding, we obtain the law of cosines, 62 + c2-a2 = 26ccos^. 28. Prove the relations in (1) of example 25 directly from a figure. Suggestion. See the figures in § 52. 61. Area of a triangle. (i) In terms of two sides and the sine of their included angle. Let F denote the area of any triangle, as ^J5C in § 52. Then F=^^ .i)C/2 = be sin A/2. [31] (ii) In terms of the three sides. By [31] and [13], we have F=bcsm(A/2)GOs(A/2). Hence, by [28], F = Vs(s-a) (s-b)(s-c). [32] (ill) In terms of one side and the sines of the angles. In [31], putting for c its value b sin C/sin .B (obtained from the law of sines), we have ^ p^b^sinAsinC 2 8inB "- -' EXERCISE XXm 1. State [31], [32], [33] in words. 2. Given A = 75°, 6 = 10, c = 40 ; to find F. 3. Find the areas of the triangles in examples 1-4 in Exercise XXI. . 4. Find the areas of the triangles in examples 1-4 in Exercise XX. 5. Find the areas of the triangles in examples 1-4 in Exercise XXII. 6. Find the areas of the triangles in examples 1-4 in Exercise XIX. THREE IMPORTANT CIRCLES 93 62. Circumscribed circle. Let D denote the diameter of the circle circumscribed about the triangle ABC. ^ ~^\C Through C draw the diameter COHj and join HB. Then Z A = Z. H, Z CBH = 90°, aY^ 7^ -r^B and CH = D. .'. sin ^ = sin ^ = a/D. .'. a/sin A = D. (1) Compare (1) with [23]. Example. Find the radii of the circles circumscribed about the triangles in examples 1-4 in Exercise XIX. 63. Inscribed circle. Let r denote the radius of the circle inscribed in the triangle ABC. Join the center with the points of contact Z>, Ej F. Draw OAj OB, OC. C FlO. 42 Then OD = OE = OF = r. A ABC = A BOC -{- A COA -\-AAOB. .-. F=ar/2 -{- br/2 + cr/2 = r(a-\-b -\-c)/2 = rs. 60 94 PLANE TRIGONOMETRY Hence r == F/s, or by [32], r = V(s -a)(s- b) {s - g)/s. (1) Compare (1) with [29]. Example. Find the radii of the inscribed circles of the triangles in examples 1-4 in Exercise XXII. 64. An escribed circle of a triangle is a circle which is tangent to one side of the triangle and to each of the other two sides produced. Ex. 1. If ra denotes the radius of the escribed circle tangent to the side a, prove that r« = F/{s -a) = Vs{s-b){s-c)/{s-a). A ABC = AACD-\-/\ ABD - A BCD. .:F=ra{b-\-c-a)/2 = ra{s-a). Fig. 44 Ex. 2. Find the radii of the three escribed circles of the triangles in examples 1-3 in Exercise XXII. CHAPTER VI RADIAN MEASURE, GENERAL VALUES, TRIGONOMETRIC EQUATIONS, INVERSE FUNCTIONS 65. A radian is an angle which, when placed at the center of a circle, intercepts between its sides an arc equal in length to the radius of the circle. That is, if arc AB is equal to the ^ ^, ^^-^^ ' radius OA, then ^^' V \^?- / X / ^^ angle AOB = a radian. / \ / \JU 66. Constant value of the radian. [ . \/ \ r i ^ In fig. 45, let o A Fig. 45 arc AB = OA = r units. Then, by Geometry, we have the following proportion : Z AOB : 180° = arc AB : semicircumference, i.e. a radian : 180° = r : irr. From this proportion, we have TT radians = 180° = 2 right angles. [34] Formula [34] expresses the relation between radians, degrees, and right angles, and should be fixed in mind. From[34],aradian=(180/7r)°, or (180/3.1416)° (1) -57°.295779, or 57°.3, approximately (2) = 57° 17' 44".8, approximately. (3) From [34], 1° = (7r/180) radian, (4) or a right angle = 90° = (7r/2) radians. (5) Hence 270° = (3 tt / 2) radians, and 360° = 2 tt radians. (6) 95 96 PLANE TRIGONOMETRY Ex. 1. Express in radians the angle 8° 15'. 8° 15' = (33/4)° = (33/4) (;r/180) radian by (4) = 0. 144 radian, approximately. Ex. 2. Express in radians the sum of the three angles of a triangle. Ex. 3. Express in degrees the angle 5/8 radian. 5/8 radian = (5/8) (57° 17' 44".8) by (3) = 35° 48' 35". 5. 67. Radian measure of angles. In fig. 45, by Geometry, we have ZAOC : ZA0B = 3,VG ABC : arc AB. (1) Let A AOC = N radians, arc ABC = s units, and arc AB = OA = r units. Substituting these values in proportion (1), we obtain N radians : 1 radian = sir. .-. N = s/r. [35] That is, the number of radians in an angle at the center of a circle is equal to the intercepted arc divided by the radius. Cor. If r = l, N = s. The number of radians in an angle is called its radian (or circular) measure. When the measure of an angle is given in terms of tt or any other numeral or numerals and no angular unit is expressed, the radian is always understood as the unit. U.g., the angle 2 is an angle of 2 radians, and the angle 7r/2 is an angle of 7r/2 radians. The fraction 22/7 gives the correct value of Tt to two places of deci- mals, and for many purposes this value is sufficiently accurate. Ex. 1. An angle at the center of a circle whose radius is 6 ft. inter- cepts an arc of 3 ft. Find the angle in degrees. The angle = (3/5) radian §67 = (3 / 5) (180 / 7t)° = 34°^. § 66 RADIANS 97 Ex. 2. An angle of 42° 20' is at the center of a circle whose radius is 10 ft. Find the length of the arc intercepted between its sides. Let X = the number of feet in the intercepted arc. Then ic/10 = the number of radians in 42° 20', or (127/3)° = (127/3) (;r/ 180). by [34] .-. X = (127 / 64) (22 / 7) = 1j%%, approximately. Hence the intercepted arc is approximately 7^'^/^ ft. long. Ex. 3. From [35] find s in terms of N and r, and r in terms of 8 and JV^. EXERCISE XXIV 1. Express in radians two positive and two negative angles each of which is coterminal with 7t/4; 5;r/4; 3;r/2; 6;r/2; 7t/S; 2n:/S; 7t/6; bit/Q, Express in degrees each of the following angles : 2. 2 7r/3. 3. 5;r/3. 4. bit. 5. 3;r/4. Express in radians each of the following angles : 6. 45°. 8. 90°. 10. 97° 25'. 12. 43° 25' 36". 7. 135°. 9. 270°. 11. 175° 13'. 13. 38° 17' 23". Find the complement and the supplement of : 14. 7r/4. 15. 2 7r/3. 16. 3;r/4. 17. 5;r/3. Find the trigonometric ratios of each of the following angles : 18. Tt/Q. 19. ;r/4. 20. ;f/3. 21. ;r/2. 22. it. 23. Two angles of a triangle are 1/2 and 2/5. Find the third angle in radians, also in degrees. 24. The difference between the two acute angles of a right-angled triangle is n/l. Find the angles in radians, also in degrees. 25. Express in radians, also in degrees, the angle subtended at the center of a circle by an arc whose length is 15 ft., the radius of the circle being 20 ft. i 26. The diameter of a graduated circle is 6 ft., and the graduations on its rim are 5' apart. Find the distance from one graduation to the next. 98 PLANE TRIGONOMETRY 27. Find the radius of a globe which is such that the distance between two places on the same meridian whose latitudes differ by 1° 10' may be half an inch. 28. The value of the divisions on the outer rim of a graduated circle is 5' and the distance between successive graduations is 0.1 in. Find the radius of the circle. 29. Assuming the earth to be a sphere and the distance between two points 1° apart to be 69i mi, , find the radius of the earth. 68. Principal values. If sin 6 = — \ 12, we know that Q is any angle which is coterminal with — 7r/6 or — 57r/6. Of this series of values the smallest in size, — 7r/6, is called the principal value of 0. The principal value of an angle having a given trigonometric i-atio is the angle, smallest in size, which has this ratio. Hence, if sin 6 or esc 6 is positive, the principal value of 6 lies between and 7r/2 ; if sin 6 or esc 6 is negative, the principal value of 6 lies between — 7r/2 and 0. If tan 6 or cot 6 is -f, the principal value of 6 lies between and 7r/2; if tan 6 or cot 6 is — , the principal value of 6 lies between — 7r/2 and 0. If cos 6 or sec 6 is +, the principal value of 6 lies between and 7r/2, preference being given to the positive value; if cos B or sec 6 is — , the principal value of B lies between 7r/2 and TT. 69. Angles having the same sine. Let Z. XOP = A, A XOP' = ir - A. Then sin ZOP' =: sin XOP, esc XOP' = esc XOP, and any angle which has the same sine or cosecant as A is, P' p or can be made, coterminal with .1 X. y^ or TT — A . ^S^-*^ >/ In this chapter n will denote M' -^^^^— j-i -L-x any positive or negative integer, Pig. 46 including zero. TRIGONOMETRIC EQUATIONS 99 When n is even, nir + A includes all the angles, and only those, which are coterminal with A. Now, when n is even, nir -[- A = TiTT + (- 1)M. (1) Again, when n is odd, n — 1 is even, and (n — 1) ir -{- (ir — A) includes all the angles, and only those, which are coterminal with IT — A. But when n is odd, {n-l)ir + {ir-A)=n'Tr -A =mr^-{- 1)M. (2) From (1) and (2) it follows that the expression nir -\- (— 1)"^ includes all the angles, and only those, which are coterminal with A or ir ~ A. Hence the general expression mr4-(— 1)'^A includes all the angles, and only those^ which have the same sine or cosecant as A. That is, the general value of in the equation sin = sin A (or esc ^ = esc ^) is nir + (— 1)"^^ E.g., if sin ^ = sin(;r/3), then d = nTt + {- l)«;r/3. Example. Find the general value of 6, if sin ^ = — 1 /2. (1) The principal value of d in (1) is — 7t/6. Substituting for — 1/2 in (1) its identical expression sin(— ;r/6), we obtain the equivalent equation sin^ = sin(-;r/6). (2) ,.,0 = n7r + {-1)-{-7t/6) = ?i7r-(-l)»-7f/6. (3) By using the principal value of d in (2), we obtain the simplest expres- sion for 6 in (3). Sometimes, however, in solving an equation it is advantageous to use some other than the principal value of the unknown angle. Observe that the unit understood with 6 and A is the radian. If the degree were the unit, we would write d = n- 180° + (— 1)"^. 100 PLANE TRIGONOMETRY Fig. 47 70. Angles having the same cosine. P Let Z XOP = A,Z. XOP' = - A, Then cos XOP' = cos XOP, ,vf and sec XOP' = sec XOP, M ' and any angle which has the same cosine or p* secant as A is, or can be made, coterminal with A or — ^. Now 2 niT ± A includes all the angles, and only those, which are coterminal with A or —A. Hence 2 rnr ± A includes all the angles, and only those, which have the same cosine or secant as A. That is, the general value of 6 in the equation cos ^ = cos ^ {or sec 6 = sec A) is 2 nir ± A. E.g., if cos e = cos(;r/9), then e = 2n7t ± 7t/9, or 2 n • 180° ± 20°. Example. Find the general value of 0, if cos ^ = — V3/2. Substituting for — •v/3/2 its identical expression cos (5 ^/6), we obtain the equivalent equation cos^ = cos(5;r/6). .-. ^ = 2 n;r ± 5 ;r/6, or 2 n • 180° ± 150°. 71. Angles having the same tangent. Let Z XOP =:A,Z XOP' = tt -\- A. Then tan XOP' = tan XOP, cot XOP' = cot XOP, and any angle which has the same tangent or cotangent as A is, or can be made, coterminal with A or TT + ^. When n is even, nir -\- A includes M_ all the angles, and only those, which are coterminal with A. When n is odd, 7^ — 1 is even, and (n — l)TT + {ir-\- A), i.e. nir -\- A, includes all the angles, and only those, which are coterminal with ir -\- A. Xl P m' 'ry^ ^0 • M ' V Fig. 48 TRIGONOMETRIC EQUATIONS 101 Hence nir -j- A includes all the angles^ and only those, which have the same tangent or cotangent as A. That is, the general value of in the equation tan B = tan A (or cot $ = cot A) is nir + A. E.g.^ if cote = cot(- ;r/5), 6 = nit -{- {- it/b), or mt — 7t/6. Example. Find the general value of d, if tan 6 = — y/S. Substituting for — V3 its identical expression tan(— ;f /3), we obtain the equivalent equation tane = tan(— ;r/3). .-. e = n7t - 7t/S, or n • 180° - 60°. 72. A trigonometric equation is an equation which involves one or more trigonometric ratios of one or more unknown angles, as tan ^ = 1 or 2 sin^ $ + V^ cos ^4-1 = 0. To solve a trigonometric equation is to obtain an expression for all the angles which will satisfy it. The first step in solving a trigonometric equation is to solve it algebraically for some trigonometric ratio of the unknown angle ; the second step is to apply one or more of the princi- ples in §§ 69-71. Some elementary types of trigonometric equations are solved below. Ex. 1. Solve the equation sin2e =1/4. First step. sin ^ = ± 1 /2. Second step. .-. sin 6 = sin ( ± tt /6). .■.d = n7t-h{-l)»{±7t/6) = n7f±7e/6. §69 Ex. 2. Solve the equation cos^ ^ = 1 /2. Denote the two values of cos d by cos di and cos 02. Then cos ^i = V2/2 = cos (7r/4), (1) and cos ^2 = - V2 /2 = cos (5 it/ 4). (2) From (1), ^1 = 2 ihtc ± 7r/4. (3) From (2), ^2 = 2 n2?r ± 5 7r/4 = (2 W2 ± 1) ;r ± ;f /4. (4) 102 PLANE TRIGONOMETRY When n is even, mt ±7t /^ includes 2 n\7t db 7^/4 in (3) ; and when n is odd, nit ±Tt /^ includes (2 n2 ± 1) tt ± 7r/4 in (4). Hence Q = mt±, it /^. Observe that, by using 5 7r/4 instead of the principal value of d^ in (2), the two expressions for 6 in (3) and (4) are more readily combined into \ one expression. Ex. 3. Solve the equation 2 sin2 ^ -f V3 cos ^ + 1 = 0. (1) Putting for sin^ d its identical expression 1 — cos^ 6^ we obtain the equivalent equation 2 cos2 6* - V3 cos ^ - 3 = 0, (2) which involves only one function of the unknown angle 6. Factor, (cos 6 - V3) (2 cos 6 + V3) = 0. (3) By Algebra, (3) is equivalent to the two equations cos^=:V3, cos^ = -V3/2. Now, by § 26, cos ^ = V^ is impossible, and cos^ = - V3/2 = cos(5;r/6) gives e = 2mt ±bTt/Q. §70 Ex. 4. Solve the equation tan bd = cot 2 6. Substituting for cot 2 its identical expression tan(7r/2 — 2 0), we obtain the equivalent equation tan50 = tan(7r/2 -2 0). .-. 50 = n7r + (;r/2 -2 0). §71 .-. = (w;r + ;r/2)/7. Ex. 5. Solve tan sec = - y/2. (1) Putting for sec0 its identical expression Vtan^ + 1, we obtain the equivalent equation tan Vtan2 0+1 = - V2. (2) Square, tan* + tan2 = 2. Factor, (tan2 + 2) (tan2 - 1) = 0. Hence tan = ± V- 2 or ± 1. (3) Now tan = ± V— 2* is impossible. Since tan sec is negative in (1), tan and sec are opposite in quality, whence is in the third or fourth quadrant. Hence tan = + 1 gives = 2ii7r + 5;r/4, and tan =: - 1 gives = 2 n7r — 7r/4. TRIGONOMETRIC EQUATIONS 103 Observe that the sohitions of (3) which do not satisfy (1) were intro- duced by squaring (2). Ex. 6. Given sin ^ = — 1/2 and tan ^ = 1 / V3 ; to find the general value of 6. Since sin ^ is — , and tan is + , must be in the third quadrant. The smallest positive angle in the third quadrant which will satisfy sin ^ = — 1/2 is 7 7r/6, and this angle satisfies also tan 6 = l/y/S. Hence d = 2n7C + 7 tc /6, or 2n- 180° + 210°. EXERCISE XXV Solve each of the following equations : 1. sin2(9 = l. 3. tan2^=l. 5. cos2^=l/4. 2. csc-2 6 = 2. 4. cot2 ^ = 3. 6. sec2 ^ = 4/3. 7. 2 sin2 ^ + 3 cos ^ = 0. 10. sin^ ^ - 2 cos 4- 1 /4 = 0. 8. cos2 ^ - sin = 1 /4. 11. sin ^ + cos ^ = V2. 9. 2 V3 cos2 d = sin 0. 12. 4 secS d-1 tan2 = 3. 13. tan + cot = 2. 14. tan2 - (1 + V3) tan + V3 = 0. 15. cot2 + (V3 + l/V3)cot = 1, 6 = tan-i \ = mt -\- it / A., and the principal value of tau-il is ;r/4. If cos ^ = 1 /2, e = cos-i (1 /2) = 2 n;r ± 7r/3, and the principal value of cos-i(l/2) is Tt/Z. If sin^ = V2/2, ^ = sin-i(V2/2) = htt + (- l)«;r/4, and the prin- cipal value of sin-i( V2/2) is ;r/4. Example. If a and c are positive and c < 1, and the inverse functions are restricted to their principal values, show that sin-^c + cos— ^c = 7r/2, tan— 'a + cot— ^a = ;r/2. INVERSE FUNCTIONS 107 76. Sum and difference of two inverse tangents. To prove tan"^ m ± tan~^ n = tan~^ [36] Let A = tan~^ m, B = tan~^ n. (1) Then tan A = m, tan B = n. (2) Using the notation in (1), [36] becomes A ±B = tan-^ T , or tan (A ± B) = (3) 1 q: 77171 ^ 1 qi mn ^ ^ To prove (3), by § 37 we have tan A ± tan B tan {A ± B) 1 qi tan A tan B (m ± n) / (1 ip mn). by (2) The proof above ilhistrates the method of procedure in establishing relations between inverse functions. Ex.1. Prove that sin- 1(3/5) -f sin- 1(8/ 17) = sin- 1(77/ 85). (1) Let ^^sin-i(3/5), 5 = sin-i(8/17). (2) Then sin ^ = 3/5, sin 5 = 8/17. (3) .-.cos ^ = 4/5, cos B= 15/17. (4) Using the notation in (2), (1) becomes ^ + 1? = sin- 1(77/ 85), or sin (^ + J?) = 77/85. (6) To prove (5), by [7] we have sin {A -\- B) = sin ^ cos J5 + cos A sin B Ex. 2. Prove that cos-i (4/5) + tan- 1 (3/5) = tan- 1 (27/11). (1) Let ^ = cos-i(4/5), B = tan-i(3/5). (2) Then cos ^ = 4 / 5, tan A = S/4, tan B = 3/5. (3) Using the notation in (2), (1) becomes ^ + B = tan- 1(27/11), or tan (^ + 5) = 27/11. (4) 108 PLANE TRIGONOMETRY To prove (4), by [11] and (3) we have 3/4 + 3/5 _27 tan {A + B) 1- (3/4) (3/5) 11 Ex. 3. Prove that 2 tan-i (1/3) + tan-i (1 /7) = 7r/4. (1) Let ^ = tan-i(l/3), 5 = tan-i(l/7). (2) Then tanJ. = l/3, tanJ5=l/7. (3) Using the notation in (2), (1) becomes 2A + B=7t/4, or tan(2^ + 5) = tan(7r/4) = 1. (4) ^ r.KH r. . 2tan^ 2/3 3 By [151, tan 2 ^ = = —^ = - • (6) ^ ^ -" l-tan2A 1-1/9 4 ^ ' To prove (4), by [11], (3), and (5) we have 3/4 + 1/7 _, tan {2A-J- B) 1- (3/4) (1/7) Ex. 4. Solve the equation tan-i 2 x + tan-i 3 ic = ;r/4. (1) 2 ic + 3 ic By §76, tan-i2x + tan-i3x = tan-i ^^^- (2) From (1) by (2), tan-i^^±?| = ^ (3) 1 — D X-^ 4 -r, /ex 2X+SX ^ 7t ^ .-. x = 1/6 or - 1. x = l/6 satisfies (1) for the principal values of tan-i 2 x and tan— ^ 3 x. X = — 1 satisfies (1) for the values tan-i(-2) = 116°33'55^ tan-i(-3) = -71°33'55'^ Ex. 6. Solve the equation tan- 1 + tan- 1 = tan- ^ - 7 ). ( 1 ) X — 1 X tan-i h tan-i = tan- ^ (2) X - 1 X 1 —X From (1) by (2), ^^^-^ + '^ = _ 7^ or x = 2. 1 — X X = 2 satisfies (1) when for tan-i (- 7) we take 98° T 48". INVERSE FUNCTIONS 109 EXERCISE XXVn 1. Find the value of vers (tt / 6) , vers (tt / 4) , vers (tt / 3) , vers (3 tt / 4), versO, vers(;r/2), vers 7f , vers (3^/2). 2. Find the value of covers (tt/G), covers (;r/4), covers (;r/3), covers (3 TT/ 4), covers 0, covers (;r/2), covers ;r, covers (3 ;r/ 2). 3. Express in radians the general value of sin-i(V2/2); sin-i(- V3/2); cos-i(V3/2); cos-i(-l/2); tan-i(V3/3); tan-i(-V3); cot-i(-l); cot-i(V3/3). Prove each of the following relations for principal values : 4. tan-12 + tan-i0.5 = 7r/2. 5. tan-i 7 + tan-i 3 = 153° 26' 6'^5, given 0.5 = tan 26° 33' 53".5. 6. tan-i tan-i — ; — = -• n m + n 4 7. tan-i(l/7) + tan-i(l/13) = tan-i(2/9). 8. 2tan-i(2/3) = tan-i(2/3) + tan-i(2/3) = tan-i(12/6). 9. tan-i(3/4) + tan-i(3/5) - tan-i(8/19) = 7r/4. Add tan- 1(3/ 5) to tan- 1 (3/ 4) and then subtract tan- 1 (8/19). 10. sin-i(3/5) + sin-i(8/17) = sin-i(77/85). 11. cos-i(4/5) + cos-i (12/13) = cos-i (33/65). 12. cos-i(4/5) + tan-i(3/5) = tan-i(27/ll). Solve each of the following equations : 13. tan-ix + tan-i {1 - x) = tan-i (4/3). 14. tan-.^-t + Un-i^±i = |. x-2 x+2 4 15. tan-ix + 2cot-ijc = 2;r/3. 16. tan-i (X + 1) + tan-i (x - 1) = tan- 1 (8/31). 17. cos-i h tan-i — = — - • x2 + 1 x2 - 1 3 18. sin-ix + sin-i2x = 7r/3. 19. sin-i (5/x) + sin-i (12 /x) = 7t/2. Suggestion. Observe that sin- 1 (5/x) and sin-i(12/x) are comple- mentary angles. CHAPTER VII PERIODS, GRAPHS, IMPORTANT LIMITS, COMPUTATION OF TABLE, HYPERBOLIC FUNCTIONS 77. Periods of the trigonometric functions. As an angle increases from to 2 7r radians, its sine first increases from to 1, then decreases from 1 to — 1, and finally increases from — 1 to 0. As the angle increases from 2 tt to 4 tt radians, the sine again goes through this same series of changes. Thus the sine goes through all its changes while the angle increases by 2 7r radians. This is expressed by saying that the period of the sine is 2 it. Similarly the cosine, secant,. or cosecant goes through all its changes while the angle increases by 2 tt radians. The tangent or cotangent, however, goes through all its changes while the angle increases from to tt radians. Hence the period of the sine, cosine, secayit, or cosecant is 217 radians; while the period of the tamjent or cotangent is ir radians. Since, as the angle increases, each trigonometric function goes through again and again the same series of values, these functions are called periodic functions. 78. Curve of sines. Let OX = OP^^^ 1, ^3 // k> ZXOPi = 7r/6, ^ \ and ZZOP2 = 7r/3. M,M, Then, if ^ = Z XOP, Fig. 49 we have, when ^=0, 7r/6, 7r/3, 7r/2, 27r/3, 57r/6, tt, 77r/6, 47 r/3 J Sin^=0, iV/iPi, M^P^, OPs, MoTo, M^P^, 0, 110 il/iPi, -il/aA, CURVE OF SINES 111 The series of values through which sin passes as the angle increases can be graphically represented by means of the points of a curve constructed as follows : Let OX and F in fig. 50 be two fixed straight lines at right angles to each other, and let an angle of one radian be repre- sented by a unit of length along OX. Take OR = ir units, i.e. about 3} units of length ; also take BX = ir units, and OZ' = — TT units. Then OR represents tt radians ; OX, 2 tt radians ; and OX', — TT radians. Fig. 50 Lay off on OX the distances representing 0, 7r/6, tt/S, it 12, 2 7r/3, etc. At N, the end of 7r/6, draw NP X OX and equal to MxP\ in fig. 49 ; at the end of 7r/3 draw a perpendicular equal to M^P^ ; and so on. Through the ends P, P', P", •of these perpendiculars draw a smooth curve. Then if from any point in this curve, as P, we draw PN ± OX, the directed line NP represents the sine of the angle whose radian measure is represented by the directed line ON. This curve, called the curve of sines, or sinusoid, consists of portions similar to OP"RQX placed one after another. This illustrates graphically that the period of the sine is 2 IT. 79. Curve of cosines. Using fig. 49, we obtain, when e=0, 7r/6, 7r/3, 7r/2, 27r/3, 57r/6, tt, 77r/6, 47r/3, cos^=+l, OMi, OM^, 0, 01/2, -OMi, -1, -OMi, -OM^, 112 PLANE TRIGONOMETRY Lay off on OX the distances representing 0, tt/G, 7r/3, 7r/2, 2 7r/3, etc. At the points thus determined erect perpendiculars equal to OX, OM^, OM^, 0, — OM^, etc., and trace a curve through the points P, P', P", etc. The curve thus obtained is called the curve of cosines. P Y Fig. 51 Observe that the curve of cosines has the same form as the curve of sines. These two curves differ only in their position with reference to the origin 0. 80. Curve of tangents. Using fig. 49, where XOP = 27r/5, we obtain, when 0=0, 7r/6, 7r/3, 27r/5, 7r/2, -7r/6, -7r/3, -27r/5, -7r/2,. • ., tan ^=0, X^i, XT^, XT, oo, -XT^, -XT^, -XT, -oo, • •. Proceeding as in § 78, we obtain the curve represented by the continuous lines in fig. 62. This graph is called the curve of tangents. Since tan(7r/2) = oo, the curve of tangents will have no point on RK, but the infinite branch P'P" will approach indefinitely near to the line RK without ever touching it. The same is true of the infinite branch PiPz with reference to the line RiKi. The curve of tangents will evidently consist of an unlimited number of similar but disconnected branches, any one of which is parallel to the branch P^OP". CURVE OF TANGENTS 113 Curve of cotangents. Using fig. 49, by § 30, we obtain, when ^=0, 7r/10, 7r/6, 7r/3, 7r/2, 27r/3, Bir/d, Qtt/IO, tt, .••, cot^=oo, XT, XT^, ZTi, 0, -XT^, -XT^, -XT, -oo,.--. .''liT Fig. 52 Proceeding as above, we obtain the curve represented by the dotted lines in fig. 52 as three of the infinite number of disconnected branches of the curve of cotangents. 81. Curve of secants. Using fig. 49, we obtain, when ^=0, 7r/6, 7r/3, 27r/5, 7r/2, -7r/6, -7r/3, -27r/5, -7r/2,. • •, sec^=+l, OTi, 07^2, OT, oo. 0^1, OTs, or. 00, Proceeding as above, we obtain BPK and B'P'K' as two branches of the curve of secants; repetitions of these two branches make up the rest of the curve. 114 PLANE TRIGONOMETRY Similarly the curve of cosecants can be constructed, three branches of which are represented by the dotted curved lines in fig. 53. 2T Fig. 53 82. *Lt(sin 0/6) = lt(tane/e) = 1, when 6 = 0, the unit of 6 being the radian. Let be the radian measure of any positive acute angle A OP. Draw the arc AP, the chord AP, PM ± OA, A r± OA. *When, according to its law of change, a variable will become and remain less in size than any assignable constant value but will never become zero, the variable is called an infinitesimal, or is said to approach zero as its limit. When a variable approaches a constant so that their difference is an infini- tesimal, the variable is said to approach the constant as its limit, or the constant is said to be the limit of the variable. Lt (sin e /9) is read the limit of sine /e; = is read e approaches as its limit or e is an infinitesimal. The reciprocal of an infinitesimal is an infinite and is denoted by co. Any number which is neither an infinitesimal nor an infinite is called n finite number LIMIT OF SIN 0/0 115 Then AOAP < sector 0.4P < A OA T. (1) By Geometry, the area of the sector OAP is OA -arc AP/2. Hence, from (1), we have OA • MP < OA . arc .4P < OA - A T. (2) Dividing each member of (2) by OA^, we obtain MP I OP <2iVGAP/0A Vl - (.0003)2 > .99 999 99^ Hence, to seven places, cos 1' = 0.99 999 99 + . (3') By (4), sin ^ > ^ cos d. .'. sin 1' > (1' in radians) cos V. 116 PLANE TRIGONOMETRY .-. sin r > 0.00 029 088 821 x 0.99 999 99 by (1'), (30 > 0.00 029 088 821 x (1 - 0.00 000 01), or sin 1' > 0. 00 029 088 820 + . (4^ From (2') and (4') it follows that to ten places sin r = 0.00 029 088 82 + , (5') the eleventh figure being or 1. This example affords a good illustration of the use of corollary 1, below. Two very important corollaries to the theorem above are the following : Cor. 1. If Q is very small, it can be substituted for either sin 6 or tan 6 in approximate calculations ; or vice versa. Cor. 2. i/* 6 is an infinitesimal and it is substituted for either sin 6 or tan 9 in any function, the limit of the function will not be changed. 83. Convenient formulas for computing the sine and cosine of any angle are 2 W,* d,« d>8 is the radian measure of the angle. For the proof of (b) and (c), see § 94 in Taylor's Calculus. The sine of any angle is arithmetically equal to the sine or cosine of some angle not greater than 7r/4 ; hence its value can be computed by taking <^ in (b) or (c) not greater than 7r/4. Likewise the cosine of any angle can be computed by taking <^ in (c) or (b) not greater than 7r/4. Observe that for <^ < 7r/4 the series in (b) or (c) converges rapidly, and only a few terms need to be computed. Having found the value of the sine and the cosine, the log- arithmic sine and cosine can be obtained from a table of logarithms of numbers. The logarithmic tangent can then be SIMPSON'S METHOD 117 found by subtracting log cos from log sin, and log cot by subtracting log sin from log cos, or log tan from 0. Example. Compute sin 11° 12' 23'' and cos 11° 12' 23". By §66, 11° = (0.01 745 329 252 x 11) radian = 0.19 198 621 772 radian. 12' = (0.00 029 088 821 x 12) " = 0.00 349 065 852 " 23" = (0.00 000 484 814 x 23) " = 0.00 Oil 150 722 " .-. = 11° 12' 23" =0.19 558 838 346 " Substituting this value for

/r = sin <^. (2) .'. a/r -{- ib /r = Qos -{- i sin <^) r. (4) (cos + i sin ) r is the trigonometric form of a complex number in which cos 4> -\- i sin <^ is the quality unit and r the modulus. Either {a/r -\- ib /r)r or (cos 4> ■\- i sin )r in (4) is called the type form of the complex number a + ib. Hence a complex number in the type form is an arithmetic multiple of a quality unit. The angle <^ is called the angle, or amplitude, of the complex number a + ib. Between — it and + tt there is one, and only one, value of <^ which will satisfy equations (2). This value of <^ is called its principal value. If <^' denotes the principal value of , the general value of <^ is 2 niT + <^', where n is any integer. Thus the angle, or amplitude, of a complex number is many-valued. When 6 = 0, a H- t^ = a, a real number ; when a = 0, a + t6 = ib, an imaginary number. 124 PLANE TRIGONOMETRY Thus a -f ih includes reals and imaginaries as particular cases. Ex. 1. Reduce the complex number — 1 — V — 3 to the trigonometric type form. Here a = — 1, & = — V3- .-. r=Vn-3 = 2. by (1) Hence cos = -1/2, sin = - V3/2. by (2) Since cos0 and sin are both — , is in the third quadrant ; hence its principal value is — 2 ;r/3. ... _i _Vir3 3::[cos(-2 7r/3) + isin(-2 7r/3)] -2. Hence the arithmetic value, or modulus, of — 1 — V— 3 is 2, and its quality unit is represented by a unit vector which makes the angle — 2;r/3 with the vector representing +1. Ex. 2. Reduce the complex number 2 — V— 5 to the algebraic type form. Here a = + 2, 6 = - V5 ; ••• r = 3. Ex. 3. Reduce the unit sin — i cos to the trigonometric type form. sin — i cos = cos (0 - 90°) + i sin (0 — 90°). Thus the angle of the quality unit sin — i cos is — 90°. EXERCISE XXIX Represent graphically and reduce to the type form : 1. i+VZTi. 5. _^3 + i 9. _6-i8. 2. 1 - V^. 6. - V3 - i. 10. - V5 - * Vll- 3. — 1 — i. 7. 3 — i4. 11. sin0 + icos0. 4. _ 1 + v^ITs. 8. - 3 + i 2. 12. - sin ^ + i cos d. 13. By constructing the sum in each member, show that (a + ih) + (« + iy) = (a + iy) + (x + i6) = (a + x) + i (6 + y), and thus prove geometrically that the commutative and associative laws of addition hold true for complex numbers. GENERAL QUALITY UNIT 125 91. General quality unit cos <|) + i sin <|>. Let ABA' be a circle whose radius is 1, and let OA = the unit +1. B Then if <^ = ^ OP, cos <^ + t sin <^ = OP ; if = A OPi', cos + ^ sin = A OPi, cos + i sin <^ = OPi ; if <^ = yI op', cos -^ism = OP'. That is, the quality unit cos -{- i sin = cos + * sin = + 1. When = ± 7r/2, cos <^ + t sin <^ = cos(± 7r/2) + i sin(± 7r/2) = *i. When = ±7r, cos <^ + t sin <^ = cos (± tt) + i sin (± tt) = ~1. These results are evident also from the figure. 126 PLANE TKIGONOMETRY 92. Product of quality units. By the distributive law of multiplication, we obtain (cos <^i + i sin <^i) (cos cf>2 + ^ sin c^g) = cos <^i cos 2 = cos 2 + i (sin i cos <^2 + cos <^i sin 2) = cos (<^i -\- cji2) + i sin (cf>i + cfj^). § 34 Hence, in general, we have (cos n + i sin <|)n) = C08((|)i + 4)2 + • ■ • + n) + i sin(, = ct>, = ... = n, then, from [39] of § 92, we obtain (cos + i sin <^)" = cos ncfi 4- i sin w<^. (1) That is, the nth power of a quality unit is a quality unit whose angle is n times the angle of the base. (ii) Taking the wth root of each member of (1), we obtain (cos ny^'^ = cos + i sin <^. Putting for nct> and therefore /n for <^, we have (cos <^ + ^ sin y^" = cos (/n) + i sin (/n). (2) DE MOIVRE'S THEOREM 127 Let 5 be a positive integer, then, from (2), we obtain (cos <^ + * sin c^)"/" = [cos {4> /n) + i sin (<^/w) J = cos (s/n) + i sin (s/7i). (3) (iii) By §92, (cos -{-i sin <^) [cos (— ) + i sin (— <^)] = cos + i sin = +1. Hence cos (— <^) + 1 sin (— <^) is the reciprocal of cos -{- i sin <^. That is, (cos (f> -{- i sin <^)-^ = cos (— <^) + i sin (— <^). (4) Let p be any positive integer or fraction. Then, from (4), (cos -{- i sin <^)~^ = [cos (— <^) + i sin (— )Y = cos (— p) + i sin (— jxf}). (5) Erom (1), (3), and (5) it follows that for any commensura- ble real value of n, we have (cos <|) + i sin (j))"^ = cos n<|) + i sin n<|). [40] Formula [40] is called De Moivre's theorem. CoR. 1. From (2), cos (<^/n) + i sin (<^/n) is one of the nth roots of cos <^ + i sin <^. CoR. 2. By § 28, cos (— <^) + i sin (— ) = cos <^ — i sin <^. Hence, by (4), cos <^—i sin <^ is the reciprocal of cos + ^ sin<^. That is, the conjugate quality ^tnits cos <^ + i sin <^ and cos — i sin <^ are reciprocals of each other. In fig. 58 observe that OP and OPi represent reciprocal, or conjugate, quality units, while OP and OP' represent opposite quality units. When OP coincides with OB, OP^ and OP' both coincide with OB'; this illustrates that the reciprocal, or con- jugate, of i is also the opposite of i. When OP coincides with OA or OA', OPi does also; that is, either +1 or ~1 is its own reciprocal. 128 PLANE TRIGONOMETRY 94. To divide by the quality unit cos + i sin <^, we multiply by its reciprocal cos (— <^) + i sin (— ), or cos . Ex. 1. — ^— ^ = (cos 01 + i sin 0i) [cos (- 02) +i sin (- 02)] cos 02 + * sin 02 = cos (01 — 02) + i sin (0i — 02). Ex. 2. (^os'^ + ^s^^'^)' = (cos + i sin 0)5 (cos ^ + i sin ey (cos^-isin^)7 ^ "^ ^ "^^ ^ ^ ' = (cos 5 + i sin 5 0) (cos 7 ^ + i sin 7 ^) = cos (5 + 7 ^) + i sin (5 4- 7 6). EXERCISE XXX Prove each of the following identities : 1. (cos0 + isin0)3 = +1, [cos(2 ;r/3)+ isin(2;r/3)]3 = +1, [cos(4;r/3)+ isin(4;r/3)]3 =: +1. Hence each of these three quality units is a cube root of +1. 2. [cos(7r/5)±isin(7r/5)]6 = -l, [cos(3 7r/5)± isin(3 7r/5)]5 =-1, (cos TT lb i sin 7t)^ = -1. Observe that cos tt + i sin tt = cos rt — ismTt. 3. [cos {7t/6)±i sin {7t/6)f = [cos {7t/2)±i sin {7t/2)Y = [cos(5;r/6)±isin(5 7r/6)]6=-l. 4. What is proved by examples 2 and 3 ? Illustrate the meaning of examples 1, 2, 3 by vectors. Express as a quality unit each of the following : _ (cos ^ + i sin 0)^ _ (cos ^ - i sin 6)^^^ 5. • o. • (cos — i sin 0)^ (cos — i sin 0)^ /s (cos ^ + i sin 6) (cos + i sin 0) (cos j3 -}- i sin /3) (cos 7 + * sin 7) [cos (7r/6) - i sin (Tr/G)]"/^ (cos + t sin 0)^ [cos(;r/6) + isin(;r/6)]i/2* * (sin ^ + i cos ^)6 ' COMPLEX NUMBERS 129 10. By De MouTe's theorem prove identities (1) and (2). cos n0 = cos«

— '• '. (2) cos 710 + i sin n0 = (cos

-\- i sin <^ = cos (2 titt + <^) + i sin (2 ^^tt + <^). .-. (cos 4> + i sin <^)i/« = [cos (2 ?i7r + (^) + i sin (2 titt + <^)]^ " = cos ^^-^H-isin ^^-^- (1) If in (1) we give to n the values 0, 1, 2,8, A,- ■ , q — 1, in suc- cession, we obtain the following q qth roots of cos <^ + i sin : 9z = 0, cos (/q) -\- i sin {(^/q), . 27r-h , . . 2 7r + (^ 71 = 1, COS + t Sin J Att -{- d> ..47r-f -{- i sin ^ has q, and only q, unequal qth roots. 97. Any complex 7iumber has q, and only q, unequal qth roots. [(cos cf> -\- isin <^) rj^'^ = (cos <^ + i sin <^) ^ /-^r^/?. (1) The quality unit cos -\- i sin <^ has q, and only q, unequal qth roots, and the arithmetic number r has only one qth root ; whence the second member of (1) has q, and only q, unequal values. Hence to find the q qth roots of any complex number, reduce the number to the type form, find the q qth roots of its quality unit, and multiply each by the ^-th root of the modulus. ROOTS OF COMPLEX NUMBERS 131 Observe that an algebraic number has q unequal qth roots because its quality unit has q unequal qth roots. Ex. 1. Find the three cube roots of — 27. -27 = (-!)• 27 = (cosTT + isin^r) -27. .-. V- 27 = (cos TT + i sin TT) 1 /-^ • 3. , , . . ,,,o 2n7t + Tt ^ . . 2n7r -\- 7t (cos 7t + ism 7t)^'^ = cos h I sin — ; ^ ' 3 3 when n = 0, =cos(;r/3) + isin(;r/3) = 1/2 + i V3/2; when 71 = 1, = cos ;r + i sin tt = — 1 ; when n = 2, =cos(5 7r/3) + isin(5 7r/3) = 1/2 -i^S/2. Hence V_ 27 = -3, (1 /2 ± i V3/2) • 3. Ex. 2. Find the four fourth roots of 8 + 8 V^. 8 + 8 V3T3 = (1 /2 + i V3/2) . 16 = [cos {tt/S) + i sin (7r/3)] • 16. ... V8 + 8 V33 = [cos(;r/3) + isin(7r/3)] i/* -.2. r / /o^ , • • / /oMi/d 2n7r + 7r/3 , . . 2n7t + 7t/3 [cos(7r/3) + tsm{7t/d)Y^* = cos — + ism — ; 4 4 when n = 0, = cos (7r/12) + isin (;r/12); when n = 1, = cos (7 ;r/ 12) -f isin (7 ;r/12); when n = 2, = cos (13 ;r / 12) + i sin (13 7t / 12) when n = 3,. = cos (19 ;r/12) + isin (19 ;r/ 12). ) (1) Multiplying each of the four units in (1) by 2, we obtain the four required fourth roots. EXERCISE XXXI Find all the values of : 1. 11/3. 5. 161/4. 9. (4V3 + i4)'/3. 2. (-1)1/3. 6. 321/5. 10. (i_ 4^3)1/4. 3. 11/6. 7. (_243)i/\ 11. (l+\/Zl)i/6, 4. (- l)i/«. 8. {-iy/^ 12. (V3-V^)2/5. 13. Solve the equation x* - x^ + x2 - x + 1 = 0. (1) Multiply by (X + 1), x^ + 1 = 0, or x = (- l)i/5. 132 PLANE TEIGONOMETRY Multiplying by jc + 1 introduces the root - 1, the other four of the five fifth roots of — 1 are the roots of equation (1). 14. Solve the equation x^^ — 1 = 0. Factor, (x^ - 1) {x^ + 1) {x» - i) {x^ + i) = 0. The twelve roots, V/% (- l)i/3, ii/3^ {-ly/^ are readily found by De Moivre's theorem. 15. Solve the equation x^ + x* + x^ + 1 = 0. Factor, (x* + 1) (x^ + 1) = 0. 16. Solve the equation x^ + 1 = 0. 98. Exponential form for cos cj) + i sin <|>. If in series (d) of § 85 we replace x by i(l>, we obtain the series (1). ^ + "l' + -\2'+-\f IT ^ (1) = COS <^ + i sin . by (b), (c), § 83 That is, series (1) is equal to a quality unit whose angle is . If, as is suggested by (d), we define c'* as an exponential symbol for the series (1), we have e'^ = cos <^ + i sin . (2) That is, €*''* is an exponential symbol for a quality unit whose angle is . Substituting — toi in (2), we obtain c"'*^ = cos <^ — i sin . (3) From (2) and (3), e'* and e"~'* denote reciprocal quality units. From (2) and §§ 92 and 94, it follows that ^i ,^-i9 =. ^ii-0)^ (c**)^ = C*^*. THE EXPRESSION e^ 133 That is, €** obeys the fundamental laws of exponents. The unit of is the radian. Why? Putting 1 for <^ in (2), .we obtain c* = cos 1 -h ^' sin 1. That is, c* denotes a quality unit whose angle is a radian. Example. What does e-* denote ? 6^ ? Observe that for all values of the arithmetic value of e'* is 1. Hence, as varies from — oo to + oo, e'* varies in quality only. Therefore, e in e»<^ cannot have the meaning it has in e^. Observe that the numerical measure of any directed line or force in a given plane can be expressed in the form e'* • a, where a is an arithmetic number which gives its length or size and

= — — (4) The values in (4) are known as Euler's exponential values of cos <^ and sin <^. Compare (4) with [37] in § 85. CHAPTEK IX MISCELLANEOUS EXAMPLES EXERCISE XXXII Express all the angles which are coterrainal with : 1. 45°. 2. 132° 3. - 35°. 4. - 100°. 5. it/^. Find all the other trigonometric ratios of A when : 6. sin ^ = 4/7. 8. cos^ = -3/8. 10. secJ. = 7/4. 7. tanJ. = 3/2. 9. cot^ = -7/5. 11. csc^ = -5/4. 12. In what quadrant is A in each of the examples 6-11 inclusive ? Construct A in each. In terms of a function of an angle less than 45° express : 13. sin 94°. 16. cot 320°. 19. cos (-175°). 14. cos 128°. 17. sec 190°. 20. tan (-200°). 15. tan 215°. 18. sin (-75°). 21. cot (-300°). In terms of each of the other functions of A find the value of : 22. sin^. 24. tan^. 26. sec ^. 23. cos J.. 25. cot J.. 27. csc^. Identities Prove each of the following identities : 28. (tan A + cot ^ ) sin ^ cos ^ = 1. 29. (sec A — tan A) (sec A + tan A) = 1. 30. (esc A — cot A) (esc A + cot A) = 1. 31. (sin B -cosB)^ = l-2 sin B cos B. 32. sin B + cosB=y/2 cos {B - 7C/4). 33. sin B - cos B = - V2 cos {B + tt /4). 134 ^ IDENTITIES 135 34. sin {A -\- 7t/S)-{- sin (A - 7t/3) = sin A. 35. cos {A + 7r/6) + cos {A - 7t/6)= V3 cos A. 36. (cot J. + tan 5) /(tan vl + cot B) = cot ^ tan B. 37. 1 - tan* A =2 sec2 ^ - sec* A. 38. sec ^/ (1 + cos B) = (tan 5 - sin B)/sm^ B. 39. sec2 A csc2 4 = tan2 A + cot2 J. + 2. 40. tanB + sec5 = tan(jB/2 + ;r/4). 41. (1 + tan B)/{1- tan B) = (cot B -\- 1) / (cot 5 - 1). 42. sin /I / (1 -f cos ^) + (1 + cos A) /sin A =2 esc ^. 43. sec -3 A. — sin^ A = (cos ^ — sin A) (1 + sin A cos tI). 44. (sin A cos E + cos A sin 5)2 -i- (cos A cos 5 — sin J. sin 5)2 = 1. 45. cot^ — tan ^ = 2cot 2-4. 46. sec 2 ^ = sec2 A / {2 - sec2 ^) . 47. 2sec2^ =sec(^ + ;r/4)sec(^ - 7r/4). 48. sin2^ = 2tan^/(l + tan2^). 49. 2 sin J. + sin 2 ^ = 2 sin^ A /{I- cos A). 50. Find sin ( J. + J5 + C7) in terms of the sine and cosine of A^ B, C. Applying formula [7] twice and [8] once, we obtain sin(^ + 5 + C) = sin(^ + 5)cos C + cos(^ + B)sin G = (sin A cos B + cos A sin B) cos C + (cos A cos 5 — sin ^ sin B) sin C = sin A cos J5 cos C + cos ^ sin B cos C 4- cos A cos 5 sin C — sin J. sin B sin C (1) It A, B, C are the angles of a triangle, sin (A + B + C) = sin 180° = 0. Hence, from (1), we obtain sin -4 cos 5 cos C + cos^sin J5cosC + cos^ cos B sin C = sin ^ sin 5 sin C 5i. Applying formula [8] twice and [7] once, prove cos {A -\- B -{- C) = cos A cos B cos C — cos A sin B sin C — sin A cos 2? sin C — sin ^ sin B cos C. (2) 136 PLANE TRIGONOMETRY 52. Find tan {A + B + 0) in terms of tan A, tan 5, tan C. Applying formula [11] three times, we obtain tan(^ + ^ + C)^^^^(-^ + ^^ + ^^"^ ^ ' 1 - tan (^ + ^) tan C tan A + tan B 1 — tan A tan B + tanC tan A + tan B ^ ^ tan C 1 — tan A tan B _ tan J. + tan B + tan C — tan A tan ^ tan C „, ~ 1 — tan A tan J5 — tan A tan C — tan B tan C If ^, B, O are the angles of a triangle, tan {A-\-B^-C) = tan 180° = 0. Hence, from (3), tan A + tan B + tan C = tan A tan B tan C 53. Puttmg ^ = 5 = in (1), (2), (3) of examples 50, 51, 52, prove sin 3 ^ = 3 sin ^ (1 - sin2 J.) - sin2 J. = 3sin J. -4sin3^, (4) cos SA = 4 cos^ J. — 3 cos J., (5) ^ o . _ 3 tan Jl - tan 3 ^ tan 3 ^ = — (6) 1 - 3 tan2^ ^ ' Identity (5) is useful in solving cubic equations. See example 112. 54. Writing SA = 2A + A, prove (4), (5), (6) in example 53 by using [7], [8], [11], [13], [14], and [15]. 55. Substituting ^ for 3 ^ in (4) and (5) of example 53, we obtain sin e = 3 sin (^/3) - 4 sin3 (^/3), cos ^ = 4 cos3 (d/Z) - 3 cos (d/S). Prove each of the following identities : 56. sin 4 J. = 4 sin ^ cos ^ - 8 sin^ ^ cos J. = 8 cos^ ^ sin ^ — 4 cos A sin A. 57. cos4^ = l-8cos2^ + 8cos4^ = l-8sin2J. + 8sin4^. 58. cos 780° = 1/2. 60. cos 2550° = y/S/2. 59. sin 1485° = V2/2. 61. sin(- 3000°) = - V3/2. 62, tan (- 2190°) = - V3/3. EQUATIONS 137 Equations In what quadrant is A in each of the following equations ? 63. sin J. cos^ = - 2/3. 65. sec ^ tan J. = - 3. 64. sin ^ tan ^ = 4. 66. cot J. + 3 sin ^ = 0. 67. Solve the equation sin 2 ^ = 2 cos 6. (1) Substituting for sin 2 d its identical expression 2 sin 6 cos 6, from (1) by Algebra we obtain the equivalent equation 2 sin ^ cos ^ = 2 cos 6, or cos (sin ^ — 1) = 0. From cos ^ = 0, e = n7t ±7t/2. §70 From sin ^ = 1, d = UTt +{- l)'^7t/2. §69 Hence mt ± it /I includes all the values of Q in (1). Solve each of the following equations : 68. cos2^ = 2sin^. 82. 2 sin-2 x - 2 = - V2 cos x. 69. cos ^ = sin 2^. 83. cos2y4-2sin2?/ — fsiny = 0. 70. sin ^ = cos 2 ^. 84. sin ^ + sin 2 ^ = 1 - cos 2 6. 71. tan A tan 2 A =2. 85. cos y — cos 2y = 1. 72. cos ^ + cos 2 ^ = 0. 86. sin (45° + z) + cos (45° -z) = l. 73. cot ^ tan 2 ^ = 3. 87. sec 2 z + 1 = 2 cos z. 74. 4 cos 2 ^ + 3 cos J. = 1. 88. cos 2 z = a (1 - cos z). 75. sin sec 2 ^ = 1 . 89. tan 2 y tan y = 1. 76. cot ^ tan 2 ^ - sec 2 d. 90. sec ^ = 2 tan ^ + 1 /4. 77. sin 2 ^ = 3 sin2 ^ - cos2 ^. 91. sin-^x + sin-i (x/2) = 120° 78. sin ^ + cos 2 ^ = 4 sin2 0. 92. sin- 1 z + 2 cos- 1 z = 210°. 79. sin 2 ^33 cos 4^. 93 tan-i ?/ + 2 cot-iy = 135°. 80. sec X + tan x = ± y/S. 9 „ 94. tan- 1 -^ — = 60°. 81. tan X + 2 V3 cos x = 0. 1 - z2 95. tan-i z + tan-i 2 z = tan-i 3 V3. 96. tan x + tan 2 x = 0. 138 PLANE TRIGONOMETRY 97. tan2 x + cot2 x = 10/3. 99. sin A + cosA= sec A. 98. 4 cos 2 ^ + 6 sin 6 = 5. 100. sin {6 + 30°) sin (d - 30°) = 1/2. Systems of Equations 101. Solve for r and 6 the system rsind = a, (1) rcos^ = 6. (2) ^ ^^' Divide (1) by (2), tan d = a/b, or 6 = tan-i(a/6). Square (1) and (2) and add, r2(sin2^ + cos2^) = a2 + 62. .-. r = Va2 4. 52. 102. Solve for r, 0, and the system r cos sin ^ = a, (1) ^ r cos cos ^ = 6, (2) i- (a) r sin = c. (3) j Divide (1) by (2), tan d = a/b, or ^ = tan-i (a/6). (4) Square (1) and (2) and add, r2 cos2 = a2 + 62. (5) From (5), r cos =: VoM^. (6) Divide (3) by (6), tan = c/Va2 + 62^ or = tan-i (c/ Va2 + 52). Square (3) and add (5), r = Va2 + &2 _,. c2. (7) 103. Solve for x and y the system sin X + sin 2/ = a, (1) cos X + cos 2/ = 6. (2) By § 40 we obtain from system (a) the equivalent system (b) 2 sin i (x + ?/) cos U^ - V) 2 cos i (x + ?/) cos I (x - 2/) = 6, Divide (3) by (4), tan i (x + 2/) = a/6. (5) Hence sin ^^{x + y) = ± a /^a^ + h\ (6) Substituting the value of sin \{x-\-y) in (3), we obtain cos \{x-y) = ± Va2 + 62/2. (7) } iral( a, (3) 1 6. (4) J (a) W From (5), From (7), Hence and 104. Solve the system sin X cosx SYSTEMS OF EQUATIONS 139 X + y = 2 tan-i {a/b). (8) | X - y = 2 cos-i (± Va2 + W/2). (9) J ^^' X = tan-i (a/b) + cos-i ( ± Va2 + 6V2), y = tan-i (a/b) - cos-i (± Va2+^/2). - sin y = a, ^ — cosy = b. J 105. Solve for x and y the system X sin ^ + y sin ^ = a, (1) ^ X cos ^ + y cos ^ = b. (2) j ' ' Since (1) and (2) are each linear algebraic equations in x and y, system (a) is solved as a linear algebraic system. 106. Solve for r and the system r sin {6 + A) = a, i{e-\-B) = b.j (a) By [7] and [8], from (a) we obtain the equivalent system (b). r sin 6 cos A + r cos d sin ^ = a, i ' J. (b) r cos ^ cos J5 — r sin ^ sin B =b. J Solve (b) as an algebraic linear system in r sin 6 and r cos 6 as the unknowns. The resulting system can then be solved for r and 6 as in example 101. 107. Solve the system cos (X + y) + cos {x-y) = 2, (1) ^ sin(x/2) + sin(y/2) = 0, (2) J for values of x and y less than 2 tt. By [8] and [10], from (1) we obtain the equivalent equation cos X cos y — sin x sin y + cos x cos y + sin x sin y = 2, or cosxcosy=:l. (3) From (3), cos x and cos y are both + 1 or both —1. Why ? Hence x and y are both coterminal with or both coterminal with 7t. (a) (2)/ 140 PLANE TRIGONOMETRY The solution x = 0, 2/ = of (3) satisfies (2) ; also the solution x — Tt, y = — 7t, or X = — 7t, y = 7t oi (S) satisfies (2). Observe that either z = 7t, y = tt, or x = — it, y = — tc is & solution of (3), but neither is a solution of (2). 108. Solve for B and F the system W- Fsinh- Rcosh = 0,^ W- Fcosh- Rsinh = 0.) Observe that this system is algebraic and linear in R and F. 109. Eliminate 6 from the system cc = r(^ -sin ^), (1) y = r{l -cose). (2) From (2), y = r vers d, or 6 = vers-i(?//r). (3) From (2), cos d = (r — y)/r. .-. sin ^ = ± ■V2ry -y^/r. (4) From (1), (3), (4), x = r vers-i {y/r)T V2 ry - y^. (5) 110. Eliminate d from the system a cos 6 -\-bsm6 = c, (1) "i dcosd -\- esme=f. (2) J Solving the system for cos 6 and sin 6, we obtain . n af-cd . ce — bf .„. sm = — , cos ^ = -^- (3) ae — bd ae — bd Squaring the members of equations (3) and adding, we obtain {ae - bdf ... (ae - bdf = {af - cdf + (ce - bff. 111. Eliminate 6 from the system acosd -\-b sin ^ = c, bcosS — a sin 8 = d. 112. Solve the cubic equation x^ -Spx + q = 0. (1) Putting x — z/n, we obtain z3 _ 2>-pn^z + gn3 = 0. (2) Now by (5) in example 53, we have the identity cos3^ = 4 cos3^ - 3cos^, pr cos3^-(3/4)cos^ -cos(3^)/4 = 0. (3) CUBIC EQUATIONS 141 Comparing identity (3) with equation (2), we see that cos A is a, root of (2) when n and A satisfy the conditions 3pn2 = 3/4, and qn^ = - cos {S A) / 4. Hence ?i = l/(2Vp), and cosSA =- 4:qn^ z=-q/{2p^^^). (4) Observe that (4) can always be solved when p is positive and q/{2p^^^) is arithmetically equal to or less than 1. If Ai is the principal value of A which satisfies (4), then the values Ai + 27t/S and ^i + 4 ;r/3 also satisfy it. Hence the roots of equation (1) are cos Ai/n, cos (^1 + 2 7t/S) /n, and cos (^i + 4 7t/S) /n, i.e. 2 Vp cos J-i, 2 VP cos (^i + 2 7r/3), and 2 VP cos {Ai + 4 tc/S). By Algebra, we know that the general cubic equation z/3 + 3 a?/2 + 6y + c = can be transformed into one of the type (1) by putting y = x — a. 113. Solve ' a;3 + 6x2 + 9x + 3 = 0. Putting X = ?/ — 2, we obtain y* — 3 ?/ + 1 = 0. Putting y = z/n, we obtain z^ — S n^z + n^ = 0. Now cos8^ -(3/4)cos^ -(1/4) cos 3^ = 0. Hence z = cos A, when n^ = 1 /4 and n^ = — cos 3 J. /4, i.e. when n = l/2, and cos 3^ =-1/2 = cos 120°. .-.3^1 = 120°, or ^1 = 40° Hence z = cos 40°, cos (40° + 120°), or cos (40° + 240°). .-. y = 2 cos 40°, 2 cos 160°, or 2 cos 280°. .-. X = - 2 + 2 cos 40°, -2 -2 cos 20°, or -2+2 cos 80°. Having given sin 15° = ( V3 - l)/(2 V2), and cos 15° = ( V3 + l)/(2 V2), solve each of the following equations : 114. a^ -24a; -32 = 0. 116. 2x3_3a;_i = o. 115. x3-6x2 + 6x + 8 = 0. 117. x3 + 3x2-l = 0. 118. x3 + 4x2 + 2x-l =0. 142 PLANE TRIGONOMETRY EXERCISE XXXm Triangles 1. Two towers are 3 mi. apart on a plain. The angle of depression of one, from a balloon directly above the other, is observed to be 8° 15'. How high is the balloon ? 2. The shadow of a tree 101.3 ft. high is found to be 131.5 ft. long. Find the elevation of the sun. 3. A rock on the bank of a river is 130 ft. above the water level. From a point just opposite the rock on the other bank of the river the angle of elevation of the rock is 14° 30' 2V'. Find the width of the river. 4. A rope 38 ft. long, when fastened to the top of a tree 29 ft. high, just reaches a point in the plane of the foot of the tree. Find the angle which the rope makes with the ground. 5. A window in a house is 24 ft, from the ground. Find the inclina- tion of a ladder placed 8 ft. from the side of the building and reaching the window. 6. A ladder 40 ft. long reaches a window 33 ft. high, on one side of a street. Its foot being at the same point, it will reach a window 21 ft. high on the opposite side of the street. Find the width of the street. 7. A lighthouse 54 ft. high is situated on a rock. The angle of eleva- tion of the top of the lighthouse, as observed from a ship, is 4° 52', and the angle of elevation of the top of the rock is 4° 2'. Find the height of the rock and its distance from the ship. 8. A man standing south of a tower, on the same horizontal plane, observes its angle of elevation to be 54° 16' ; he goes east 100 yd. , and then finds its angle of elevation to be 50° 8'. Find the height of the tower. 9. A pole is fixed on the top of a mound, and the angles of elevation of the top and the bottom of the pole are 60° and 30° respectively. Prove that the length of the pole is twice the height of the mound. 10. Given that the radius of the earth is 3963 mi., and that it sub- tends an angle of 57' 2" at the moon. Find the distance of the moon from the earth. TRIANGLES X43 11. Given that the radius of the earth is 3963 mi., and that it sub- tends an angle of 9" at the sun. Find the distance of the sun from the earth. 12. Solve example 1 in § 57 by the principles of right triangles. The given parts are a side and two angles. In fig. 38 draw AH ± BC. In the right triangle AHB, compute the sides c and BR. Then compute c in the right triangle HCA. 13. Solve the first four examples in Exercise XIX by the principles of right triangles. 14. Solve example 1 in § 58 by the principles of right triangles. The given parts are two sides and an angle opposite one of them. In fig. 39 draw CH ± AB. Compute the sides CH and AH in A ABH. Then compute B and HB in A HBC. 15. Solve examples 1, 3, 5, and 7 in Exercise XX by the principles of right triangles. 16. Solve the example in § 59 by the principles of right triangles. The given parts are two sides and their included angle. In A ABC draw BH ± CA. Compute CH and BH in A CHB. Compute A and c in triangle BHA. 17. Solve the first four examples in Exercise XXI by the principles of right triangles. 18. Solve example 6 in § 60 by the principles of right triangles. The given parts are the three sides. In the A ABC draw AH ± BC and let x = HC. Then b^ - x'^ = AH^ = c^ - {a - x)\ Hence h^ - x^ = c^ - a^ - x^ i- 2 ax. .: cc = (a2 + 62_c2)/(2a). Whence HC and b are known in A HAC, and c and BH in A BAH. 19. Solve the first four examples in Exercise XXII by the principles of right triangles. 144 PLANE TRIGONOMETRY 20. A tree stands at a distance from a straight road and between two milestones. At one milestone the line to the tree is observed to make an angle of 25° 15' with the road, and at the other an angle of 45° 17'. Find the distance of the tree from the road. 21. From the decks of two ships at C and D, 880 yd. apart, a cloud A, in the same vertical plane as C and D and between them, is observed. Its angle of elevation at C is found to be 35°, and at D 64°. Find the height of the cloud above the surface of the sea, the height of the eye in each case being 21 ft. 22. To determine the distance between two ships at sea, an observer noted the interval between the flash and report of a gun fired on board each ship, and measured the angle which the two ships subtended. The intervals were 4 seconds and 6 seconds respectively, and the angle 48° 42'. Find the distance between the ships, the velocity of sound being 1142 ft. per second. 23. In order to find the breadth of a river a base line of 600 yd, was measured in a straight line close to one side of it, and at each extremity of the base the angle subtended by the other end and a tree upon the opposite bank was measured. These angles were 53° and 79° 12' respec- tively. Find the breadth of the river. 24. A straight road leads from a town ^ to a town 5, 12 mi. distant ; another road, making an angle of 77° with the first, goes from ^ to a town C, 7 mi. distant. Find the distance between the towns B and C. 25. Two lighthouses A and ^ are 11 mi. apart. A ship C is observed from them to make the angles BAC = 31° 13' 31" and ABC = 21° 46' 8". Find the distance of the ship from A. 26. Two posts A and B are separated by a swamp. To find the dis- tance between them a point C is so taken that both posts are visible from it. By measurement, AC = 1272.5 ft., BC = 2012.4 ft., and ZACB = 41° 9' 11". Find the distance AB. 27. Two buoys A and B are one half mile apart. Find the distance from ^ to a point C on the shore if the angles ABC and BAC are 77° 7' and 67° 17' respectively. 28. The elevation of the top of a spire at one station. A, was 23° 50' 15", and the horizontal angle at this station between the spire and another PROBLEMS 145 station, B, was 93° 4' W. The horizontal angle at B was 54° 28' 30", and the distance between the stations 416 ft. Find the height of the spire. 29. In order to find the distance of a battery at B from a fort at F, distances BA and AC were measured to points A and C from which both the fort and the battery were visible, the former distance being 2000 and the latter 3000 yd. The following angles were then measured : Z BAF = 34° 10', ZFAC= 74° 42', and Z FCA = 80° 10'. Find the distance of the fort from the battery. 30. The distances of two islands from a buoy are 3 and 4 mi. respec- tively. The islands are 2 mi. apart. Find the angle subtended by the islands at the buoy. 31. Two rocks in a bay are c yd. apart, and from the top of a cliff in the same vertical plane with the rocks their respective angles of depression are A and 3 A. Show that the height of the cliff is c sin 3 J./ (2 cos A). 32. A person wishes to find the distance between two places A and B on opposite sides of a brook. He walks from E to a bridge 2 mi. away. Crossing this he continues his walk 6 mi. in the same direction to C, which he knows to be 3 mi. from A. If .4 is 4 mi. from the bridge, show that AB = 5.86 mi., nearly. 33. A person at the top of a mountain observes the angle of depression of an object in the horizontal plane beneath to be 45°; turning through an angle of 30° he finds the depression of another object in the plane to be 30°. Show that the distance between the objects is equal to the height of the mountain. 34. From a window on a level with the bottom of a steeple the angle of elevation of the top of the steeple was 40°. At another window 18 ft. vertically above the former, the angle of elevation was 37° 30'. Find the height of the steeple. 35. Find what angle a tower will subtend at a distance equal to six times the height of the tower. Find where an observer must station himself that the angle of elevation may be double the former angle. 36. Two ships are a mile apart. The angular distance of the first ship from a fort on the shore, as observed from the second ship, is 35° 14' 10"; the angular distance of the second ship from the fort, observed from the first ship, is 42° 11' 53". Find the distance in feet from each ship to the fort. 146 PLANE TKIGONOMETRY 37. The sides of a triangle are 17, 21, 28. Prove that the length of a line bisecting the greatest side and drawn from the vertex of the opposite angle is 13. 38. Along the bank of a river is drawn a base line of 500 ft. The angular distance of one end of this line from an object on the opposite side of the river, as observed from the other end of the line, is 53° ; the angular distance of the second extremity from the same object, observed from the first extremity, is 79° 12'. Find the breadth of the river. 39. Two forces, one of 410 lb. and the other of 320 lb., make an angle of 51° 37'. Find the size and direction of their resultant. 40. An unknown force combined with one of 128 lb. produces a resultant of 200 lb., and this resultant makes an angle of 18° 24' with the known force. Find the size and direction of the unknown force. Areas and Regular Polygons 41. Two sides of a parallelogram are 59.8 ch. and 37.05 ch., and the included angle is 72° 10'. Find the area. 42. The three sides of a triangle are 49 ch., 50.25 ch., and 25.69 ch. Find the area. 43. One side of a regular pentagon is 25. Find the area. 44. One side of a regular decagon is 46. Find the area. 45. In a circle with a diameter of 125 ft. find the area of a sector with an arc of 22°. 46. In a circle with a diameter of 50 ft. find the area of a segment with an arc of 280°. 47. A building is 37.54 ft. wide and the slope of the roof is 43° 36'. Find the length of the rafters. 48. What angle at the center of a circle does a chord which is 4/7 of the radius subtend ? 49. The side of a regular pentagon is 2. Find the radius of the inscribed circle. 50. The side of a regular decagon is 23.41 ft. Find the radius of the inscribed circle. REGULAK POLYGONS 147 51. The perimeter of a regular polygon of 11 sides is 23.47 ft. Find the radius of the circumscribed circle. 52. The perimeter of a regular heptagon inscribed in a circle is 12. Find the radius of the circle. 53. Find the perimeter of a regular decagon circumscribed about a unit circle. 54. Find the perimeter of a polygon of 11 sides inscribed in a unit circle. 55. The perimeter of an equilateral triangle is 17.2 ft. Find the area of the inscribed circle. FORMULAS {sin A CSC A = 1. cos A sec A = 1. tan A Got A = 1. 2. tan ^ = sin ^ /cos J. 3. cot ^ = cos ^/sinyl. 4. sin^^ H- cos^^ = 1. 5. tan^ A -^1 = sec^ A . 6. cot^^ H- 1 = csc^^. 7. sin (^ + j5) = sin ^ cos 5 + cos A sin 5. 8. cos (A -]- B) = cos ^ cos B — sin ^1 sin B. tan ^ + tan B Paob 1 — tan A tan B cot A cot 5 — 1 9. tan(^ -i-B) 10. cot U + 5) ^ ^ cot 5 + cot ^ 11. sin ( J — 5) = sin ^ cos 5 — cos A sin B. 12. cos (^ — 5) = cos ^ cos 5 + sin A sin 5. . „ , . . „. tan ^ — tan 5 13. tan(^ -B) = rj— — — -• ^ ^ 1 + tan A tan B 14. cot (A - B) cot ^ cot J3 + 1 cot B — cot A 148 31 53 66 57 55 56 67 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. FORMULAS sin 2^ = 2 sin ^ cos A. cos 2.4 = cos^ ^ — sin^ A (i) = l-2sinM (ii) = 2cosM-l. (iii) 2 tan A tan 2^ = cot 2A = sin 1 - tan2 A cot^ yl - 1 2 cot A — cos .4 2 H- cos ^ 2 cos A -h cos ^ ' A _ 11 + cos A "^^2^\l-cos^ sm C + sm Z) = 2 sin — - — cos — ^ C - D C + D . sm C — sm D = 2 cos — - — sm C ^- D cos C 4- cos 2) = 2 cos — - — cos 2 C- 7) cos C — cos Z> = — 2 sm — - — sm — - — sin C + sin Z) _ tan^(C + i>) sin C — sin D ~ tan -J- (C — D) a h c sin ^ ~ sin j5 ~" sin C 149 Paoe 57 57 58 68 59 60 61 75 150 PLANE TRIGONOMETRY 29. a2 = &2_|_c2_2^>ccos^ ^>2 + c^ - a^ 30. cos .4 31. 2 be a-\-b tan ^ {A + B) a — b tan -I- (^ — B) 3^. taii^- = ^-p^cot- 33. sin 34. cos 35. tan ^) ^ /5(g-a) 2 \ be A_ Us -b)(s- c) 2~\ s(s-a) Page 76 79 88 36. r = V(5 -a)(s- b) (s - c)/s. 37. tan(yi/2)=r/(5-a). 38. F=Jcsin^/2. 39. F = -s/s(s-a)(s-b)(s-c). ,^ „ ft^sin A sin C 40. F = ^r—. — 2 sin 5 41. TT radians = 180° = 2 right angles. 42. N = s/r. 43. sin(270°±^) = cos(180°±^) = -sin(90°±^) = -cos^. 44. cos (270° ± J) = - sin (180° ±^) = - cos (90° ± ^) = ± sin A. 45. tan (270° ± ^) = - cot (180° ± A) = tan (90° ± ^) = q: cot ^. , 89 92 95 96 '44 FORMULAS 151 46. sin3^ = 3sin^ -4sin»^. 47. cos 3^ = 4 cos*^ — 3 cos A. 3 tan A — tan^^ 48. tan 3.4 1 -3tanM Page 136 49. In the following table any two expressions in the same line are equal arithmetically. Whether they are like or oppo- site in quality is known by § 22, when it is known in what quadrant A is. See pages 32, 33. 8\nA tan^ 1 V^8ec2y(-1 sec^ 1 sec^ 1 sin^ ^/l-cos2^ cos^ v^H-tan2^ 1 COt^ csc^ V^C8C2^-1 cos^ "^l-sin^A sin A tan^ 1 tan^ "^l + cot'A 1 cot^ cot A C8C^ V^l-C082.4 cos A cos A 1 tan A ^^sec2^-l 1 ^sec^A-1 sec A sec A ^^0802^-1 sin^ 1 cot A ^C8C2^-1 "^l-cos^A cos A 1 sec^ ^l+COt2^ cot^ CSC^ >^l-8in^A 1 sin^ Vcsc2^-1 v^l + tan2^ tan^ CSC A ^^l + cot*^ CSC A Vl-COS2vl ^sec^A-1 50. If sin = sin A,e = n7r-\-(- 1)M. 51. If cos = cos A, = 2 TiTT ± A. 52. If tan^ = tanyl, ^ = n7r + ^. m :tn 53. tan~^?>i ± tan~^7i- = tan" 1 q= mn Page 99 100 101 107 ANSWERS EXERCISE I 1. sin ^ = 2/5, csc^ = 5/2, cos^ =V21/5, sec^ = 5/V21, tan ^ = 2/V21, cot A = V21 /2 ; A=2S° 34' 40^ . 2. sin ^ = 4/5, esc ^1 = 5/4, cos ^ = 3/5, sec A = 6/3, tan^ = 4/3, cot^ = 3/4; A = 63° 8'. 3. sin A = V7/4, esc ^ = 4/V7, cos J. = 3/4, sec ^ = 4/3, tan A =^7 /3, cot ^ = 3/ V7 ; ^ = 41° 24' 30". 4. sin A = V8/3, esc A = 3/V8, cos ^ = 1/3, sec^ = 3, tan ^ = V8, cot A = 1/ V8 ; ^ = 70° 32'. 5. sin A = 1/V17, esc ^ =V17, cos^ =4/V17, sec ^ =V17/4, tan ^ = 1/4, cot ^ = 4 ; ^ = 14° 2' 15". 6. sin A = 4/5, csc^ = 6/4, cos^ = 3/6, sec A = 6/3, tan A = 4/3, cot ^ = 3/4 ; A = 63° 7' 46". 7. sin ^ = 2/v'29, esc A = V29/2, cos A = 5/V29, sec A = V29/5, tan^ = 2/5, cot^ = 6/2; A = 21° 48' 6". 8. sin A = 3/VlO, esc A = VlO/3, cos A = 1/VlO, sec A = VIO, tan ^ = 3, cot ^ = 1 /3 ; ^ = 71° 34'. 9. sin A = 4/6, esc A = 5/4, cos .4 = 3/5, sec A = 5/3, tan ^ = 4 /3, cot ^ = 3/4 ; A = 53° 8'. 10. sin A = V7/4, esc ^ = 4/V7, cos J. = 3/4, sec^ = 4/3, tan A = V7/3, cot^ = 3/V7 ; ^ = 41° 24' 30". 11. sin A = 2/5, esc^ = 6/2, cos^ =V21/6, sec ^ = 5/V21, tan A = 2/V21, cot A = V21/2 ; ^ = 23° 35'. 12. sin A =2/3, csc^ = 3/2, cos^ = V5/3, sec^ = 3/V5, tan ^ = 2/ V5, cot ^ = V5/2 ; ^ = 41° 48' 30". 153 154 PLANE TRIGONOMETRY 13. sin JL = 4/V17, csc^=V17/4, cos^ = 1/V17, 8ecA=y/n, tan J[ = 4, cot ^ = 1 /4 ; J. = 75° 57' 49''. 14. sin ^ = 1 / V50, CSC A = V50, cos J. = 7 / V^O, sec A = V50/7, tan ^ = 1/7, cot ^ = 7 ; ^ = 8° 7' 48". 15. sin ^ = 9/V82, esc 4 = V82/9, cos A = 1/V82, sec A = V82, tan ^ = 9, cot ^ = 1 /9 ; ^ = 83° 39' 35". 16-18. For answers see table on page 151. EXERCISE n 1. sin 60° ; cos 30° ; tan 55° ; cot 75° ; esc 5° ; sec 14° ; cos 16° 46' ; sin 24° 17'. 2. 45°. 4. 18°. 6. 5°. 8. 90°/(m + n). 3. 30°. 5. 36°. 7. 5°. 9. 60°/(c - 1). EXERCISE m 1. 5 = 65°, 6 = 64.335, c = 70.98. 2. A = 35°, a = 14.281, c = 17.434. 3. 5 = 25°, a = 63.441, 6 = 29.582. 4. ^ = 75°, a = 74.64, c = 77.28. 5. ^ = 55°, 6 = 35.01, c = 61.05. 6. ^ = 35°, 6 = 49.152, a = 34.416. r. -4 = 20°, B = 70°, c = 106.4. 8. ^ = 25°, 5 = 65°, c = 55.15. 9. ^ = 20°, B = 70°, a = 34.2. 10. A = 20°, B = 70°, 6 = 46.985. 11. A = 15°, a = 10.352, 6 = 38.636. 12. B = 80°, a = 5.289, c = 30.45. 13. 5 = 70°, a = 27.36, 6 = 75.176. ANSWERS 155 14. A = 65°, b = 13.989, c = 33.09. 15. A = 65°, B = 25°, b = 14.086. 16. ^ = 15°, ^ = 75°, c = 51.75. - EXERCISE rV 1. 160.7 yd. 5. 42.68 ft. 9. 81.98 ft. 2. 50°. 6. 107.22 ft. 11. 1174.6 ft. 3. 96.06 ft. 7. 136.63 ft. 12. 3770 ft. 4. 122.02 ft. 8. 44.78 ft., 37.58 ft. 13. 98.097 ft., 68.69 ft. 14. 36.08 ft., 154.23 ft. 15. 33.51 ft., 28.12 ft. to nearest tower. 16. 74.24 ft., 51.98 ft. 17. 378.21 ft., 417.17 ft. 18. h = altitude = 32.14 ft., c = base = 76.6 ft., Q = area = 1231 sq. ft. 19. ZC = 40°, ZA = 70°, h = 93.97 ft., Q = 3213.8 sq. ft. 20. 61.04 ft., ZA = 35°, ZC= 110°. 21. h = 62.836 ft., r = 76.71 ft., Q = 2764.8 sq. ft. 22. h = 107.23 ft., r = 118.3, Q = 5361.3 sq. ft. 23. c = r=: 57.74 ft., Q = 1443.5 sq.ft. 28.492.4 ft. 24. 20°. 29. 2515 ft. 25. R = 274.75 ft., area = 237150sq. ft. 30. 125 (V3 + 3) ft. 26. 25°; 14.09 ft. 31. 30°. 27. 917.136 ft. EXERCISE V 1. 2d qdt. 4. 2d qdt. ; 4th qdt. 7. 2d qdt. ; 1st qdt. 2. 4th qdt. 5. 2d qdt.; 1st qdt. 8. 4th qdt. ; 3d qdt. 3. 2d qdt. 6. 4th qdt. ; 3d qdt. 9. 3d qdt. ; 3d qdt. 156 PLANE TRIGONOMETRY 10. 847° is coterminal with 127°; 1111° with 31°; -225° with 135°; - 300° with 60° ; 942° with 222° ; - 1174° with - 94° or 266°. 11. 405° and 1125°; - 315° and - 675°; 390° and 750°; - 330° and - 690° ; 460° and 820° ; - 260° and - 620° ; 560° and 1280° ; - 160° and - 520°; 350° and 710°; - 370° and - 730°; 260° and 620° ; - 460° and - 820°. 12. -75°; 15°. 15. - 224° 22' 17''; -134° 22' 17". 13. - 138° ; - 48°. 16. 122° 14' 21" ; 212° 14' 21". 14. -205° 17' 14"; -115° 17' 14". 17. 255° 28' 42"; 345° 28' 42". 18. 60°. 19. 175°. 20. 30°. 21. 340°. 22. 317°. EXERCISE VI 1. MP = - 2, OP =3, OJlf = T V5 ; CSC ^ = -3/2, cos ^= T \/5/3, sec J. =T 3/V5, tan J. =± 2/V5, cot^ =± V5/2. 2. 3f P = ± 5, OJIf = ± 2, OF = V29 ; cot ^ = 2/5, sin ^ = ± 5/V29, esc ^ = ± V29/5, cos ^ = ± 2/ V29, sec^ =± V29/2. 3. JWP = ± 3, Oilf = =F 1, OP = VIO ; cot JL = - 1 /3, smA=± 3/VlO, CSC ^ = ± VlO/3, cos ^ = T 1 /VIO, sec J. = ^ VIO. 4. OM = 2, OP = S, MP = ±y/5', sec A = 3/2, sin J. = ± V5/3, esc J. = ± 3 / V5, tan J. = ± V5/2, cot ^ = ± 2 / V5. 5. csc^ =- 8/7, cos^ ==F V15/8, sec^ =T 8/V15, tan^ = ± 7/V15, cot J. =:± V15/7. 6. cot^ = 1/7, sin J. =± 7/(5 V2), csc^ =± 5 V2/7, cos ^ = ± 1 /(5 V2), sec J. = ± 5 V2. 7. sec A =-7/3, sin ^ =±2 VlO/7, csc^ =±7/(2 VIO), tan^ =T 2 VlO/3, cot^ =T 3/(2 VIO). 8. JfP = ±3, OJlf=± 5, OP = V34; tan^ = 3/5, sin ^ = ± 3/V34, csc^ = ± V34/3, cos ^ = ± 5/V34, sec ^ = ± V34/5. 9. sec^ =- 5/4, sin^ =± 3/5, esc ^ = ± 5/3, tan ^ =^ 3/4, cot^ =T 4/3. 10. OM =1, OP = 2, ifP = ± V3 ; cos ^ = 1/2, sin A=± V3/2, CSC J. = ± 2 / V3, tan A=± V3, cot ^ = ± 1 / V3. ANSWERS 167 11. cos ^ = - 2/3, sin ^ = ± V5/3, esc ^ = ± 3/ V5, tan^ =qF V5/2, cot^ =^2/y/b. 12. sin A =-3/5, cos^ =t4/6, sec4 =:f 5/4, tan ^ =±3/4, cot J. =±4/3. EXERCISE Vn 1. csc^ =-3/2, cos^ =T^1 -4/9 = T V5/3, sec^ = rp 3/ V5, tan J. =± 2/V5, cot^ =± V5/2. 2. sec ^ = 3, sin^ = ± 2 V2/3, csc^ = ± 3/(2 V2), tan^ = ± 2 V2, cot J. =± 1/(2 V2). 3. CSC ^ = 5, cos ^ = ± 2 V6/5, sec ^ = ± 5/(2 V6), tan ^ = ± 1 /(2 V6), cot ^ = ± 2 y/Q. 4. sec J. =- 4/3, sin^ =± V7/4, esc ^ =± 4/V7, tan^ ==F V7/3, cot^ =T 3/V7. 5. cot J. =- 3/4, sin^ =± 4/5, esc ^ = ± 5/4, Cos^ =q:3/5, sec A = T 5/3. 6. tan J. = - 1/2, sin ^ = ± 1/ V5, esc ^ = ± y/b, cos J. = =F 2/ V5, sec^ =T V5/2. 7. tan^ = 2/3, sin ^ =±2/V13, esc ^ =±V13/2, cos^ =±3/V13, sec^ =±V13/3. 8. cot^ = 2/5, sin^ =± 5/V29, esc ^ =± V29/5, eos^ =±2/V29, see^ =±V29/2. 9. sin ^ =- 1/V3, eos^ =T\/(2/3), sec^ =T V(3/2), tan ^ = ± 1 / V2, cot J. = ± V2. 10. cos^ = 1/4, sin^ = ±V15/4, esc J. =±4/V15, tan^ =± V15, cot^ =± 1/V15- 11. cot^ =_i/V7, sin A=±V14/4, ese^=±4/V14, eos^ =± V2/4, see^ =t4/V2. 12. sec A = —t sin ^ = ± '— , esc A =± m c Vc^ Vc2 — m2 ^ m tan A=± , cot J. = ± m Vc2 13-17. See table on page 151. 158 PLANE TRIGONOMETRY EXERCISE IX 1. sin 12°. 5. -cos 25° 9. cos 30°. 13. esc 36°. 2. -tan 43°. 6. sin 6°. 10. cot 25°. 14. -tan 16° 54'. 3. sin 17°. 7. -cot 24° 11. -cot 26°. 15. -tan 33° 39'. 4. cos 24°. 8. sin 22°. 12. -esc 23°. EXERCISE X 1. The cosine of the sum\ _ j cos first ■ cos second of any two angles J ~ \ — sin first • sin second. 2. (V2+V6)/4. 6. 1; 0. 3. (V6-V2)/4. 7. 0; 1. 4. (V6-V2)/4; (V6 + V2)/4. 8. ^'^(1 + V42); tV(V21 -4 V2). 5. (V6-V2)/4; (V6 + V2)/4. 9. ^V(8 + 5V3); ^^(2^6-^16). EXERCISE XI 1. The cosine of the difference') _ f cos first - cos second of any two angles J ~ I + s,m first • sin second. 2. (V6-V2)/4; (V6 + V2)/4. 3. (V6-V2)/4; (V6 + V2)/4. 4. (2 V2 -V15)/12; (2V30 + 1)/12. 5. (3V5-2V7)/12; (6 + V35)/12. EXERCISE Xn 1. The tangent of the difference 1 _ f the difference of their tangents of any two angles / ~~ 1 1 + product of their tangents 2. (V3 + 1)/(V3-1). 4. 1; 7. 3. (V3-1)/(V3 + 1). 5. 1; 1/7. ,_ ^ / ^ . TJX cot^ + cot^ .^ .. _. cot B- cot ^ 12. tan {A + B) = ; tan {A — B) = cot J. • cot J5 - 1 cot ^ • cot J5 + 1 -. o ^ / >• . T>x 1 — tan ^ • tan J5 . ^ , . _, 1 + tan ^ • tan B 13. cot (A + B)= -— ; cot (A-B) = —^ tan ^ + tan J5 tan -4 - tan B ANSWERS 169 EXERCISE Xm - ^, , 4. X * . , twice the tangent of the angle 1. The tangent of twice an angle = — 1 — (the tangent of the angle)^ 2. V3/2; 1/2; V3. 3. V3/2; - 1/2; -V3. 4. 2 sin 3^ cos 3 ^ ; cos2 3 ^ - sin2 3 ^, 1 - 2 sin^ 3 A, or 2cos2 3^ - 1; 2 tan 3^ /(I - tan2 3^). 5. 2 sin (3^/2) cos (3^/2); cos2(3^/2) - sin2(3^/2), or l-2sin2(3^/2), or 2 cos2(3^/2) - 1 ; 2tan(3^/2)/[l-tan2(3^/2)]. EXERCISE XIV 1 11^ T X J. 1 + COS angle 1. cos half an angle = square root of ^— ; tan TiaJf an angle = square root of — • 1 + cos angle 2. V2-V2/2; V2 +V2/2; ^ P ~ ^^ =^S - 2 ^2. \2 + V2 3. V2-V3/2; V2TV3/2; V? - 4 V3. 4. V3/3; V6/3; V2/2. - / I -g 11 + a /I - a _ /I - cos 2 ^ /I + cos 2 ^ /I - ®- V 2 ' V ~2 • VlT COS 2^ 1 — cos 4 ^ /I + COS 4 ^4 /I — cos 4 J. . /I - cos 4 ^ /I + cos 4 .4 11 - V 2 ' \ 2 ' Vrr , /I - cos 6 ^ /I + cos 6 ^ /I ^- \ — ^ — ' V — 2 — ' Vi cos 4^ 1 — cos 6 J. + cos 6 J. 160 PLANE TRIGONOMETRY EXERCISE XV }■ { The difference of the sines of any two angles The sum of the cosines of any 1 _ two angles j ~ The difference of the cosines^ _ of any two angles J ~ twice cos half sum into sin half difference. J twice cos half sum into \ cos half difference. r twice sin half sum into \ sin half difference. 15. (1) sin {A-]- B) = (2 V2 + V3)/6 : cos(^ + J5) = (2V6-l)/6; sin 2 A =:V3/2; cos 2^ = 1/2; sin {A-B)z= (2 V2 - V3)/6 ; cos (J. - 1?) = (2 V6 + l)/6 ; sin2E = 4 V2/9; cos2JS = 7/9; (2) sin ( J. + B) 3z - (2 V2 - V3)/6 ; sin (^ - ^) = - (2 V2 + V3)/6 ; cos U + -B) = -(2 V6 + l)/6 ; cos {A - B)=- (2 V^ - l)/6 ; sm2^ = V3/2; sin 2B = - 4 V2/9; cos2^ = l/2; cos2B = 7/9. 16. (1) (2) tan {A-\- B) : tan {A - B) cot (^ + B) cot U - -S) sec {A + B) esc ( J. + ^) tan 2 4 cot 2 ^ sec 2B CSC 2 B tan (^ + B) tan (^ - ^) cot {A + B) cot (^ - B) ; sec (^ + B) CSC (A + B) tan 2 ^ cot 2^ sec 2 B CSC 2 B (2V2+V3)/(2V6-1): (2 V2 - V3)/(2 V6 + 1) (2V6-1)/(2V2+V3) (2V6 + 1)/(2V2-V3) 6/(2 V6-1); 6/(2V2 + V3); V3; 1/V3; 9/7; 9/(4 V2). (2V2-V3)/(2V6 + 1) (2V2+V3)/(2V6-1) (2V6 + 1)/(2V2-V3) (2V6-1)/(2V2+V3) -6/(2V6 + l); -6/(2V2-V3); V3; 1/V3; 9/7; -9/4 V2. ANSWERS 161 EXERCISE XVI 9. sin(3a;/4) = ±V[l - cos(3x/2)]/2 ; cos(3x/4)=±V[l + cos(3x/2)]/2; tan(3x/4)=±Vl - cos(3x/2)/ Vl + cos(3x/2). 10. sin (3x/4) = 2sin(3x/8)cos(3x/8); cos (3x/4) = cos2(3x/8) - sin2(3ic/8) ; tan(3x/4) = 2taii(3x/8)/[l - tan2(3x/8)]. 16. (1) sin {A + B) = {2+^l5)/6; sin {A - B) = {2 - V15) /6 ; cos(^+ B) = {^/6 - 2 V3)/6; cos(^ - B) = (V5 + 2 V3)/6 sin2^=4V5/9; cos2^ = l/9; sin2J5=V3/2; cos25 = -l/2. tan2^=:4V5; cot 2^ = •v/5/20 ; tan2B = -V3; cot2B = -V3/3. EXERCISE XVn 1.^=23% 6 = 11.779, c= 12.796. 2. B = 52°, b = 10.355, c = 13.14. 3. J5 = 75°, 6 = 6.7614, a = 1.8117. A. A= 40°, a = 16.782, c = 26.108. 5. A= 34° 22' 9'', B = 55° 37' 51", 6 = 0.51176. 6. ^ = 33° 8' 56'', B = 56° 51' 4", c = 499.26. 1. A= 39° 49' 22", B = 50° 10' 38", a = 48.863. S. B = 81°, a = 148.41, c = 948.68. 9. ^ = 49° 53' 53", B = 40° 6' 7', c = 4.4632. 10. 5 = 43° 37', a = 3821.5, 6 = 3641.3. 11. ^ = 35° 53' 56". 5, 5 = 54°6'3".5, 6=731.23. 12. A = 66° 51', a = 176.53, c = 191.99. 162 PLANE TRIGONOMETRY 13. A = 71° 22', a = 2.4099, b = .81268. 14. B = 58° 15', b = 77.632, c = 91.294. 15. ^ = 32° 10' 15'', J5 = 57° 49' 45", a = 388.45. 16. -4 = 7° 53' 42", 6 = 644.11, c = 650.27. EXERCISE XVin 1. C=180°-2^, r = c/ {2 -cos A), h = c-tSinA/2. 2. J. = 90° - 0/2, r = ^/cos(C/2), c = 2 ^ • tan(C/2). 3. ^ = tan-i(2/i/c), C = 2 • tan-i(c/2/i), r = V4¥T^/2. 4. r = 2.055, 71=1.6853, ^ = 55° 5' 30", Q = 1.9819. 5. r = 7.706, c = 3.6676, C = 27° 32', Q= 13.7253. 6. Let X = length of rafter and Q = area of roof ; then, since the roof projects one foot over the side of the barn, we have : a; = 21 /sin 45°, Q = 2-x-82; x = 29.698, Q = 4870.44. 7. r = 1.61804, 7i = 1.53882, F= 7.69417. 8. r = 11.2692, 71 = 10.8852, 2?^= 380.99. 9. r = 1.0824, c = 8284, F= 3.3136. 10. r = 1.5994, 7i = 1.441, p = 9.715. 11. h = 28.971, r = 31.357, A = (difference between polygon and in- scribed circle) = 144.51, Be = 307.8. 12. A =14.536, r = 16.134, A = 48.48, Dc = 105.47. 15. 99.64 sq. ft. 16. 2.393. EXERCISE XIX 1. ^ = 65° 15', 6 = 95.6025, c = 89.648. 2. B = 72°14', a = 75.132, c = 92.788. 3. ^ = 31° 20', 6 = 184.896, c = 191.978. 4. C = 60°, a = 255.38, 6 = 282.56. 5. ^ = 15° 43', 6 = 222.1, c = 321.08. ANSWERS 163 6. J5 = 66°, a = 765.43, c = 1035.4. 7. 1253.2 ft. 9. 300 yd. 11. 294.77 ft. 8. 1116.6 ft. 10. 12296 ft., 13055 ft. 12. 4211.8 ft. EXERCISE XX 1. ^ = 32° 25' 36", C= 106° 24' 24'', c = 259.4. 2. A= 28° 20' 48", C = 39° 35' 12", a = 293.56. 3. J5 = 32°36'9", C = 83° 33' 5", c = 6.621. 4. Bi = 51° 18' 22", Ci = 88° 41' 38", Cx = 218.525 ; Bi = 128° 41' 38", Ca = 11° 18' 22", Cg = 42.853. 5. ^1 = 31° 57' 46", Ai = 120° 44' 14", ai = 120.31 ; B2 = 148° 2' 14", A2 = 4° 39' 46", 03 = 11.379. 6. Impossible. 7. Ci = 46° 18' 40", ^1 = 93° 9' 13", ai = 69.4567 ; C2 = 133° 41' 20", A2 = 5° 46' 33", ag = 7.0005. 8. ^1 = 51° 18' 27", d = 98° 21' 33", Ci = 43.098 ; A2 = 128° 41' 33", C2 = 20° 58' 27", d = 15.593. 9. ul = 54° 31' 13", C = 47°44'7", c = 50.481. 10. ^1 = 24° 57' 54", Ci = 133° 47' 41", Ci = 616.67 ; ^2 = 155° 2' 6", C2 = 3°43'29", C2 = 55.41. 11. ^1 = 16° 43' 13", ^1 = 147° 27' 47", ai = 35.519; B2 = 163° 16' 47", Ai = 0° 54' 13", Oa = 1.0415. EXERCISE XXI 1. ^ = 42° 50' 58", 5 = 64° 9' 2", c = 374.06. 2. 5 = 132° 18' 28", C= 14° 34' 23", a = 67.75. 3. A = 109° 15' 30", B = 45° 4' 30", c = 440.45. 4. ^ = 60° 44' 39", B = 47° 21' 21", c = 966.28. 5. B = 54° 1' 13", C = 63° 49' 9", a = 44.825. 6. 902.94 ft. 8. 1331.2 ft. 10. 10.532 mi. 7. 13.27 mi. 9. 4.8112 mi. 11. 9.6268 mi. 164 PLANE TRIGONOMETRY EXERCISE XXn 1. A = 74° 40' 18'', B = 47° 46' 38", C = 57° 33' 4". 2. A = 59° 19' 14", B = 68° 34' 8", C = 52° 6' 40". 3. A = 45° 11' 50", B = 101° 22' 12", C = 33° 25' 58". 4. A = 11° 33' 52", B = 49° 8' 3", C = 59° 18' 5". 5. A = 54° 3' 10", B = 30° 47' 22", C = 95° 9' 24". 6. Bi = 41° 41' 26", Bi = 138° 18' 34", Ci = 111° 52' 34", C2 = 15° 15' 26", ci = 177.2; C2 = 50.248. 7. A = 17° 16' 11", B = 28° 43' 49", c = 14.424. 1 8. A = 18° 12' 22", B = 135° 50' 46", C = 25° 56' 52". ' 9. Since c>a, C>A> 90°, which is impossible, as a , triangle cannot have two obtuse angles. , 10. B = 48° 34' 44", A = 49° 38' 16", a = 76.015. 11. B = 145° 35' 24", C = 7° 11' 36", a = 104.57. 12. A = 57° 52' 44", B = 70° 17' 24", C = 51° 49' 50". 13. Ai = 70° 12' 48", A2 = 109° 47' 12", Bi = 57° 22' 56", ^2 == 17° 48' 32", 61 = 28.79; 62 = 10.454. \ 14. A = 32° 44' 40", C = 63° 48' 28", b = 137.39. ; 15. 6=143.52, B = 146° 43' 10", C = 14° 4' 7". i 16. A = 67° 55' 15", B = 54° 4' 45", c = 85.36. i 17. 44° 2' 9", 51° 28' 11' , 84° 29' 40". 18. 44° 2' 56". 20. N. 4° 23' 2" W., or S. 4°23'2" W. 19. 60° 51' 8". 21. 60°. EXERCISE XXm 2. 1931.8. 3. 44770; 781.617; 149689; 314543. 4. 123794 ; 53596.3 ; 10.6665 ; Fi = 11981.7, F2 = 2349.63. 5. 1016.23; 30.858; 1430.3; 170346. 6. 3891.64; 3319.38; 9229.4; 31246.4. ANSWERS 165 EXERCISE XXIV 1. 9;r/4, 17;r/4, -7 7r/4, -15;r/4; 13;r/4, 21;r/4, -3 7r/4, ll7r/4; 7;r/2, 11 7r/2, - ;r/2, - 5 7r/2 ; 9 7r/2, 13;r/2, -3;r/2, 7 7zr/2; 7;r/3, 13 7r/3, - 5 7r/3, - 11 7r/3; 8;r/3, 14 7r/3, -4;r/3, 10;r/3; 137r/6, 25;r/6, -ll7r/6, -237r/6; 177r/6, 29;r/6, 7 7r/6, -19;r/6. 2. 120°. 4. 900°. 6. ;r/4. 8. ;r/2. 3. 300°. 5. 135°. 7. 3 7r/4. 9. 3 7r/2. 10. 12859/7560, or 1.7. 14. ;r/4, 3;r/4. 11. 115643/37800, or 3.06. 15. - 7r/6, 7r/3. 12. 35827/47250, or 0.76. 16. - ;r/4, ;r/4. 13. 1516273/2268000, or 0.67. 17. - 7 ;r/6, - 2 7r/3. 18. sin(7r/6) = l/2, cos{7r/6) = V3/2, tan(;r/6) = V3/3. 19. sin(7r/4)=V2/2, cos(7r/4) = V2/2, tan(;r/4)=l. 20. sin(7r/3) = V3/2, cos(;r/3) = 1/2, tan(;r/3) = V3. 21. sin(;r/2) = 1, cos(;r/2) = 0, tan(7r/2) = oo. 22. sin TT = 0, cos TT = — 1, tan tc = 0. 23. 157/70 or 2.24; 157 (57° 17' 44^8)/ 70. 24. 55/98; 55 (57° 17' 44^8)/ 98. 25. 3/4; 3(57°17'44''.8)/4. 26. 11/210 in. 29. 3980^^ mi. EXERCISE XXV 1. e = UTt -\- {-l)n{±7r /2) = n7t ±7t/ 2. 2. n7t±7t/i. 6. n7t±7t/Q. 10. 2n7r±7t/3. 3. n7r±7r/4. 7. 2n7t±2 7t/S. 11. 2n7t-\-7t/4. 4. n7r±7t/e. 8. n7r + (-1)« • 7r/6. 12. n7t±7t/6. 5. nit±7t/3. 9. n;r + (-l)« • (7r/3). 13. titt + 7r/4. 166 PLANE TRIGONOMETRY 14. nTC + ;r/4, mt + 7t/3. 16. utc ± ;r/4. 15. nje + 5 7t/6, uTt -^ 2 7r/3. 17. n • 180° + 53° 7' 45''. 18. (2 w + 1) . 180°, 2 n • 180° + 53° 8'. 19. [n;r + (-!)« ;r/4]/5. 22. {2n7t + 7t/2)/d, 2 mt- 7r/2. 20. 2n7r/9, 2n;r. 23. (2n7r ± 7r/2)/(m ± r). 21. (nnr + 7r/2)/(r + l). 24. 2w7r + ;r/4, 2n7r - 3 7^/4. ^ must be in the first or third quadrant. Why ? 25. 2n7r-;r/6, 2n7r + 6;r/6, 2w7r-7r/3, 2M;r + 2;r/3. d must be in the second or fourth quadrant. Why ? 26. 2n7r- 7r/4, 2n7r + 3 7r/4 28. (2n + l)7r + ;r/4. 27. n;r/3 + 7r/6. 29. 2n;r-;r/6. 30. 7 7r/12 and ;r/4; (2 m + n) 7r/2 ± tt/G + (- l)«(;r/12) and (n — 2m);r/2 + (— l)«;r/12 T 71^/6, where m and n are any integers. 31. 25;r/24 and 19;r/24; {m + 2n) tc /2 + 7t /S ± 7t /12 and (2n — m)7t/2±7t/12 — 7i/S, where m and n are any integers. EXERCISE XXVI 1. nn/^, {2n7t ±7t/S)/3. 8. UTt ± 7t/4, 2n7t ±27C /S. 2. TiTi: ± 7t / 4:, 2 uTt ± 7t / S. 9. 2ii7r + 7r/6 ± 7r/4. 3. n;r ± ;r/4, n7r-(-l)»;r/6. 10. 2n;r + ;r/4. 4. »i7r/2 ± 7r/8, 2n7r/3 ± 7r/9. 11. 2n7t - tc/S ± 7t/4. 5. 2n;r/3±7r/6, n;r4-(-l)'»7r/6. 12. 2?i;r + 7r/4 ± ;r/5. 6. n;r/3, (2w + l/3)7r/4. 13. 2 n • 180° + 68° 12'. 7. n7t/S,n7C±7t/S. 14. 2w • 180°±60°-33°4r24". EXERCISE XXVn 1. (2-V3)/2; (2-V2)/2; 1/2; (2+V2)/2; 0; 1; 2; 1. 2. 1/2; (2-V2)/2; (2-V3)/2; (2-V2)/2; 1; 0; 1; 2. ANSWERS 167 3. riTT +(-!)» •7r/4; n;r -(-1)« • 7r/3 ; 2n7r±^/6; 2n;r±2;r/3; mt-\-7t/Q\ nit- It /Z; mt-it/^\ n7t+7t/o. 13. 1/2. 15. V3. 17. V3. 19. 13. 14. ±V2/2. 16. -8,1/4. 18. ±V21/14. EXERCISE XXIX 1. (1/V2 + i/V2) • V2, or (cos ;r/4 + isin 7r/4) • V2. 2. [l/V2 + i(-l/V2)]- V2, or [cos(- ;r/4) + isin(- ;r/4)] • V2. 3. [-l/V2 + i(-l/V2)]- V2, or [cos(-3;r/4) + isin(-3 7r/4)]. y/2. 4. (_ 1/2 + i V3/2)-2, or (cos2 7r/3 + isin 2 ;r/3) • 2. 5. (- V3/2 + i/2) -2, or (cos5 7r/6 + isin 5 7r/6) • 2. 6. [- V3/2 + i(- 1/2)] .2, or [cos(- 5 ;r/6) + i sm(- 5;r/6)] • 2. 7. [3/5+i(-4/5)].5, or [cos(- 53°80 + isin(- 53°80] • 5. 8. (- 3/ V13 + i ' 2/ V13) • V13, or (cos 146° 19' + isin 146° 19') • V13. 9. [-3/5 + i(-4/5)].10, or[cos(-126°52') + isin(-126°520]-10. 10. [_V5/4 + i(-Vll/4)]-4, or [cos (- 123° 590 + i sin (- 123° 59')] • 4. 11. cos (90° -^ ^"i REC'D U REC'D LD DEC - 1 1958 JAN 2l95iRt^C^i:D LD 21-100m-9,;47(A5702sl6)476 l3Oct'60lC SEP29l$60 UCT211960 f^. aAb33 T3 THE UNIVERSITY OF CALIFORNIA UBRARY