UC-NRLF B M EMT Dbl s 01 MCTMF1 PV ev\ C|\V) aifyatraitcai ^Itfararg primer ^ScJpigl&r ^rooklgn, jc ^ Stechert-Hafn Inc. 31 E. 10 St C{pv h s #* ELEMENTS OF PLANE TRIGONOMETRY By D. A, MURRAY, Ph.D, Professor of Applied Mathematics in McGill University. INTRODUCTORY COURSE IN DIFFERENTIAL EQUA- TIONS, for Students in Classical and Engineer- ing Colleges. Pp. xvi + 236. A FIRST COURSE IN INFINITESIMAL CALCULUS. Pp. xvii + 439. DIFFERENTIAL AND INTEGRAL CALCULUS. Pp. xviii + 491. PLANE TRIGONOMETRY, for Colleges and Second- ary Schools. With a Protractor. Pp. xiii + 212. SPHERICAL TRIGONOMETRY, for Colleges and Secondary Schools. Pp. x + 114. PLANE AND SPHERICAL TRIGONOMETRY. In One Volume. With a Protractor. Pp. 349. PLANE AND SPHERICAL TRIGONOMETRY AND TABLES, in One Volume. Pp. 448. PLANE TRIGONOMETRY AND TABLES. In One Vol- ume. With a Protractor. Pp. 324. LOGARITHMIC AND TRIGONOMETRIC TABLES. Five- place and Four-place. Pp. 99. NEW YORK: LONGMANS, GREEN, & CO, ELEMENTS OF PLANE TRIGONOMETRY BY DANIEL A. MURRAY, Ph.D. Professor of Applied Mathematics in MgGill University LONGMANS, GREEN, AND CO FOURTH AVENUE & 30TH STREET, NEW YORK LONDON, BOMBAY, AND CALCUTTA 1912 REPLACING ? G2 3 o I Copyright, 1911, BY LONGMANS, GREEN, & CO. First Edition, July, 1911 Reprinted, August, 1912 October, 1912. THE SCIENTIFIC PRESS HOBERT DRUMMOND AND COMPANV BROOKLYN, N- Y. QAssz /In PREFACE This text-book is shorter than my former text-book, entitled " Plane Trigonometry," by the omission of many of the notes and several of the topics in that book and by the more condensed treatment of other topics. There is also a marked difference in arrangement. Thus, radian measure, the periodicity of the trigonometric functions, their general values, and their graphs, and the inverse trigo- nometric functions, which are discussed in the later chapters of the "Plane Trigonometry/ ' are treated in the earlier chapters of the "Elements of Plane Trigonometry." The line definitions of the functions are explained more fully, and the unit circle is used to a greater extent, in this book than in the former one. D. A. Murray. May 1, 1911. M552649 CONTENTS CHAPTER I Trigonometric Functions of Acute Angles ART. PAGE 1. Angle denned 1 2. Degree measure 2 3. Trigonometric functions denned for acute angles 2 4. Problems 5 5. Trigonometric functions of 45°, 60°, 30°, 0°, 90°. Variation of functions 8 6. Relations between the trigonometric functions of an acute angle and those of its complement 12 7. Relations between the trigonometric functions of an acute angle . 13 CHAPTER II Solution of Right-angled Triangles 8. Solution of a triangle 18 9. Cases in the solution of right-angled triangles 19 10. Projection of a straight line upon another straight line 24 11. Measurement of heights and distances 25 12. Solution of isosceles triangles 28 13. Related regular polygons and circles 29 14. Problems requiring a knowledge of the points of the mariner's compass 32 15. Examples in the measurement of land 33 vii vili CONTENTS CHAPTER III Angles in General and Their Trigonometric Functions ART. PAGE 16. General definition of an angle 36 17. Measurement of angles 39 18. Value of a radian 40 18a. Motion of a particle in a circle. 42 19. The convention for signs of lines in a plane 43 20. Trigonometric functions defined for angles in general 44 21. Line definitions of the trigonometric functions. Geometrical representation of the functions 48 22. Changes in the values of the functions when the angle varies. Limiting values of the functions 51 23. Periodicity of the trigonometric functions 55 24. Graphs of the functions 56 25. Relations between the trigonometric functions of an angle 60 26. Functions of -A, 90°, TA, 180, TA in terms of functions of A . 62 27. Reduction of trigonometric functions of any angle to functions of acute angles 66 CHAPTER IV General Values. Inverse Trigonometric Functions. Trigonometric Equations 28. General values 68 29. Inverse trigonometric functions 72 30. Trigonometric equations. Trigonometric identities 75 CHAPTER V Trigonometric Functions of the Sum and Difference of Two Angles general formulas 31. To deduce sin (A + B), cos (A + £) 78 32. To deduce sin (A -B), cos (A -B) 81 33. Fundamental formulas «, 83 34. To deduce tan (A+B), tan (A-B), cot (A + £), cot (A-B) 84 35. To deduce sin 2 A, cos 2 A, tan 2A 85 36. Transformation formulas 88 CONTENTS ix CHAPTER VI Relations between the Sides and Angles of a Triangle ah¥. page 37. Notation. Simple geometrical relations 94 38. The law of sines ■ 94 39. The law of cosines ' 96 40. The law of tangents 97 41. Functions of the half -angles of a triangle in terms of its sides. ... 98 CHAPTER VII Solution of Oblique Triangles 42. Cases for solution. General remarks on methods of solution 101 43. Case I. Given one side and two angles 102 44. Case II. Given two sides and an angle opposite to one of them. . 103 45. Case III. Given two sides and their included angle 107 46. Case IV. Given three sides 108 47. Use of logarithms in the solution of triangles 109 48. Cases I, II, logarithms used 109 49. Case III, logarithms used Ill J>0. Case IV, logarithms used 112 51. Problems in heights and distances 113 CHAPTER VIII Miscellaneous Theorems 52. Area of a triangle 117 53. Area of a circular sector 118 54. Circles connected with a triangle 119 55. Relations between the radian measure, the sine, and the tangent of certain angles 123 Answers to the Examples ...t 128 ELEMENTS OF PLANE TRIGONOMETRY CHAPTER I TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES 1. Angle defined. An angle XOP is the amount of turning which a line makes when it revolves about from Fig. 1. the position OX into the position OP (Figs. 1, 2). Accord- ingly (Fig. 2), one-fourth of a revolution = a right angle; a complete revolution = four right angles. 2 ELEMENTS OF PLANE TRIGONOMETRY When the amount of turning is less than one-fourth a revo- lution (i.e., less than right angle XOY), the angle is called an acute angle. 2. Degree (or sexagesimal) measure of angles. Since all right angles are equal, a right angle may be chosen as the unit of measurement. A right angle, however, is incon- veniently large as a unit. Accordingly a ninetieth part of a right angle, called a degree, is taken for unit. Degrees are divided into minutes, and minutes into seconds, according to the following table of angular measure : 60 seconds = 1 minute, 60 minutes = 1 degree, 90 degrees = 1 right angle. Degrees, minutes, and seconds are denoted by symbols: thus, 23 degrees 17 minutes 20 seconds is written 23° 17' 20". 3. Trigonometric functions (defined for acute angles). From any point P in one of the lines bounding an angle A (Figs. 3, 4, 5) draw a perpendicular PM to the other bound- ing line. (The angles A in Figs. 3, 4, 5 are equal.) In any of these triangles AMP there can be formed six ratios with the lines AM, MP, AP, viz.: MP AM MP_ AM AP_ AP^ ~AP' AP' AM' MP' AM' MP' TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES Since all the triangles AMP above are similar, each of these ratios has the same value whatever be the position of P on a bounding line of the angle. These six ratios are called trigonometric functions of the angle A, and are given names as follows: MP . is called the sine of the angle A ; AP AM . -7-p is called the cosine of the angle A ; -j-jn? is called the tangent of the angle A ; -rjrp is called the cotangent of the angle A ; AP AM is called the secant of the angle A ; AP (1) -TTp is called the cosecant of the angle A. Short symbols for these functions and definitions appli- cable for any right-angled triangle are given in (2) : opposite side hypotenuse ' • A I MP \ sin A {=jp) COS tan A cot A sec / AM\ adjacent side \ AP/~ hypotenuse ' / MP\ opposite side \ AM/ ~ adjacent side' / AM\ adjacent side \ MP) ~ opposite side ' / AP\ hypotenuse \ AM/ ~ adjacent side' cosed \ = mp) hypotenuse opposite side* (2) 4 ELEMENTS OF PLANE TRIGONOMETRY The symbol esc A is also used for cosecant A. From the definitions of these functions and the proper- ties of similar triangles the following properties are easily deduced : (1) To each value of an angle there corresponds but one value of each trigonometric function. (2) To each value of a trigonometric function there corre- sponds but one value of an acute angle. (3) Two unequal acute angles have different values for each trigonometric function. The values of the trigonometric functions for angles from 0° to 90° are arranged in tables. These values, which may be given to four, five, six or seven places of decimals, are called the Natural sines, Natural cosines, etc. The logarithms of these values of sines and cosines (with 10 added) are called Logarithmic sines, Logarithmic cosines, etc. In addition to functions (1), (2), the following are occa- sionally used: versed sine of A = vers A = l — cos A ; coversed sine of A = covers A = \— sin A. EXAMPLES 1. Suppose that the line OP (Fig. 2) revolves about in a counter-clockwise direction, starting from the position OX; show that, as the angle XOP increases, its sine, tangent, and secant increase, and its cosine, cotangent, and cosecant decrease. Test this conclusion by an inspection of a table of Natural functions. 2. Find by tables, sin 17° 40', sin 76° 43', cos 18° 10', cos 61° 37', tan 79° 37' 30", cot 72° 25' 30". Log sin 37° 20', log cos 71° 25', log tan 79° 30' 20". TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES 5 3. Find the angles corresponding to the following Natural and Logarithmic functions: sine= .15327, sine= .62175, cosine =.85970, cosine =.6 1497, tangent = .42482, tangent = .60980, Log sine =9.79230, Log cosine =9.96611, Log tangent = 9.82120. 4. Problems. The student is recommended to try to solve Exs. 1, 2, 3, 4 without help from the book. EXAMPLES 1. Construct the acute angle whose cosine is § . its other trigonometric functions? Find the number of degrees in the angle. The required angle is equal to an angle in a right-angled triangle, in which "the side adjacent to the angle is to the hypot- enuse in the ratio 2:3." Construct a right-angled triangle AST which has side AS=2, and hypotenuse A T= 3. The angle A is the angle required, for cos A = § . What are Now ST= V3 2 -2 2 = V5 = 2.2361. Fig. 6. Hence, the other functions are VE V5 sin A = —-=.7454, tan A = -^-=1.1180, cot A = -7= =.8944, v5 sec A = -=1.5000, cosec A = — 7== 1.3416. V5 The measure of the angle can be found in either one of two ways, viz.: (a) by measuring the angle with the protractor; (b) by finding in the table the angle whose cosine is j or .6667. The latter method shows that A = 48° 11.4'. [Compare the result obtained by method (a) with the value given by method ELEMENTS OF PLANE TRIGONOMETRY a 2. A right-angled triangle has an angle whose cosine is §, and the length of the hypotenuse is 50 ft. Find the angles and the lengths of the two sides. By method shown in Ex. 1, construct C ; an angle A whose cosine is f. On one boundary line of the angle take a length AG to represent 50 ft. Draw GK perpen- dicular to the other boundary line. 7. cos A = l =.6666. . . , .-. A =48° 11.4', .\ J3= 90 -A = 41° 48.6'. AK cos A = — = .6666. . ., AG ' .*. Ai£=50X.6666. . ., = 33.333 sin A = (Ex. 1) KG Vb AG~~T' V5 KG= — X 50 = 37.27. o The problem may also be solved graphically as follows: Measure angles A, G with the protractor. Measure AK, KG directly in the figure. 3. A ladder 24 ft. long is leaning against the side of a building, and the foot of the ladder is distant 8 ft. from the building in a horizontal direction. What angle does the ladder make with the wall? How far is the end of the ladder from the ground? Graphical method. Let AC represent the ladder, and BC the wall. Draw AC, AB, to scale, to represent 24 ft. and 8 ft. respectively. Measure angle ACB with the protractor. Fig. 8. Measure BC directly in the figure. TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES ^ Method of computation. 5C= x/ IC 2 -ZB 2 =V576-64=V512= 22.63 ft. smACB=j^= ^=.33333, .-. ACB=19°28.2'. 4. Find tan 40° by construction and measurement. With the protractor lay off an angle SAT equal to 40°. From any point P in AT draw PR perpendicularly to AS. Then measure AR, RP, and substitute the values in the ratio, -pp tan 40° = -Tp- Compare the result thus ob- tained with the value given for tan 40° in the tables. 5. Construct the angle whose tangent is f. Find its other functions. Measure the angle approximately, and compare the result with that given in the tables. Draw a number of right-angled, obtuse-angled, and acute-angled triangles, each of which has an angle equal to this angle. 6. Similarly for the angle whose sine is $; and for the angle whose cotangent is 3. 7. Similarly for the angle whose secant is 2£; and for the angle whose cosecant is 3J. 8. Find by measurement of lines the approximate values of the trigonometric functions of 30°, 40°, 45°, 50°, 55°, 60°, 70°; compare the results with the values given in the tables. // any of the following constructions asked for is impossible, explain why it is so. 9. Construct the acute angles in the following cases: (a) when the sines are %, 2, f ; (6) when the cosines are i, £, 3; (c) when the tangents are 3, 4, § ; (d) when the cotangents are 4, |; (e) when the secants are 2, 3, i; (/) when the cosecants are 3, 4, 8 ELEMENTS OF PLANE TRIGONOMETRY 10. Find the other trigonometric functions of the angles in Ex. 9. Find the measures of these angles, (a) with the pro- tractor, (b) by means of the tables. 11. What are the other trigonometric functions of the angles: (1) whose sine is — ; (2) whose cosine is — ; (3) whose . a .a a tangent is r-; (4) whose cotangent is =-; (5) whose secant is -; a (6) whose cosecant isr? 12. A ladder 32 ft. long is leaning against a house, and reaches to a point 24 ft. from the ground. Find the angle between the ladder and the wall. 13. A man whose eye is 5 ft. 8 in. from the ground is on a level with, and 120 ft. distant from the foot of a flag pole 45 ft. 8 in. high. What angle does the direction of his gaze, when he is looking at the top of the pole, make with a hori- zontal line from his eye to the pole? 14. Find the functions of 45°, 60°, 30°, 0°, 90°, before reading the next article. 5. Trigonometric functions of 45°, 60°, 30°, 0°, 90°. Varia- tion of functions. J? A. Functions of 45°. Let AMP be an isosceles right- angled triangle, and let each of the sides about the right angle be equal to a. Then Fig. 1 1 . angle A = 45°, and AP = a\/2 . sin 45° = sin A = MP 1 AP a V2 V2* Thus, by definitions Art. 3 and Fig. 10, TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES sin 45°=-/-=, tan 45° = 1, sec 45°= V^ cos 45° = —=, cot 45° = 1 , cosec 45° = V2. V2 The sides of triangle AMP are proportional to 1, 1, a/2. Hence, in order to produce the ratios of 45° quickly, it is merely necessary to draw Fig. 11; from this figure the ratios of 45° can be read off at once. B. Functions of 30° and 60°. Let ABC be an equi- A°\ lateral triangle. From any -4r ~ a D vertex B draw a perpen- dicular BD to the opposite side AC. Then angle DAB = 60°, angle ABD = SQ°. If AB=2a, then AD = a, and DB = V4a 2 -a 2 = ax / 3. ,. S in60°=sinZ)^=5|^^. AB 2a 2 Thus, from Fig. 12, \/3 sin 60° =-2", tan60° = \/3, sec 60° =2, cos 60° =4, cot 60°=- ~j=. 2^ a/5' Also, Thus, sin30°=sm^D=^=| a 4 112 sin 30°=-, tan 30°=-^, sec 30° =-7=, 2' V3 V3 a/3 r cos 30° - — , cot 30° - V3, cosec 30° = 2. 10 ELEMENTS OF PLANE TRIGONOMETRY In ADB the sides opposite to the angles 30°, 60°, 90°, are respectively proportional to 1, V3, 2. Hence, in order to produce the functions of 30°, 60°, at a moment's notice, it is merely necessary to draw Fig. 13, from which these func- tions can be immediately read off. C. Functions of 0° and 90°. Let the hypotenuse in each of the right- angled triangles in Fig. 14 be equal to a. MP AP' AM sin MAP cos MAP" AP It is apparent from this figure that if the angle MAP approaches zero, then the perpendicular MP approaches zero, and the hypotenuse AP approaches to an equality with AM; so that, finally, if MAP = 0, then MP = 0, and AP = AM. Therefore, when MAP = 0, it follows that sin 0°=-=0, tan 0°=-=0, cos0°=- = l, sec0°=- = l. a a (AM approachmg a\ :. Also cot = I , Trt t~t- --?{ ) = unlimited number = \MP approaching 0/ X approaching _ Q / AP equal too \ ,. . •, cosec — l ttd r~' ?, ) = unlimited number = qo . \MP approaching 0/ As MAP approaches 90°, AM approaches zero, and MP TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES 11 approaches to an equality with AP. Therefore, when MAP = 90°, it follows that sin90° = - = l, tan90° = oo, sec90° = oo, cos 90°=- =0, cot 90°=- = 0, cosec90°=- = l. Thus as the angle increases from 0° to 90°, its sine increases from to 1; its cosine decreases from 1 to 0; its tangent increases from to oo ; its cotangent decreases from oo to 0; its secant increases from 1 to oo ; its cosecant de- creases from oo to 1. EXAMPLES N.B. Exponents. In trigonometry (sin x) n is usually written sin n x, and similarly for the other functions. This is not done when n= —1. The reason for this exception is given in Art. 29. Find the numerical value of 1. sin 60° + 2 cos 45°. 2. sec 2 30° + tan 3 45°. 3. sin 3 60° + cot 3 30°. 4. cos 0° sin 45° + sin 90° sec 2 30°. 5. 4 cos 2 30° sin 2 60° cos 2 0°. 6. 3 tan 3 30° sec 3 60° sin 2 90° tan 2 45°. 7. 10 cos 4 45° sec 6 30°. 8. 2 sin 5 30° tan 3 60° cos 3 0°. 9. x cot 3 45° sec 2 60°= 11 sin 2 90°; find x. 10. x(cos30 o + 2sin90 o +3cos45 o -sin 2 60 o ) = 2sec0°-5sm90 o ; find x. 12 ELEMENTS OF PLANE TRIGONOMETRY 6. Relations between the trigonometric functions of an acute angle and those of its comple- ment. When two angles added to- gether make a right angle, the two angles are said to be complementary, and each angle is called the comple- ment of the other. Thus, in Kg. 15, P = 90°-A and P is the complement of A. Now MP Fig. 15. sin ^ == Tp" == cos P= cos (90°- A) AM cos A =-jTp= sin P = sin (90°- A) MP tan A = -^ = cot P = cot (90°- A) cot A = -^p - tan P- tan (90°- A) AP sec A =-7-Tr = cosec P= cosec (90°— A) cosec A — AP MP secP = sec (90°- A) These six relations can be expressed briefly: Each trigonometric function of an angle is equal to the corresponding co-function of its complement. EXAMPLES 1. Compare the functions of 30° and 60°; of 0° and 90°. 2. Express the following as functions of angles less than 45°: sin 78° 20', cos 80° 30', tan 50°, cot 65°, sec 71°, cosec 80°. TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES 13 3. If ski x=cos (2z + 40°), find a value of x. 4. If cot 2x = tan (x— 30°), find a value of x. 6. Show that in a triangle ABC, sin JB = cos i(A + C). 7. Relations between the trigonometric functions of an acute angle. A. Reciprocal relations between the functions. Inspection of the definitions (2), Art. 3, shows that: (a) sinA = r. cosec A=— — r, or, sin A cosec A=l: v ' cosec A smi ' ' (i) (b) cos A = t, sec A= 7, or, cos A sec -4=1; v ' sec A cos A 7 ' 1 1 (c) tan A= — —r, cot A=- j, or, tan A cot -4=1. cot ^1 tan ./L 2?. 7%6 tangent and cotangent in terms of the sine and cosine. In the triangle AMP (Fig. 3), MP MP AP sin A tanA ~AM~AM = ^A ; (2) AP AM AM AP cos A cot A = MP = MP = stiA- (3) AP C Relations between the squares of certain functions. In the triangle AMP (Fig. 3), indicating by WP 2 the square of the length of MP, MP +AM =AP . On dividing each member of this equation by AP , AM , MP , in turn, there is obtained 14 ELEMENTS OF PLANE TRIGONOMETRY (mp\ 2 (AMy /Apy \APj + \APj ~\AP) ' (mp\ 2 (AMy /ap\ \am) + \am) ~\am) (mpv /amv /apv \MP, V /AMy (AP\ 7 + \MPj ~\MP) In reference to the angle A, these equations can be written : sin 2 A + cos 2 A = l 9 tan 2 .4 + 1 =sec 2 A, (4) 1 + cot 2 A = cosec 2 A. Note. An equation involving trigonometric functions is a trigonometric equation. Thus, for example, tsaiA = l. One angle which satisfies this equation is the acute angle A = 45°. Other solutions can be found after Art. 28 has been taken up. EXAMPLES In the following examples, the positive values of the radicals are to be taken. The meaning of the negative values is shown in Art. 20. 1. Given that sin A = J, find the other trigonometric ratios of A by means of the relations shown in this article. V3 cosec A =— — r=2: cos A = \ x l — sin 2 A=— ~ : sin A ' 2 3 sec A = 1 2 sin A 1 tan A— .— ,_ ; cos A V3 cos A V3 ' cot A = — \=V3. tan A TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES 15 These results may be verified by the method used in solving Exs. 1, 5-7/ Art. 4. 2. Express all the ratios of angle A in terms of sin A. sin A = sin A; cos A = v. 1 — sin 2 A ; sin A sin A 1 V 1 —sin 2 A tanA = r= , : cotA = - - = : — ; cos A Vl —sin 2 A tan A sin A sec^l = -:-= . — . ; cosecA = — cosA Vl —sin 2 A ' sin A 3. Prove that ; ; — t + t-— : — 7= 2 sec 2 A. 1— sin A 1 + smA 1 _1 _2 2 1 — sin A 1 + sinA 1 — sin 2 A cos 2 A 4. Prove that sec 4 A -1 = 2 tan 2 A+tan 4 A. sec 4 A-l=(sec 2 A) 2 -l = (l+tan 2 A) 2 -l = 2tan 2 A+tan 4 A. 5. Solve the equation 4 sin 6—3 cosec 0=0. 3 4sin0 — =— z=0. sin # .*. 4 sin 2 0-3 = 0. 3 V3 . V3 .*. sin 2 0= j, .*. sin^=+-^-, and sin 0= —. On taking the pZws sign, one solution is the acute angle 0=60°; other solutions will be found later. For the minus sign there is also a set of solutions; these will be found later. 16 ELEMENTS OF PLANE TRIGONOMETRY 6. Solve 2 sin 2 cosec 0-5 + 2 cosec 0=0, 2 sin 2 2 sin sin 2 sin 2 0-5 sin + 2 = 0, (2sin0-l)(sin0-2)=O. .*. sin0=J, and sin 0=2. The acute angle whose sine is \ is 30°; hence 0=30° is one solution. The sine cannot exceed unity; hence sin 0=2 does not afford any solution. 7. Given cos A = j, sin B= J, tan C=2, cot D = %, sec E=S, cosec ^=2.5; find the other trigonometric ratios of A, B, C, D, E, F, by the algebraic method. Verify the results by the method used in Art. 4. 8. Find by the algebraic method the ratios required in Exs. 1, 5-7, 10, 11, Art. 4. 9. Express all the trigonometric ratios of an angle A in terms of: (a) cos A; (6) tan A) (c) cot A; (d) sec A; (e) cosec A. Arrange the results and those of Ex. 2 neatly in tabular form. Prove the following identities: 10. (sec 2 A — 1) cot 2 A = 1 ; cos A tan A = sin A ; (1 — sin 2 A) sec 2 A = l. 11. sin 2 sec 2 = sec 2 0-1; tan 2 0- cot 2 0= sec 2 0- cosec 2 0, sin A cos A )sec A ; l + sin0 sec 2 A cosec 2 A ' cosec A sec A. (tan 0+ sec 0) 2 = 1 -sin 0' TRIGONOMETRIC FUNCTIONS OF ACUTE ANGLES 17 13. sec 2 A + cosec 2 A = tan 2 A + cot 2 A + 2 ; 1 + tan 2 A sin 2 A cosec A 1 + cot 2 A cos 2 A ' cot A + tan A cos A. cos A sin A 14. - — -r+: — r=sin A+cos A; 1 —tan A 1 —cot A 1 ■= sec A + tan A ; sec 4 A —sec 2 A =tan 4 A +tan 2 A. sec A —tan A Solve the following equations: 15. 2 sin 0=2 -cos 6. 16. tan + cot 0=2. 17. tan + 3 cot = 4. 18. 6 sec 2 0-13 sec 0+5 = 0. 19. 8 sin 2 0-10 sin 0+3 = 0. 20. sin + 2 cos 0=2.2. CHAPTER II SOLUTION OF RIGHT-ANGLED TRIANGLES Applications 8. Solution of a triangle. There are two methods for finding the unknown parts of triangle (sides and angles) when a sufficient number of parts are given, viz. : (a) The graphical method; (6) The method of computation. The graphical method consists in drawing a triangle which has angles equal to the given angles, and sides proportional to the given sides, and then measuring the remaining parts directly from the drawing. The method of computation consists in writing the formulas by which the unknown parts can be found and doing the arithmetical work. The latter method is the more exact. The graphical method affords a rough check on the results obtained by computation and leads to the detection of large errors. General suggestions for solving problems. (1) Make an off-hand estimate as to what the magnitude required may be, and write this estimate down; (2) Solve the problem by the graphical method; (3) Solve the problem by the method of computation; (4) Check the accuracy of the results arithmetically, using formulas not used in process (3). 18 SOLUTION OF RIGHT-ANGLED TRIANGLES 19 9. Cases in the solution of right-angled triangles. All the possible sets of two elements that can be made from the three sides and the two acute angles of a right-angled triangle are the following: (1) The two sides about the right angle. (2) The hypotenuse and one of the sides about the right angle. (3) The hypotenuse and an acute angle. (4) One of the sides about the right angle, and an acute angle. (5) The two acute angles. (An unlimited number of triangles can have two given acute angles.) Relations (2), Art. 3, for the triangle AMP (Fig. 3) show that if any two sides of a right-angled triangle be given, or any side and an acute angle be given, the remaining parts of the triangle can be found. In solving a triangle, the general method of procedure, after making an off-hand estimate and finding an approxi- mate solution by the graphical method, is as follows: First : Write all the relations (or formulas) which are to be used in solving the problem. Second : Write the check formulas. Third: In making the computations arrange the work as neatly as possible. This last is important, because, by attention to this rule, the work is presented clearly, and mistakes are less likely to occur. The computations may be made either with or with- out the help of logarithms. The calculations can generally be made more easily and quickly by using logarithms. 20 ELEMENTS OF PLANE TRIGONOMETRY EXAMPLES 1. In the triangle ABC, right-angled at C, a=42 ft., 6=56 ft. Find the hypotenuse and the acute angles. I. Computation without logarithms. [Four-place tables.] a 42 tanA = r = — = .7500. .*. A = 36° 52'.2. o 56 B=90-A. - .'. £=53°7'.8. c=Va 2 + 6 2 = V1764 + 3136. .'. c=70 ft. Check: a=c cos B = 70 X cos 53° 7'.8=70X.6000 = 42 ft. II. Computation with logarithms. Given: a=42 ft. To find: * A = 6=56 ft, B= a c= Formulas: tan A = ^. (1) 5=90° -A. (2) ^ecfcs; tan £=-. , a a 2 = c 2 -b 2 C_ sinA' «(c+6)(c-fc). Logarithmic formulas: log tan A = log a —log 6. log c = log a —log sin A . log a = 1 .62325 log a = 1 .62325 log 6=1.74819 log sin .4 = 9.77815-10 .*. log tan A = 9.87506 -10 .\ log c= 1.84510 .*. A = 36° 52' 12" .'. c=70 .'. 5=53° T 48" * This is to be filled after the values of the unknown quantities have been found. It is advisable to indicate the given parts and the unknown parts clearly. SOLUTION OF RIGHT-ANGLED TRIANGLES 21 The work can be more compactly arranged, as follows: Checks: log a= 1.62325 log tan B= 10.12494-10 log 6=1.74819 /. B = 53° 7' 48" ,\ log tan A = 9.87506 -10 c + 6=126 .'. A = 36°52' 12" c-6= 14 /. 5 = 53° T 48" log (c + 6) = 2.10037 log sin A = 9.77815 -10 log (c -6) = 1.14613 .*. log c= 1.84510 . , 2 0/fncn & _ 7n .. loga 2 = 3.24650 *'' ° ;. log a =1.62325 Note. In every example it is advisable to make a complete skeleton scheme of the solution, before using the tables and proceeding with the actual computation. In this exercise, for instance, such a skeleton scheme can be seen on erasing all the numerical quantities in the equations that follow the logarithmic formulas. 2. In a triangle ABC right angled at C, c=60 ft., 6=50 ft.; find side a and the acute angles. I. Computation without logarithms. 6 50 * o cos A = - =—=.8333. .'. A=33°33'.75. c 60 B=90°-A. .'. 5=56°26'.25. A &-»**• G a=c sin A = 60X.5528 =33.17 ft. FlG " 17 ' Check: a=6 tan A = 50 X. 6635= 33.17. II. Computation with logarithms. Given: c=60 ft. . To find: A = 6=50 ft. B= 6 <*= c' Checks: a 2 = c 2 — 6 2 =(c+6)(c— 6). B=90°-A. a=6tanA. a = c sin A. 22 ELEMENTS OF PLANE TRIGONOMETRY Logarithmic formulas : log cos A = log b —log c. (If necessary.) log a = log c + log sin A . log 6= 1.69897 log c= 1.77815 (1) (2) log cos A = 9.92082 -10 (3) = (D-(2) /. A = 33° 33' 27" .-. £=56 p 26' 33" log sin A = 9.74255 -10 (4) .'. log a= 1.52070 (5) = (2) + (4) .*. a = 33.16 log tan A = 9.82173 - .*. log a= 1.52070 = (l) + (6) c+6=110 c-b = 10 log (c+6) = 2.04139 log(c-6) = l .-. log a 2 = 3.04139 .'. log a =1.52070 •10 (6) (7) Note. There is a slight difference between the results obtained by the two methods. This is due to the fact that the calculations have been made with a four-place table in one case, and with a five-place table in the other. A four- place table will give an angle correctly to within one minute; a five-place table will give it correctly to within six seconds, and sometimes to within a second. 3. In a triangle right angled at C, the hypotenuse is 250 ft., and angle A is 67° 30 r . Solve the triangle. I. Computation without logarithms. B = 90° -A = 90° -67° 30 r = 22° 30'. a = c sin A = 250 Xsin 67° 30 , = 250X. 9239=230.98. 6 = ccosA = 250Xcos67°30 , = 250X.3827= 95.68. Checks: a 2 =c 2 —b 2 , or a=6tanA. II. Computation with logarithms. Given c=250 ft. To find: B= A = 67° 30'. a= b= Formulas: B = 90° -A. Checks: a 2 =c 2 -b 2 a = c sin A. =(c+6)(c— 6). 6 = ccosA. SOLUTION OF RIGHT-ANGLED TRIANGLES 23 Logarithmic formulas: .*. B=22° 30' log c= 2.39794 log sin A = 9.96562 -10 log cos A = 9.58284 -10 .-. log a = 2.36356 .-. log 6= 1.98078 .'. a=230.97 .'. 6= 95.67 log a=log c+log sin A. log 6= log c+log cos A. c + 6 = 345.67 c -6= 154.33 log (c + 6) = 2.53866 log (c -6) = 2.18845 .*. loga 2 =4.72711 .'. log a =2.36356 4. In a triangle ABC right angled at C, 6=300 ft. and A = 37° 20'. Soive the triangle. I. Computation without logarithms. B= 90° -A = 90° -37° 30' = 52° 40'. 6 300 c= =377.3. cos A .7951 a=b tan A = 300 X. 7627 = 228.8. Checks : a 2 = c 2 —b 2 , a = c sin A . II. Computation with logarithms. Given Formulas: A = 37° 20'. 6=300 ft. B= 90° -A. 6 cos A a=6 tan A. .*. £=52° 40' log 6=2.47712 log cos A = 9.90043 -10 log tan A = 9.88236 -10 .'. logc=2.57669 .'. loga = 2.35948 .'. c=377.3 .-. a=228.8 To find: B= c= a~ Checks : a 2 =c 2 —b 2 = (c+6)(c-6) c+6 = 677.3 c-6= 77.3 log (c+6) = 2.83078 log (c -6) = 1.88818 .*. loga 2 = 4.71896 .'. log a =2.35948 24 ELEMENTS OF PLANE TRIGONOMETRY N.B. Check all results in the following examples. The given elements belong to a triangle ABC which is right angled at C. From the given elements solve the following triangles: 5. c=18.7, a= 16.98. 6. a= 194.5, 6 = 233.5. 7. c=2934, ,4 = 31° 14' 12". 8. a=36.5, J3=68°52'. 9. a = 58.5, 6=100.5. 10. c=45.96, a= 1.095. 11. c=324, A = 48°17'. 12. 6 = 250, ^ = 51° 19'. 13. c=1716, ^ = 37° 20' 30". 14. a = 2314, 6=1768. 15. 6 = 3741, A = 27°45'20". 16. c = 50.13, a = 24.62. Solve Exs. 17-24 by two methods, viz. : (1) with logarithms; (2) without logarithms. 17. a =40, 5=62° 40'. 18. c=9, a =5. 19. a =4.5, 6=7.5. 20. c=15, ,4 = 39° 40' 21. c=12, £=71° 20'. 22. c=12, a=8. 23. 6=15, 5=42° 30'. 24. a=8, 6=12. 10. Projection of a straight line upon another straight line. Let AB, of length I, be inclined at an angle a to LR. If per- pendiculars AM, BN, are drawn to LR, MN is called the (orthog- i/> onal) projection of AB on LR. is N Through A draw AD parallel FlG 20 to LR. Then Projection = MN = AD = AB cos DAB = I cos a. That is, the projection of one straight line upon another straight line is equal to the product of the length of the first line and the cosine of the angle of inclination of the two lines. SOLUTION OF RIGHT-ANGLED TRIANGL V ES 25 EXAMPLES In working these examples use logarithms or not, as appears most convenient. Check the results. 1. A ladder 28 ft. long is leaning against the side of a house, and makes an angle 27° with the wall. Find its projections upon the wall and upon the ground. 2. What is the projection of a line 87 in. long upon a line inclined to it at an angle 47° 30'? 3. What are the projections: (a) of a line 10 in. long upon a line inclined 22° 30' to it? (i) of a line 27 ft. 6 in. long upon a line inclined 37° to it? (c) of a line 43 ft. 7 in. long upon a line inclined 67° 20' to it? (d) of a line 34 ft. 4 in. long upon a line inclined 55° 47' to it? 11. Measurement of heights and distances. When an object is above the observer's eye, the angle between the line from the eye to the object, and the horizontal line through Fig. 21. the eye and in the same vertical plane as the first line, is called the angle of elevation of the object, or simply the elevation of the object. When the object is below the observer's eye, this angle is called the angle of depression of the object, or simply the depression of the object. 26 ELEMENTS OF PLANE TRIGONOMETRY EXAMPLES 1. At a point 150 ft. from, and on a level with, the base of a tower, the angle of elevation of the top of the tower is observed to be 60°. Find the height of the tower. Let AB be the tower, and P the point of observation. By the observations, AP=150ft., APB=W°. Fig. 22. AB=AP tan 60°= 150X^3 = 150X1.7321 = 259.82 ft. 2. In order to find the height of a hill, a line was measured equal to 100 ft., in the same level with the base of the hill, and in the same vertical plane with its top. At the ends of this line the angles of elevation of the top of the hill were 30° and 45°. Find the height of the hill. Let P be the top of the hill, and AB the base line. The vertical line through P will meet, AB produced in C. AB= 100 ft. CAP=S0°, CPP=45°; the height CP is required. Let BC=x, and CP=y. [ In triangle CAP, CP Fig. 23. in CBP % 2^=tan30°; CP BC = tan 45°. Hence, V z+100 = tan 30° =.57735, and -=tan45°=l. x a) (2) From (2), x=y. Substitution in (1) gives --(»/+• 100) X. 57735. SOLUTION OF RIGHT-ANGLED TRIANGLES 27 .\ 2/(1 -.57735) =57.735. 3. A flagstaff 30 ft. high stands on the top of a cliff, and from a point on a level with the base of the cliff the angles of elevation of the top and bottom of the flagstaff are observed to be 40° 20' and 38° 20', respectively. Find the height of the cliff. Let BP be the flagstaff on the top of the cliff BL, and let C be the place of observation. PP=30 ft., LCB=S8° 20', LCP=40° 20'. Let CL = x, LB=y. P 7 / J30J.FY. In LCB, LB y^=tan38°20 r ; /X0^ i.e., -=.7907. In LCP, LP -^=tan40°20'; J4. QQ +,Jj Fig. 24. i.e., y+30 -.8491. Hence, on division, y 7907 y + 30 8491' On solving for y, LB=y=406.18ft. 4. At a point 180 ft. from a tower, and on a level with its base, the elevation of the top of the tower is found to be 65° 40/5. What is the height of the tower? 5. From the top of a tower 120 ft. high the angle of depression of an object on a level with the base of the tower is 27° 43'. What is the distance of the object from the top and bottom of the tower? 28 ELEMENTS OF PLANE TRIGONOMETRY 6. From the foot of a post the elevation of the top of a column is 45°, and from the top of the post, which is 27 ft. high, the elevation is 30°. Find the height and distance of the column. 7. From the top of a cliff 120 ft. high the angles of depres- sion of two boats, which are due south of the observer, are 20° 20' and 68° 40'. Find the distance between the boats. 8. From the top of a hill 450 ft. high, the angle of depres- sion Of the top of a tower, which is known to be 200 ft. high, is 63° 20'. What is the distance from the foot of the tower to the top of the hill? 9. From the top of a hill the angles of depression of two consecutive mile-stones, which are in a direction due east, are 21° 30' and 47° 40'. How high is the hill? 10. For an observer standing on the bank of a river, the angular elevation of the top of a tree on the opposite bank is 60°; when he retires 100 ft. from the edge of the river the angle of elevation is 30°. Find the height of the tree and the breadth of the river. 11. Find the distance in space travelled in an hour, in con- sequence of the earth's rotation, by an object in latitude 44° 20'. [Take earth's diameter equal to 8000 mi.] 12. At a point straight in front of one corner of a house, its height subtends an angle 34° 45', and its length subtends an angle 72° 30'; the height of the house is 48 ft. Find its length. 12. Solution of isosceles triangles. An isosceles triangle can often be solved on dividing it into two equal right- angled triangles. SOLUTION OF RIGHT-ANGLED TRIANGLES 29 EXAMPLES 1. The base of an isosceles triangle is 24 in. long, and the vertical angle is 48°; find the other angles n and sides, the perpendicular from the vertex and the area. Only the steps in the solution will be indicated. Let ABC be an isosceles triangle having base Ai?=24 in., angle (7=48°. / Draw CD at right angles to base; then CD I bisects the angle ACB and base A B. Hence, .Ajf-^. in the right-angled triangle ADC, AD= Fig 25 iAB=12, ACD=iACB=24°. Hence, angle A, sides AC, DC, and the area, can be found. 2. In an isosceles triangle each of the equal sides is 363 ft., and each of the equal angles is 75°. Find the base, per- pendicular on base, and the area. 3. In an isosceles triangle each of the equal sides is 241 ft., and their included angle is 96°. Find the base, angles at the base, height, and area. 4. In an isosceles triangle the base is 65 ft., and each of the other sides is 90 ft. Find the angles, height, and area. 5. In an isosceles triangle the base is 40 ft., height is 30 it. Find sides, angles, area. 6. In an isosceles triangle the height is 60 ft., one of equal sides is 80 ft. Find base, angles, area. 7. In an isosceles triangle the height is 40 ft., each of equal angles is 63°. Find sides and area. 8. In an' isosceles triangle the height is 63 ft., vertical angle is 75°. Find sides and area. 13. Related regular polygons and circles. Among the 30 ELEMENTS OF PLANE TRIGONOMETRY problems on regular polygons which can be solved with the help of right-angled triangles are the following: (a) Given the length of the side of a regular polygon of a given number of sides, to find its area; also, to find the radii of the inscribed and circumscribing circles of the polygon; (6) To find the length of the sides of regular polygons of a given number of sides which are inscribed in, and circum- scribed about, a circle of given radius. For example, in Fig. 26, let AB be a side, equal to 2a, of a regular polygon of n sides, and let C be the centre of the inscribed circle. Draw CA, CB, and draw CD at right angles to AB- Then D is the middle point of AB. 360° 180° By geometry, angle ACD = %ACB = h Also, angle DAC= 90°- ACT). n n Thus, in the triangle ADC, the side AD and the angles are known; therefore CD, the radius of the circle inscribed in the polygon, can be found. On making similar constructions, the solution of the other problems referred to above will be Fig. apparent. The perpendicular from the centre of the circle to a side of the inscribed polygon is called the apothem of the polygon. SOLUTION OF RIGHT-ANGLED TRIANGLES 31 EXAMPLES 1. The side of a regular heptagon is 14 ft. : find the radii of the inscribed and circumscribing circles; also, find the difference between the areas of the heptagon and the inscribed circle, and the difference between the area of the heptagon and the area of the circumscribing circle. 2. The radius of a circle is 24 ft. Find the lengths of the sides and apothems of the inscribed regular triangle, quadri- lateral, pentagon, hexagon, heptagon, and octagon. Compare the area of the circle and the areas of these regular polygons; also compare the perimeters of the polygons and the circum- ference of the circle. 3. For the same circle as in Ex. 2, find the lengths of the sides of the circumscribing regular figures named in Ex. 2. Compare their areas and perimeters with the area and cir- cumference of the circle. 4. If a be the side of a regular polygon of n sides, show that R, the radius of the circumscribing circle, is equal to 180° iacosee ; and that r, the radius of the circle inscribed, n 180° is equal to \a cot -. 5. If r be the radius of a circle, show that the side of the • 180 ° , , regular inscribed polygon of n sides is 2r sin ; and that 180° the side of the regular circumscribing polygon is 2r tan f» 6. If a be the side of a regular polygon of n sides. R the radius of the circumscribing circle, and r the radius of the 180° circle inscribed, show that area of polygon =Jna 2 cot . . 360° Q 180° = hnR 2 sin = nr z tan . 2 n n 32 .ELEMENTS OF PLANE TRIGONOMETRY 14. Problems requiring a knowledge of the points of the Mariner's Compass. The circle in the Mariner's Compass is divided into 32 equal parts, each part being thus equal to 360° ■*■ 32, i.e., 11J°. The points of division are named as indicated on the figure. Fig. 27. It will be observed that the points are named with refer- ence to the points North, South, East, and West, which are called the cardinal points. Direction is indicated in a variety of ways. For instance, suppose C were the centre of the circle; then the point P in the figure is said to bear E.N.E. from C, or, from C the bearing of P is E.N.E. Similarly, C bears W.S.W. from P, or, the bearing of C from P is W.S.W. The point E.N.E. is 2 points North of East, and 6 points East of North. Accordingly, the phrases E. 22J° N., N. 67J° E., are sometimes used instead of E.N.E. s Fig. 28. If LA is 6 mi., SOLUTION OF RIGHT-ANGLED TRIANGLES 33 EXAMPLES 1. Two ships leave the same dock at 8 a.m. in directions S.W. by S., and S.E. by E. at rates of 9 and 9^ mi. an hour respectively. Find their distance apart, and the bearing of one from the other at 10 /& a.m. and at noon. 2. From a lighthouse L two ships *?(£"-'-- A and B are observed in a direction N.E. and N. 20° W. respectively. At the same time A bears S.E. from B. what is LB1 15. Examples in the measurement of land. In order to find the area of a piece of ground, a surveyor measures distances and angles sufficient to provide data for the com- putation. An account of his method of doing this belongs to works on surveying. This article merely gives some examples which can be solved without any knowledge of professional details. In solving these problems, the student should make the plotting or mapping an important feature of his work. The Gunter's chain is generally used in measuring land. It is 4 rods or 66 feet in length, and is divided into 100 links. An acre = 10 square chains = 4 roods = 160 square rods or poles. EXAMPLES 1. A surveyor starting from a point A runs S. 70° E. 20 chains, thence N. 10° W. 20 chains, thence N. 70° W. 10 chains, thence S. 20° W. 17.32 chains to the place of beginning. What is the area of the field which he has gone around? 34 ELEMENTS OF PLANE TRIGONOMETRY Fig. 29. Make a plot or map of the field, namely, ABCD. Here A B represents 20 chains, and the bearing of B from A is S. 70° E. BC represents 20 chains, and the bearing C from B is N. 10° W., and so on. Through the most west- erly point of the field draw a north- and-south line. This line is called the meridian. In the case of each line measured, find the distance that one end of the line is east or west from the other end. This easting or westing is called the departure of the line. Also find the distance that one end of the line is north or south of the other end. This northing or southing is called the latitude of the line. For example, in Fig. 29, the departures of AB, BC, CD, DA, are B X B, BL } CH, DD\, respectively; the latitudes of the boundary lines are AB X , BiCi, CiD h D X A, respectively. The following formulas are easily deduced: Departure of a line = length of line X sine of the bearing; Latitude of a line = length of line X cosine of the bearing. By means of the departures, the meridian distance of a point (i.e., its distance from the north-and-south line) can be found. Thus the meridian distance of C is C\C, and C\C= DtD + HC. Hence in Fig. 2.9, AB U B X B, B X C U dC, C X D U D\D can be computed. Now area ABCD = trapezoid D1DCC1 + trapezoid C\CBBi —triangle ADD\ —triangle ABB% m The areas in the second member can be computed; it will be found that area ABCD = 26 acres. Note. Sometimes the bearing and length of one of the lines enclosing the area is also required. These can be com- puted by means of the latitudes and departures of the given lines. SOLUTION OF RIGHT-ANGLED TRIANGLES 35 2. In Ex. 1, deduce the length and bearing of DA from the lengths and bearings of AB, BC, CD. 3. A surveyor starts from A and runs 4 chains S. 45° E. to B, thence 5 chains E. to C, thence 6 chains N. 40° E. to D. Find the distance and bearing of A from D; also, the area of the field ABCD. Verify the results by going around the field in the reverse direction, and calculating the length and bear- ing of BA from the lengths and directions of AD, DC, CB. 4. A surveyor starts from one corner of a pentagonal field, and runs N. 25° E. 433 ft., thence N. 76° 55' E. 191 ft., thence S. 6° 41' W. 539 ft., thence S. 25° W. 40 ft., thence N. 65° W. 320 ft. Find the area of the field. Deduce the length and direction of one of the sides from the lengths and directions of the other four. 5. From a station within a hexagonal field the distances of each of its corners were measured, and also their bearings; required its plan and area, the distances in chains and the bearings of the corners being as follows: 7.08 N.E., 9.57 N. J E., 7.83 N.W. by W., 8.25 S.W. by S., 4.06 S.S.E. 7° E., 5.89 E. by S. 3^° E. CHAPTER III ANGLES IN GENERAL AND THEIR TRIGONOMETRIC FUNCTIONS 16. General definition of an angle. The amount of rota- tion that a line OP makes in turning about a point from an initial position OX until it comes to rest in a terminal position OQ is said to be the angle described (or generated) by the line OP. Angles unlimited in magnitude. Since the revolving line may make any number of complete revolutions before com- Fig. 30. Fig. 31. Fig. 32. ing to rest, angles can be of any magnitude and can be unlimited in magnitude. When necessary the number of revolutions can be indi- cated as in Fig. 32, which represents the angle 60° and the 36 ANGLES AND THEIR TRIGONOMETRIC FUNCTIONS 37 angle 3.360° + 60°. The lines in this figure may represent any angle ft -360° + 60°, n denoting any whole number. Ex. Taking the initial line in the same position as OX in Fig. 30 draw the terminal lines of the angles 40°, 160°, 220°, 325°, 437°, 860°, 1020°, 180°, 360°, 720°. Positive and negative angles. When the turning line revolves in the counterclockwise direction (as in Fig. 30), the angle described is called positive and is given the plus sign; when the turning line revolves in the clockwise direc- tion (as in Fig. 31), the angle described is called negative and is given the minus sign. Ex. Taking the initial line in the same position as OX in Fig. 30 draw the terminal lines of the angles 140°, —200°, -430°, 335°, -850°, 820°. Coterminal angles. Angles having the same initial and terminal lines respectively are called coterminal angles. Ex. Show that, if the same initial line be taken, the angles 40°, -320°, 400°, 760°, -1040° are coterminal. Congruent angles. Angles differing by multiples of 360° are called congruent angles. Ex. Show that the angles in the preceding exercise are congruent. Complementary angles. Supplementary angles. Angles whose sum is 90° are called complementary angles; angles whose sum is 180° are called supplementary. In other words, the complement of angle A = 90°— A ; the supplement of angle ^4 = 180°— A Angles in the various quadrants. In Fig. 33, XX f and BN s Y >PX / V X 7 \« p. (,>* 38 ELEMENTS OF PLANE TRIGONOMETRY YY f are at right angles and OX is the initial line. In this figure XOY, YOX', X'OY', Y'OX, are called the first, second, third, and fourth quadrants, respectively. When the turning line ceases its x'- revolution at some position be- tween OX and OY, the angle described is said to be an angle in the first guadrant; when the final position of the turning line is between OY and OX' the angle described is said to be in the second quadrant) and so on for the third and fourth quadrants. EXAMPLES 1. Lay off the following angles: In the case of each angle name the least positive angle that has the same terminal line. Name the quadrants in which the angles are situated. In the case of each angle name the four smallest positive angles that have the same terminal line. (a) 137°, 785°, 321°, 930°, 840°, 1060°, 1720°, 543°, 3657°. (6) -240°, -337°, -967°, -830°, -750°, -1050°, -7283°. 2. What are the complements and supplements of 40°, 227°, -40°? complement of 40°= 90°- 40°= 50°; supplement of 40°= 180°— 40°= 140°. complement of 227°= 90° -227°= -137°; supplement of 227°= 180°-227°= - 47°. complement of - 40° = 90° - ( - 40°) - 130° ; supplement of -40°- 180°- (-40°) =220°. 3. What are the complements of -230°, 150°, -40°, 340°, 75°, 83°, 12°, -295°, -324° ; 200°, 240°, -110°, -167°? 4. What are the supplements of the angles in Ex. 3? ANGLES AND THEIR TRIGONOMETRIC FUNCTIONS 39 17. Measurement of angles. There are three systems of angular measure, the sexagesimal, the centesimal, and the circular. In the sexagesimal system the unit angle is one-ninetieth of a right angle, called a degree, and the subdivisions are minutes and seconds, as described in Art. 2. This measure is used in the solution of triangles and in elementary mathe- matics generally. In the centesimal system the unit angle is one-hundredth of a right angle, called a grade. The table of centesimal measure is 1 right angle = 100 grades. 1 grade = 100 minutes. 1 minute = 100 seconds. This system, which is used in France, has not come into general use. In the circular system the unit angle is the angle which is subtended at the centre of a circle by an arc equal in length to the radius. This angle is called a radian. The radian Fig. 34. Fig. 35. measure (often called the circular measure) of an angle is the number of radians it contains. Radian measure is the measure used in higher mathematics. It is also used in various problems and discussions in elementary mathematics. 40 ELEMENTS OF PLANE TRIGONOMETRY 18. Value of a radian. Relation between radian measure and degree measure. In Fig. 35 arc AB = radius, and thus angle AOB = a radian. By geometry, angle AOB arc A B complete angle about length of circle' a radian radius 1 Le *' 360° = 2ttX radius = 2k' 180° /. a radian = =57° 17' 44".8. (1) IT Thus a radian has a constant value, and accordingly can be used as a unit. From (1) ir radians = 180°. (2) Also, from (1) or (2), l° = ^r radians. By means of relation (2) an angle expressed in the one measure can be expressed in the other. Notation. An angle 2 radians is expressed 2 r or 2 C (r from radian, c from circular). The symbol r, or c, is often omitted. For instance, angle n denotes the angle n radians i.e., 180°; angle «r denotes the angle ~ radians, i.e., 90°; angle 6 denotes' an angle containing 6 radians. Radian measure as ratio of subtended arc to radius. Length of arc. Since the arc of a circle equal to the radius subtends at the centre an angle equal to a radian, the number of times an arc contains the radius is the radian measure of its subtended angle; i.e., _ ■„ . . length of subtended arc Number of radians in angle = - — : — — 3 — -r. . (3) length of radius v ANGLES AND THEIR TRIGONOMETRIC FUNCTIONS 41 If a = length of arc, r = length of radius, and 0= radian measure of the angle, (3) may be written e = a v a=rQ, (4) (5) From this In words, arc = radius X radian measure of angle. Formula (4) derived otherwise. In Fig. 36 AOB is a radian. Let r = length of radius, a = length of arc AP, 6 = radian measure of angle AOP. By geometry, angle AOP _ arc AP angle A OB ~ arc AB } angle AOP a 1 radian ~ r' number of radians in AOP, 0=2. i.e., Fig. 36. o-7r, — 107T, in o EXAMPLES 1. How many degrees are there in 2.5 radians? 2. How many radians are there in 231°? 3. Express J*", 4 r , J r in degrees. t-i x-l * X 7C X K 2 ' 4. Express the angles -, -, j, -, ^rr, 3tt, degrees. 5. Express in radians 83° 20', 142° 30'. 6. Express in radians (in terms of n) 45°, 210°, 300°, 120°, 225°. 7. What is the radian measure of the angle which at the centre of a circle of radius 1J yds. subtends an arc of 8 in.? Also express the angle in degrees. 42 ELEMENTS OF PLANE TRIGONOMETRY 8. Find the radius of the circle in which an arc 12 in. subtends an angle 2 radians at the centre. 9. The radius of a circle is 10 in. How long is the arc sub- tended by an angle of 4 radians at the centre? 10. How long does it take the minute hand of a clock to turn through — If radians? 18a. Motion of a particle in a circle. Let r ft.* = radius of the circle on which the particle is moving; s ft. = length of arc traversed ; and 6 radians = angle swept over by the radius joining the moving point to the centre of the circle. Then, by Art. 18, Eq. *(5), s = rQ. It should be noted that this equation is not true unless the radius and the arc are measured in the same unit of length and the angle is measured in radians. Again, suppose the particle is moving at a uniform rate on a circle of radius r ft.* Let v ft. per sec. = the linear velocity of the particle in its path, and (o radians per sec. = its angular velocity. f Then v=ro>; and thus « = ^» * Or whatever may be the unit of length. f The angular velocity of the particle is the number of radians swept over in a unit of time — in this instance a second — by the radius joining the moving point to the centre of the circle. ANGLES AND THEIR TRIGONOMETRIC FUNCTIONS 43 Suppose the angular velocity =N revolutions per minute. Since 1 revolution = 2n radians, N revolutions per minute = licN radians per min. irJV ,. =-sr radians per sec. EXAMPLES 1. A wheel of a carriage which is travelling at the rate of 8 miles per hour is 3 ft. in diameter. Find the angular velocity of any point of the wheel about the axle, in radians per second. 2. Compare the angular velocities of the hour, minute, and second hands of a watch. 3. Express in terms of radians per minute an angular velocity of 25° per second. 4. A wheel makes 300 revolutions per hour. Express its angular velocity (a) in radians per sec. ; (6) in degrees per min. 5. An automobile having a 28-in. wheel travels at a speed of 15 miles an hour. What is the angular speed of the wheel about the axle (a) in revolutions per minute? (6) in radians per second? 6. A belt travels over a 30-in. pulley which is carried on a shaft making 300 revolutions per minute. Find the linear speed of the belt in feet per second. 19. Convention for signs of lines in a plane. In Fig. 37 X'OX is a horizontal line to which YOY' is drawn at right 44 ELEMENTS OF PLANE TRIGONOMETRY Pi angles. Lines measured horizontally towards the right are taken as positive, and accordingly lines measured horizontally towards the left are negative. Thus OM 1} OM±, M 2 are positive, OM 2 , OM 3 , M x O are negative. Lines measured vertically upward, as M X P\, M2P2, are taken as positive, and lines measured verti- cally downward, as M 3 P S , M±P 4 , are negative. M, X M* P 3 O M 1 MtX Y' Fig. 37. 20. Trigonometric functions defined for angles in general. In Figs. 38-41 the horizontal line OX is taken as the initial line, and angles in the first, second, third, and fourth quad- Fig. 38. Fig. 39. ft the cosine of angle A =cos -^ = ~ OP the secant of angle A = sec A = q^ > the cosecant of angle A = esc ^4. = jjbfp • Two other trigonometric functions are occasionally used, viz.: the versed sine of angle A = vers A = 1 — cos A, the cover sed sine of angle A = covers A = l — sin A. Inspection of Figs. 38-41, in connection with the defini- tions, shows that the trigonometric functions of angles in the various quadrants have the signs stated in the following table : Quadrant. sin A cos A tan A cot A sec A cosec A I. + + + + + + II. + - - - - + III. - - +. + - - IV. - + - - j_ - 46 ELEMENTS OF PLANE TRIGONOMETRY EXAMPLES 1. Tell the signs of the following functions: (a) sin 100°; (b) cos 220°; (c) tan 230°; (d) sec 340°; (e) cot (-130°); (/) esc 560°. 2. Describe the angle 480°, and find the values of its sine, cosine, and tangent. 3. Describe the angle 945° ; and find the values of its sine, cosine, and tangent. 4. In which quadrants may an angle lie, if : (a) its sine is positive; (b) its cosine is negative; (c) its tangnet is positive; (d) its tangent is negative. 5. In what quadrant must an angle lie, if: (a) its sine and cosine are negative; (b) its sine is positive and tangent negative; (c) its sine is negative and tangent positive; (d) its cosine is positive and tangent negative. 6. Construct angle A, when sin A = f . Find the remaining functions of A. The definition of the sine, Art. 20, shows that for this angle M?=+3, OP=4. The construction then is as indi- cated in Fig. 42. There are thus two sets of angles whose sines are J, viz., the angles whose terminal line is OP, and the angles whose terminal line is OPi. Each set of angles is unlimited in number, and the angles in it differ from one another by mul- tiples of 360°. Fig. 42. ANGLES AND THEIR TRIGONOMETRIC FUNCTIONS 47 For an angle having OP for terminal line, cos A = Vf tan A sec A = - vf csc A cot A = VI vT ~~ 3 For an angle having OPi for terminal line, Vf . 3 cos A= — tan A = VT cot A = — Vf 3 ' X l M 4 A 4 secA= =, cscA = — . Vf 3 3 r 3 —3 3 1 7. Construct angle A, given tan A=— —. Note — j = "4 _== 34- The definition of the tangent, Art. 20, shows that for this angle, MP=-3 and 0M= *4, or MP=3 and 0Af=-4. The construction then is as indicated in Fig. 43. There are thus two sets of angles whose tangents are — }, viz., the angles whose terminal line is OP and the angles whose terminal line is OP x . The angles in each set are congruent. 8. Calculate the remaining functions of the angles in Ex. 7. 9. Construct the angles which have the following functions, and calculate the remaining functions: (a) cos x= - s-; (&) sin#=— -p-; (c) cos#=-r-; Fig. 43. 48 ELEMENTS OF PLANE TRIGONOMETRY (d) sec x=p (.) tan,=i. [Note. J-±j-=£] 10. Construct the angles and find their remaining functions in the following cases: 3 2 (a) sin x = =■ and cos x negative ; (jb) cos x = ^ and sin a; negative ; 3 (c) tan x = ^- and sin # negative ; ((f) sec a; = 2 and tan x negative. 21. Line definitions of the trigonometric functions. Geo- metrical representation of the functions. In Fig. 44 XOP is an angle of x radians. A circle, whose radius is a unit in length, is described about 0. This circle is called a unit circle. Angle XOF = 90°. From A and B tan- gents AT, BS are drawn to meet the .terminal line OP. From P, the intersection of the terminal line with the circle, PM is drawn at right angles to OX. Then 7 B cot X SA t N sseex I S 8 C cosX M A.X Angle in First Quadrant. Fig. 44. sm a> = cos cc = tan »i MP OP(-l) Oflf OP(-l) 'Oi(-l) = MFy = OM y =AT, cot ^- M p- 0jB( = 1) -^» * Since the triangles OMP, OBS, are similar, f Thus also, vers x = l — cos £ = OA — OM — M A ; covers x = 1 — sin x = OB — MP = NB. ANGLES AND THEIR TRIGONOMETRIC FUNCTIONS 49 OT sec * = oXC=T) =ot ' OP OS* csc —Ep-OTj^-oi*. . , arc AP Also ^"" n A ( — n =arc AP * Since the radius is of unit length the lengths of these lines are expressed in the same numbers as the ratios which are the trigonometric functions. Accordingly the lines can represent the functions. Also, on the same scale, the arc represents the radian measure of the angle. (Compare definition (3), Art. 18.) The line definitions of the trigonometric functions of an angle may be thus expressed, the angle being at the centre of a unit circle : The sine is the length of the perpendicular drawn from the extremity of the terminal radius to the initial radius. The cosine is the length of the line from the centre of the circle to the foot of this perpendicular. The tangent is the length of the tangent drawn at the extrem- ity of the initial radius from this extremity to where the tangent meets the terminal radius produced. The secant is the length from the centre along the terminal line to where it meets the tangent drawn at the extremity . of the initial radius. The cotangent is the length of the tangent drawn at the extremity of the radius which makes an angle + 90° with the initial radius, from this extremity to where this tangent meets the terminal radius produced. The cosecant is the length from the centre, along the terminal line to where it meets the tangent drawn at the extremity of the radius which makes an angle +90° with the initial radius. ELEMENTS OF PLANE TRIGONOMETRY Figs. 45, 46, 47 show the lines which represent the trigo- nometric functions of angles in the second, third, and fourth quadrants respectively. The lines are named as in Fig. 44. The line representing the tangent is always drawn from A, \S cotx B Y ~\cosxM A 3 1 M /t \-2 Angle in Second Quadrant. Fig. 45. B cot X S/J 'T mf R- | 1 M fi C08X)>S ]ax \l -^^_ Angle in Third Quadrant. Fig .46. and the line representing the cotangent is always drawn from B, to the terminal line of the angle. The signs of OM, MP, AT, BS follow the convention in Art. 19, and the * Angle in Fourth Quadrant. Fig. 47. Fig. 48. direction from towards P is taken as positive. Inspection of Figs. 44-47 shows that the functions of angles in the, various quadrants have the signs set down in the table ♦in Art. 20 * * The line definitions of the trigonometric functions were employed before the ratio definitions were suggested. ANGLES AND THEIR TRIGONOMETRIC FUNCTIONS 51 22. Changes in the values of the functions when the angle varies. Limiting values of the functions. In Fig. 48 AOP is a variable angle x, and a unit circle is described about 0. MP The sine. Now sin x = 7jpTZT\ = M?- When angle x approaches 0° as a limit, MP passes through the values M2P2, M1P1 to zero as a limit. Thus sin 0° = 0. When OP revolves positively from OA toward OB as a limit, x increases from 0° to 90° as a limit, and MP increases from zero through MxP 1} M 2 P 2 , M^P^ to OB( = l) as a limit. Thus sin 90° = 1. When OP revolves from OB to OA lf x increases from 90° to 180°, and MP decreases from 0£( = 1) through M&P B , MqP 6 to zero. Thus sin 180° = 0. When OP revolves from OAi to OBi, x increases from 180° to 270° and MP changes from zero through M 7 P 7 , M 8 P 8 to OJBi(- -1), and thus sin 270°= -1. When OP revolves from OBi to OA, x varies from 270° to 360°, and MP changes from 0#i(=-l) through M 9 P 9 , ^10^10 to zero. Thus sin 360° = 0. The cosine. Now 0M rw cos x — A)p/_-i\ = OM. 52 ELEMENTS OF PLANE TRIGONOMETRY When angle x approaches 0° as a limit, OM passes through the values OM 2 , OM x to OA ( = 1) as a limit. Thus cos 0° = 1. When OP revolves positively from OA to OB, x increases from 0° to 90° and OM decreases from OA( = l) through OM h OM 2 , OM s , OM 4 , to zero. Thus cos 90° = 0. When OP revolves from OB to OA u x increases from 90° to 180° and OM changes from zero through OM 5 , OM 6 to OA^-1). Thus cos 180°- -1. When OP revolves from OA i to OB 1} x increases from 180° to 270° and OM changes from OAi( = -l) through OM 7 , OM 8 to zero. Thus cos 270° = 0. When OP revolves from OB x to OA } x increases from 270° to 360° and OM changes from zero through OM 9} OM XQj toOA = l. Thus cos 360° = 1. The tangent. First method, using the ratio definitions. Now MP tan * = OM' When angle x approaches 0° as a limit, MP approaches zero and OM approaches A ( = 1 ) . Thus tan0° = 0. ANGLES AND THEIR TRIGONOMETRIC FUNCTIONS 53 When x increases from 0° to 90°, MP is approaching MP OB=(l) and OM is approaching zero. Thus jyjur, which is positive during this change, is constantly increasing and becomes unlimited in value. Thus tan 90° = oo. When x is increasing from 90° to 180°, MP is changing from OJ5( = l) to zero and OM is changing from zero to MP OA i ( = — 1 ) . Thus Y^Tj, which is negative during this change, changes from — oo to 0. Thus tan 180° = 0. N.B. When x is increasing from 0° to 180°, tanz= +oo when x reaches 90°, and tan x = — oo when x leaves 90°. Thus tan x changes sign when x is passing through the value 90°. If x is decreasing from 180° to 0°, tan x = — oo when x reaches 90°, and tanz=+oo when x leaves 90°. Thus the sign of tan x when z = 90°, depends on the way in which x has approached the value 90°. When x is increasing from 180° to 270°, MP is changing from zero to OBi ind OM is changing from OAi to zero. MP Thus :~ T7, which is positive during this change, changes from to an unlimited value. So tan 270° = oo . When x is increasing from 270° to 360°, MP is changing from OB i to zero, and OM is changing from zero to OA. 54 ELEMENTS OF PLANE TRIGONOMETRY MP Thus :~ T7, which is negative during this change, changes from — oo to 0. So tan 360° =0. Tan 270°, like tan 90°, has an unlimited value whose sign is positive or negative, according to the way in which the angle has approached the value 270°. f The tangent. Second method, using the line definitions. In Fig. 49, the variable angle AOP = x. A unit circle is drawn and the tangent drawn at A. Now tanz AT Fig. 49. OA(-l) = AT. When x approaches 0°, AT passes through AT 2y AT\ to zero; and thus tan 0°=0. When x changes from 0° to 90°, AT passes through ATi, AT 2 , ATzj AT 4, A T 5 to an unlimited value; and thus tan 90° = oo. The student can trace by means of Figs. 45, 46, 47 the further changes in tan x when x increases from 90° to 360°. The tracing of the changes in cot x, sec x, esc x, as x changes from 0° to 360°, is left as an exercise to the student. ANGLES AND THEIR TRIGONOMETRIC FUNCTIONS 55 The changes for the above six functions when the angle is increasing from 0° to 360° are indicated in the following table: In the second quadrant the Y In the first quadrant the sine decreases from lto sine increases from Oto 1 cosine decreases from Oto- -1 cosine decreases from lto tangent increases from- -oo to tangent increases from Oto 00 cotangent decreases from Oto- - oo cotangent decreases from oo to secant increases from- -ooto- - 1 secant increases from lto 00 cosecant increases from lto 00 • cosecant decreases from oo to 1 y V A\ A In the third quadrant the In the fourth quadrant the sine decreases from Oto- - 1 sine increases from- - lto cosine increases from- - lto cosine increases from Oto 1 tangent increases from Oto 00 tangent increases from- -oo to cotangent decreases from qo to cotangent decreases from Oto - -00 secant decreases from - - lto- ■ oo secant decreases from oo to 1 cosecant increases from- ■ ooto- - 1 cosecant decreases from - - lto - -00 Yx 23. Periodicity of the trigonometric functions. When the angles 360° to 720° are described the sine goes through all its changes in the same order as when the angles 0° to 360° are described. Also the sine repeats the same value every time the angle changes by 360°, i.e., 2£ That is, n denoting a whole number sin (n • 360° + x) = sin x, i.e. sin (2mr +x) = sin x. Accordingly the sine is said Fig. 50. to be a periodic function, and its period is 2x. 56 ELEMENTS OF PLANE TRIGONOMETRY Similarly, the cosine, secant, cosecant have the period 2n. When the angles 180° to 360° are described the tangent goes through all its changes in the same order as when the angles 0° to 180° are described. Also the tangent repeats the same values every time the angle changes by 180°, i.e., it. That is, n denoting any whole number, i.e., tan (n • 180° + x) = tan x, tan {nn + x) = tan x. Accordingly the tangent is a periodic function, and is period is it. Similarly, the cotangent has the period n. 24. Graphs of the functions. Graph of sin x. From the equation, y = smx form corresponding pairs of x and y. Convenient values to TZ TZ TZ tz 2n take for x are 30°, 45°, 60°, 90°, 120°, . Thus: . ,i.e. '6'4 ; 3' 2' 3 X it "6 Tt T It it ~2 27t 3 3tt 4 5tt 6 it 7 6* 3 2* 7 I* 2tt 9ar 4 Tt y .5 .71 .87 1 .87 .71 .5 -.5 -1 -.71 .71 -.71 Choose any convenient length to represent the angle 2n (360°) ; * mark the points corresponding to angles ^, j t TZ ^, . . . ; at these points erect ordinates representing the o * E.g., one such length that may be taken is the length of a unit circle. ANGLES AND THEIR TRIGONOMETRIC FUNCTIONS 57 corresponding values of y; draw a smooth curve through the ends of these ordinates. This curve, Fig. 51, is called "The curve of sines 11 or "the sine curve. 11 A very convenient way to draw the ordinates corre- sponding to values of x is to take the sines from the unit circle (Art. 21), as indicated in Fig. 52. Graph of cos x. From the equation 2/ = cos x form corresponding values of x and y, and proceed as in the case of the sine. The ordinates can also be obtained readily by using a unit circle. The cosine curve is in Fig. 53. Y Fig. 51. — Graph of sin x. rr Fig. 53. — Graph of cos x. Graph of tan oc. From the equation y = tan x 58 ELEMENTS OF PLANE TRIGONOMETRY form corresponding values of x and y; thus X 30° Tt "6 45° St 4 60° Tt ~3~ 90° it ~2 120° 2 3* 135° 3 4* 180° Tt 210° 7 6* 270° ~2 300° 5 360° 2it y .58 1 1.73 CO -1.73 -1 .58 00 -1.73 Then, on proceeding as in the case of the sine the graph in Fig. 54 is obtained. Y Fig. 54. — Graph of tan x. By using a unit circle, see Fig. 49, the ordinates can be drawn quickly. Graphs of cot as, sec ac y cosec oc. These can be obtained by proceeding as in the case of the preceding graphs. They are shown in Figs. 55, 56, 57. Fig. 55. — Graoh of cot x. ANGLES AND THEIR TRIGONOMETRIC FUNCTIONS 59 F 1 I Fig. 56. — Graph of sec x. Fig. 57. — Graph of esc x. EXAMPLES 1. Draw various graphs for sin 6, cos 6, tan 0, by varying the scales used in representing radians and trigonometric functions. 2. Draw graphs of the following: (a) sin# + cos#; (b) sin 6— cos 6) (c) sin 26; (d) cos 30; 0) sin -; (/) tan 2« GO ELEMENTS OF PLANE TRIGONOMETRY 3. Draw, with the same axes of reference, graphs of sin 6 and cos 6; and from your figure obtain values of 6 between 0° and 360° for which (1) sin 0=cos 0; (2) sin 6+ cos 6 = 0. Also with the help of this figure draw the graph of sin 0-j-cos 6. 25. Relations between the trigonometric functions of an angle. The relations between the trigonometric functions of an acute angle were set forth in Art. 7. These relations hold for the ratios of any angle. A. Inspection of the definitions, Art. 20, shows the reciprocal relations, namely: tan^cot^i = l; cos A sec -4 — 1; sin A esc A — 1, (1) * A 1 1 A 1 i.e., cot A =.- t: secA = 7 ; esc A=- — r. tan A' cos A' sin A B. In each of the figures in Art. 20, MP OM MP OP smA OM OP cos A t&nA OM~OM~cosA> cotA ~MP~MP~smA- (2) OP OP C. In each of the figures in Art. 20, MP 2 + OM 2 = OP 2 . On dividing both members of this equation by, OP . 2 ■ 2 OM , MP , in turn, and following the same process as that adopted in Art. 7, it results that sin 2 A + cos 2 A = l] sec 2 . 1 = 1 + tan 2 .4; cosec 2 ^4=1 + cot 2 A. (3) Relations C also follow directly from the line definitions, Art. 21, as shown in Fig. 44 or Figs. 45-47. ANGLES AND THEIR TRIGONOMETRIC FUNCTIONS 61 , EXAMPLES 1. Given that sinA = f; find the other functions of A by means of the relations shown in this article. [In Ex. 6, Art. 20, this problem is solved geometrically; here it is solved algebraically.] ±V7 . 1 4 cos A = ± Vl — sin 2 A = — - — : sec A = cos A -j- \/y ' 1 4 . sin A 3 cosecA=— — 7 = 5-; tanA = sin A 3 ' cos A ± V7 ' cot A = - j= — — . tan A 6 Since the given sine is positive, the corresponding angles are in the first and second quadrants. Hence the double values of the calculated ratios may be paired as follows: sin A cos A tan A cot A sec A cosec A 3 +VY 3 VT 4 4 4 4 a/7 3 v7 3 3 -V7 3 -VT 4 4 4 4 V7 3 Vf 3 Find the other ratios algebraically, and verify the results geometrically, when: 2. cos A =— f. 3. tanA=-f. 4. cosec A=— 5. 5. cotA= — 3. Find the other ratios algebraically, and verify the results geometrically, when angle A satisfies the following pairs of conditions : 6. sin A = J and tan A negative. 7. tan A = \/3 and sec A negative. 8. cosA=— J and sin A positive. 9. sin A = — f and tan A positive. 62 ELEMENTS OF PLANE TRIGONOMETRY 10. If sin 0= a . „ ' . „ „ , prove that m 2 + 2mn + 2n 2 ' tan 0= ± m 2 + 2mn 2mn+2n 2 ' a 2 —b 2 11. If cos 6= , rs t find sin and tan 0. Verify each of the following relations by reducing the first member to the second: 12. cos x tan x= sin x. 13. sec x— tan x> sin z= cos x. cot 2 A 14. = cos 2 A. l + cot 2 A 16. (l + tan 2 z) cos 2 0=1. 16. cos 4 0-sin 4 0+1 = 2 cos 2 0. 1 1 17 + ■ = 1. tan 2 B+l ' cot 2 £+l 18. sec 4 0-tan 4 6=2 sec 2 0-1. 26. Functions of — A, 90° + A, 180° + A in terms of func- tions of A, A being any angle. Functions of —A. Describe the angles A, — A, OP being p P- Pn R Fig. 59 Fig. 60. Fig. 58. the terminal line of A, and OPi the terminal line of —A. In Figs. 58, 59, 60, 61, A is in the first, second, third, fourth quadrants respectively. ANGLES AND THEIR TRIGONOMETRIC FUNCTIONS 63 Take OP l = OP, and draw PM, P X M X at right angles to OX. Then in each figure OM^OM, M,P, MP. Thus, for an angle A in any quadrant, sin (--4) = -gfp-=-^p=-sin^; cos {—A) So also tan (— A) 0M l OM = cos A. OPi OP -tan A; cot (—A) = —cot A; sec (—A) = sec A; esc (—A) = —esc A. Functions of 90° — ^. Describe the angles A, 90°— A, OP being the terminal line of A, and OP\ the terminal line of 90°- A. In Figs. 62, 63, 64, 65, A is in the first, second, third, fourth quadrants respectively. / p ' ^ M Mi 1 A * * H7 x iv Jf iy\ M 0^ Fig. vtf*' V 63. Fig. 64. If, 0^>v Fig. 6 i X O M x M X Fig. 62. 5. Take OP, = OP, and draw Pikf, P,ilf, at right angles to OX. Then in each figure OMx=MP, MxPx^OM. Thus, for A in any quadrant, and so for all angles A, sm (90°-A)==-^p- =Qp=cos A; /rtrtQ „ OM, MP . A cos (90°-A)=-gp-=-gp =sm A; 64 ELEMENTS OF PLANE TRIGONOMETRY fMA ix M X P X OM tan (90°-A)= m ^- =]^p = cot A- cot (90°- A) = w ^ i = m = tan i; sec (90°- A) =-qj^=mp = ^ M esc (90°- A) OiPi OP = sec A. M l P l OM Hence, the function of any angle is the same as the co- function of its complement. Functions of 90° + A. Describe the angles A, 90° + A, OP being the terminal line of A and OPi the terminal line of 90° + A. In Figs. 66, 67, 68, 69, A is in the first, second, third, fourth quadrants respectively. Fig. 66. Fig. 67. Fig. 68. Fig. 69. Take OPx^OP, and draw PM, P X M X at right angles to OX. Then in each figure OM x - - MP, MxPx = OM. Hence, for any angle A 7 MiPi _ OM ~ OP MP ~~OP ' So also, tan (90° + A) - - cot A ; cot (90° + A) = - tan A ; sec (90° + A) = - esc A ; esc (90° + A)= sec A. sin (90°+ A) = : OP cos(90°+A)=-^p- cos A; sin A. ANGLES AND THEIR TRIGONOMETRIC FUNCTIONS 65 Functions of 180° -X Describe the angles A, 180°- A, OP being the terminal line of A, and OPi the terminal line of 180°- A. In Figs. 70, 71, 72, 73, A is an angle in the first, second, third, fourth quadrants respectively. p x p p p x ' x f\^^^1x M O Mi Mi o M M x M Fig. 70. Fig. 71. Fig. 72. Fig. 73. Take OP x = OP, and draw PM, P x Mi at right angles to OX. Then in each figure OM x = - OM, MtPi = MP. Hence, for any angle A, MiP t sin (180°- 4) = cos (180°- A) OPi 0M X MP OP OM = sin A = —cos A. OPx OP So also, tan (180°- A) = -tan A; cot (180°- A) = -cot A; sec (180°-A) = -sec A; esc (180°- A) = esc A. Thus, a function of the supplement of an angle and the same function of the angle itself are numerically equal; the sines and cosecants respectively of supplementary angles have the same sign, but the cosines and the remaining functions respectively have opposite signs. Functions of 180° +A. On proceeding as in the pre- ceding cases it will be found that sin (180° +A) - - sin A, cos (180° + A) = - cos A, tan(180° + A)= tan A, cot(180° + A)= cot A, sec (180° + A) = -sec A, esc (180° + A) = -esc A. 66 ELEMENTS OF PLANE TRIGONOMETRY Since, OX being the initial line, all angles having the same terminal line have the same value for a function, it follows that any function of 360° + A, and of rc-360° + A (n = any whole number) is equal to the same function of A. Also, any function of 360°— A, and of n-360°—A (n = any whole number) is equal to the same function of —A. EXAMPLES 1. Express the functions of 270°— A in terms of functions otA. 2. Express the functions of 270° + A in terms of functions of A. 27. Reduction of trigonometric functions of any angle to functions of acute angles. By means of the relations in Art. 26 the functions of any angle can be expressed in terms of the functions of an angle between 0° and 90°, and in terms of the functions of an angle between 0° and 45°. EXAMPLES 1. sin 700° = sin (360° + 340°) = sin 340° = sin (-20°) = -sin 20°= -.3420. 2. tan975° = tan (2- 360° + 255°) = tan 255° = tan (180° + 75°) = tan75° = cot 15° = 3.7321. 3. esc (-1160°)= -esc (1160°) = -esc (3-360° + S0°) = -esc 80°= -sec 10°. Hence esc (-1160°)= -sec 10°= -_i_--^.= -1.015. ANGLES AND THEIR TRIGONOMETRIC FUNCTIONS 67 4. Express the following as functions of acute angles: (a) sin 287°. (d) sec 925° 10'. (g) tan (-1055°). (6) cos 332°. (e) sin 2150°. (h) cos (-2055°). (c) tan 218° 30'. (/) cot (-487°). (i) esc 310° 30'. 5. Find the values of the following: (a) sin 346° 10'. (d) sec (-310°) (g) esc 876°. (6) cos231 o 30 / . (e) cot 950°. (h) cos (-1131°). (c) tan 174° 15'. (/) sin (-2830°) (i) tan (-1487°). 6. Prove the following: (a) cos 240° cos 120°-sin 120° cos 150°- 1. (6) tan 675° sec 540° + cot 495° esc 450° =0. CHAPTER IV GENERAL VALUES. INVERSE TRIGONOMETRIC FUNCTIONS. TRIGONOMETRIC EQUATIONS 28. General values. All angles having the same initial and terminal lines have the same values for each trigono- metric function. The general value of an angle having a given trigonometric function is an expression, or formula, which includes all the angles that have that function. General expression for all angles having the same sine and cosecant. Let a be an acute angle having a given sine, a, say. Let its terminal line be OP. All angles having OP o Fig. 74. for terminal line have the same sine a. These angles are the angles in the expression, ra-360° + a, i.e., 2m -180° + a, (ra = any integer). {%) Also sin (180°— a) = sin a = a. In Fig. 74, OPi is the terminal line of 180°- a. All angles having OPi for terminal line have the same sine a. They are the angles n the expression, m-360 o + (180 o -a), i.e., (2m + l)180°-a, (m an integer), (it) 68 GENERAL VALUES 69 Expressions (i) and (it) are both included in the expres- sion, nl80°+(— l) n a, i.e., mr-f(— l) n a, n an integer. For, if n is even, a has the sign +, as in (i); if n is odd, a has the sign — , as in (u). Since esc a = i , this expression includes all angles that have the same cosecant as a. Otherwise expressed: sin a -sin [rc-180° + (— l) n a] = sin [n7r + (— l) n a]; esc a=csc [tt-180° + (— l) n a] = csc [ft7r + (— l) n a]. 'Hie proof is similar for a in any quadrant. Vs Ex. 1. Give the general expression for a if sin a = -^~. The least positive value of a is 60°. .*. The general value of a is n-180°+(-l) n 60°, i.e., ™r+(-l)|. On giving n the values 0, 1, 2, 3, ... , particular values of a are obtained, viz., 60°, 120°, 420°, 480°, . . . General expression for all angles having the same cosine and secant. Let a be an acute angle having a given cosine, a, say. Let its terminal line be OP. All angles having OP for terminal line have the same cosine a. These angles are the angles in the expression, n - 360° + a , {n = an integer. ) (m) Also cos (— a) = cos a = a. 70 ELEMENTS GF PLANE TRIGONOMETRY In Fig. 75, OP i is the terminal line of —a. All angles having OPi for terminal line have the same cosine a. They are the angles in the expression, n • 360° — a j (n = an integer. ) (iv) Expressions (Hi) and (iv) are both included in the expres- sion, n • 360° ± a, i.e., 2wir ± a, (n an integer.) Since sec a = , this expression includes all angles that have the same secant as a. Otherwise expressed: cos a = cos (n • 360° ± a) = cos (2m: ± a) ; sec a = sec (n • 360° ±a)= sec (2tt7r ± a) . The proof is similar for a in any quadrant. Fig. 75. Ex. 2. Give the general expression for a when cos a=J. The least positive value of a is 60°. .'. The general value of a is n- 360° ±60°, i.e., 2nn± n -. o On giving n the values 0, 1, 2, 3, ... , particular values of a are obtained, viz., ±60°, 420°, 300°, 780°, 660°, . . , GENERAL VALUES 71 General expression for all angles having the same tangent and cotangent. Let a (Fig. 76) be an acute angle having a given tangent, a, say. Let its terminal line be OP. All angles having OP for terminal line have the same tangent a. These angles are the angles in the expression, ra-360 o + a, i.e., 2m -180° + a, (m an integer.) (v) Also tan (180° + a) = tan a. In Fig. 76, OP i is the terminal line of 180°+ a. All angles having OPi for terminal line have the same tangent a. They are the angles in the expression m- 360° + (180° + a) i.e., (2m + l)180° + a, (m an integer) (vi) 180°+ a \a M, -1 \CL 1 y^o M -a{ Fig. 76. Expressions (v) and (vi) are both included in the expres- sion, w-180° + a, i.e., mr + a, n an integer. Since cota=- , this expression includes all angles tan a' r that have the same cotangent as a. Otherwise expressed : tan a = tan (n • 180° + a) = tan (un + a) ; cot a=cot (tt-180° + a:)=cot (nn+a). The proof is similar for a in any quadrant. 72 ELEMENTS OF PLANE TRIGONOMETRY Ex. 3. Give the general expression for a when tan a = \/3. The least positive value of a is 60°. .'. The general value of a is n-180° + 60°, i.e., nn +—. o On giving n the values 0, 1, 2, 3, ... , particular values of a are obtained, viz., 60°, 240°, 420°, 600°, . . . EXAMPLES 4. Given sin A = i, find the general value of A . Also find the four least positive values of A. 5. Given cos A = — J, find the general value of A and the three least positive values of A. 6. Find the general value of which satisfies each of the following equations : (a) sin = 0. (g) sin#= — (6) cos = 0. (h) cos 6= ]=. 2 ' V2 (c) tan = 0. (i) tan 0=1. 1 Vs' (d) cot0=-^. (/) cot0=-l. 2 (e) sec 0=1. (k) sec 0=— -=^. V3 (/) csc0=\/2. (I) tan0= — ~ V3 29. Inverse trigonometric functions. It has been seen that the sine is a function of the angle. On the other hand, the value of the angle depends on the value of the sine, and thus the angle is a function of the sine. This function of GENERAL VALUES 73 the sine is called the inverse sine or the anti-sine. Similarly, the angle is a function of each of the other trigonometric functions. Thus there arise other inverse trigonometric functions or anti-trigonometric functions, as they are some- times called, viz., the inverse cosine or the anti-cosine, the inverse tangent or the anti-tangent, etc. E.g., since sin 30° = J, 30° is an angle whose sine is J. This line is also expressed thus : 30° is an inverse sine of \ (or an anti-sine of J), and more briefly, on using a symbol invented for the inverse sine, 30° = sin" 1 J. In general : the two statements, " the sine of an angle 6 " is m, 6 is " an angle whose sine is ra," are briefly expressed, 6 = sin -1 m = arc sin m * The expressions, sin -1 m, cos -1 m, tan -1 m, etc., . . . , are the symbols for the inverse trigonometric functions. The expression tan -1 2, for instance, is read " the inverse * This notation is the one used on the Continent. 74 ELEMENTS OF PLANE TRIGONOMETRY tangent of 2," "the anti-tangent of 2," and may also be read " an angle (or the set of angles) whose tangent is 2." N.B. The trigonometric functions are ratios, and thus are pure numbers; the inverse trigonometric functions are angles. For instance (see Art. 28, Exs. 1, 2, 3), \/3 sin" 1 — -=n- 180°+ (- l) n 60°, n an integer; cos -1 - = tt-360°±60°, n an integer; tan" 1 Vs=n- 180° + 60°, n an integer. Thus, while each of the trigonometric functions has a single definite value, each inverse trigonometric function has an infinite number of values. The angle having the smallest numerical value in an inverse function is called the principal value of the inverse function. E.g. The principal value of tan -1 Vs is 60°; The principal value of sin _1 ( y= J is —45°. When the smallest numerical value has both positive and negative signs the positive sign is taken; thus The principal value of cos -1 \ is 60°. N.B. It should be noted that the " — 1 " in the symbol for an inverse trigonometric function is not an algebraic exponent. See Art. 5, N.B. in Examples. Thus: sin -1 m does not denote (sin m)~ l , i.e., -. ; v ' ' sin m (tan x)~ l , i.e., t , should not be written tan -1 x. tan x GENERAL VALUES 75 EXAMPLES 1. Prove that sin -1 £=cos -1 Vl—x 2 . Let 6 = sin -1 x. Then sin#=£. Hence cos 6 = Vl — sin 2 d= Vl — x 2 . Thus 0=cos _1 Vl — x 2 ; i.e., sin -1 x = cos _1 Vl — x 2 . 2. Construct the following: (a) sin" 1 !. (c) tan" 1 (-2). (e) sec" 1 2. (6) cos-i(-|). (iQ cot-i(I). (/) csc-i(-0. Read the following identities, and prove them : 3. sin(tan- £)= ±± 7. «rt|-o«n(±^. 4.ta„(sin- 1 g)=±|. 8. 8 m-»A=tan-«( ± |). 6. cot" 1 1 = nn* + -. 9. sin ( cos" 1 - J = tan I sin" 1 — = J . 6. cos" 1 ! — K )=2n7t*±—. 10. tan -1 m=cos _1 — r=r-. V 2 / 3 Vl + m 2 30. Trigonometric equations. Trigonometric identities. Trigonometric equations are equations in which one or more trigonometric functions or inverse trigonometric functions are involved. These equations are true only under certain conditions, viz., for certain values of the angles. To solve these equations is to find these values. * n a whole number. 76 ELEMENTS OF PLANE TRIGONOMETRY E.g. The equation tan x=l. Here x = tan" 1 1 = 45,° 225,° etc. The general, or complete, solution of this equation is z=n-180 o + 45 o , i.e., vm+j. Trigonometric identities are in the form of equations and are unconditionally true, i.e., are true for all values of the angles involved. E.g. sin 2 x + cos 2 x = 1, and the other relations in Art. 25, (1), (2), (3), are identities. EXAMPLES 1. Solve the equation sin 2 x — 2 cos x-\- } = 0. Here 1 — cos 2 x — 2 cos £-f-J = 0. .'. 4 cos 2 x+S cos x— 5 = 0; i.e., (2 cosx + 5) (2 cosz-l)=0. .*. 2 cos £ + 5 = 0, or 2 cos x— 1 = 0. .'. cosz= — f, or cos x =J. There is no solution for cosx=— f, since the cosine of an angle lies between — 1 and +1.* From cos x = J comes the solution, z=tt-360 o ±60°. In solving an equation containing several functions, the general method is to reduce the equation to a form in which only one function appears. * In other words: The solution is impossible. GENERAL VALUES 77 2. Prove the identity cos 4 A — sin 4 .4 = 1 — 2 sin 2 A. cos 4 A — sin 4 A = (cos 2 A + sin 2 A) (cos 2 ^4 — sin 2 A) = 1 • (1-sin 2 A -sin 2 4) = 1-2 sin 2 A. Solve each of the following equations: 3. 2 cos 2 z + 5 sinx— 4 = 0. 8. sin z=tan 2 x. 4. sin z + csc x = 2. 9. 2 sin 2 # + V3 cos rr+l = 0. 5. sin y + cos y = V2. 10. 4 sec 2 5-7 tan 2 5=3. 6. sin 2 a;=l. 11. 4 tan x— cot z = 3. 7. 2 cos A + sec A = S. 12. sec 2 ?/— 5 tan y+5 = 0. 13. Given 4 sin 2 = 3, find the values of which are between 0° and 500°. 14. Given sin z+cos x cot x = 2, find the values of x which are between 0° and 360°. Prove the following identities: 15. sin 3 + cos 3 = (sin + cos 0) (1 — sin cos 0) . 16. sin z (cot rr+2) (2 cot x+l)=2 cosec x+5 cos a:. 17. cos 6 A + sin 6 A = 1 — 3 cos 2 A sin 2 ^4 . 18. cos 6 x + 2 cos 4 x sin 2 z-f-cos 2 # sin 4 # + sin 2 #=1. sec0 + csc0 l + cot0 tan 0+1 19. sec — esc 1 — cot tan 0—1' sec 2 + esc 2 20. tan0 + cot = sec esc CHAPTER V TRIGONOMETRIC FUNCTIONS OF THE SUM AND DIFFERENCE OF TWO ANGLES General Formulas 31. To deduce sin (A + B), cos (A + B). N.B. In Arts. 31, 32 the conventions (Arts. 16, 19, 20) regarding the signs of angles and lines are followed. The formulas can be derived, however, for Figs. 77-79 by using the definitions in Art. 3. Case L A and B both acute. In Figs. 77, 78, O M N Fig. 77. M O N Fig. 78. XOL = A, LOT = B, XOT = A+B. (In Fig. 77, A +B is acute; in Fig. 78, A +B is obtuse.) From any point P in OT, PM, PQ are drawn at right angles to OX, OL respectively; QN is drawn at right angles to OX ; VQR is drawn parallel to OX. Now VQL = XOQ = A. VQP - VQL + LQP = 90° + A. 78 THE SUM AND DIFFERENCE OF TWO ANGLES 79 • /ixln • yn P MP N Q+KP NQRP NQ Now, QQ=sinA; ,\ NQ = OQ sin A. 7?P Also OP =sin H>^ = sin (90° + A) = cos A. .*. RP = QP cos A. • / ^ i z>\ OQ sin A QP cos A .-. sm(A+B)= — Qp — + — Qp — . But ~ - cos QOP = cos B; ^p = sin QOP = sin B. .\ sm (A+B) =sm A cos B + qos Asm B. (1) fjj-n* ynp 0M ON+QR* ON OR cos(A+B)=cosXOP=Qp= — ^p — = qp+qP' ON Now -fvj = cosA; .-. ON = OQcos A Also, 7^ = cos FQP = cos (90°+A) = -sin A. .-. Qfl=-QPsinA. .. OQ cos A QP . .'. cos(A+£)= — qp — -Q-psmA; .'. cos (A + B)= cos A cos B— sin A sin B. (2) Case II. A and B any angles. First suppose B acute and A obtuse, viz., 90° + A', A' thus being acute. Since A -90°+ A', A'=-(90°-A). * Note that OiV, Qi2, are in opposite directions. 80 ELEMENTS OF PLANE TRIGONOMETRY /. sinA'=-sin(90°-.A)*=-cosA-, cos A' = cos (90°- A) * - sin A. Then sin (A+£) = sin (90°+A'+B)=coa (A'+B)* = cos A' cos B— sin A' sin B = sin A cos 5 +cos A sin B. Also, cos (A +^)= cos (90° + A'+£) = -sin (A'+£)* = —sin A f cos B—cos A' sin 5. = cos A cos #— sin A sin B. The above procedure shows that formulas (1), (2) still hold true when one of the angles A, B, in Case II, is increased by 90°, and that these formulas will continue to hold true as the angles continue to be increased by 90°. Hence these formulas are true for angles in any quadrants, and thus for all angles. Note. Formulas (1), (2) can also be derived from a figure, as in Case I, for A and B in any quadrants. EXAMPLES 1. Derive sin 75° = sin (30° + 45°)= ^t} . ^ . ^ rn V3-1 2. Derive cos 75° = =-. 2V2 3. Given sin#=i, siny=$, and x and y both acute, find sin (x+y), cos (x+y). 4. If tan x= J and tan y = -fa, find sin (x+y) and cos (x + y) when x and y are acute angles. 5. Prove sin (60° + x) -cos (30° + z)=sin x. * See Art. 26. THE SUM AND DIFFERENCE OF TWO ANGLES 81 6. Prove cos (60°+J5)+sin (30° +B) = cos B. 7. Prove cos (x + y) cos z + sin (x + y) sin #=cos y. 8. Find sin (sin -1 J-fsin -1 J) when the angles are between 0° and 90°. 9. Prove sin (sin -1 ra+sin -1 n) = mv / l- n 2 ±nVl — m 2 . 10. Prove cos (sin -1 ra + sin -1 n)=Vl — m 2 Vl — n 2 ± mn. 32. To deduce sin (A—B), cos (A—B). First method. Formulas (1), (2), Art. 31, are true for all angles. On taking — B instead of B, there results : sin (A — B)=sinA cos (— B)+cos A sin (— B); .'. sin (A —B) = sin A cos B —cos A sin B. (1) Also, cos (A — B)= cos A cos (— B)~ sin A sin (-B). .'. cos (^1 — U) =cos ^L cos B + sin A sin ^. (2) Second method. Case I. A and B, both acute, J. >B. In Fig. 79, From any point P in 07 7 , PM, PQ are drawn at right angles fe to OX, OL, respectively; QN is at right angles to OX; VPR is drawn parallel to OX. Now XOL = A, LOT=-B, XOT = A-B. N M Fig. 79. RQP = 90°-OQN = A; . QPR = 90°-RQP = 90°-A; 82 ELEMENTS OF PLANE TRIGONOMETRY .-. VPQ = 1$0°-QPR = 90° + A. sin (A - B) = sin XOP = ~gp= —gp — = Qt > - ~jp. But jvpsin A; .*. J\TQ= 30 sin A Also ~ - sin VPQ - sin (90° + A) - cos A. .'. RQ = PQ cos A. ,a m OQsinA PQcosA /. sin (A -5)- — ^p— ^p — . But pyp = cos QOP = cos ( — #)=cos 5; PQ QP — YYp = ^p = sin QOP = sin (—B) = —sinB; .'. sin (A — B) = sin A cos B—cos A sin 5. sj m yn P 0M ON+RP ON PR* cos (A - B)= cos X0P=Qp =—jjp — =s oP~'5F' ON Now 7)7) = cos ^;* •'• ON = 0Q cos A. P7? Also pQ=cos 7PQ = cos (90° + A) = -sin A. .'. PR=-PQ sin A. /, DX OQ cos A PQsinA .-. cos(A-£)= — ^p — + — ^p — . .*. cos (A — 5) = cos A cos B+ sin A sin B. * Since ltP=-PR. THE SUM AND DIFFERENCE OF TWO ANGLES 83 Case II. A and B any angles. The methods of proof are similar to those shown in Art. 31, Case II, and Note. EXAMPLES V3-1 1. Derive sin 15°= sin (45° -30°) = 2. Derive cos 15°= cos (60° -45°) = 2\/2 * V3+1 2V2 ' 3. If cos A = |f and cos B = %, find sin (A — B) and cos (A — B) when A and B are acute angles. 4. If tan x=j and tan y = ^, find sin (x — y) and cos (x — y) \ ] m x and y are acute angles. 5. Find sin (sin -1 ^ — sin -1 i) when the angles are acute. Verify the following identities: 6. cos (^+45°)+ sin (A- 45°)= 0. 7. cos (30° + x)~ cos (30° - x) = - sin z. 8. cos (x + y) cos (z — y) — cos 2 z — sin 2 y. 9. sin (x + y) sin (x — 2/) = cos 2 2/ — cos 2 x. „ rt sin (4 +B) . , . D 10. 1 ^ = tan A±tan J5. cos A cos B 33. Fundamental formulas. Formulas (1), (2), Art. 31, are called the addition formulas or theorems in trigonometry ; formulas (1), (2), Art. 32, are called the subtraction formulas or theorems. These four formulas are also called the funda- mental formulas of trigonometry. For convenience they are brought together: 84 ELEMENTS OF PLANE TRIGONOMETRY sin (A + B) = sin A cos B+ cos A sin B. (1) sin (A —B) =sin A cos B —cos ^ sin B. (2) cos (-4 + B) = cos ^t cos B —sin ^4 sin B. (3) cos (4— B)=cos A cos JB+sin ^4 sin B. (4) 7n words : Of any two angles, sine sum = sine first • cosine second + cosine first - sine second sine difference =sine first - cosine second — cosine first - sine second cosine sum = cosine first- cosine second— sine first- sine second cosine difference = cosine first- cosine second+sine first -sine second 34. To deduce tan (A + B), tan(A-B), cot (A + B), cot (A-B). sin (A + B) sin A cos B + cos A sin B tan(A+£) = cos (A +2?) cos A cos B— sin A sin i? On dividing each term of the numerator and the denom- inator of the second member by cos A cos B there is obtained tan A + tan B tan (.*+*) = r _ tan ^ tanj? . (1) In a similar way it can be shown that tan A —tan B s v tan (A -B) = -— — -. (2) 1 + tan A tan B v J • t . ^ cos (A +B) cos A cos B— sin A sin B COt (A + J5) = , a t r>\ = — A d"! J—' D- sin (A +B) sin A cos 5 +cos A sin B On dividing each term of the numerator and the denom- inator of the second member by sin A sin B there is obtained cot A cot B— cot B + cot In a similar way it can be shown that co t (A + i?)= cotjS+cotA . (3) THE SUM AND DIFFERENCE OF TWO ANGLES 85 j. , A r»N COt ^4 COt 5 + 1 cot ^-^° cotg-cot^ - (« EXAMPLES 1. If tanP=2, tanQ = $, show that tan(P+Q) = 7, tan(P-Q) = l. 2. Derive tan 75° from tan 45° and tan 30°. 3. Derive tan 15° from tan 60° and tan 45°. 4. Prove: tan (45° + x) = \ + ^ n X ; tan (45°-x) = t 1 ~ tan x . ' 1-tanz' 7 1 + tanz 5. Prove: cot (45° + x) = ^t^li ; cot (45°-:r) = Cot * +1 cotx+T COtx— 1 6. If tan A = £, tan 5= J, find tan (A + 5) ani tan (A — B). 7. Show that tan" 1 m+tan" 1 n = ta.nr l m+ ^ . 1 — mn Let x = tan _1 ra, and 2/=tan -1 n. Then tanx = ra, tany=n. „ / , \ tanx + tan?/ m-\-n Now tan (x + y)= - — — ■*- = - — — . 1 — tan x tan y 1 — mn .'. £ + v=tan *- ; i.e., tan 1 m+tan 1 n=tan~ 1 - . 9 1 — ran' 1 — mn 8. Show that tan -1 m — tan -1 n = tan -1 1 + mw 9. Find tan -1 7 + tan -1 3, and tan -1 7 — tan -1 3. 10. Find tan -1 2 + tan -1 -5, and cot -1 2 + cot -1 -5. 35. To deduce sin 2-4, cos 2A, tan2A. On putting B = A in formulas (1), (3), Art. 33, there results: 86 ELEMENTS OF PLANE TRIGONOMETRY sin 2A = 2 sin A cos A; (1) cos 2A = cos 2 A - sin 2 A ; (2) i.e., =1-2 sin 2 A*; (3) i.e., = 2 cos 2 .4-1* (4) On putting B = A in formula (1), Art. 34: 2 tan A tan2A = l^U^A- ® On putting 2^4. = z ; then A = $x; and these formulas are expressed : sin x = 2 sin Jz-cos \x, cos z = cos 2 Jo;— sin 2 }x, etc. In words : sine any angle = 2 sine half-angle • cosine half-angle, cosine any an#Ze = (cosine half-angle) 2 — (sine half-angle) 2 , = 1 — 2 (sine half-angle) 2 , •= 2 (cosine half-angle) 2 — 1 . . 2 tangent half-angle tangent any «^_ T _^____. EXAMPLES 1. Find cos 22£° from cos 45°. 2cos 2 22£°=l + cos45°. • co 3 2 22A°-Vli M- 1 + V? 1 + 1-4142 .. cos 22* - 2 ^ + V 2J— 2VF"2X1.4142 = - 8536 ' .\ cos 22i°=.9239. ♦Sinoecos 2 ^ + sin 2 A = l. THE SUM AND DIFFERENCE OF TWO ANGLES 87 2. (a) Deduce sin 22^° from cos 45°. (b) Deduce tan 22^° from tan 45°, (c) From functions of 180° derive sin 90°, cos 90°, tan 90°. cot 2 A-l 3. Derive cot 2A 2 cot A 3 4. Express cos 3x and sin 3x in terms of functions of —x. 5. Express cos 3x and sin 3x in terms of functions of 6x. 6. Express cos 6x and sin Qx in terms of functions of 3x. 7. Derive the following: . , - /l —cos oc /l +COS 00 sin §« - \ — — ; cos fcc = \ ; , /l -cos oc tan Jos — \— . 11 1 + COS oc 8. Derive the following: sin 3A = 3 sin A —4 sin 3 .4; cos 3A =4 cos 3 ^4 —3 cos A\ ' A 3 tan A — tan 3 ^4 tan 3^4 = — - — — — x— - — . 1— 3 tan 2 A [Suggestion : sin 3 A = sin (2 A + A) = sin 2A cos A + cos 2 A sin A = etc] 9. Verify the following identities: (1) cot A-cot2A=cosec2A. (2) 1 +tan2A tan A = sec 2A. ^ / . A , M 2 1 , • , /.x sin2A . , V 3) (sm-±cos-j =l±smi. (4) 1 + CQs2 ^ tan A 88 ELEMENTS OF PLANE TRIGONOMETRY /KX 1 — cos 24 . /fi ,.l + cos4 x 4 ( 5 ) gl - oj = tan4. (6) - — -. — -r— = cot-. sin 24. N smi 2 _ 2 tan 4 . /ox 2— sec 2 4 _ . (7) rTT — 2-r = sm 2 ^- ( 8 ) 9-r-=cos 24. 1 + tan 2 4 sec 2 4 (9) cos 4 0-sin 40= cos 20. (10) cot 0-tan =2 cot 20. /1ix sin20 cos 20 a ,• . sin3x cos 3a; _ (11) — r—z ^- = sec 0. (12) — s = 2. sin cos sin # cos x 1-tan 2 ^ /io\ /• ^ fiA\ j. x smi 1 — cos a: (13) cos 4 = r . (14) tan— = - = — : . 1 , 2 4 2 1 + cos a; sini ^.v J . 4 tan a: — 4 tan 3 x (15) tan4x=- — — — = — — — j— . 1 — 6 tan 2 x + tan 4 a; (16) sin 4a: = 4sina:cosa: — 8 sin 3 a: cos a; = 8 cos 3 x sin a; — 4 cos x sin x. (17) cos 4x=l — 8 sin 2 z + 8 sin 4 2=1 — 8 cos 2 a: +8 cos 4 a;. 4 10. (a) If sin 4=-, calculate cos 4, sin 24, cos 24, tan 24. 2 1 (6) If cos 24=—, prove tan 4=— \/~b. o o 1 4 (c) If tan 4 =-, prove cos 24 = + -. . o 5 36. Transformation formulas. From formulas (l)-(4), Art. 33, there follow, on making the add tions and sub- tractions indicated : sin (^ + -B)+sin (A—B) = 2 sin A cos B. (1) sin (A + B) -sin (A -B) = 2 cos A sin B. (2) cos (4 + J3) + cos (A—B)= 2 cos A cos B. (3) cos (A + B) -cos (A -B) = -2 sin A sin ^. (4) THE SUM AND DIFFERENCE OF TWO ANGLES 89 By this set of formulas products of sines and cosines can be transformed into sums and differences. Thus: In words (reading the second members first): Of any two angles, 2 sin one • cos the other = sin sum + sin difference,* (1/) 2 cos one • sin the other = sin sum — sin difference, (2') 2 cos one • cos the other = cos difference + cos sum, (3') 2 sin one • sin the other = cos difference — cos sum. (4') On putting A+B = P, A~B = Q, and solving for A and B, A = UP + Q), B = i(P-Q). ■ Formulas (l)-(4) then take the forms: sin P + sin Q= 2 sin — r~ cos . (5) sin -P— sin Q= 2 cos 9 sin . (6) COS -P + COS Q - 2 COS ^2 cos *!z2 (7) cos 1* -cos £ = -2 sin ^-^ sin ^2 (8) By this set of formulas sums and differences of sines and cosines can be transformed into products. In words: Of any two angles, the sum of the sines =2 sin half sum -cos half difference,* (5') the difference of the sines =2 cos half sum- sin half difference, (6') * The difference is taken: first angle -the second. 90 ELEMENTS OF PLANE TRIGONOMETRY the sum of the cosines =2 cos half sum • cos half difference, (70 the difference of the cosines = — 2 sin half sum ■ sin half difference. (8') EXAMPLES a ai_ xi. x COS X— COS?/ 1. Show that ■ ^= —tan J (z+y) tan J(s— y). COS Z + COS i/ cos a: — cos ?/_ —2 sin \(x-\-y) sin J(^~2/) cos z + cos y 2 cos J(x + ?/) cos %(x — y) = —tan J(^ + 2/) tan %(x— y). 2. Show that 2 sin (A +45°) sin {A -45°)= sin 2 A -cos 2 A. 2 sin (A + 45°) sin {A -45°) =. cos ( A + 45° - A - 45°) - cos (A + 45° + A - 45°) , = cos 90° — cos 2 A = sin 2 A — cos 2 A . « ot_ xu .sin A + sin 3A. 3. Show that r - — - t = tan 2A. cos A + cos 3A si n A + sin 3A Jl sin^(3A + A) cosK3A-A) cos A + cos 3A ~2 cos i(3A + A) cos i(3A - A) 1 sin 2A • = pr-r=tan 2A. cos2A 4. Solve the equation sin 50 + sin = sin 30. .'. 2 sin 30 cos 20 = sin 30. .'. sin 30(2 cos 20-1)= 0. /. (a) sin 30=0; (6) 2 cos 20-1 = 0. From (a), 30=0°, 180°, etc.; the general value of 30 is nn (n being any integer). .*. = 0°, 60°, etc.; the general value of is — . o From (b), cos 20 = £. .\ 20= ±60°, etc.; its general value is 2rwr±— . o .*. 0= ±30°, etc.; its general value is nn±— . o THE SUM AND DIFFERENCE OF TWO ANGLES 91 5. Transform each of the following sums and differences into a product: (1) sin 3x + sin bx. (2) sin 7A — sin 5A. (3) cos2z— cos 6z. (4) cos bx + cos 9x. (5) sin mA + sin nB. (6) cos mx— cos ny. (7) sin Sx + cos bx. (8) cos 4x— sin 2x. 6. Transform each of the following products into a sum or a difference : (1) sin 5x cos Sx. (2) cos 7x sin bx. (3) sin 2z sin Qx. (4) cos bx cos 9x. (5) sin mA sin nB. (6) cos nx cos my. (7) sin 4z sin 2x. (8) cos 7z cos Sx. 7. Show that the value of sin (n + 1)5 sin (n-l)£+cos (w+ 1)5 cos (n-l)B is independent of n. Verify the following identities: 8. (a) sin (w+l)A + sin (n— 1)A = 2 sin nA cos A. (6) cos (n+l)A + cos (n— 1)A=2 cos nA cos A. 9. cos (A + 5) cos A + sin (A + 5) sin A = cos 5. 10. esc 2 A + cot 2 A = cot A . 11. sin 5 A sin A = sin 2 3 A — sin 2 2A. .. sin Sx— sin x „„ sinA + sini? tan £(A + Z?) 12. 5— = tanz. 13. -^ — rX-, — 5 =- \) . ' cos 3x+cos x sin A — sin 5 tan i(A — 5) sin (a?+y) _ tan x + tan ?/ cos (x + y) __\ — tan x tan ?/ sin (x— y) tana;— tan?/' ' cos (x— y) l + tanxtany' 16. sin Sx + sin bx = 8 sin x cos 2 x cos 2x. 17. cos 20° cos 40° cos 80° = .125 (without use of tables). 720 1681 92 ELEMENTS OF PLANE TRIGONOMETRY 18. Given tan A=-, t&nB=-, tanC=^, find tan (A+B+C), Prove the following: 19. sin \2 tan-!^j = 20. sin[isin-i(-|)] = ±| and ±i 21. 3 sin -1 a:=sin _1 (3x— 4x 3 ). 22. 3 cos -1 x=cos _1 (4:X 3 -3x). 23. sec -1 3 = 2 cot -1 V2. 24. tan" 1 * + tan" 1 y + tan" 1 * = tan" 1 5±y±£zfSL. l — xy — yz—zx 25. sm -i- + sm-i- + sini- = -. -1 a — b , x -1 b — c - c—a 26. tan 1 — — r + tan 1 1 , , +tan J — — =0. 1 + ao 1 + 6c 1 -f ca n „ T - A A a , ^ u . . 2a6 . _ . ±4a6(a 2 -6 2 ) 27. If tan £- = r , snow that sin A — , , 9 . sm 2 A = — , „ , LOx0 . 2 6 a 2 + o2 7 (a 2 + 6 2 ) 2 Solve the following equations: 28. cos -cos 70 = sin 40. 29. sin 20 + sin 40 = V2 cos 0. 30. sin 2a + cos 2a =1. 31. sin 2a + 2 cos 2a = 1. 32. sin 20 + 2 sin 40 + sin 6(9 = 0. 33. 4 sin cos 20=1. 34. tan -1 2^ + tan -1 3x=— . 35. cos -1 £— aia -1 x=cos~ 1 x\/3. o 36. tan" 1 (z+l)+tan -1 (x— l)=tan _1 — . ol 37. Find tan (A + B+C) in terms of tan A, tan B, tan C. Thence show: (a) If A+£+C= 90°, THE SUM AND DIFFERENCE OF TWO ANGLES 93 tan A tan 2? + tan B tan C+tan Q tan A = l; (6) If A + B + C= 180°, tan A + tan B+ tan C=tan A tan B tan C. 38. Prove the following, given that A + B + C= 180°: A B C (a) cos A + cos B + cos C=l+4 sin— sin — sin—; A B C (b) sin A+sin 5+sin C=4 cos — cos — cos — . CHAPTER VI RELATIONS BETWEEN THE SIDES AND ANGLES OF A TRIANGLE 37. Notation. Simple geometrical relations. Notation: In stating and deriving the relations in Arts. 37-41 the triangle is denoted as ABC, and the sides opposite the angles A, B, C, as a, b, c, rsepectively. Simple geometrical relations : (a) A +B + (7=180°. (6) The greater side is opposite the greater angle, and conversely. 38. The law of sines. From 1 C in the triangle ABC draw CD at right angles to opposite side AB, and meeting AB or AB produced in D. (In Fig. 80 B is acute, in Fig. 81 o ~ — ^o D BV Fig. 80. A c BD V Fig. 82. B is obtuse, and in Fig. 82 B is a right angle.) Produce AB to V. In what follows, AB is taken as the positive direction. 94 THE SIDES AND ANGLES OF A TRIANGLE 95 In CD A, DC = b sin A. In CDB (Figs. 80, 81), DC = a sin VBC = a sin B. In Fig. 82, DC=BC=a=a sin B * Therefore, in all three triangles, a sin B = b sin A. Tj a ° , a sini Hence, - — -r = - — B , and r = ~ — ™- (lj sin A sin B' b sm B w Similarly, on drawing a line from B at right angles to AC, it can be shown that a c ^ a sin A - — T=-r— 7?, and -— - — jy. (2) sm A sm C c sm C v Hence, in any triangle ^4.5(7, a b c -^— = — — . (3) sin A sin B sin O* v ' In words: The sides of any triangle are proportional to the sines of the opposite angles. The circle described about ABC. Each fraction in (3) gives the diameter of this circle. Let (Fig. 83) be its centre and R its radius. Draw OD at right angles to any side, say AB. Then AD = \c, AOD = C, and AD = AO sin AOD; i.e., \c = R sin C. Hence < IB = -^—. (4) sin C v * V £ = 90°, and sin 90°= 1. 96 ELEMENTS OF PLANE TRIGONOMETRY 39. The law of cosines. The angle A is acute in Fig. 84, obtuse in Fig. 85, right in Fig. 86. From C draw CD at right angles to AB. The direction AB is taken as positive. In Figs. 84, 85, BC 2 = DC 2 +DB 2 . In Fig. 84, DB = AB-AD; in Fig. 85, DB=DA+AB=-AD+AB. Hence, in both figures, ~BC 2 =DC 2 + (AB-AD) 2 = DC 2 +AD 2 +AB 2 -2AB.AD. In Fig. 84, AD - AC cos B AC; in Fig. 85, ^ AD = AC cos BAG (Art. 20). Also, DC 2 + AD 2 = AC 2 . c c r^ C xh \ *v^ b ^S^rt X • ! \ i \ ^ Sfc jk A D B DA Fig. 84. Fig. 85. Hence, in both figures, B Fig. 86. (1) BC -AG +AB -2AG-ABco8A; that is, a 2 = b 2 + c 2 — 2bc cos A. This formula also holds for Fig. 86; for there, cos 4 = cos 90° =0. Similar formulas for b, c, can be derived in like manner, or can be obtained from (1) by symmetry, viz. : b 2 = c 2 +a 2 -2ca cos B, c 2 = a 2 -\-b 2 -2ab cos C. THE SIDES AND ANGLES OF A TRIANGLE 97 In words: In any triangle, the square of any side is equal to the sum of the squares of the other two sides minus twice the product of these two sides multiplied by the cosine of their included angle. Relation (1) may be expressed as follows: b 2 +c 2 -a 2 C0SA = -^^' » Similarly: ■ n c 2 +a 2 -b 2 „ a 2 + b 2 -c 2 COS B = = , COS C = tr-r . 2ca ' 2ab 40. The law of tangents. In any triangle ABC, for any two sides, say a, b, a __ sin A b ~ sin B' From this, on composition and division, a—b sin A — sin B a + &~sin^4.+sini? 2 cos j(A+B) sin j(A-B) = 2smi(A+B)cosi(A-By ^ 36 ' (5) ' ^ a _ b tanjU-B) *./ «+ft"tanJU+B) - [Art. 25, A, B.] (1) In words : The difference of two sides of a triangle is to their sum as the tangent of half the difference of the opposite angles is to the tangent of half their sum. Now A+B = 180°-C. .\ }(A+5)=90°~ /. tan i(A +B) = tan f 90°- ^ ) = cot ^ . Hence, relation (1) may be expressed: 98 ELEMENTS OF PLANE TRIGONOMETRY taniU-B) = ^cotl C . 41. Functions of the half -angles of a triangle in terms of its sides. Let s denote half the sum of the sides of the triangla. Then 2s = a + 6+c, and 2s— 2a = 2(s— a) = — a + 6+c. Similarly, 2(s-b)=a-b + c, 2(s-c)=a+b- c. By Vrt. 39 (2), cos A = — ^ -a 2 B> \rt. 35, 2 sin 2 |^ = 1 -cos A, 2 cos 2 iA = l +cos A. .*. 2 sin 2 J.4 ".-. 2cos 2 JA i 6 2 +c 2 — a 2 26c , 6 2 +c 2 -a 2 ~ 1+ 26c 26c-6 2 -c 2 +a 2 26c+6 2 +c 2 -a 2 26c 26c a 2 -(6-c) 2 (6+c) 2 -a 2 26c 26c (a— 6+c)(a + 6 — c) (6+c+a)(6+c- •a) 26c 26c 2(s-6)-2(s-c) 2s-2(s-a) 26c * 26c *" be (1) cos2L4- 8(8 - a >. * be (2) Since tan 2 \A = sin 2 JA-f-cos 2 \A, it follows that THE SIDES AND ANGLES OF A TRIANGLE 99 lwt»i*- .»-,ft / «-«> . ( 3 ) 2 s K s-a) v u ' ■- sia^=V^^-' ""^^Fl (4) Similar formulas hold for %B, %C, viz.: sin 2 \B = : cos 2 \B = ; ac ac ' • 2 in ( s-a) (s-b) s(s-c) sin 2 J(7 = - 1 • cos 2 W = -^—r — . 2 ab 2 ab (s—a)(s—c) tan 2 J£ = tan 2 iC = s(s-6) ; (s— a)(s— 6) s(s — c) Formula (5) can be given a more symmetrical form. For, on multiplying the numerator and denominator in the second member of (3) by (s—a), (s—a)(s—b)(s—c) _ s(s— a)- tan 2 \A — whence tan \A = - 1 - J.-"'-*><—\ (6) •* s - a \ 8 T , Is— a c 100 ELEMENTS OF PLANE TRIGONOMETRY The circle .nscribed in ABC. The r in the formulas above is equal to the radius of this circle. In Fig. 87 is the centre of the circle, and L, M, N, its points of contact with the sides. Let r denote the radius. By geometry, AN = MA, BL = NB, CM = LC. .-. AN+BL+LC = s, i.e., AN+a = s. .'. AN = s—a. NO Now tan }A = tan NAO^-r^^ AN s—a' (9) Comparison of (8) and (9) shows that the r in (8) must have the same value as the r in (9). Accordingly, radius r -V— — a)(8 — b)(8 — c) EXAMPLES 1. Prove: (a) with a figure, (6) without a figure, that in any triangle ABC, a=b cos C +c cos B ; and write corresponding formulas for b and c. [Suggestion: Draw the perpendicular to BC or BC pro- duced.] 2. If the sines of the angles of a triangle are in the ratios of 13:14:15, prove that the cosines are in the ratios 39:33:25. 3. In ABC, if a : b : c = 8 : 7 : 5, find the angles. ( 4. The sides of a triangle are proportional to the numbers 4, 5, 6; find the least angle. 5. Prove that a cos B — b cos A = ir~ CHAPTER VII SOLUTION OF OBLIQUE TRIANGLES 42. Cases for solution. General remarks on methods of solution. In order that a triangle may be constructed, three elements, one of which must be a side, are required. Hence, there are four cases for construction and solution, namely, when the given parts are as follows: I. One side and two angles. II. Two sides and the angle opposite to one of them. III. Two sides and their included angle. IV. Three sides. Careful attention should now be paid to the remarks in Art. 8 on the methods of solution of triangles and to the general suggestions made there for solving problems and checking results. These remarks and suggestions apply also to the problems in this chapter. Oblique triangles can be solved (by computation) in the following ways : (a) By dividing them conveniently into right-angled triangles, solving these triangles, and combining the results. This method is not discussed here, but is left as an exercise for the student. Full details are in Murray, Plane Trigonometry, Art. 34. (6) By means of certain relations in Chap. VI, logarithms not being used. 101 102 ELEMENTS OF PLANE TRIGONOMETRY (c) By means of certain relations in Chap. VI, logarithms being used. Cases I-IV are solved in manner (b) in Arts. 43-46; these cases are solved in manner (c) in Arts. 48-50. 43. Case I. Given one side and two angles. In triangle ABC, suppose that A, B, a are known; it is required to find C, b, c. In this case (see Fig. 80, Art. 38), C = 1S0°-(A+B); a a - — n = - — j, whence b = - — T -smB: sin B sin A' sin A ' _ — — =--. — - whence c = - — j -sin C. sin C sin A' sin A Checks: a 2 = b 2 +c 2 — 2bc cos A ; — — ^ = - — j ; or other ' sin B sin A ' results in Chap. VI which have not been used in the solution. EXAMPLES 1. Solve the triangle PQR, given: PQ= 12 in., Solution: * R = Q = 40°, PR= P=75°. RQ = R= 180°- (P+Q) = 180°- (40° +75°) = 65°. _PR_ = _PQ_ RQ PQ sin Q sin R' sin P sin R' /. P#=-^L s inQ, /eQ = -^%.sinP, sin R smB sin 65° sin 65° * Results to be written here. SOLUTION OF OBLIQUE TRIANGLES 103 = .9 1 063 X - 6428 ' =i& x - 9659 > = 13.24 X. 6428, = 13.24 X -9659, = 8.51 in. = 12.8 in. Check: Take some relation of Chap. VI not involving PQ, P, Q, as above; e.g.: PR _ sin Q RQ~ sin P' According to this, 8.51 sin 40° ^ . 64^8 . 12.8 " sin 75° .9659' which gives, on multiplying up, 8.2198 ... . =8.2281. This shows that the results are very nearly accurate. They are as accurate as can be obtained with four-place tables. Another check: Ex. Use relation (1), Art. 40, as a check. 2. In ABC, vl = 50 o , B=75°, c = 60 in. Solve the triangle. 3. In ABC, ,4 = 131° 35', 5 = 30°, 6 = 5£ ft. Find a. 4. In ABC, B = 70° 30', C=78° 10', a=102. Solve the tri- angle. 5. In ABC, B = 9S° 22', (7=41° V, a = 5.42. Solve the tri- angle. 44. Case II. Given two sides and an angle opposite to one of them. In the triangle ABC let a, b, A be known, and C, B, c be required. The triangle will first be constructed 104 ELEMENTS OF PLANE TRIGONOMETRY with the given elements. At any point A of a straight line LM, unlimited in length, make angle MAC equal to angle A, and cut off AC equal to b. About C as a centre, and with a radius equal to a, describe a circle. This circle will either: (1) Not reach to LM, as in Fig. 88. (2) Just reach to LM, thus having LM for a tangent, as in Fig. 89. (3) Intersect LM in two points, as in Figs. 90, 91. b a b, A\ a B z> c a / /k\ V a ' c ~, _,_. Fig. 88. Fig. B M B, --B-'" B M Fig. 91. Each of these possible cases must be considered. In each figure, from C draw CD at right angles to AM; then CD = b sin, A. In case (1), Fig. 88, CBCD; that is, a>b sin A. If a>b, then the points B, Bi, in which the circle inter- sects LM, are on opposite sides of A, as in Fjg. 90, and there is one triangle which has three elements equal to the given elements, namely, ABC. If a 100+16-49 67 0 2 ' VT' 2' + VJ' _VW __2_ VT _2_ _3^ 3 3 ' 3' + 2 ' + Vr 2' vT () 5' 5' V 3' 4' V 5' 5' 4' 3' 4' 3' 132 ANSWERS M -u^H 1 -u^l -A- A _A_ W + 5 ' 5' + 2 ' Tv^i' 2' V2T \/21 2 V21 2 5 5 5 ' 5' 2 ' V2l' 2 ' V2l' A A A A A A _A A _A _ 3 A _A 5' 5' 3' 4' 3' 4 ' 5' 5' 3' 4 ' 3 ' 4 A A i A A A. _A _A A A _A _A 5 ' 5 ' 3 ' 4 ' 3 ' 4 ' 5' 5 ' 3 ' 4 ' 3' 4* 3 2\/J0 3 2VlO 7 _7 10. (a) - ? , - ? , 2ViQ , 3 , avl _ > 3. V21 2^ _^H _2_ J[ ^5_ () 5 ' 5' 2 ' V2l' 2' V2l' 3 2 j5 2_ VJ3 Vl3 (C) VI3' Vfs' 2' 3' "" 2 ' ~ 3 ' (<*) 2 • 2' "' VT "' V3" V3, * 2, -^ Art. 25, Page 62 a 2 + 6 2 ' a 2 — 6 2 Art. 26, Page 66 1. sin (270°-.4) = -cos4, cos (270°-, 4) = -sin A, tan (270° -.4) =cqt A, cot (270° -.4) =tan A } ' sec (270°-vl)= -csc A, esc (270°-4) - -sec A. 2. sin (270° + A)=-cos4, cos (270° + A) -sin A, tan (270° + 4) = -cot4, cot (270° + A) = -tan 4, sec (270° + 4)=csc4, esc (270° + 4) = -sec A. Art. 27, Page 67 4. (a) -sin 73°, -cos 17°. (e) -sin 10°, -cos 80°. (6) cos 28°, sin 62°. (/) cot 53°, tan 37°. (c) tan 38° 30', cot 51° 30'. (g) tan 25°, cot 65°. (d) -sec 25° 10', -esc 64° 50'. (h) -sin 15°, -cos 75°. (0 -esc 49° 30', -sec 40° 30'. 5. (a) -.2391. (d) 1.555. (g) 2.458. (6) -.6225. (e) .8391. (h) .6293. (c) -.1007. (/) .7660. (i) -1.0724 ANSWERS 133 Art. 28, Page 72 4. n-180° + (-l) n 30°, nn + (-l) n ^; 30°, 150°, 390°, 510°. 5. n- 360° ±120°, 2nsr±J-«r; 120°, 240°, 480°. 6. (a) n-180°, nn. (g) n- 180°-(-l) w 60°, nx-i-l?^-. ~. (h) n.360°±135°, 2mz±^ TV 4 *(fc) n-360°±90°, 2nn±\. (h) n-360 o ±135°, 2nK± 371 (c) n-180°, rnz. (i) n- 180° +45°, nn + n (d) n-180° + 60, n^+^-. (?) n.l80° + 135°, rwr+^p (e) n.l80°±90°, mz±\. Ck) n.360°±30°, 2nn±~. A 6 (/) n.l80°+(-l) n 45°, nn + (-l) n j. (I) n.l80° + 150°, nn + ^. Art. 30, Page 77 3. x = n7z + ( — l) n —; x = sin -1 2, impossible. o 4. x = 2n7r+— . 5. y = 2ri7t+j-. A 4 6. x=mz±^r. 7. A=2mz± 7 ^, A=2mz. 8. st-wr, z = n-360° + (-l)»38°10'.3; a; ^sin-^- 1.618), impossible. 9. x = 2nn±— -; a; = cos -1 >/3^ impossible. o 10. B = mt±%-. 11. x = n7r+^, z=n-180 o +165 o 47'.7. b 4 12. 2/ = n-180° + 71°33 , .8, */ = n-180 o + 63° 26'. 13. 60°, 120°, 240°, 300°, 420°, 480°. 14. x = 30°, x = 150°. Art. 31, Pages 80, 81 + V21+2V15 +V / 3l5-V / 2' 4 i. I c 2\/2 + V / 3~ 20 ' 20 ' 4 ' 5 ' 5 ' 6 Art. 32, Page 83 84 . 187 44 m 117 2V / 2-V / 3~ 205 ' 205' 125 ' 125* 6 * In this instance the general value may be expressed; m- 180°±90°, mn±— . 134 ANSWERS Art. 34, Page 85 6. 1; y. 9. tan-i(— i-) ; tan-^. 10. tan- 1 oo, i.e., (2n+ !)-£-; cot" 1 0, i.e., (2n + l)-£-. Art. 35, Pages 87, 88 3 3 3 3 ^ 3 4. cos 2 —a: — sin 2 —a;, 2 cos 2 — x — 1, 1—2 sin 2 — x; 2 sin -^-x cos —a;. /l+cos6x ll 5- V^2— ; \- cos 6a; 2 6. cos 2 3a:— sin 2 3x, 2 cos 2 3a; — 1, 1—2 sin 2 3x; 2 sin 3a; cos Sx. 10. M 4; *£, -1; ,$ Art. 36, Pages 91, 92, 93 5. (1) 2 sin 4a; cos x. (2) 2 cos 6A sin A. (3) 2 sin 4x sin 2a;. (4) 2 cos 7x cos 2a;. . mA+nB mA—nB . mx + ny . mx — ny (5) 2 sin g cos 2 ' ^ ~ 2sm ^sin ^-2. (7) 2 sin (45° -x) cos (45° -4x). (8) 2 cos (45° -x) sin (45° -3x). 6. (1) J(sin8x+sin2x). (2) \ (sin 12a; -sin 2x). (3) £(cos4z— cos 8a;). (4) £(cos 14a; + cos 4x). (5) $[cos (m A — nB)— cos (mA+nB)}. (6) ijcos (nx + my) + cos (nx— my)}. (7) £(cos2x— cos 6a;). (8) £( cos 10a;+cos4a;). 18. 1. 29. ^ = (2n + l)90°, = {4rc+(-l) n }15°. 30 a-y, a=n-90° + (-l) n 45°. 31. a = n.90° + (-l) n 45°, a-* sin- 1 (-|)=n.90 o + (-l) n (-18° 26'. 1). 32. 0=n^r, 0=n.l8O°±9O°. 4 ' 33. ^ = {6n+(-l)»}30°,