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About Google Book Search Google's mission is to organize the world's information and to make it universally accessible and useful. Google Book Search helps readers discover the world's books while helping authors and publishers reach new audiences. You can search through the full text of this book on the web at |http: //books .google .com/I V ^ i" '' . * > < 1 1 V SeU^Ht^^Mr^ y // Jl^ ^%/u/»-^^^. ^, 1^ ' ;* ^ > ^ ^ i::* v'"^ ^^ ^^ " ^ 4 2. r^^ *-♦ -> U./2 '. t ,^ ^ * r *x t--' / CV <._ \ \ L^ r '\ A V ' <■;/,,:/•- PLANE AND SPHERICAL TRIGONOMETRY BY ELMER A. LYMAN I MICHIGAN STATB NORMAL COLLBaS AND EDWIN C. GODDARD UNITBRSITT OF MICHIGAN U.^ ALLYN AND BACON ISoston BVit Ci)tcajQ[o \C\C ' I • • • » » • !.. •-.. •• •• • . THE NEW YOr.K PUBLIC LIBRARY 551730 A A:;iOR, LENOX AND TILDEN FOUNDATIONS COPTEIGHT, 1899, 1900, BY ELMEB A. LYMAN AND EDWIN 0. GODDARD. » • * * • • • ' • • ♦ • » ♦ • 1 • • • « » •* « • • • t .•• • « * « • • • - r • • « • • • * • • • • ' * Norfaiootr $ret(0 3, B. duhing ft Co. — Berwick & Smitb Norwood Mail. U.8.A. PREFACE. Many American text-books on trigonometry treat the solution of triangles quite fully ; English text-books elaborate analytical trigonometry; but no book available seems to meet both needs adequately. To do that is the first aim of the present work, in the preparation of which nearly everything has been worked out and tested by the authors in their classes. The work entered upon, other features demanded attention. For some unaccountable reason nearly all books, in proving the formulae for functions of a ± p, treat the same line as both posi- tive and negative, thus vitiating the proof ; and proofs given for acute angles are (without further discussion) supposed to apply to all angles, or it is suggested that the student can draw other figures and show that the formulae hold in all cases. As a matter of fact the average student cannot show anything of the kind ; and if he could, the proof would still apply only to combi- nations of conditions the same as those in the figures actually drawn. These difficulties are avoided by so wording the proofs that the language applies to figures involving any angles, and to avoid drawing the indefinite number of figures necessary fully to establish the formulae geometrically, the general case is proved algebraically (see page 58). Inverse functions are introduced early, and used constantly. Wherever computations are introduced they are made by means of logarithms. The average student, using logarithms for a short gQ time and only at the end of the subject, straightway forgets what manner of things they are. It is hoped, by dint of much prac- tice, extended over as long a time as possible, to give the student a command of logarithms that will stay. The fundamental for- ^ mulae of trigonometry must be memorized. There is no substi- tute for this. For this purpose oral work is introduced, and there are frequent lists of review problems involving all prin- ciples and formulae previously developed. These lists serve the • • • m 00 00 iv PREFACE. further purpose of throwing the student on his own resources, and compelling him to find in the problem itself, and not in any model solution, the key to its solution, thus developing power, instead of ability to imitate. To the same end, in the solution of triangles, divisions and subdivisions into cases are abandoned, and the student is thrown on his own judgment to determine which of the three possible sets of formulas will lead to the solu- tions with the data given. Long experience justifies this as clearer and simpler. . The use of checks is insisted upon in all computations. For the usual course in plane trigonometry Chapters I-VII, omitting Arts. 26, 27, contain enough. Articles marked * (as Art. * 26) may be omitted unless the teacher finds time for them without neglecting the rest of the work. Classes that can accom- plish more will find a most interesting field opened in the other chapters. More problems are provided than any student is ex- pected to solve, in order that different selections may be assigned to different students, or to classes in different years. Do not assign work too fast Make sure the student has memorized and can use each preceding formula, before taking up new ones. No complete acknowledgment of help received could here be made. The authors are under obligation to many for general hints, and to several- who, after going over the proof with care, have given valuable suggestions. The standard works of Levett and Davison, Hobson, Henrici and Treutlein, and others have been freely consulted, and while many of the problems have been prepared by the authors in their class-room work, they have not hesitated to take, from such standard collections as writers gen- erally have drawn upon, any problems that seemed better adapted than others to the work. Quality has not been knowingly sacri- ficed to originality. Corrections and suggestions will be gladly received at any time. E. A. L., Ypsilanti. E. C. G., Ann Arbor. October, 1900. CONTENTS. Chapter I. Angles — Measurement of Angles. PAGS Angles; magnitude of angles . . • . ... . . . 1 Rectangalar axes ; direction 2 Measurement; sexagesimal and circular systems of measurement; the radian 3 Examples 6 Chapter II. The Trigonometric Functions. Function defined .8 The trigonometric functions . • . . ... • . 9 Fundamental relations • . 11 Examples 14 Functions of 0°, 30% 45°, 60% 90** 15 Examples • • • • 18 Variations in the trigonometric functions . . • • • 19 Graphic representation of functions ....••• 22 Examples 27 Chapter HE. Functions of any Angle — Inverse Functions. Relations of functions of - ft 90** i ft ISO*' ± ft 270'* ±6 to the functions ot$ 29 Inverse functions 35 Examples ,36 Review . . . . • 38 Chapter IV. Computation Tables. Natural functions 40 Logarithms . • • • 40 Laws of logarithms • . 42 Use of tables 45 Cologarithms . • • • • • • • • • .49 Examples • • • 50 V vi CONTENTS. Chapter V, Applications. PAOK Measurements of heights and distances •••••• 51 Common problems in measurement • • 52 Examples 54 Chapter VI. General FoRMULiE — Trigmjnometric Equations and Identities. Sine, cosine, tangent oia± P 56 Examples 59 Sin $ ± sin ff>, $ ± cos <l> . 61 Examples 62 Functions of the double angle 63 Functions of the half angle 64 Examples 64 Trigonometric equations and identities ..•••• 66 Method of attack 66 Examples 67 Simultaneous trigonometric equations 69 Examples 70 Chapter VII. Triangles. Laws of sines, tangents, and cosines i 72 Area of the triangle 76 Solution of triangles 76 Ambiguous case 78 Model solutions ; 80 Examples 83 Applications 84 Review 86 Chapter "Viil. Miscellaneous. Incircle, circumcircle, escribed circle 92 Orthocentre, centroid, medians 94 Examples 96 Chapter IX. Series. Exponential series 97 Logarithmic series 99 Computation of logarithms 100 De Moivre's theorem 103 Computation of natural functions 104 Hyperbolic functions 109 Examples 110 CONTENTS. vii Chapter X. Spherical Trigonometry. PAGE Spherical triangles 112 General formulae 114 Right spherical triangles 123 Area of spherical triangles 125 Examples 128 Chapter XL Solution of Spherical Triangles. General principles 129 Formulae for solution 130 Model solutions 131 Amhiguous cases 132 Right triangles 134 Species 135 Examples 137 Applications to Geodesy and Astronomy 138 PLANE TRIGONOMETRY. >>»Co<i- CHAPTER I. ANQLISS— MEASUREMENT OF ANQLES. 1. Angles. It is difficult, if not impossible, to define an angle. This difficulty may be avoided by telling how it is formed. If a line 'revolve ahovi one of its points^ an angle is generated^ the magnitude of the angle depending on the amount of the rotation. Thus, if one side of the angle d, as OjR, be originally in the position OX^ and be revolved about the point to the position in the figure, the angle XOB is generated. OXia called the initial line, and any position of OB the. terminal line of the angle ,v^^ r x formed. The ansfle is , ,--'' considered positive if gener- -'' ft i ated hy a counter-clockwise rotation of OR^ and hence negative if generated hy a clockwise rotation. The magnitude of depends on the amount of rotation of Oi2, and since the amount of such rotation may be unlimited, there is no limit to the possible magnitude of angles, for, evidently, the revolving line may reach the posi- tion OR by rotation through an acute angle ^, and, likewise, by rotation through once, twice, •••, w times 860®, plus the acute angle 0, So that XOR may mean the acute angle 0,0 + 360^ + 720^ ..., ^ + n • 360^ 1 2 PLANE TRIGONOMETRY. In reading an angle, read first the initial line, then the terminal line. Thus in the figure the acute angle XOR^ or 2T, is a positive angle, and ROX^ or nr, an equal negative angle. Ex. 1. Show that if the initial lines for ), ), ^, — }, right angles are the same, the terminal lines may coincide. 2. Name four other angles having the same initial and terminal lines as I of a right angle ; as } of a right angle ; as ) of a right angle. 2. Rectangular axes. Any plane surface may be divided by two perpendicular straight lines XX' and TY' into four _ portions, or qttadrants. XX' is known as the axiods^ TV as the y-axis^ and the two together are called axes of refer- X ' L X ence. Their intersection is the origin^ and the four portions of the plane surface, XOY^ TOX\ y" X'0T\ Y'OX, are called respec- Yyq,, 2. tively the firsts second^ thirds and fourth quadrants. The position of any point in the plane is determined when we know its dis- tances and directions from the axes. 3. Any direction may be considered positive. Then the opposite direction must be negative. Thus, if AB represents any positive line, BA is an equal nega- tive line. Mathematicians usually consider lines measured in the same direction as OX or OT (Fig. 2) as positive. Then lines measured in the same direc- tion as OX' or OT' must be negative. The distance of any point from the y-axis is called the abscissa^ its distance from the 2;-axis the ordinate^ of that point ; the two together are the cod'rdinates of the point, usually denoted by the letters x and y respectively, and written (re, y). p" o N \"t Y' Fio. 3. ANGLES — MEASUREMENT. 3 When taken with their proper signs, the codrdinates define completely the position of the point. Thus, if the point P is + a units from YT , and + h units from XX' ^ any convenient unit of length being chosen, the position of P is known. For we have only to measure a distance ON equal to a units along OX, and then from N measure n distance h , units parallel to OF, and we arrive at the position of the point P, (a, 6). In like manner we may locate P', ( — a, ft), in the second quadrant, P", (—a, — 6), in the third quadrant, and P'", (a, — 6), in the fourth quadrant. Ex. Locate (2, -2); (0,0); (-8, -7); (0, 5); (-2, 0); (2, 2); (m, n). 4. If OX is the initial line, 6 is said to be an angle of the firnt^ second^ thirds or fourth quadrant^ according as its ter- minal line is in the first, second, third, or fourth quadrant. It is clear that as OR rotates its quality is in no way affected, and hence it is in all positions considered positive^ and its ex- tension through 0, 0R\ negative. The student should notice that the initial line may take any position and revolve in either direction. While it is customary to consider the counter-clockwise rotation as forming a positive angle, yet the condi- \R X X* li *^®"^ ^^ ^ figure may be such •^ \ • • / ^hat a positive angle may be generated by a clockwise rota- tion. Thus the angle XOR in each figure may be traced as a positive angle by revolving the initial line OX to the posi- tion OR. No confusion can result if the fact is clear that when an angle is read XOR, OX is considered a positive line revolving to the position OR. OX' and OR' then are negative lines in whatever direc- tions drawn. These conceptions are mere matters of agreement, and the agreement may be determined in a particular case by the conditions of the problem quite as well as by such general agreements of mathema- ticians as those referred to in Arts. 3 and 4 above. Fio. 4. 5. Measurement. All measurements are made in terms of some fixed standard adopted as a unit. This unit must 4 PLANE TRIGONOMETRY. be of the same kind as the quantity measured. Thus, length is measured in terms of a unit length, surface in terms of a unit surface, weight in terms of a unit weight, value in terms of a unit value, an angle in terms of a unit angle. The measure of a given quantity is the number of times it contains the unit selected. Thus the area of a given surface in square feet is the number of times it contains the unit surface 1 sq. ft. ; the length of a road in miles, the number of times it contains the unit length 1 mi. ; the weight of a cargo of iron ore in tons, the number of times it contains the unit weight 1 ton ; the value of an estate, the number of times it contains the unit value $1. The same quantity may have different measures, according to the unit chosen. So the measure of 80 acres, when the unit surface is 1 acre, is 80, when the unit surface is 1 sq. rd., is 12,800, when the unit surface is 1 sq. yd., is 887,200. What is its measure in square feet ? 6. The essentials of a good unit of measure are : 1. That it be invariable^ i.e. under all conditions bearing the same ratio to equal magnitudes. 2. That it be convenient for practical or theoretical pur- poses. 3. That it be of the same kind as the quantity measured. 7. Two systems of measuring angles are in use, the sexa- gesimal and the circular. The sexagesimal system is used in most practical applica- tions. The right angle, the unit of measure in geometry, though it is invariable, as a measure is too large for con- venience. Accordingly it is divided into 90 equal parts, called degrees. The degree is divided into 60 minutes^ and the minute into 60 seconds. Degrees, minutes, seconds, are indicated by the marks ^ ' ", as 36^ 20' 15". The division of a right angle into hundredths, with subdivisions into hundredths, would be more convenient. The French have proposed such MEASUREMENT OF ANGLES. a centesimal system, dividing the right angle into 100 grades, the grade into 100 minutes, and the minute into 100 seconds, marked ^^ ^\ as 50s 70^ 28^\ The great labor involved in changing mathematical tables, instruments, and records of observation to the new system has prevented its adoption. 8. The circular system is important in theoretical con- siderations. It is based on the fact that for a given angle the ratio of the length of its arc to the length of the radius of that arc is constant, i,e. for a fixed angle the ratio arc : radius is the same no matter what the length of the radius. In the figure, for the angle ^, OA OB 00 AA' BB' 00' That this ratio of arc to radius for a fixed angle is constant follows from the established geometrical principles : 1. The circumference of any circle is 2 tt times its radius. 2. Angles at the centre are in the same ratio as their arcs. The Radian. It follows that an angle whose arc is equal in length to the radius is a constant angle for all circles, since in four right angles, or the perigon, there are always 27r such angles. Thi% constant angle^ whose arc is equal in length to the radius, is taken as the unit angle of circular measure, and is called the radian. From the definition we have 4 right angles = 360® = 2 tt radians, 2 right angles = 180® = tt radians, 1 right angle = 90® = - radians. Fio. 6. TT is a numerical quantity, 3.14159+, and not an angle. When we speak of 180° as tt, 90° as ^, etc., we always mean v radians^ ^ radians, etc. 6 PLANE TRIGONOMETRY. 9. To change from one system of measurement to the other we use the relation, 2 TT radians = 360®. 1QAO . •. 1 radian = =^ = 57^2958- ; TT i.e. the radian is 57®.3, approadmately. Ex. 1. Express in radians 75"" W. 75° 30' = 75°.5 ; 1 radian = 57^3. .% 75** 30' = ^ = 1.317 radians. 67.3 2. Express in degree measure 3.6 radians. 1 radian = 57*^.3. .-. 3.6 radians = 3.6 x 57^3 = 206° 16' 48". EXAMPLES. 1. Construct, approximately, the following angles ; 50®, — 20°, 90°, 179°, -135°, 400°, -380^ 1140°, ^radians, ^radians, -^radians, 4 3 6 3 TT radians, — -^ I'^dians, — ^ radians. Of which quadrant is each angle? ^ ^ 2. What is the measure of : (a) f of a right angle, when 30° is the unit of measure ? (6) an acre, when a square whose side is 10 rds. is the unit ? (c) m miles, when y yards is the unit ? 3. What is the unit of measure, when the measure of 2} miles is 50? 4. The Michigan Central K.R. is 535 miles long, and the Ann Arbor R.R. is 292 miles long. Express the length of the first in terms of the second as a unit. 5. What will be the measure of the radian when the right angle is taken for the unit ? Of the right angle when the radian is the unit ? 6. In which quadrant is 45°? 10°? -60°? 145°? 1145°? -725°? Express each in right angles ; in radians. 7. Express in sexagesimal measure 2[, Z., 1, 6.28, i, I^, - i2[, radians. 3' 12' ' TT 3 3 ' EXAMPLES, 7 8. Express in each system an interior angle of a regular hexagon ; an exterior angle. 9, Find the distance in miles between two places on the earth's equator which are 11° 15' apart. (The earth's radius is about 3963 miles.) 10. Find the length of an arc which subtends an angle of 4 radians at the centre of a circle of radius 12 ft. 3 in. 11. An arc 15 yds. long contains 3 radians. Find the radius of the circle. 12. Show that the hour and minute hands of a watch turn through angles of 30' and 6° respectively per minute ; also find in degrees and in radians the angle turned through by the minute hand in 3 hrs. 20 mins. 13. Find the number of seconds in an arc of 1 mile on the equator ; also the length in miles of an arc of 1' (1 knot). 14. Find to three decimal places the radius of a circle in which the arc of 71° 36' 3".6 is 15 in. long. 15. Find the ratio of - to 5°. 6 16. What is the shortest distance measured on the earth's surface from the equator to Ann Arbor, latitude + 42° 16' 48" ? 17. The difference of two angles is 10°, and the circular measure of their sum is 2. Find the circular measure of each angle. 18. A water wheel of radius 6 ft. makes 30 revolutions per minute. Find the number of miles per hour travelled by a point on the rim. 4 / I / ' ^v CHAPTER II. THE TRiaONOMETRIC FUNCTIONS. 10. Trigonometry, as the word indicates, was originally concerned with the measurement of triangles. It now includes the analytical treatment of certain functions of angles^ as well as the solution of triangles by means of cer- tain relations between the functions of the angleg of those triangles. U. Function. If one quantity depends upon another for its value, the first is called a function of the second. It always follows that the second quantity is also a function of the first ; and, in general, functions are so related that if one is constant the other is constant, and if either varies in value, the other varies. This relation may be extended to any number of mutually dependent quantities. Illustration. If a train moves at a rate of 30 miles per hour, the distance travelled is a function of the rate and time, the time is a function of the rate and distance, and the rate is a function of the time and distance. Again, the circumference of a circle is a function of the radius, and the radius of the circumference, for so long as either is constant the other is constant, and if either changes in value, the other changes, since circumference and radius are connected by the relation 0=2 irR. Once more, in the' right triangle NOP^ the ratio of any two sides is a function of the angle a, because all the right triangles of which a is one angle are similar, Le. the ratio 8 THE TRIGONOMETRIC FUNCTIONS. 9 of two corresponding sides is constant so long as a is con- stant, and varies if a varies. Thus, the ratios JVP N'F' N"P" OP OP' OP" and ON 0N> ON" ;> etc., NP N'P' N"P"' depend on a for their values, i.e. are functions of a. 12. The trigonometric functions. In trigonometry six functions of angles are usually employed, called the trigono- metric functions. By definition these functions are the six ratios between the sides of the triangle of reference of the given angle. The triangle of reference is formed ly drawing from some point in the initial line^ or the initial line produced^ a perpendicular to that line meeting the terminal line of the angle. X N —X —X Fig. 8. Let a be an angle of any quadrant. Each triangle of reference of a, NOP^ is formed by drawing a perpendicular to OJf, or OX produced, meeting the terminal line OR in P. 10 PLANE TRIGONOMETRY. If a is greater than 360°, its triangle of reference would not differ from one of the above triangles. It is perhaps worthy of notice that the triangle of reference might be defined to be the triangle formed by drawing a perpendicular to either side of the angle, or that side produced, meet- ing the other side or the other side produced. In the figure, NOP is in all cases the triangle of reference of a. The principles of the fol- lowing pages are the same no matter which of the triangles is considered the triangle of FiQ. u. reference. It will, however, be as well, and perhaps clearer, to use the triangle defined under Fig. 8, and we shall always draw the triangle as there described. 13. The trigonometric functions of a (Fig. 8) are called the sine^ co%ine^ tangent^ cotangent^ secant^ and cosecant of a. These are abbreviated in writing to sin a, cos a, tan a, cot a, sec a, CSC a, and are defined as follows : sin a = 2?3^ = M, whence y ^rsina; nyp. r ^ CM a = i:^ = —9 whence x^^rcosa; nyp. r taiia = ^^ = ~» whence y = xtana; cot tt = = —J whence x=i y cotct; sec a = ^^^ = —9 whence r = x8eca; base 0? CSC a = ^^ = -9 whence r^yc%c a. perp. y ^ 1 — cos a and 1 — sin a, called versed-sine a and coversed-sine a, respec- tively, are sometimes used. Ex. 1. Write the trigonometric functions of j8, NPO (Fig. 8), and .• compare with those of a above. The meaning of the prefix co in cosine, cotangent, and cosecant appears from the relations of Ex. 1. For the sine of an angle equals the cosine, i.e. the complement-siney of the complement of that angle ; the tangent THE TBIGONOMETRIC FUNCTIONS. 11 of an angle equals the cotangent of its complementary angle, and the secant of an angle equals the cosecant of its complement' ary angle. 2. Express each side of triangle ABC in terms of another side, and some function of an angle in all possible ways, as a = & tan Ay etc. Fio. 10. 14. Constancy of the trigonometric functions. It is impor- tant to notice why these ratios are functions of the angle^ i.e. are the same for equal angles and different for unequal angles. This is shown by the principles of similar triangles. Fig. 11. In each figure show that in all possible triangles of refer- ence for a the ratios are the same, but in the triangles of reference for a and a', respectively, the ratios are different. The student must notice that sin a is a single symbol. It is the name of a number, or fraction, belonging to the angle a ; and if it be at any time convenient, we may denote sin a by a single letter, such as o, or x. Also, sin^a is an abbreviation for (sin a) 2, Le. for (sin a) x (sin a). Such abbreviations are used because they are convenient. Lock, Ele- mentary Trigonometry. 15. Fundamental relations. From the definitions of Art. 13 the following reciprocal relations are apparent : sina = cosa = CSC a 1 seca^ tana = -4— > cot a Also from the definitions, cos a C8Ca=: 8eca = cota = cota = sinci 1 CO8 a 1 tan a COS a sin a 12 PLANE TRIGONOMETRY. From the right triangle NOP, page 9, j,2 + a^ = r2; whence (1) ^ + ^-1 ^ + ^-1' • (2) (3) 1 + '^='*. y* y* From (1) 8iii*a+co8*a=lj nn «= VI — coi? a ; eosass? (2) ta]i^a+l=sec^a| tow «= Vwc^ a— 1 ; «e<?a=s? (S) l-f-eot*a=C8C*a; cot a=^^/c8e^ a-^1; esca=i? The foregoing definitions and fundamental relations are of the highest importance, and must be mastered at once. The student of trigonometry is helpless without perfect familiarity with them. These relations are true for all values of ce, positive or negative, but the signs of the functions are not in all cases positive, as appeal's from the fact that in the triangles of reference in Fig. 8 x and y are sometimes negative. The equations sin a = ± VI — cos^ a, tan a = ± Vsec^ a — 1, cot a = ± Vcsc^ a — 1, have the double sign ±, Which sign is to be used in a given case depends on the quadrant in which a Ues. 16. The relations of Art. 15 enable us to express any function in terms of any other, or when one function is given, to find all the others. Ex. 1. To express the other functions in terms of tangent : r.;« ^1 1 tan a . ^^x « 1' . sm a = = = — ; cota= ; CSC a VI + cot2 a VI + tan^ a ^^ana 1 1 ^^^^^TITZ- /. . =' seca = Vl + tan2a; sec a Vl + tana « ^ tani = tan«; csc« = ^^ + ^^^ > tana THE TRIGONOMETRIC FUNCTIONS. 18 In like manner determine the relations to complete the following table : sma cosce tana cot a seca CSC a sma cos a tana tan a VI + tan* a 1 VI + tan* a tana 1 tana VI + tan*a VI 4- tan* g tana cot a A ^. sec a CSC a 2. Given sin a = f ; find the other functions. co8a=vT^^ = jV7; tana =-*-= — = f\/7; jVr V7 1 /- 14/- 14 cot a = -^=- = i V7 ; sec a = = -=- = 4 V7 ; esc a = i = ^« ^V7 Ja/7 V7 t 3 3. Given tan ^ + cot ^ = 2 ; find sin ^. 1 tan^ + tan^ = 2, tan* ^ - 2 tan ^ + 1 = 0, tan ^ = 1. ... sin <^= , ^°^ - = }V2. VI + tan* ^ Or, expressing in terms of sine directly, 5HL$+ S21$^ = 2, cos^ sm^ I sin* ^ + cos* ^ = 2 sin ^ cos ^ sin* ^ — 2 sin ^ cos ^ + cov ^ = ; whence sin ^ — cos ^ = 0, sin ^ = cos ^. .*. sin ^ = ^VS. 4. Prove sec* x — sec* x = tan* x + tan* a?. sec** — sec*x = sec* a; (sec* x — 1) = (1 + tan* x) tan* ar = tan*x + tan*x. 5. Prove sin* y + cos* y = 1 — 3 sin* y cos* y. sin* y + cos* y = (sin* y + cos* y) (sin* y — sin* y cos* y + cos* y) s= (sin* y + cos* y)* — 3 sin* y cos* y = 1 — 3 sin* y cos* y. 14 PLANE TRIGONOMETRY. ,6. Prove -i5^^ + -52ii_=8ec«c8c^ + l. 1 — cot z 1 — tau z sinz C08Z tau z cot z cos z sin z + i 7- = V 1 — cot z 1 — tan z • __ cos 2 ^^ _ sing sin 2 cos z sin* 2; _^ cos* z cos 2 (sin 2 — cos z) sin 2 (cos z — sin 2) _ sin' 2 — cos' 2 _ sin* 2 4- sin 2 cos 2 4- cos* 2 sin 2 cos z (sin 2 — cos 2) sin 2 cos 2 ^l+8iD»C08»^_^ +l=8ec*C8C^ + l. Sin 2 cos 2 sm 2 cos 2 In solving problems like 3, 4, 5, and 6 above, it is usually safe, if no other step suggests itself, to express all other functions of one member in terms of sine and cosine. The resulting expression may then be re- duced by the principles of algebra to the expression in the other member of the equation. For further suggestions as to the solution of trigono- metric equations and identities see page 66. \ ^ EXAMPLES. y 1. Find the values of all the functions of a, if sin « = | ; if tan ce = f ; if sec a = 2 ; if cos a = J V3 ; if cot a = | ; if esc a = v^. 2. Compute the functions of each acute angle in the right triangles whose sides are : (1) 3, 4, 5 ; (2) 8, 15, 17 ; (3) 480, 31, 481 ; (4) a,h,c\ >,ex 2a:y a:* + y^ (5) — ?-, . ^ ^ , X + y. x-y x-y 3. n cos a = JV, find the value of ^^"""^^^"" cos a— cot a y 4. If 2 cos a = 2 — sin a, find tan a. 5. If sec* a esc* a — 4 = 0, find cot a. 6. Solve for sin p in 13 sin j8 + 5 cos* P = 11. Prove -^ 7. sin* ^ ~ cos* ^ = 1 — 2 cos* ^. V 8. (sin a + cos a) (sin a — cos a) = 2 sin* a — 1. '' V ^ 9. (sec a + tan a) (sec a — tan a) = 1. 'y^ 10. cos* p (sec* )8 ~ 2 sin* j8) = cos* p + sin* p. ^ ^ J. cos v 11. tan t; 4- sec r = / 1 — sin V 12 sin w _ 1 4- cos w 1 ~ cos w sin w 13. (sectf + l)(l-costf) = tan*tfcosA / FUNCTIONS OF CERTAIN ANGLES. 15 14. sin* t — sin^ t = cos* t — cos^ U 1 — sin p sin p 16. (tan A + cot il)^ = sec^il csc« A. 17. sec^x — sin* x = tan^x + cos^ar. In the triangle ABC, right angled at C, 18. Given cos A = ^^ BC = 45, find tan B, and AB. •2 _ ♦,2 ' ,19. If cos A = ^, " ^ . and AB = m^ + n^, find ^ C and 5a fn* 4- n* , 20. If -4 C = m + n, BC zzm—n, find sin il, cos 5. 21. In examples 18, 19, 20, above, prove sin^il + cos^il = 1; ' 1 + tan^il =8ec2^. 17. Functions of certain angles. The trigonometric func- tions are numerical quantities which may be determined for any angle. In general these values are taken from tables prepared for the purpose, but the principles already studied enable us to calculate the functions of the following angles. 18. Functions of 0^. If a be a very small angle, the value of y is very small, and decreases as a diminishes. , ,t7 ■ [« ^ Clearly, when a approaches fig 12 ^ 0^ as a limit, y likewise ap- proaches 0, and X approaches r, so that when a = 0®, y = 0, and x = r. .'. «t«0» = ^ = 0, cot 0" = —^ = CO, r tan co,0° = -=l, «ecO° = -^ = l, r cos 0° to„0°=2^ = 0, c»cO° = -r^==cc. X sm U In the figure of Art. 18, by diminishing a it is clear that \?e can make y as small as we please, and by making a small enough, we can make the value of y less than any assignable quantity, however small, so that sin a ap- proaches as a limit 0. This is what we mean when we say sin 0° = 0. In like manner, it is evident that, by sufficiently diminishing a we can make cot a greater than any assignable quantity. This we express by saying cot QP = co. 16 PLANE TRIGONOMETRY. 19. Functions of 30**. Let NOP be the triangle of refer- ,R ence for an angle of 30^. Make triangle NOP' = NOP. Then POP' is an equilateral triangle (why?), and ON bisects PF. Hence 9%n FiQ. 13. 30<> = ^ = f = J. r 2y 2 Also a; = Vr2 - y2 = V3y2 = y VS. c.«tf 30^ = 2, r 2y * •etf 30° = f VS, «a« 80° = ^ = -^ = 4= « i^. <»«30°=V3. X yV3 V8 * 20. Functions of 45°. Let NOP be the triangle of refer> ence. If angle NOP = 45°, OPN= 46°. Fio. U. ,♦. jf = x, and r = Va:* + y* = V2 a? = a; V2. Then «iw 46° = « = -^ = J V2, «iw46° = ^ = CO* 46° = - = -^ = i V2, «an45° = ^ = - = l. a; a; Find cot 46°, «ec 46°, cse 46°. FUNCTIONS OP CERTAIN ANGLES. 17 2L Functions of 60^. The functions 'of 60^ may be com- puted by means of the figure, or they may be written from the func- tions Off the complement, or 30®. Let the student in both ways ^how that sin 60° = J V3, cos 60® = J, Compute also the other functions of 60®. 22. Functions of 90' x2r Fio. 16. . If a be an angle very near 90®, the value of x is very small, and de- creases as a increases toward 90®. Clearly when a approaches 90® as a limit, X approaches 0, and t( ap- proaches r, so that when a = 90®, a;=0, y = r. .-. «m90®=l, c?o8 90®=0, eaw90®=Qo. Compute the other functions. Also find the functions of 90® from those of its complement, 0®. 23. It is of great convenience to the student to remember the functions of these angles. They are easily found hy recalling the relative values of the sides of the triangles of reference for the respective angles^ or the values of the other functions may readily be computed by means of the funda- mental relations, if the values of the sine and cosine are remembered, as follows : a 0» 80° 45» 60° 90° sine cosine iVi ivI iVs JV2 Jvl iVi 1V5 18 PLANE TRIGONOMETRY. ORAL WORK. 1. YOiich is greater, sin 45** or i sin 90** ? sin 60° or 2 sin 30'' ? 2. From the functions of 60^ find those of 30°; from the functions of 90°, those of 0°. Why are the functions of 45° equal to the co-functions of 45°? 3. Given sin A =:\, find cos A ; tan A» 4. Show that sin ^ esc ^ = 1 ; cos C sec C = 1 ; cot x tan x = 1, 6. Show that sec« - tan^ e = c8c^0- cot^ $ = sin« $ + cos^ 0. 6. Show that tan 30° tan 60° = cot 60° cot 30° = tan 45°. 7. Show that tan 60° sin^ 45° = cos 30° sin 90°. 8. Show that cos a tan a = sin a; sin j3 cot j3 = cos j3. 9. Show that l-*ang30° ^ ^^ g^o = i cos 0°. l + tan2 30° ' 10. Show that (tan y + cot y) sin y cos y = 1. * EXAMPLES. ^ *^ 1. Show that sin 30° cos 60° + cos 30° sin 60° = sin 90°. Y^^ ' 2. Show that cos 60° cos 30° + sin 60° sin 30° = cos 30°. 3. Show that sin 45° cos 0° - cos 45° sin 0° = cos 45°. 4. Show that cos2 45° - sina45° = cos 90°. 5. Show that tan 45° + tan 0° ^ ^^^ ^^o. / 1 ~ tan 45° tan 0° If il = 60°, verify a o^r, A jl — c os A 6. 8m_ = ^ 7. tan4 = A^ 2 ^1 + — cos^ / 2 ^1 + coSil 8. coSi4 = 2cos24~l = l-2sin«:i. 2 2 If a = 0°, )8 = 30°, y = 45°, 8 = 60°, c = 90°, find the values of 9. sin p + cos 8. / 10. cosjS + tan8. 11. sin j3 cos 8 + cos j3 sin 8 — sin e. 12. (sin P + sin f ) (cos « + cos 8) — 4 sin a (cos y + sin c) . VARIATIONS IN THE FUNCTIONS. 19 24. Variations in the trigonometric functions. Signs. Thus far no account has been taken of the signs of the functions. By the definitions it appears that these de- pend on the signs of re, y, and r. Now r is always positive, and from the figures it is seen that x is positive in the fii*st Ml. + C90, + (r-) (r+) Cot. 4^ Sin. + C09. + Tan,+ CM, + See, + C!MX + %-) Co».-h See. -»- Fig. 17. and fourth quadrants, and y is positive in the first and second. Hence For an angle in the first quadrant all functions are positive^ since re, y, r are positive. In the second quadrant x alone is negative^ so that those functions whose ratios involve 2?, viz. cosine^ tangent^ co- tangent^ secant^ are negative; the others, sine and cosecant^ are positive. In the third qvxidrant x and y are both negative^ so that those functions involving r, viz. sine^ cosine^ secant^ cosecant^ are negative; the others, tangent and cotangent^ rtb positive. In the fourth quadrant y is negative^ so that sine^ tangent^ cotangent^ cosecant are negative^ and cosine and secant^ positive. Valves. In the triangle of reference of any angle, the hypotenuse r is never less than x or y. Then if r be taken of any fixed length, as the angle varies, the base and perpen- dicular of the triangle of reference may each vary in length X u from to r. Hence the ratios - and - can never be greater r r than 1, nor if x and y are negative, less than —1; and -^ - X y 20 PLANE TRIGONOMETRY. cannot have values between + 1 and — 1. But the ratios ^ and - may vary without limit, i.e. from + oo to — oo. X y Therefore the possible values of the functions of an angle are : sine and cosine between + 1 and — 1, i.e. nnt and cosine cannot be numerically greater than 1; tangent and cotangent between + oo and — oo, i.e. tangent and cotangent may have any real valite; secant and cosecant between + oo and + 1, and — 1 and — oo, i.e. secant and cosecant may have any real values^ except values between + 1 and — 1. These limits are indicated in the following figures. The student should carefully verify. Ate iy>to Bin, M**-! Cm »0°«10 Tan, W-ioD Cm IW^a^i Tan UO-70 Ml. 41 Cm. -0 2kra. —flo F * t * • 9S 6 £ sm o-±o o /♦« -1 -0 Cm^. Tom o'^-tO Bin, -1 Cm. -0 +1 +0 QQ O K •i-o. -t-i. -t-0^ 0' O -0. +1, -0-^380" -1 + S/II8T0--1 (M« 170^^70 Tom STO'£ao Tan, +«>| -00 If' 270" FiQ. 18. 25. In tracing the changes in the values of the functions as a changes from 0® to 360°, consider the revolving line r as of fixed length. Then x and y may have any length between and r. Sine, At 0°, sin a = ^ = - = 0. As a increases through r r the first quadrant, y increases from to r, whence - increases from to 1. In passing to 180® sin a decreases from 1 to 0, VARIATIONS IN THE FUNCTIONS. 21 since y decreases from r to 0. As a passes through 180°, y changes sign, and in the third quadrant decreases to nega- tive r, so that sin a decreases from to — 1. In the fourth quadrant y increases from negative r to 0, and hence sin a increases from — 1 to 0. Co%ine depends on changing values of x. Show that, as a increases from 0° to 360°, cos a varies in the four quadrants as follows: 1 to 0, to — 1, — 1 to 0, to 1. Tangeid depends on changing values of both y and x. At 0°, y = 0, a; = r, at 180°, y = 0, a; = - r, at 90°, 2? = 0, y = r, at 270°, a; = 0, y = - r. Hence tan 0° = -2^ = - = 0. As a passes to 90°, y increases X r to r, and x decreases to 0, so that tan a increases from to oo. As a passes through 90°, x changes sign, so that tan a changes from positive to negative by passing through oo. In the second quadrant x decreases to negative r, y to 0, and tan a passes from — oo to 0. As a passes through 180°, tana changes from minus to plus by passing through 0, because at 180° y changes to minus. In the third quadrant tana passes from to oo, changing sign at 270° by passing through 00, because at 270° x changes to plus. In the fourth quadrant tan a piasses from — oo to 0. Cotangent. In like manner show that cot a passes through the values oo to 0, to — oo, oo to 0, to — oo, as a passes from 0° to 360°. Secant depends on x for its value. Noting the change in X as under cosine, we see that secant passes from 1 to oo, — 00 to — 1, — 1 to — 00, 00 to 1. Cosecant passes through the values oo to 1, 1 to oo, — 00 to — 1, — 1 to — 00 . The student should trace the changes in each function fully, as has been done for sine and tangent, giving the reasons at each step. 22 PLANE TRIGONOMETRY. a 0° to 90° 90° to 180° 180° to 270° 270° to 360*» sin to 1 1 to -0 to -1 - 1 to -Q cos 1 to - to - 1 -1 to -0 to 1 tan to 00 - 00 to - to 00 -00 to -0 cot 00 to - to -00 00 to - to -00 sec 1 to 00 — 00 to — 1 — 1 to —00 00 to 1 CSC 00 to 1 1 to 00 — 00 to — 1 — 1 to —00 * 26. Graphic representation of functions. These variations are clearly brought out by graphic representations of the functions. Two cases will be considered : I, when a is a constant angle ; II, when a is a variable angle. I. When a is a constant angle. The trigonometric functions are ratios, pure numbers. By so choosing the triangle of reference that the denomi- nator of the ratio is a side of unit length, the side forming the numerator of that ratio will be a geometrical representa- tion of the value of that function, e.g. if in Fig. 19 r = 1, then sin a = ?f = |.= y. This may be done by making a a r 1 central angle in a circle of radius 1, and drawing triangles of reference as follows : FiQ. 19. GRAPHIC REPRESENTATION OF FUNCllONS. 23 In all the figures AOP = a, and SP SP T>r> OB OB ^j. \j ,^' * ;^ ,. /• . BP AD AD .J. tan a = = = = AD^ OB OA 1 . OA HO EC Tpr, OP OD OB ^n OP 00 00 f,n ''"' = BP = OE = -=^^- It appears then that, by taking a radius 1, sine is represented by the perpendicular to the initial line, drawn from that line to the terminus of the arc sub- tending the given angle; cosine is represented by the line from the vertex of the angle to the foot of the sine; tangent is represented by the geometrical tangent drawn from the origin of the arc to the terminal line, produced if necessary; cotangent is represented by the geometrical tangent drawn from a point 90° from the origin of the arc to the terminal line, produced if necessary; secant is represented by the terminal line, or the terminal line produced, from the origin to its intersection with the tangent line; cosecant is represented by the terminal line, or the terminal line produced, from the origin to its intersection with the cotangent line. 24 PLANE TRIGONOMETRY. The9e lines are not the functions^ but in triangles drawn as explained their lengths are equal to the numerical values of the functions, and in this sense the lines may be said to represent the functions. It will be noticed also that their directions indicate the signs of the functions. Let the student by means of these representations verify the results / of Arts. 24 and 25. II. WTien a is a variable angle. Take XX* and TT' as axes of reference, and let angle units be measured along the 2;-axis, and values of the func- tions parallel to the y-axis, as in Art. 3. We may write corresponding values of the angle and the functions thus : a-(f, 30°, 46^ 60», W, 120**, 136^ 150°, 180°, 210°, 226°, Bina=0, J, 1V2,JV3,1, iV3, Jv^, J, 0, - J, - Jv^, «= 240°, 270°, 300°, 316°, 330°, 360°, -30°, -46°, -60°, -90°, etc., Bina=-J>/3, -1, -jV3, -jV2, -J, 0, -J, -jV2, -i>/a, -1, etc. These values will be sufi&cient to determine the form of the curve representing the function. By taking angles between those above, and computing the values of the function, as given in mathematical tables, the form of the curve can be determined to any required degree of accuracy. Reduc- ing the above fractions to decimals, it will be convenient to make the y-units large in comparison with the a;-units. In the figure one a?-unit repre- sents 15°, and one y-unit 0.25. Measuring the angle values along the a;-axis, and from these points of division measuring the corresponding values of sin a parallel to the y-axis, as in Art. 3, we have, approximately. Curves of Sine and Cosecant, Sine Cosecant Fio.20. GRAPHIC REPRESENTATION OF FUNCTIONS. 25 C>Xi = 30° = 2 units, 0X^ = ^5'' « 3 units, Xi Fi = J =2 units, Xg Fg = 0. 71 = 2. 84 units, OX^ = 60° =4 units, etc., -y8F8 = 0.86 = 3.44 units, etc. We have now only to draw through the points F^, Fg, Fg, etc., thus determined, a continuous curve, and we have the sine-curve or sinusoid. The dotted curve in the figure is the cosecant curve. Let the student compute values, as above, and draw the curve. In like manner draw the cosine and secant curves, as follows : I 1 Curves of Cosine and SecarU, Cosine Secant "" Fig. 21. Tangent curve. Compute values for the angle a and for tan a, as before : a = 0°, 30°, 46°, 60°, 90°, 120°, 186°, 160°, 180°, 210°, 226°, 240°, 270% tan a = 0, jV3, 1, VS, ±«, -V3, -1, -jVS, 0, JV3, 1, V8, ±oo, a = - 30°, - 46°, - 60°, - 90°, etc., tan a = — J V3, — 1, — V3, ± «, etc. Then lay off the values of a and of tan « along the a?, and parallel to the y-axis, respectively. It will be noted that, 26 PLANE TRIGONOMETRY. as a approaches 90^, tan a increases to oo, and when a passes 90^, tan a is negative. Hence the value is measured parallel Curves of Tangent and Cotangent Tangent Cotangent Fio. 22. to the y-axis downward, thus giving a discontinuous curve, as in the figure. * 27. The following principles are illustrated by the curves : 1. The sine and cosine are continuous for varying values of the angle, and lie within the limits + 1 and — 1. Sine changes sign as the angle passes through 180°, 360®, •••, n 180**, while cosine changes sign as the angle passes through 90*=; 270*^, — , (2w + l) 90°. Tangent and cotangent are discontinuous, the one as the angle approaches 90°, 270°, ..., (2 w+1) 90°, the other as the angle approaches 180°, 360°, —, 71180°!, and each changes sign as the angle passes through these vetoes. The limiting values of tangent and cotangent are + oo and — oo. 2. A line parallel to the y-axis cuts any of the curves in but one point, showing that for any value of a there is but one value of any function of a. But a line parallel to the a;-axis cuts any of the curves in an indefinite number of points, if at all, showing that for any value of the function there are an indefinite number of values, if any, of a. GRAPHIC REPRESENTATION OF FUNCTIONS. 27 8. The curves afford an excellent illustration of the varia- tions in sign and value of the functions, as a varies from to 360^, as discussed in Art. 25. Let the student trace these changes. 4. From the curves it is evident that the functions are periodic^ i.e. each increase of the angle through 360® in the case of the sine and cosine, or through 180° in the case of the tangent and cotangent, produces a portion of the curve like that produced by the first variation of the angle within those limits. 6. The difference in rapidity of change of the functions at different values of a is important, and reference will be made to this in computations of triangles. (See Art. 64, Case III.) A glance at the curves shows that sine is chang- ing in value rapidly at 0®, 180°, etc., while near 90°, 270°, etc., the rate of change is slow. But cosine has a slow rate of change at 0°, 180°, etc., and a rapid rate at 90°, 270°, etc. Tangent and cotangent change rapidly throughout. Ex. Let the student discuss secant and cosecant curves. ORAL WORK. 1. Express in radians ISO'*, 120°, 45°; in degrees, ^ radians, 2ir, fir, iir. 2. If } of a right angle be the unit, what is the measure of } of a right angle? of 90°? of 135°? 3. Which is greater, cos 30° or 4 cos 60°? tan^ or cot J? sin ? or cos- ? ^ ' ' 6 3 4 4 4. Express sin a in terms of sec a ; of tan a ; tan a in terms of cos a ; of sec a, 5. Given sin a = {, find tan a. If tan a = 1, find sin a, esc a, cot a ; also tan 2 a, sin 2 a, cos 2 a. 6. If cos a = 4, find sin ^, tan ^< ' 2 2 7. In what quadrant is angle f, if both sin t and cos t are minus ? if sinf is plus and cos< minus? if tan< and cot< are both minus? if sint and CSC f are of the same sign ? Why ? 8. Of the numbers 3, f, — 5, — 4> ^> "" ^> °°» ^> which may be a value of sin;) ? of Bocp ? of tan p ? Why ? \ 28 PLANE TRIGONOMETRY. EXAMPLES. 1. If 8in26''40' = 0.44880, find, correct to 0.00001, the cosine and tangent. 2. If tan a = V3, and cot p = | V3, find sin a cos /3 — cos a sin p, 3 j,^^!^^^ sin 30^ cot 300^ cos 60O tan 00° sin 90° cos 0° Prove the identities : 4. tan^(l -cot2>4) + coti4(l -tan2^) = 0. 5. (sin A + sec i4 )2 + (cos A + esc .4)2 = (1 + sec ^ esc Ay. 6. sin^ X cos a; esc re — cos' x esc x sin^ x + cos^ x sec a; sin :e = sin* x cos x + cos'z sinx. 7. tan^ w + cot^ w = sec^ to esc* to — 2. 8. sec* V + cos* u = 2 + tan* i; sin* v, 9. cos* r + 1 = 2 cos« r sec « + sin*«. 10. CSC* t — sec* ^ = cos* t esc* ^ — sin* t sec* f. 11. The sine of an angle is ; find the other functions. 12. If tan il + sin .4 = m, tan A — sin -4 = n, prove to* — n* = ^Vmi, Solve for one function of the angle involved the equations : 13. sintf + 2costf = 1. 16. 2sin*x + cosx - 1 = 0. , ^ cos a 3 ^^' 8ec*x — 7 tan x — 9 = 0. 14. = -. tan a 2 18. 3 esc y + 10 cot y - 35 = 0. 15. V3csc*^ = 4cotft 19. sin*i;-}cost;-l = 0. 20. a sec* to + b tan to + c — a = 0. 21. If ?HLii=v^, *!£ii = v^, find .4 and B. sm B tan B 22. Find to five decimal places the arc which subtends the angle of 1° at the centre of a circle whose radius is 4000 miles. 23. If CSC il = }\/3, find the other functions, when A lies between ~ and w. 2 24. In each of two triangles the angles are in G. P. The least angle of one of them is three times the least angle of the other, and the sum of the greatest angles is 240^ Find the circular measure of each of the angles. CHAPTER III. FXJITCTIOITS OF ANY ANGLE — INVERSE FUNCTIONS. 28. B}*^ an examination of the figure of Art. 24 it is seen that all the fundamental relations between the functions hold true for any value of a. The table of Art. 16 expresses the functions of a, whatever be its magnitude, in terms of each of the other functions of that angle if the ± sign be prefixed to the radicals. The definitions of the trigonometric functions (Art. 12) apply to angles of any size and sign, but it is always possible to express the functions of any angle in terms of the func- tions of a positive acute angle. The functions of any angle ^, greater than 360°, are the same as those of ^ ± n • 360°, since and ±n' 360° have the same triangle of reference. Thus the functions of 390°, or of 750°, are the same as the functions of 390°— 360°, or of 750°— 2-360°, i.e. of 30°, as is at once seen by drawing a figure. So also the functions of — 315°, or of — 675° are the same as those of - 315° + 360°, or of - 675° + 2-360°, i.e. of 45°. For functions,of angles less than 360° the relations of this chapter are important. 29. To find the relations of the functions of — 0, 90° ± ^, 180° ± 0, and 270° ±0to the functions of 0^ being any angle. Four sets of figures are drawn, I for an acute angle, II for obtuse. III for an angle of the third quadrant, and IV for an angle of the fourth quadrant. In every case generate the angles forming the compound angles separately, i.e. turn the revolving line first through 29 80 PLANE TBIGONOMETBY. W (») (0) -e i80°±e ^:5 3KF ^^ j:^T FUNCTIONS OF ANY ANGLE. W 82 PLANE TRIGONOMETRY. O*', 90°, 180°, or 270°, and then from this position through 0^ or — 0^ as the case may be. Form the triangles of refer- ence for (a) the angle 0, (b) - 0, ((?) 180° ± 0, (rf) 90° ± 0, (6)270°±^. The triangles of reference (a), (6), ((?), (c?), and (e), in each of the four sets of figures, I, II, III, IV, are similar, being mutually equiangular, since all have a right angle and one acute angle equal each to each. Hence the sides x, y^ r of the triangles (d) are homologous to x\ y\ r^ of the cor- responding triangles (6) and ((?), but to y\ a;', r', of the corresponding triangles (c?) and (e). For the sides x of triangle (a) and aj' of the triangles (6) and ((?) are opposite equal angles, and hence are homologous, but the sides y^ are opposite this same angle in triangles (rf) and (e), and there- fore sides y' of (df) and («) are homologous to x of (a). Attending to the signs of x and x', y and y in the similar triangles (a) and (6), sin(-^) = ^ = -^ =-sin^. a?' X r r cos (— ^)=-r = - = cos ^, tan (- ^) = ^= - ^ = - tand. ^ ^ aj' a; Also in the similar triangles (a) and ((?), sin (180° - ^) = ^ = I = sin ^, cos (180° - ^) = ^ = - ? = - cos d, tan (180° - ^) = ^ = - ^ = - tand. X X In like manner show that sin(180° + ^) = -sin^, cos(180° + ^) = -cos^, tan (180° + 0)= tan 0. FUNCTIONS OF ANY ANGLE. 83 Again, in the similar triangles (a) and (d"), 8in(90° + d) = ^ = - =cosd, cos (90° + 0)=^ = -^=-ava0, tan(90° + ^)=^ = --=-cot^. • X y Show that sin (90*^ - ^) = cos 0, cos (90^ - ^) = sin 0, tan (90^-^) = cot ^. .; ' ' Finally, from the similar triangles (a) and (e), show that sin (270°±^)=-cos^, cos (270° ± ^) = ± sin 0, tan (270° ± ^) = T cot 0. From the reciprocal relations the student can at once write the corresponding relations for secant, cosecant, and cotangent. 30. Since in each of the four cases x\ y^ of triangles (6) and (js) are homologous to x, y of triangle (a), while x\ y* of the triangles (d) and (g) are homologous to y, x of triangle (a), we may express the relations of the last article thus : The functions of -[qm . n correspond to the same functions of 0, while those of oyno . n correspond to the co-functions of 0j due attention being paid to the signs. The student can readily determine the sign in any given case, whether be acute or obtuse, by considering in what quadrant the compound angle, 90° ± 0^ 180° ± 0^ etc., would 84 PLANE TRIGONOMETRY. lie if were an acute angle, and prefixing to the correspond- ing functions of the signs of the respective functions for an angle in that quadrant. Thus 90° + ^, if ^ be acute, is an angle of the second quadrant, so that sine and cosecant are plus, the other functions minus. It will be seen that sin (90° + ^) = +COS 0, cos (90° + ^) = - sin 0, etc., and this will be true whatever be the magnitude of 0. It will assist in fixing in the memory these important relations to notice that when in the compound angle is measured from the y-axis, as in 90° ± 0^ 270° ± 0^ the functions of one angle correspond to the co-functions of the other, but when in the compound angle is measured from the 2;-axis, as in ± d, 180° ± ^, then the functions of one angle correspond to the same functions of the other. These relations, as has been noted in Art. 28, can be extended to angles greater than 360°, and it may be stated generally that function ^ = ± function (2 w • 90° ± ^), function ^ = ± co-function [(2 n + V) 90° ± ^]. Computation tables contain angles less than 90° only. The chief utility of the above relations will be the reduction of functions of angles greater than 90° to functions of acute angles. Thus, to find tan 130° 20', look in the tables for cot 40° 20', or for tan 49° 40'. Why? Ex. 1. What anglea less than 360° have the same numerical cosine as 20°? cos 20° = - cos (180° ± 20°) = cos (360° - 20°). .-. 200°, 160°, 340° have the same cosine numerically as 20°. 2. Find the functions of 135° ; of 210°. sin 135° = sin (90° + 45°) = cos 45° = J V^, cos 135° = cos (180° - 45°) = - cos 45° = - i V2, etc. sin 210° = sin (180° + 30°) = - sin 30° = - J. Let the student give the other functions for each angle. INVERSE FUNCTIONS. 85 ORAL WORK. 1. Determine the sine and tangent of each of the following angles : 30°, 120°, - 30°, - 60°, J IT, 2} ir, - 135°, - ir. 2. Which is the greater, sin 30° or sin (-30°)? tan 135=» or tan 45° ? cos 60° or cos( - 60°) ? sin 22° 30' or cos 67° 30' ? 3. What positive angle has the same tangent as -? the same sine as 50°? ^ 4. If tantf = -1, findsinft 5. Find sin 510°, cos(- 60°), tan 150° 6. Reduce in two ways to functions of a positive acute angle, cos 122° tanl40°30',sin(-60°). 7. Find all positive values of x, less than 360°, satisfying the fol- lowing equations : cos x = cos 45°, sin 2 x = sin 10°, tan 3 a: = tan 60°, sin a? = sin 30°, tan x = tan 135°. 8. What angles are determined when (a) sine and cosine are + ? (6) cotangent and sine are — ? (c) sine + and cosine — ? (<f) cosine — and cotangent + ? INVERSE FUNCTIONS. 31. That a is the sine of an angle 6 may be expressed in two ways, viz., sin ^ = ^, or, inversely, = sin""^ a, the latter being read, 6 equals an angle whose sine is a, or, more briefly, is the anti-sine of a. The notation sin*^ a, cos"^ a, tan*^ a, etc., is not a fortunate one, but is so generally accepted that a change is not probable. The symbol may have been suggested from the fact that it ax = b, then x = a'^ 6, whence, by analogy, if sin 6 = a, 6 = sin"^ a. But the likeness is an analogy only, for there is no similarity in meaning. Sin~^ a is an angle 6, where sin = a, and is entirely different from (sin a)-^ = -: — . In Europe the symbols arc sin a, arc cos a, etc., are employed. 32. Principal value. We have found that in sin ^ = a, for any value of ^, a can have but one value ; but in = sin"-^ a, for any value of a there are an indefinite number of values of (Art. 27, 2). Thus, when sin ^ = a, if a = J, ^ may be 80°, 150°, 890°, 610°, - 330°, etc., or, in general, wtt +(- 1)«80°. In the solution of problems involving inverse functions. 86 PLANE TRIGONOMETKY, the numerically least of these angles, called the principal value^ is always used ; i.e. we understand that sin~i a, tan""-^ a, are angles between + 90® and — 90®, while the limits of cos-^a are 0® and 180®. Thus, sin-i^ = 30®, sin-i(- J)=:- 30®, cos-^J^GO®, cos--i(-^)=120®, ORAL WORK. How many degrees in each of the following angles? How man} radians ? 1. COS' 2 2. tan-il? 3. cot-i(-V3)? 4. 8in-i(-jV2)? 5. cos-i(-i>/2)? 7. tan-iV5? 8. cos-iQ? 9. sin-il? 10. tan-iQ? 11. tan-i(-l)? 12. sin-i(-l)? Find the values of the functions : 13. sin(tan-i jVa). 14. tan(cos"^ 1). 15. tan(cot"^[— «]). 16. cos(tan~ioo). 17. sin(am-ijV2). 18. tau(tan~^x). 3. K tf = csc-i a, prove = cos 19. cos(sin-iO). 20. 8in(cos-i[- 1]). 21. cos(cot-i>/3). 22. tan(sin-i[- 1]). 23. 8in(tan-i[-l]). Ex. 1. Construct cot~i |. Construct the right triangle xyr, so that 2 = 4, y = 3, whence angle xr = cot"^ |. 2. Find co8(tan~^ -j^). Let = tan"i -j^, whence tan 6 = ^y and cos = ^. .'. cos tf = cos(tan-i -fg) = J^. a C8C$ = a ; .'. sin tf = -> a and cos 0=^1^, = ^^^. or 6i = co8-»:i^iHI EXAMPLES- 87 EXAMPLES. 1. Construct sin*i}, tan-^^, cos~i(-- }). /i/*^ 2. Find tan(sin-lA)»si"(tan-lA)• / . 3. K tf = sin-i a, prove B = tan-^ » 4. Show that sm"^ a = 90® — cos"^ a. # 5. Prove tan-i V3 + cot-iV3 = ^. 6. Prove tan-if sin ^^ = cos-i Jv^. / 7. What angles, less than 360°, have the same tangent numerically as 10°? 8. Given tan 143° 22' = - 0.74357 ; find, correct to 0.00001, sine and cosine. 9. If cot2(90° + i8) + csc(90° - )8) - 1 = 0, find tan j3. 10. Find all positive values of a:, less than 360°, when sin a:=8in 22° 30'; when tan 2 ar = tan 60°. 11. When is sin x = " ^ ^ possible, and when impossible ? 12. Verify sin'i i + cos-i^ + tan-i>/3 = sin-i ^. 13. What values of x will satisfy sin-i(a:2 - x)- 30°? 14. If tan^ 6 - sec« a = 1, prove sec B + tan« tf esc ^ = (3 + tan« a)*. 15. Prove sin -4 (1 + tan ^ ) + cos ^ (1 + cot A) = sec il + esc A. 16. Solve the simultaneous equations : sin-i(2 a: + 3 y) = 30° and 3ar + 2y = 2. 17. Verify (a) tan 60° = ^^ (J) cos 60° = ~ cos 120° . cos 120°* 1 - tan^ 30° , 1 + tan2 30°' (c) 2 sina 60° = 1 - cos 120°. 18. Show that the cosine of the complement of ^ equals the sine of the supplement of 2- 88 PLANE TRIGONOMETRY. REVIEW. Before leaving a problem the student should review and master all principles involved. 1. Construct cos"^ ^ ; 8in-i( — j) ; tan"i 2. 2. Find cos (sin-i f ) ; tan (cos"^ [ — i] ) • 1/ 3. Prove cot"^ a = cos"^ — ^^ » 4. Given a = cot-i f , find tan a + sin (90** + a), 5. Find tan (sin-ij + cos-i^). 6. State the fundamental relations between the trigonometric func- tions in terms of the inverse functions. Thus, sin*^a = csc*^-, sin"^a = cos"^Vl — a^, etc. a 7. Find all the angles, less than 360**, whose cosine equals sin 120^ 8. Given cot'^ 2.8449, find the sine and cosine of the angle, correct tb 0.0001. 9. K tan2 (180° - tf) - sec (180'' + tf) = 5, find cos 0. 10. K sin 61 = I, find J-^B!|±co8?| Y^ ' tan^tf-cos^^ 11. Is sin 2? — 2 cos a; + 3 sin 2? — 6 = a possible equation ? 12. Verify (a) sin 60*' = ^ ^^^ ^Q° . ^ ^ ^ l + tan230° (b) 2 cos^ 60' = 1 + cos 120*». (c) cos 60° - cos 90° = 2 cos« 30° - 2 cos« 46®. 13. If sin a: = ^ ^^^ "^, — ^^— -, find sec x and tan x. 14. Prove 1 + sing - cos^ 1 + sin^-f cosg^ ^^^g^ ^ 1 + sing + cosg 1 + sing - cosg 15. Prove cos 45° + cos 135° + cos 30° + cos 150° - cos 210° + cos 270° = sin 60°. - 16. If tan = prove that sin g(l + tan 6) + cos g(l + cot 6) - sec g = ?• 17. Solve sin^x + sin^ (x + 90°) + sin^ (x + 180°) = 1. EXAMPLES. 89 18. Given cos^ a = m sin a — n, find sin a. 19. If sin2i8=— 2--r,findj3. ^ 20. Given tan 238° = 1.6, find sin 148°. 21. Prove tan-i m + cot'i m = 90^ 22. Find sin (sin"^ jo + co8~^/>) . 23. Solve cot2 ^ (2 esc ^ - 3) + 3 (esc tf - 1) = 0. 24. Prove sin^ a sec^ )8 + tan^ j3 cos^ a = sin^ a + tan^ )3. 25. Prove cos« V + sin« F=l-3sinaF+3 sin* F. 26. What values of A satisfy sin 2 il = cos 3 /I ? 27. ntanC = ^ ^-"^^ and tan Z> ==VLil22L2^ find tan D in terms -^ m 'l+cosC Ol 171. 28. If sin a: — cos ar + 4 cos^a: = 2, find tan x ; sec ar. '^^^ i j: ]^ 2 cos* ar 29. Does the value of sec a?, derived from sec* a: = — — , give a possible value of a:? "" 30. Prove [cot (90° - ^ ) - tan (90'' + il )] [sin (180° - il) sin (90° + ^)] = 1. 31. Prove (1 + sin -4)* [cot il + 2 sec il (1 — esc -4)] + esc A cos« ^ = 0. 32. Given sin a: = m sin y, and tan x = n tan y, find cos x and cos y. 33. Given cot201° = 2.6, find cos 111°. 34. Find the value of * cos-i J + sin-i J V2 + csc-i( - 1) + tan'i 1-2 cot-^ V3. ^35. Solve 2cos*^ + llsintf -7 = 0. 36. Prove cos* B + cos* (J5 + 90°) + cos* (J5 + 180?) + cos* (J5. + 270°) = 2. CHAPTER IV. COMPUTATION TABLES. 33. Natural functions. It has been noted that the trigo- nometric functions of angles are numherB^ but the values were found for only a few angles, viz. 0°, 30°, 45*^, 60°, 90°, etc. In computations, however, it is necessary to know the values of the functions of any angle, and tables have been prepared giving the numerical values of the functions of all angles between 0° and 90° to every minute. In these tables the functions of any given angle, and con- versely the angle corresponding to any given function, can be found to any required degree of accuracy ; e,g. by look- ing in the tables we find sin 24° 26'= 0.41363, and also 1 .6415 = tan 58° 39'. These numbers are called the natural functions^ as distinguished from their logarithms, which are called the logarithmic functionB of the angles. Ex. 1. Find from the tables of natural functions : sin 35° 14'; cos 54" 46'; tan 78° 29'; cos 112° 58'; sin 135°. 2. Find the angles less than 180° corresponding to : sin-i 0.37865 ; cos'i 0.37865 ; tan-i 0.58870 ; cos'i 0.00291 ; sin-i 0.99999. 34. Logarithms. The arithmetical processes of multi- plication, division, involution, and evolution, are greatly abridged by the use of tables of logarithms of numbers and of the trigonometric ratios, which are numbers. The principles involved are illustrated in the following table : Write in parallel columns a geometrical progression having 'the ratio 2, and an arithmetical progression having the dif- ference 1, as follows : 40 LOGARITHMS. 41 It will be perceived that the numbers in the second column are the indices of the powers of 2 producing the corresponding numbers in the first column, thus : 2^ = 64, 2" = 2048, 218 = 262144, etc. The use of such a table will be illustrated by examples. Ex. 1. Multiply 8192 by 128. From the table, 8192 = 2i«, 128 = 2^ Then by actual multiplication, 8192 x 128 = 1048576, or by the law of indices, 2i» x 2^ = 2» = 1048576 (from table). Notice that the simple operation of addition is sub- stituted for multiplication by adding the numbers in the second column opposite the given factors in the first column. This sum corresponds to the number in the first column which is the required product. 2. Divide 16384 by 512. 16384 -T- 512 = 32, which corresponds to the result obtained by use of the table, or 2" -j- 2« = 2« = 32. The operation of subtraction takes the place of division. 3. Find v^262144. v^262l44 = v^ = 2^ = 2« = 8. In the table, 262144 is opposite 18. 18 -r- 6 = 3, which is opposite 8, the required root ; i.e, simple division takes the place of the tedious process of evolution. ' 6. Find \^^2768. G. p. A. p. 1 2 1 4 2 8 3 16 4 32 5 64 6 128 7 256 8 512 9 1024 10 2048 11 4096 12 8192 13 16384 14. 32768 15 65536 16 131072 17 262144 18 524288 19 1048576 20 4. Cube 64. 5. Multiply 256 by 4096. 7. Divide 1048576 by 32768. 35. The above table can be made as complete as desired by continually inserting between successive numbers in the first column the geometrical mean, and between the opposite numbers in the second, the arithmetical mean, but in prac- tice logarithms are computed by other methods. The num- bers in the second column are called the logarithms of the numbers opposite in the first column. 2 is called the base of this system, so that the logarithm of a number is the exponent by which the base is affected to produce the number. 42 PLANE TRIGONOMETRY. Thus, the logarithm of 612 to the base 2 is 9, since 29 = 512. Logarithms were invented by a Scotchman, John Napier, early in the seventeenth century, but his method of constructing tables was different from the above. See Encyc, Brit., art. " Logarithms,^^ for an exceedingly interesting account. De Morgan says that by the aid of logarithms the labor of computing has been reduced for the mathematician to about one-tenth part of the previous expense of time and labor, while Laplace has said that John Napier, by the invention of logarithms, lengthened the life of the astronomer by one-half. Columns similar to those above might be formed with any other number as base. For practical purposes, however, 10 is always taken as the base of the system, called the common system, in distinction from the natural system, of which the base is 2.71828 •••, the value of the exponential series (^Higher Algebra^. The natural system is used in theoretical discus- sions. It follows that common logarithms are indices, positive or negative, of the powers of 10. Thus, 108 = 1000 ; i.e, log 1000 = 3 ; 10-2 = ^=0.01; i.g. log0.01 = -2. 36. Characteristic and mantissa. Clearly most numbers are not integral powers of 10. Thus 300 is more than the second and less than the third power of 10, so that log 300 = 2 plus a decimal. Evidently the logarithms of numbers generally consist of an integral and a decimal part, called respectively the charac- teristic and the marvtissa of the logarithms. 37. Characteristic law. The characteristic of the loga- rithm of a number is independent of the digits composing the number, but depends on the position of the decimal point, and is found by counting the number of places the first significant figure in the number is removed from the units' place, being positive or negative according as the first significant LOGARITHMS. 48 figure is at the left or the right of units^ place. This follows from the fact that common logarithms are indices of powers of 10, and that 10**, n being a positive integer, contains n-\-l places, while lO'** contains w — 1 zeros at the right of units' place. Thus in 146.043 the first significant figure is two places at the left of units' place ; the characteristic of log 146.048 is therefore 2. In 0.00379 the first significant digit is three places at the right of units' place, and the charac- teristic of log 0.00379 is - 3. To avoid the use of negative characteristics, such charac- teristics are increased by 10, and — 10 is written after the logarithm. Thus, instead of log 0.00811 = 3.90902, write 7.90902 — 10. In practice the — 10 is generally not written, but it must always be remembered and accounted for in the result. £x. Determine the characteristic of the logarithm of : 1; 46; 0.009; 14796.4; 280.001; 10« x 76; 0.525; 1.03; 0.000426. 38. Mantissa law. The mantissa of the logarithm of a number is independent of the position of the decimal point, but depends on the digits composing the number, is always positive^ and is found in the tables. For, moving the decimal point multiplies or divides a number by an integral power of 10, i.e. adds to or subtracts from the logarithm an integer, and hence does not affect the mantissa. Thus, log 225.67 = log 225.67, log 2256.7 = log 225. 67 X 101 = log 225.67 + 1, log 22567.0 = log 225.67 x 102 = log 225.67 + 2, log 22.567 = log 225.67 x lO-i = log 225.67 + (- 1), log 0.22567 = log 225.67 x 10-8 = log 225.67 4- (- 3), so that the mantissse of the logarithms of all numbers com- posed of the digits 22567 in that order are the same, .35347. Moving the decimal point affects the characteristic only. The student must remember that the mantissa is always positive. 44 PLANE TRIGONOMETRY. Log 0.0022567 is never written -3 +.35347, but 3.35347, the minus 8ign being written above to indicate that the characteristic alone is nega- tive. In computations negative characteristics are avoided by adding and subtracting 10, as has been explained. 39. We may now define the logarithm of a number as the index of the power to which a fixed number^ called the base^ mu8t be raised to produce the given number. Thus, a' = 6, and x = logjb (where \ogJb is read logarithm of b to the base a) are equivalent expressions. The relation between base, logarithm, and number is always (base)^** = number. To illustrate: log28=3 is the same as 2^=8; logg81=4and 3*= 81 are equivalent expressions ; and so are logiQlOOO = 3 and 103 _ 1000, and logio0.001= -3 and 10-8= q.OOI. Find the value of : log^ei; log5l25; log8243; log«(a)*; log27 3; log^l. 40. From the definition it follows that the laws of indices apply to logarithms, and we have : I. The logarithm of a product equals the sum of the loga- rithms of the factors, II. The logarithm of a quotient equals the logarithm of the dividend minus the logarithm of the divisor, III. The logarithm of a power equals the index of the power times the logarithm of the number. IV. The logarithm of a root equals the logarithm of the number divided by the index of the root. For if a* = w and a^ = w, then w X m = a^"'"'', .•. log nm = x + y = log n + log m; and n-*-w = a*"*', .'.log— =a: — y = logw — logm; m also n^= Qa^y= a***, .*. log n*" =rx = r x log n ; finally, Vn = Vo* = a**, .'. logVw = ? = - log n. r r LOGARITHMS. 45 EXAMPLES. Given log 2 = 0.30103, log 3 = 0.47712, log 5 = 0.69897, find : 1. log 4. 4. log 9. 7. logl5». 10. logVjf. 2. log 6. 5. log 25. 8. logf. f 3. log 10. 6. log>/3. 9. log 15x9. ^- ^^g^/; 92 X 5« 2*xl0 USE OF TABLES. 41. To find the logarithm of a number. First, Find the characteristic, as in Art. 37. Second. Find the mantissa in the tables, thus : (a) When the number consists of not more than four figures. In the column N" of the tables find the first three figures, and in the row iV the fourth figure of the number. The mantissa of the logarithm will be found in the row opposite the first three figures and in the column of the fourth figure. Illustration. Find log 42.38. The characteristic is 1. (Why ?) In the table in column N" find the figures 423, and on the same page in row iV the figure 8. The last three figures of the mantissa, 716, lie at the interseetion of column 8 and row 423. To make the tables more compact the first two figures of the mantissa, 62, are printed in column only. Then log42.38 = 1.62716. Find log 0.8734 = 1.94121, log 3.5 = log 3.500 = 0.54407,- log 36350 =4.56050. (6) When the number consists of more than four figures. Find the mantissa of the logarithm of the number com- posed of the first four figures as above. To correct for the remaining figures we interpolate by means of the principle of proportional parts, according to which it is assumed that^ for differences small as compared with the numbers, the differences 46 PLANE TRIGONOMETRY. between several numbers are proportional to the differences be^ tween their logarithms. The theorem is only approximately correct, but its use leads to results accurate enough for ordinary computations. Ex. 1. To find log 89.4562. As above, mantissa of log 894500 = 0.95158, mantissa of log 894600 = 0.95163, .-. log 894600 - log 894500 = O.OOOOb^called the tabular difference. Let log 894562 - log 894500 = x hundred-thousandths. Now, by the principle of proportional parts, log 894562 - log 894500 ^ 894562 - 894500 log 894600 - log 894500 894600 - 894500' or - = :^, whence ar = .62 of 5 = 3.1 5 100 .-. log 89.4582 = 1.95158 + 0.00003 = 1.95161, all figures after the fifth place being rejected in five-place tables. If, however, the sixth place be 5 or more, it is the practice to add 1 to the figure in the fifth place. Thus, H x = 0.0000456, we should call it 0.00005, and add 5 to the mantissa. 2. Find log 537.0643. To interpolate we have x : 9 = 643 : 1000, i.e. x = 5.787; .-. log 537.0643 = 2.72997 + 0.00006. 3. Find log 0.0168342 = 2.22619. 4. Find log 39842.7 = 4.59816. 42. To find the number corresponding to a given logarithm. The characteristic of the logarithm determines the posi- tion of the decimal point (Art. 37). (a) If the mantissa is in the tables, the required number is found at once. Ex. 1. Find log"^ 1.94621 (read, the number whose logarithm is 1.94621). The mantissa is found in the tables at the intersection of row 883 and column 5. .-. log-n .94621 = 88.35, the characteristic 1 showing that there are two integral places. LOGARITHMS. 47 (h) If the exact mantissa of the given logarithm is not in the tables, the first four figures of the corresponding num- ber are found, and to these are annexed figures found by interpolating by means of the principle of proportional parts, as follows : Find the two successive mantissae between which the given mantissa lies. Then, by the principle of proportional parts, the amount to be added^ to the four figures already found is such a part of 1 as the difference between the successive mantissse is of the difference between the smaller of them and the given mantissa. 2. Find log-U.43764. Mantissa of log 2740 = 0.43775 of log2739 = 0.43759 Differences 1 16 Mantissa of log required number = 0.43764 of log2739 = 0.43759 Differences x 5 By p. p. a? : 1 = 5 : 16 and x = ^ = 0.3125. Annexing these figures, log-* 1.43764 = 27.393+. 3. Find log-i T.48762. The differences in logarithms are 14, 6. .'. x=-^ = .428+, 14 and log-i 1.48762 = 0.30744+. 4. Find log 891.59; log 0.023; logi; log 0.1867; log V2. 5. Find log-i 2.21042 ; log'i 0.55115 ; log-i 1.89003. 43. Logarithms of trigonometric functions. These might be found by first taking from the tables the natural func- tions of the given angle, and then the logarithms of these numbers. It is more expeditious, however, to use tables showing directly the logarithms of the functions of angles less than 90° to every minute. Functions of angles greater than 90° are reduced to functions of angles less than 90° by <./?' S^-^S'^ 48 PLANE TRIGONOMETRY. the formulsB of Art. 29. To make the work correct for seconds, or any fractional part of a minute, interpolation is necessary by the principle of proportional parts, thus : Ex. 1. Find log sin 28° 32' 21". In the table of logavithms of trigonometric functions, find 28° at the top of the page, and in the minute column at the left find 32'. Then -under log sin column find log sin 28° 32' = 9.67913 - 10 log sin 28° 33 = 9.67936 - 10 Differences 1' 23 By p. p. a: : 23 = 21" : 60", t.e. x = — x 23 = 8.06. •^ ^ '^ 60 .-. log sin 28° 32' 21" = 9.67913 + 0.00008 - 10 = 9.67921 - 10. Whenever functions of angles are less than unity, i,e, are decimals f(as sine and cosine always are, except when equal to unity, and as tan- :gent is for angles less than 45°), the characteristic of the logarithm will ibe negative, and, accordingly, 10 is always added in the tables, and it must be remembered that J.0 is to be subtracted. Thus, in the example above, the characteristic of the logarithm is not 9, but I, and the log- ;arithm is not 9.67913, as written in the tables, but 9.67913 — 10. 2. Find log cos 67° 27' 50''. In the table of logarithms at the foot of the page, find 67°, and in the iminute column at the right, 27'. Then computing the difference as ;above, x = 25. But it must be noted that cosine decreases as the angle increases itoward 90°. Hence, log cos 67° 27' 50" is less than log cos 67° 27', t.c. ithe difference 25 must be subtracted, so that log cos 67° 27' 50" = 9.58375 - 0.00025 - 10 = 9.58350 - 10. 44. To find the angle when the logarithm is given, find the .^successive logarithms between which the given logarithm ilies, compute by the principle of proportional parts the seconds, and add them to the less of the two angles corre- sponding to the successive logarithms. This will not neces- sarily be the angle corresponding to the less of the two logarithms ; for, as has been seen, the number, and, therefore, the logarithm, may decrease as the angle increases. LOGARITHMS. 49 Ex. 1. Find the angle whose log tan is 9.88091. log tan 37° 14' = 9.88079 - 10 log tan 37*» 15' = 9.88105 - 10 Differences 60" 26 log tan 37° 14' = 9.88079 - 10 ■ log tan angle required = 9.88091 — 10 Differences a:" 12 .-. X : 60 = 12 : 26, or a:" = ij x 60" = 28", approximately, and the angle is 37° 14' 28". 2. Find the angle whose log cos = 9.82348. We find a: = A X 60" = 26", and the angle is 48° 14' 26". 3. Show that log cos 25° 31' 20" = 9.95541 ; log sin 110° 25' 20" = 9.97181 ; log tan 49° 52' 10" = 0.07417. 4. Show that the angle whose log tan is 9.92501 is 40° 4' 39" ; whose log sin is 9.88365 is 49° 54' 18" ; whose log cos is 9.50828 is 71° 11' 49". 45. Cologarithms. In examples involving multiplications and divisions it is more convenient, if n is any divisor, to add log - than to subtract log n. The logarithm of - is n n called the cologarithm of n. Since log - = log 1 — log n = — log w, it follows that colog w = — log «, i.e. logw subtracted from zero. To avoid negative results, add and subtract 10. Ex. 1. Find colog 2963. log 1 = 10.00000 - 10 log 2963= 3.47173 .-. colog 2963 = 6.52827 - 10 2. Find colog tan 16° 17'. log 1 = 10.00000 - 10 log tan 16° 17' = 9.46554 - 10 .-. colog tan 16° 17' = 0.53446 50 PLANE TRIGONOMETRY. By means of the definitions of the trigonometric functions, the parts of a right triangle may be computed if any two parts, one of them being a side^ are given. Thus, ■B given a and A in the rt. triangle ABC. Then c =^ a -t- sin A, b = a -t- tan Ay and B = 90°-A, Again, if a and b are given, then tan A = -, c = a -h sin ^, and B = 90° — -4- b 3. Given c = 25.643, B = 37** 25' 20", compute the other parts. ^ = 90° - 37° 25' 20" = 52° 34' 40". a = c cos B, & = a tan B. log c = 1.40897 log a = 1.30889 log cos B = 9.89992 log tan J5 = 9.88376 log a = 1.30889 log b = 1.19265 .-. a = 20.365. .-. b = 15.583. Check: c^ = a^ -^ b^ = 20.3652 + 15.5832 = 657.57 = 25.6432. 4. Given b = 0.356, B = 63° 28' 40", compute the other parts. A = 26° 31' 20". a = sin B tan B log b = 9.55145 log b = 9.55145 colog sin B = 0.04829 colog tan B = 9.69816 log c = 9.59974 log a = 9.24961 c = 0.3979 a = 0.1777 Check: c^ - a^ = 0.1583 - 0.03157 = 0.12673 = bK EXAMPLES. Compute the other parts : ^ 1. Given a = 9.325, A = 43° 22' 35". .. ' ^2. Given c = 240.32, a = 174.6. -^ • >- 3. Given jB = 76° 14' 23", a = 147.53. y 4. Given a = 2789.42, b = 4632.19. '^ 5. Given c = 0.0213, il = 23° 14". S. Given 6 = 2, c = 3. CHAPTER V. APPLICATIONS. 46. Many problems in measurements of heights and dis- tances may be solved by applying the preceding principles. By means of instruments certain distances and angles may be measured, and from the data thus determined other distances and angles compnted. The most common instru- ments are the chain, the transit, and the compasi. The chain is used to measure distances. Two kinds are in use, the engineer's chain and the Q-unter*! chain. They each contain 100 links, each link in the engineer's chain being 12 inches long, and in the Gunter's 7.92 inches. The trarmt is the instrument moat used to measure hori- zontal angles, and with certain attachments to measure verti- cal angles. The figure shows the form of the instrument. 62 PLANE TRIGONOMETRY. The mariner''» compass is used to determine the directions, or bearings, of objects at sea. Each quadrant is divided into 8 parts, making the 32 points of the compass, so that each point contains 11° 15'. ^yx^'^ip> |>8 ^ s FiQ. 27. Fig. 28. 47. The angle between the horizontal plane and the line of vision from the eye to the object is called the angle of elevation^ or of depression^ according as the object is above or below the EievaUoiry"-.^ obscrvcr. It is evident that the elevation angle of B^ as seen from A^ is equal to the depression angle of A^ as seen from jB, so that in the solution of examples the two angles are interchangeable. PROBLEMS. 48. Some of the more common problems met with in practice are illustrated by the following: To find the height of an object when the foot is accessible. The distance jB(7, and the eleva- tion angle B are measured, and x is determined from the relation b X=BOt^nB. Fig. 29. / APPLICATIONS. 53 Ex. 1. The elevation angle of a cliff measured from a point 300 ft. from its base is found to be 30®. How high is the cliff? Then BC = 300, B = 30°. a: = 300 . tan 30° = 300 • i V3 = 100 V3. 2. From a point 175 ft. from the foot of a tree the elevation of tho top is found to be 27° 19'. Find the height of the tree. The problem may be solved by the use of natural functions, or of logarithms. The work should be arranged for the solution before the tables are opened. Let the student complete. 5C = 175. jB = 27°19'. Then x = BC tan B, Or by natural functions, logjBC= BC = 175 log tan jB = tan iB = 0.5165 log a: = .-. X = 90.39. To find the height of an object when the foot is inaccessible. Measure BB\ and ^'. .-. X = 90.3875. Then x = BO BB'-^-B'O cot cot But B^O = x cot ^', whence substituting, ^^ BB^ cot - cot 0'' which is best solved by the use of the natural functions of and 0'. 3. Measured from a certain point at its base the elevation of the peak of a mountain is 60°. At a distance of one mile directly from this point the elevation is 3Q°. Find the height of the mountain. BB' = 5280 ft., e = 30°, & = 60°. X = IL±^?§5. But y = arcot 60°. cot 30° ^ » m X — • 5280 cot 30° - cot 60° = 4572.48 ft. 54 PLANE TRIGONOMETRY. In surveying it is often necessary to make measurements across a stream or other obstacle, too wide to be spanned by a single chain. To find the distance from O to a point B on the opposite side of a stream. At C measure a right angle, and take CA a convenient distance. Measure angle A^ then FiQ. 31. ~ 5(7= (7^. tan A. 4. Find CB when angle A = 47° 16', and CA = 250 ft. 5. From a point due south of a kite its elevation is found to be 42° 30'; from a point 20 yds. due west from this point the elevation is 36° 24'. How high is the kite above the ground ? ^J5 = x.cot42°30', ^C = ar.cot36°24', AC^-AB^ = BC^ = 400. ^ .-. x^ (cota 36° 24' - cot^ 42° 30') = 400, whence EXAMPLES. 4-^\ Fia. 32. 1. What is the altitude of the sun when a tree 71.5 ft. high casts a shadow 37.75 ft. long ? 2. What is the height of a balloon directly over Ann Arbor when its elevation at Ypsilanti, 8 miles away, is 10° 15' ? 3. The Washington monument is 555 ft. high. How far apart are two observers who, from points due east, see the top of the monument at elevations of 23° 20' and 47° 30', respectively? 4. A mountain peak is observed from the base and top of a tower 200 ft. high. The elevation angles being 25° 30' and 23° 15', respec- tively, compute the height of the mountain above the base of the tower. 5. From a point in the street between two buildings the elevation angles of the tops of the buildings are 30° and 60°. On moving across APPLICATIONS. 55 / the street 20 ft. toward the first building the elevation angles are found to be each 45^ Find the width of the street and the height of each building. 6. From the peak of a mountain two towns ar6 observed due south. The first is seen at a depression of 48® 40', and the second, 8 miles farther away and in the same horizontal plane, at a depression of 20° 50'. What is the height of the mountain above the plane ? 7. A building 145 ft. long is obseiTed from a point directly in front of one corner. The length of the building subtends .tan-i 3, and the height tan-i 2. Find the height. 8. An inaccessible object is observed to lie due N.E. After the ob- server has moved 8.£. 2 miles, the object lies N.N.E. Find the distance of the object from each point of observation. 9. Assuming the earth to be a sphere with a radius of 3963 miles, find the height of a lighthouse just visible from a point 15 miles distant at sea. 10. The angle of elevation of a tower 120 ft. high due north of an observer was 35° ; what will be its angle of elevation from a point due west from the first point of observation 250 ft. ? Also the distance of the observer from the base of the tower in each position ? 11. A railway 5 miles long has a uniform grade of 2° 30' ; find the rise per mile. What is the grade when the road rises 70 ft. in one mile ? (The grade depends on the tangent of the angle.) 12. The foot of a ladder is in the street at a point 30 ft. from the line of a building, and just reaches a window 22J ft. above the ground. By turning the ladder over it just reaches a window 36 ft. above the ground on the other side of the street. Find the breadth of the street. 13. From a point 200 ft. from the base of the Forefathers* monument at Plymouth, the base and summit of the statue of Faith are at an eleva- tion of 12° 40' 48" and 22° 2' 53", respectively; find the height of the statue and of the pedestal on which it stands. 14. At a distance of 100 ft. measured in a horizontal plane from the foot of a tower, a flagstaff standing on the top of the tower subtends an angle of 8°, while the tower subtends an angle of 42° 20'. Find the length of the flagstaff. 15. The length of a string attached to a kite is 300 ft. The kite's elevation is 56° 6'. Find the height of the kite. 16. From two rocks at sea level, 50 ft. apart, the top of a cliff is ob- served in the same vei'tical plane with the rocks. The angles of eleva- tion of the cliff from the two rocks are 24° 40' and 32° 30'. What is the height of the cliff above the sea ? CHAPTER VI. OXSNERAL FORMUIt^l— TRiaONOMBTRIC EQUATIONS AND IDENTITIES. 49. Thus far functions of single angles only have been considered. Relations will now be developed to express functions of angles which are sums, differences, multiples, or sub-multiples of single angles in terms of the functions of the single angles from which they are formed. First it will be shown that, sin (a±p) = 8lnaco8p±co8asinp, cos (a ± p)= cosacosp T sinasinp tan (a ± p) = *!_. 1 q=tanatanp The following cases must be considered : 1. a, /8, a + /8 acute angles. 2. a, )8, acute, but a + 13 an obtuse angle. 3. Either a, or )8, or both, of any magnitude, positive or negative. The figures apply to cases 1 and 2. P) •Q Let the terminal line revolve through the angle a, and then through the angle /3, to the position OB^ so that angle 56 X GENERAL FORMULA. 67 XOB=: oi + 13. Through any point P in OB draw perpen- diculars to the sides of a, DP and OP, and through O draw a perpendicular and a parallel to OX, MO and IfO. Then the angle QOA = a (why?), and OJ^P is the triangle of reference for angle QOP = 90^ + a. CNP is sometimes treated as the triangle of reference for angle CPN. The fallacy of this appears when we develop cos (a + j8), in which PC would be treated as both plus and minus. Now sin(« + /3)=sinX05 = ^ = ^+^ or expressing in trigonometric ratios, ^MO 00^, NP OP 00' OP op' op = sin a cos )8 + sin (90** + a) sin /8. Hence, since sin (90*^ + a) = cos a, we have sin (a + p^=z sin a cos/8 + cos a sin /8-__^ """ In like manner cos(a + ^) = cosXOB = — = — + -^ or expressing in trigonometric ratios, ^OM 00 ON OP 00 ' op'^ op' op = COS a COS 13 + cos (90** + a) sin /3. And since cos(90*' + «)= — sin a, we have cos (^u + I3^=s cosa cos/8 — sin a sin/8. It will be noted that the wording of the demonstration ap- plies to both figures, the only difference being that when a+jS- is obtuse OD is negative. ON is negative in each figure. 50. In the case, when a, or /8, or both, are of any magni- tude, positive or negative, figures may be constructed as- before described by drawing through any point in the terminat line of 13 a perpendicular to each side of a, and through the foot of the perpendicular on the terminal line of a a perpendicular and a parallel to the initial line of a. Noting negative lines. 58 PLANE TRIGONOMETRY. the demonstrations already given will be found to apply for all values of a and /3. To make the proof complete by this method would require an unlim- ited number of figures, e.g. we might take a obtuse, both a and p obtuse, either or both greater than 180% or than 360% or negative angles, etc. Instead of this, however, the generality of the proposition is more readily shown algebraically, as follows : Let a' = 90^ + « be any obtuse angle, and a, /3, acute angles. Then sin («' + /3) = sin (90** + a + )8) = cos (a + /8) = cosacos/3— sinasin/3 = sin (90^ + a) cos /3 + cos (90° + a) sm /8(why?) = sin a' cos 13 + cos a' sin fi. In like manner, considering any obtuse angle /8' = 90*^ + /8, it can be shown that sin (a' + /8') = sin a' cos /8' + cos a' sin^S'. Show that cos (a' + /8') = cos a' cos /8' — sin a' sin /8'. By further substitutions, e.ff. a" = 90*^ ± «', /8" = 90° ± /8', etc., it is clear that the above relations hold for all values, positive or negative, of the angles a and /3. Since a and 13 may have any values, we may put — /8 for /8, and sin (a + [ — /8]) s= sin (a — /8) = sin a cos(— /8) + cosasin ( — /8) 5= sin a cos )8 — cos a sin /8 ( why ?) . Also cos (a — /3)= cos a cos(— /8) — sin a sin (— /8) = cos a cos /8 + sin a sin /3. Finally, tan C ±S^^ sin (« ± /8) _ sin « cos /3 ± cosce sin /3 cos (a ± /8) cos a cos /8 :f sin a sin )8 sin a cos /3 , cos a sin /3 COS a cos iS cos a cos iS tan (x ± tan B cos « cos 13 sin a sin /8 l^tanatan/S cos tt cos 13 cos a cos )3 EXAMPLES. 59 ORAL WORK. By the above f onnulse develop : 1. sin(2A+SB). 7. sin 90« = sin (45<» + 45*>). 2. cos (90° -5). a cos 90°. 3. tan (45° + ^). 9. tan 90°. 4. sin2 ^ = sin (A + A). 10. sin(90° + jS + y). 5. cos 2 ft 11. cos (270° - m - n). 6. tan (180° + C). 12. tan (90° + m + n). Ex. 1. Find sin 76°. sin 75° = sin (45° + 30°) = sin 45° cos 30° + cos 45° sin 30* = -L.v2^.J_.UL±^=0.9659. V2 2 v^ 2 2V5 2. Find tan 15°. tan 45° - tan 30° 1--L = ^ =:5^i^= 2 --v/3 = 0.2679. 1 + J_ V3 + 1 V3 3. Prove 2ii^-S2ilii=2. Sin A cos A Combining si^^^ cos^— co8 3^ sin^ _ sin(3^ — vi) sin A cos A sin A cos A s= 8"^2i4 _ sin (A + il) __ sin ^ cos i4 + cos ^4 sin A _ g sin ^ cos A sin ^1 cos^ sin A cos ^1 4. Prove tan-* a + tan-^ b = tan-^ ° "^ » 1 — ao Let a = tan-^o, fi = tan-^J, y = tan-^ Z' "*" * 1 — ab Hence, tan a^^a, tan B = bj tan y = "^ * 1—00 Then a + /3 = y, and hence tan (a + /3) = tan y. Expanding, tan « + tan ^ ^ ^^^ ^ ^ l-tanatanj8 ' Substituting, ^L±\ = :^±A. 1 — ao 1 — ao 60 PLANE TRIGONOMETRY. r6 EXAMPLES. 1. Find COB 15% tan 75**. 2. Prove cot (a i /8) = cotacot^Tl . ^ '^^ cot /8i cot a 3. Prove geometrically sin (a + j3) = sin a cos /3 + cos a sin j9, and cos (a + j3) = cos a cos /3 — sin a sin ^ given (a) a acute, /3 obtuse ; (6) a, j3, obtuse ; (c) a, /3, either, or both, negative angles. 4. Prove geometrically tan (a + iS) = ^^^ "*" ^^ ^^ ' 1 — tan OL tan p Verify the formula by assigning values to a and j3, and finding the values of the functions from the tables of natural tangents. 5. Prove cos (a + /8) cos (a — /3) = cos' a — sin'jS. 6. Show that tan a + tan j8 == 5^5J[«±^ cos a cos p 7. Given tan a = J, tan /8 = f, find sin (a + /8) a Given sin 280° = «, find sin 170^ 9. If a = 67° 22', /8 = 128° 40', by use of the tables of natural func- tions verify the formulae on page 56. 10. Prove tan-i^5^^JtJ^ = tan- V« + tan-i>/a. 1 — Vai U. Prove tan-i?^^ + tan-i2Ajl£^^^j^.i^ 12. Prove sec-* — ^ = sin-^ ^ Va« - a:' a 13. If a + /3 = 0), prove cos'oe + cos'jS — 2 cos a cos j9 cos » = sin*c0. 14. Solve i sin = 1 -. cos 0. y^ 15. Prove sin {A + -B) cos A — cos (-4 + 5) sin il = sin B, 16. Prove cos(il +B)cos(i4--B) + sin(il +-B)sin(-4--B) = cos2A 17. Prove sin (2 a - /8) cos (a - 2 j8) - cos (2a - )8)8in(a - 2/8) = sin(a + /8). 1& Prove sin(n'— l)acos(n+l)a+cos(n-l)asin(n+l)a = 8in2na. ~. 19. Prove sin(135° - tf) + cos(135° + tf ) = 0. ADDITION— SUBTRACTION FORMULA. 61 20. Provel-tan«atan«i8 = 528!fzJI^ 21. Prove ^"^ + ^^" ^ = tan « tan ^. cot a + cot /3 22. tan« (? - «U 1^4^^««2i«. \4 / 1 + 2 sm a cos a 51. The following formiilse are very important and should be carefully memorized. They enable us to change sums and differences to products, i.e, to displace terms by factors. sin e + sin ♦ = 2 sin^i^cos^^ sine - 8iii+ = 2 CM^±^9ln^, cose + COS+ = 2C08-^^€08-^^> cose - C08+ = - 2 sin -i-*sin -=^« Since sin (« + /8) = sin a cos jS + cos a sin )8, and sin (« — /8) = sin a cos /8 — cos a sin )8, then sin (^a + fi^-^- sin (a — )8) = 2 sin a cos )8, (1) and sin (a + /8) — sin (a — )8) = 2 cos a sin /8. (2) Also since cos (« + /8) = cos a cos /8 — sin « sin /8, and cos (a "13^ = cos a cos fi + sina sin )8, then cos C^ + 13^ + cos (a — /8) = 2 cos a cos )8, (3) and cos (« + /8) — cos (a — /8) = — 2 sin « sin /8. (4) Put a + l3 = and a-— ^ = <f> 2a = ^ + 0, anda = ^i:-^, 2/8 = ^-^, and /3 = ^^. Substituting in (1), (2), (3), (4), we have the above formulae. 62 PLAICE TRIGONOMETRY. EXAMPLES. 1. Prove "»^^ + ^^"; = tan^. CO82 + COS0 2 By formuke of last article the first member becomes 2 sm -e cos ^ 2 2 ^ se 2cos--ecos^ * 2 2 2 Prove sin « + 2 sin 3 « + sin 5 a __ sin3« sin 3 a + 2 sin 5 a + sin 7 a sin 5 a (sin tt + sin 5 tt) + 2 sin 3 cc __ 2 sin 3 « cos 2 « + 2 sin 8 « (sin 8 a -h sin 7 a) + 2 sin 5 a 2 sin 5 a cos 2 a + 2 sin 5 a _ (cos 2 « + 1) sin 8 « _ sin 8 a (cos 2 a + 1) sin 5 ce sin 5 a cos(4i4 -25)+ 008(45-2-4) \ ^ ^ o»:^ 4^ -25 + 45-2^ ^^, 4^ -25 -45 + 2.4 If sm cos L__ ^ 2 2 a^^.4il-25 + 45-2^_, 4^-25-45 + 2"T SJ cos rr cos ^ 2 2 ;=eHLii^ = tan(^+5). cos(^+5) ^ ^ 4. Prove sin 50° - sin 70^ + sin 10° = 0. 2 cos ^° ^ "^^^ sin ^''"•'^^ = 2 cos 60° sin (- 10°) = - sin 10° 5. Prove '^'^«°*»^«-°°^^''°"»^'*+'^«'^^"«=cot6«cot5tt. sin 4 a sin 3 a— sin 2 a sin 5 a+sin4a sin 7 a By (3) and (4), p. 61, cos 5 « + cos « — cos 9 ft — cos 5 « + cos 11 « + cos 9 a cos a — cos 7 a — cos 8 a + cos 7 a + cos 8 a — cos 11 a ^costt + cosll«^ 2co86«cos5« ^^^g^^^g^ cos a — cos 11a 2 sin 6 a sin 5 a ORAL WORK. By the formulae of Art. 51 transform : 6. cos 5 a + cos a. 8. 2 sin 3 cos 0. 7. cos a — cos 5 a. 9. sin 2 a — sin 4 a. FUNCTIONS OF THE DOUBLE ANGLE. 68 10. cos 9 cos 2 0. 16. cos (30°+ 2<^) sin (30°- <^). 11. sintf + sin^. 17. sin(2r + ») + sm (2r-»). 12. sin 75° sin 15°. 18. cos (2 /S ^ a) - cos 3 a. 13. cos7i>--cos2;). >c ». sin 36° + sin 54°. 14. co8(2p + 8,)8in(2p-35). ^ ««, ««» + cos 20-. 15. sin^-sin^. ^ 8m30» + co8m Prove: 22. !JE«±«]l| = tan«±^cot^. Sin a — sm /3 2 2 23 cos « + cos /3 ^ cotgL+^cot^M^ cos j8 — cos a 2 2 ^^ COS a: + cos y 2 ^^ COS X — cos y 2 .^ 26. cos 55° + sin 25° = sin 85°. Simplify: 27. 8ini? + sin2^ + 8in3^ ^ "^ C08 5+C08 2jB +C08 3iJ 23 sin C- sin 4 C + sin 7 C - sin 10 C cos C — COS 4 C + cos 7 C — cos 10 C 52. Functions of an angle in terms of those of tKe half angle. If in sin (« + /8) = sin a cos /3 + cos a sin /3^ a = fi^ then sin (a + a) = sin 2 a = 2 sin acos a*. In like manner cos (a + a) = COS 2 a = cos* a - sin^a = 2cos«a-l = l-2sin*a; 2 tan a and tan 2 a 1-tanSa 64 PLANE TRIGONOMETRY. ORAL WORK. Ex. Express Ib terms of functions of half the given angles : 1. sin 4 m. 4. cos x, 6. sin (2p — q), 2. co83/>. . . iS ^« cos (30° + 2 «^). 3. tan 5 1. ^ 8. sin {x + y), 9. From thefimctious of 30° find those of 60°; from the functions of 45°, those of 90°. 53i Functions of an angle in terms of those of twice the angle. By Art. 62, co9«=:l -2sin2?= 2cos2^- 1. /. 2 sin*^ = 1 — cos a, and 2 cos^^ = 1 + cos «• . a 8in;r .•. taii- = =±\r — cosa 2 ^^„a ^l + cosa cos- Explain the significance of the ± sign before the. radicals. Express in terms of the double angle the functions of 120®; SO*'; 90*^, with proper signs prefixed. Ex. 1. Express in terms of functions of twice the given angles each of the functions in Examples 1-8 above. • 2. From the f auctions of 45° find those of 22° 30^ ; from the functions of 36°, those of 18° (see tables of natural functions). 3. Find the corresponding functions of twice and of half each of the following angles, and verify results by the tables of natural functions : Given sin 26° 42' = 0.4493, tan 62° 24' = 1.9128, cos 21° 34' = 0.9300. 4. Prove tan-i Vt*^^^ = ^ 5-2 tan-i x = tan-^ -^^ ^1 + cosx 2 l-a;« (^^y^ 6. Ji A^ B, C toe angles of a triangle, prove ARC sin il 4 sin C + sin jB = 4 cos —^ cos — cos ^. 7. K cos^a + cos« 2 a + cos^ 3 a = 1, then cos a cos 2 a cos 3 a = 0. 8. Proye cot il — cot 2 il = esc 2 il. 9. Prove fS n \ ^ .i^'^, f'-) >-:i tanf^-*^ fl-tan*" V4 2/ 2 (M) ' 10. tan tana / /• c V / ) tan (a + ^) = 1- 1 + tani 2 J 2 sin ^ sin (2 a + ^) + sin ^ 11. If V = tan-i2 g T^' + ^1 ~ ^' . prove :ca = sin 2y. 12. Prove tan^i 21,+ ^^" 1 + tan^i ^^ , = -tan-V a; 13. Ify = sin-i JT ^ prove X = tan ^. 14. Prove cos^ a + cos' j8 - 1 = cos (a + j8) cos (a - j8). 15. Prove V(cos a - cos py + (sin a - sin py = 2 sin ^^^- 16. Prove sin-i -J-^ = tan-i a/- = ^ cos"* ^^-^• ^a + a; 'a 2 a + a; 17. Prove cos« B - cos^ <^ = sin (<^ + 9) sin (<^ - S). la Provetanil+tan(il + 120°)+tan(il -120«)=8taii3il 19. Prove tan a — tan ^ = tan ^ sec a. 2 2 20. 8tan-ia = tan-i5iLr«?. l-3a2 21. cos* 3 A (tan« 3 il - tan« ^) = 8 sin« J cos 2 il. 22. 1 + cos2(i4 - 5) cos 25 = cos* il +cos*(il -2B). 23. cot«f2: + ?\ = 2c8c2^-jecl V4 2/ 2csc20 + sec0 •■'./' 66 PLANE TKIGONOMETRY. TRIGONOMETRIC EQUATIONS AND IDENTITIES. 54. Identities. It was shown in Chapter I that sin»d + 008^^=1 is true for all values of 0^ and in Chapter YI, that sin (a + )8)= sin a cos fi + cos a sin )8 is true for all values oi a and fi. It may be shown that sin 2 A . A r t;— 7= tan -4. 1 + cos 2 J. is true for all values of J., thus : sin 2 J. _ 2 sin A cos A (by trigonometric transf orma- l + cos2A"l + 2 cos* J. - 1 tion) = ^^^^ (by algebraic transformation^ cos A = tan A (by trigonometric transformation). Such expressions are called trigonometric identities. They are true for all values of the angles involved, 55. Equations. The expression 2 cos* a — 3 cos a + 1 = is true for but two values of cos a, viz. cos a= ^ and 1, i.e. the expression is true for a = 0®, 60®, 300®, and for no other positive angles less than 360®. Such expressions are called trigonometric equations. They are true only for particular v(ilu£S of the angles involved. 56. Method of attack. The transformations necessary at any step in the proof of identities, or the solution of equa- tions, are either trigonometric^ or algebraic; i.e. in prov- ing an identity, or solving an equation, the student must choose at each step to apply either some principles of algebra, or some trigonometric relations. If at any step no algebraic operation seems advantageous, then usually the expression METHOD OF ATTACK. 67 should be simplified by endeavoring to state the different functionB involved in terms of a Bingle function of the angle, or if there are multiple angles^ to reduce all to functions of a single angle. f Algebraic Transformations Trigonometric, f Single function to change to a 1 Single angle No other transformations are needed, and the student will be greatly assisted by remembering that the ready solution of a trigonometric problem consists in wisely choosing at each step between the possible algebraic and trigonometric transformations. Problems involving trigonometric func- tions will in general be simplified by expressing them entirely in terms of sine and cosine. EXAMPLES. 1. Prove 8in3^_cos3^^2 T> 1 V sin 3 ^ cos 3 il sin 3 ^ cos ^ — cos 3 ^ sin il By algebra, — : — -r- = : — -z -z sin ^ cos ^ sin A cos A V . . ,, sin (3 -4 — i4 ) sin 2 ^ by trigonometry, = — A— -r^ = -, — j •^ ° "^ sinilcosil smAcosA 2 sin i4 cos A sin A cos A Or, by trigonometry, = 2. sin 3^ . cos3i4 _ 3sinii —isin^A _ 4cos»^ ~3coSil sin^ cosil sin J. cos^ by algebra, = 3 - 4sin«^ -icoa^A + 3 = 6 - 4(sinM + cos2^)= 2. 8ec80-l tan 8^ 2. Prove sec4^-l tan 2^ No algebraic operation simplifies. Two trigonometric changes are needed. 1. To change the functions to a single function, sine or cosine. 2. To change the angles to a single angle, 8^, 4^, or 2^1. 68 PLANE TRIGONOMETRY. By trigonometry and algebra, 1 -cos8^ sinSg cos 8 ^ __ cos 8 B ^ 1 -cos4g "" sm2g ' cos 4 COB 2 ^ by algebra, co84g(l - cos8g) ^ 8in8gcos2^ . 1 — COS 4 tf sin 2 ^ by trigonometry, cos4^(l~l +28in«4^) _ 2sin4^cos4gco82g , l-l + 28in82d 8in2fl byalgebra, 5ilLi| = 2 cos 2 tf ; sin 2 9 and sin40=:28in20co820, which is a trigonometric identity. 3. Solve 2 cos^ ^ + 3 sin = 0. By trigonometry, 2(1 - sin« tf) + 3 sin tf = 0, a quadratic equation in sin B, By algebra, 2sin30 - 3sin^ - 2 = 0, and (sin tf - 2)(2 sin tf + 1) = 0. .% sin ^ = 2, or — J. Verify. The value 2 must be rejected. Why ? .*. B = 210% and 330° are the only positive values less than 860° that satisfy the equation. 4. Solve sec — tan ^ = 2. Here tan ^ = — 0.75, .*. from the tables of natural functions, B = 143° 7' 48", or 323° V 48". Find sec B, and verify. 5. Solve 28in0sin30-8mS20 = O. By trigonometry, cos 2B — cos 4 ^ — sin* 20 = 0, also oos20 - 008*20 + 8in*20 - sin*20 = 0. By algebra, cos 2 B(\ - cos 2 0) = 0. .•. cos 20 = or 1, and 2 = 90°, 270°, 0°, or 360°, whence = 45°, 135°, 0°, or 180°. Verify. ^ o "-, (^ d- o .a TRIGONOMETRIC EQUATIONS. 69 Or, by trigonometry, 2 Bin 0(3 sin - 4 sin« 0) - 4 sin^ co8> = ; by trigonometry and algebra, esin^d - 88in4 tf - 48in2fl + 48in*fl = 0; by algebra, 2 sin^ 0-4 sin^ = 0, and 2 sin^ 0(1 - 2 sin^ 0) = 0. .*. sin = 0, or ± Vj, and = (F, 180°, 45°, 135°, 225°, or 315°. The last two values do not appear in the first solution, because only angles less than 360° are considered, and the solution there gave values of 2 0, which in the last two cases would be 450° and 630°. X Solve \\y 6. tan = cot 0. ..SJ^ aLo^2^ ^ ^- 8in«0 + CO80 = l. / 7 T^- Prnv^ 10. 2cot2i4 =cot-4 -tan-4 8. 2cos20-2Bin0=l. 9. sin 2 cos = sin 0. 11. cos 2a; + cos 2y = 2 cos(x + y)cos (a: — y). 12. (cos a + sin a)^ = 1 + sin 2 a. 57. Simultaneous trigonometric equations. 13. Solve cos(a: + y)+cos(a:-y) = 2, sin ? + sin ? = 0. 2 2 By trigonometry, cos a; cos y — sin ar sin y + cos x cos y + sin a; sin y = 2^ so that also, and Substituting, cos a: cos y = 1 ; ^l-cosa: ^^ l-cosy ^^^ and 2 • ^ 2 .% cos X = COS y. cos' a: = 1, COS 0? = db 1. .% X = 0°, or 180% y = a: = 0°, or 180°. Verify. / - c^ 5 tnI ^^ 70 PLANE TRIGONOMETBT 14. Solve for R and F. ^nJ W " Fain t - iSoost s 0^ 1F+ Fooet - /Jam t =s 0. To eUminate F, IF" cos i — Fsin t cos t — ^ cos*t = 0, IFsin t + Fcos itani" R sin* t = 0. Adding, 1F(8in t + cos t) - /{(sin* t + oos* t) = 0. .'. R = 1F(8in t + cos 0. ^4^ Sabstitating, IF — Fsin t — Tr(sin t + cos t)cos t = . « IF — 1F(8in t + cos cos t sint If TF= 3 tons, and t = 22^3(y, compute Fand R. /? = 3(0.3827 + 0.9239) = 3.9198. >;;;^ p ^ 3 - 3(0.3827 + 0.9239)0.9239 « _ j g24 0.3827 Solve: <^ 15. 472cot0-263cot<^ = 49O, 3O7cot0-379cot^ = O. 16. sin 2 X + 1 = cos x + 2 sin x. 17. co8«fl + 8ind = l. la If 2A(cosS0-8in3^-2a8in0co80 + 2ft8in0cos0 = O, provo tf=:jtan-i-i^. a — Prove : 19. tany =(1 + secy) tan ^ 20. 2 cot-i X = csc-i i^-^ 2x 21. 8in(^ + 45°) + sin (^ + 135°) = \/§ cos ^. 22 cos p 4- cos 3 p 1 ^ cos 3tf + co6 5t; 2cos2tf — 8ec2r 23. cos 3 X — sin 3 X = (cos x + sin x) (1 — 2 sin 2 x). Solve: 24. sin 2 ^ + sin = cos 2 + cos $. 25. 4co8(tf+60°)-V2=V6-4co8(tf+30°). 26. cot 20 = tan 0-1. 27. cos0 + cos20 + cos30 = O. \' V. ) 1 N v-^ \ "•-rs K. Vj TRIGONOMETRIC EQUATIONS. 71 28. sin 2 a; + v^ cos 2 x = 1. 29. 3 tan«p + 8 cos^p = 7. 30. Determine for what relative valaes of P and W the following equation is true: CO82^-^COS^-i=0. 2 IF 2 2 -or TMT 31. Compute N from the equation iV+ -J^ cos a — ~- sin a — IF cos a = 0, 3 3 when W = 2000 pounds and a satisfies the equation 2 sin a = 1 + cos a. 32. sin — tan ^(cos + sin 9) = cos 0, sin $ — tan ^ cos = 1. Prove: 33. cot(« + 15°)-tan(/-15°)= ^^^^^^ 28in2< + l 34. sin-* J — sin"* ^ = sin"* J}. 35. ^^J^+^)=^l\±^^ \4 2/ ^1— sin 01 36. 2 sin-* } = cos"* J. 37. If siuii is a geometric mean between sinB and cosB, prove cos 2 ^ = 2 sin(45 - B) cos (45 + B). 38. Prove 8in(a + j8 + y) = sin a cos j3 cos y + cos a sin j3 cos y + cos a cos j3 sin y — sin a sin fi sin y. Also find cos(a + jS + y). 39. Prove tan(« + ff + y) = ^ V^ ^""'^ ^o*" ^"^ V ^" " !''" ^^'^ ^ ' 1— tanatanjS— tanj3tany-*tanytance lift, Pi and y are angles of a triangle, prove 40. tan a + tan fi + tan y = tan a. tan fi tan y. 41. C0t^+C0t^+ COt^rrCOt^COtccot?* 2 2 2 2 2 2 If a + j8 + y = 90**, prove 42. tan a tan fi + tan j3 tan y + tan y tan a = 1. Prove : 43. sin na = 2 sin (n — 1) a cos a — sin (n — 2)ce. 44. cos na = 2 cos (n — 1) a cos a — cos (n — 2)a. A- 4. tan fn — 1) a 4- tan a 45. tan na = - f — ^ " ^ ^^ ^ 1 — tan (n — 1) a tan a CHAPTER VIT. TRIANOIiEa. 58. In geometry it has been shown that a triaiigle is determined, except in the ambiguous case, if there are given any three independent parts, as follows : I. Two angles and a side. II. Two sides and an angle, (a) the angle being included by the given sides, (6) the angle being opposite one of the given sides (am- biguous case). III. Three sides. The angles of a triangle are not three independent parts, since they are connected by the relation A + B + C = 180°. The three angles of a triangle will be designated J., jB, (7, the sides opposite, a, 5, c. But the principles of geometry do not enable us to compute the unknown parts. This is accomplished by the following laws of trigonometry: sin A sin B sin I. Law of Sines^ a 11. Law of Tangents, tan|(^-^) ^a-^ ^^^ ^ tanJ(J. + ^) a + b III. Law of Cosines, cos -4. = — ^^r^ , etc. 2 00 59. Law of Sines. In any triangle the sides are propor- tional to the sines of the angles opposite. Let ABO be any triangle, p the perpendicular from B on 6. In I (Fig. 84), O is an acute, in II, an obtuse, in III, 72 LAW OF SINES — OF TANGENTS. 78 a right angle. The demonstration applies to each triangle, but in II, sin^CB=sin2)CB (why?); in III, sin (7=1 (why?). B b D C A b C D A I. II. Fia. 34. 6 III. Now sin -4.= P P /. I? = {? sin A. or sin (7= ^t .*. » = a sin (7. a ^ Equating values of j?, ^ {? sin J. = a sin (7, sin A sin a c By dropping a perpendicular from A, or (7, on a ^ ^, show that whence sin B sin C sin ^ sin B — — - = ^, or = — — , he a sin J. _ sin ^ _ sin _ - , a e 60. Law of Tangents. The tangent of half the difference of two angles of a triangle is to the tangent of half their sum^ as the difference of the sides opposite is to their sum. BA^ii'Ck ^ sin A QinB By composition and division, g — 6 _ sin ^ — sin ^ _ 2 cos ^ (J. + -B) sin ^(A — B) a + 6 " sin^ + sin^" 2 sin J(^ + ^)co8 J(J. - B) ^ tan^(^-^) , tanJ(^ + -B)' ^j. tanK^-^) ^«"S ' tanj(^-|-^) a + b' 74 PLANE TRIGONOMETRY. 61. Law of Cosines. The cosine of any angle of a triangle is equal to the quotient of the sum of the squares of the adjacent sides less the square of the opposite side^ divided by twice the product of the adjacent sides. In each figure a^^p^-\-I)C^ (in Fig. 84, II, 2) (7 is negative; in III, zero) = c2 - ^2)2 + J2_ 26 . ^2) + ^i>2 = J2_|_^_26.^2). But AD=e cos A cos J.= 2 be 2 be • cos A; Prove that d (ioaB = fioaCs a' + c^- 2ao a2 + 62_ 2ah 62. Though these formulsB may be used for the solution of the triangle, they are not adapted to the use of loga- rithms (why?). Hence we derive the following: Since cos ^ = 2 cos^ ^-1 = 1-2 sin^- , 2 2 we have 2 cos^:^ == 1 + cos A, and 2 sin^^ = 1 - cos A. LAW OF COSINES. 75 From the latter 2 2bc 2be 2 be 2 be Let a+6+(?=2«,thena+6— {?=a + 6+c— 2i?=2«— 2tf; *•«• a + J-<?=2(«-.c), In like manner, a -- 6 + c = 2(« — J). "a-»-6 + (?=2(«-a). Substituting, 2 sin2^ = 2(<^- ft) '2rg-<?-) 2 2be 2 ^ be Show that sin-=? also sin-^== ? From 2 cos?— = 1 + cos J, show that cos - = ■yj"^"-'^), also cos — = ? and cos 1^ = ? AJso derive the formulae V >v.^ tanf = ? tan~ = ? 2 76 PLANE TBIGONOMETRY. 63. Area of the triangle. In the figures of Art. 59 the area of the triangle ABC=^ A = \'pf>. But ^scsinA .'. A = J6(?8inil. (i) Again, by law of sines, * = — : Substituting, A = sin (7 g^sinjlsinJ? 2sin(7 c^sinJ.sinjB 2sin(^+^) (why?). (ii) A A Finally, since sin J. = 2 sin — cos ~, we have from (i) A 1 iL a ' A A T ^ /«(« — d)(s — h)(s — c) A = * 6<? • 2 sm— ■ cos — = oc\-^ ^ — - ^^ ^ ^ 2 2 ^ be ' be or A = V« (« — a)(« — 6)(« — e). (iii) Find A ; (1) Given a = 10, 5 = 12, C = 45*». (2) Given a = 4, J = 6, c = 6. (3) Given a = 2, B = 45% C = GO*. SOLUTION OF TRIANGLES. 64. For the solution of triangles we have the following formulae, which should be carefully memorized : J gja A __ 9in B _ gin C II. tiui|(^-B) = ^^i|taii|(^ + B). * a + 6 * III. ««- = V^^ g ^' w ««_ = ^-L^ 2 ' »(»-a) ' , . IV. A = I dc 8to ^ = ^ «j^°^^yg = V«(* -aK»- b) (5 - g). SOLUTION OF TRIANGLES. 77 Which of the above formulae shall be used in the solution of a given triangle must be determined by examining the parts known, as will appear in Art. 69. It is always pos- sible to express each of the unknown parts in terms of three known parts. In solving triangles such as Case I, Art. 58, the law of sines applies; for, if the given side is not opposite either given angle, the third angle of the triangle is found from the relation A + B + 0=: 180®, and then three of the four quantities in — — = ^^ — being known, the solution gives a the fourth. In Case II (6) the law of sines applies, but in II (a) two only of the four quantities in — — = ^^ — are known. a Therefore, we resort to the formula tanJ(^-5) = 5L::^tanJ(^ + 5), in which all the factors of the second member are known. In Case III, tan4 = \^ ""^^^*"7^^ is clearly applicable, and is preferred to the formulae for sin— and cos — ; for, first, it is more accurate since tangent varies in magnitude from to 00, while sine and cosine lie between and 1. (See Art. 27, 6.) Let the student satisfy himself on this point by finding, correct to seconds, the angle whose logarithmic sine is 9.99992, and whose loga- rithmic tangent is 1.71668. Does the first determine the angle ? Does the second? And, second, it is more convenient, since in the complete solution of the triangle by sin — 8ix logarithms must be taken A A from the table, by cos — seven^ and by tan — but four. The right triangle may be solved as a special case by the law of sines, since sin (7=1. 78 PLAKE TRIGONOMETRY. 65. Ambiguous case. In geometry it was proved that a triangle having two sides and an angle opposite one of them of given magnitude is not always determined. The marks of the undetermined or ambiguous triangle are : 1. The parts given are two sides and an angle opposite one. 2. The given angle is acute, 3. The side opposite this angle is less than the other given side. When these marks are all present, the number of solutions must be tested in one of two ways : (a) From the figure it is apparent that there will be no solution when the side opposite is less than the perpendicular p ; one solution when side a equals p ; and two solutions when. a is greater than p. A b c No Solution, A b C A b C C' One Solution, Two Solutions, Fig. 35. And since sin ii = — » it follows that there will be no solt^ c ^ tion^ one solution^ two solutions^ according as sin A =-, < c (6) A good test is found in solving by means of loga- rithms ; and there will be no solutions, one solution, two solu* tions, according as log sin proves to be impossible, zero, possible, i,e, as log sin Q is positive, zero, or negative. This results from the fact that sine cannot be greater than unity, whence log sine must have a negative characteristic, or be zero. G6. In computations time and accuracy assume more than usual importance. Time will be saved by an orderly arrange- ment of the formulae for the complete solution, before open- ing the book of logarithms, thus : SOLUTION OF TRIANGLES. 79 Given -4., By a. Solve completely. (7=180<^-(A+5), 6 = 554!]^, c:=^Lmf^ A^labsinO. am A am A ^ ISC'* log a = log a = A -\-B = log sin 5 = log sia C = ••• C = colog sin A = colog sin A = log 6 = log c = .'. 6= .'. c = log a = log (« — 6) = log 6 = log (« — c) = log sin C = colog 8 = colog 2 = colog (« — a) = logA= 2 ) .-. A = log tan — = .-. A = 67. Accuracy must be secured by checks on the work at every step ; e.g. in adding columns of logarithms, first add, up, and then check by adding down. Too much care can- not be given to verification in the simple operations of addition, subtraction, multiplication, and division. A final check should be made by using other formulae involving the parts in a different way, as in the check above. As far as. possible the parts originally given should be used through- ' out in the solution, so that an error in computing one part: may not affect later computations. 6& The formulae should always be solved for the unknown part before using^ and it should be noted whether the solu- tion gives one value, or more than one, for each part ; e.g.^ the same value of sin 5 belongs to two supplementary angles,, one or both of which may be possible, as in the ambiguous, case. 69. Write formulae for the complete solution of the fol-^ lowing triangles, showing whether you find no solution, one solution, two or more solutions, in each case, with reasons for your conclusion : 80 PLANE TRIGONOMETRY. a h c A B C 1. 8P 26' 28" 44** 11' 20" 540 22' 12" 2. 78.54 63° 18' 20" 41** 30' 18" 3. 135.82 26.89 63*^ 28' 30" 4. 0.75 0.85 0.95 "• 5. 243 562 36** 15' 40" 6. 38.75 25.92 63° 50' 10" 7. 0.058 78^5' 83^46' 8. 2986 1493 30° 9. 48 50 26n5' MODEL SOLUTIONS. 1. Given a = 0.785, b = 0.85, c = 0.633. Solve completely. Check: A + B + C =z 180°. A = V«(« - a)(«- 6)(» - c). a = 0.735 h = 0.85 c = 0.633 2)2.268 s = 1.134 s-a = 0.349 5-6=0.284 « - c = 0.501 Check: A = 61° 53' 38" J?= 72° 46' 4" C= 45° 20' 20" 180° 0' 2" log(«-6)= 9.45332 log(«-c)= 9.69984 oology = 9.94539 colog («-«)= 0.45717 2)19.55572 logtanJ^= 9.77786 ^A =30° 56' 49" A = 61° 53' 38" log(»-a)= 9.54283 log(«-6)= 9.45332 colog«= 9.94539 colog (« - c) = 0.30016 2 )19.24170 logtanJC= 9.62085 J C = 22° 40' 10" C = 45° 20' 20" log(»-a)= 9.54283 log (« - c) = 9.69984 colog 9= 9.94539 colog(»-6)= 0.54668 2 )19.73474 log tan i 5= 9.86737 J5 = 36°23'2" B = 72° 46' 4" log«= 0.05461 log («-«)= 9.54283 log(«-6)= 9.45332 log (« - c) = 9.69984 2)18.75060 logA= 9.37530 A = 0.2373 (1) Given a = 30, 6 = 40, c = 50. (2) Given a = 2159, b = 1431.6, c = 914.8. (3) Given a = 78.54, 6 = 32.56, c = 48.9. SOLUTION OF TRIANGLES. 81 2. Given A = 57° 23' 12", C= 68° 15' 30", c = 832.56. Solve completely. a — csinii sin C 5 = 180O-(.l +C) = 54° 21' 18". j^ciBin^ ^ = j5c8in^. sin C Check: taxi^A -4 logc = 2.92042 log sin A = 9.92548 colog sin C = 0.03204 log a = 2.87794 a= 754.98 Check: a= 754.98 6= 728.38 c= 832.56 log c = 2.92042 log sin B = 9.90990 colog sin C = 0.03204 log b = 2.86236 6= 728.38 9-a= 402.98 « - 6 = 429.58 8-c= 325.40 2 )2315.92 5 = 1157.96 « (« — a) log 6= 2.86236 logc= 2.92042 log sin ^ = 9.92548 log 2 A = 5.70826 A = 515811 = 255405.5 2 log(«-6)= 2.63304 log(«-c)= 2.51242 colog «= 6.93634 colog(«-a)= 7.39471 2 )19.47651 logtani^= 9.73826 i il = 28° 41' 38' A = 57° 23' 16' Solve : (1) Given a = 215.73, B = 92° 15', C = 28° 14'. (2) Given b = 0.827, A = 78° 14' 20", B = 63° 42' 30". (3) Given b = 7.54, c = 6.93, if = 54° 28' 40". 3. Given a = 25.384, c = 52.925, B = 28° 32' 20". Solve completely. (Why not nse the same formulffi as in Example 1, or 2?) tan C-A = £^tan.^ + ^ 2 c -^a 2 180° - 5 = C + ^ = 151° 27' 40". .-. i(C + A)=: 75° 43' 50". , c sin ^ J. , ^^ .^ o sinC Check: b = asinB sin^ ' c= 52.925 log (c- a) = 1.43998 .-. i(C-A)= 54° 7'38" a= 25.384 colog (c+ a) =8.10619 l(C+.4)= 75°43'50" c+a= 78.309 log tan KC+^)= 0.59460 ^^^^^^^ C=129°51'28" c-a= 27.541 log tan i(C-i4) =0.14077 subtracting, ^4= 21°36'12'' log a =1.40456 log c= 1.72366 log sin 5=9.67921 colog sin C =0.11484 log ft =1.51771 b= 32.939 Check: log a =1.40456 log sin 5=9.67921 colog sin i4 =0.43395 log 6= 1.51772 log c= 1.72366 log sin 5=9.67921 log 2 A =2.80743 ^^641^^ 320.92 82 PLANE TRIGONOMETRY. Solve : (1) Given a = 0.325, c = 0.426, B = 48° 50' 10". (2) Given h = 4291, c = 3194, A = 73° 24' 50". (3) Given b = 5.38, c = 12.45, A = 62° 14' 40". 4. Ambiguous cases. Since the required angle is found in terms of its sine, and since sin a =* sin (180° — a), it fol- lows that there may be two values of a, one in the first, and the other in the second quadrant, their sum being 180®. In the following examples the student should note that all the marks of the ambiguous case are present. The solutions will show the treatment of the ambiguous triangle having no solution, one solution, two solutions. (a) Given 6 = 70, (? = 40, (7=47° 32' 10". Solve. Why ambiguous ? o;« » 6 sin (7 log b = 1.84610 sin JL> = . '^ logsin (7= 9.86788 colog c = 8.39794 log sin B = 0.11092 .*. 5 is impossible, and there is no solution. Why? Show the same by sin (7 > -• (J) Given a = 1.5, c = 1.7, A = 61° 66' 38". Solve. . /Y csinA loff <? = 0. 23045 sin ^ . ° a log sin A = 9.94564 colog a = 9.82391 logsin (7= 0.00000 (7=90° * and there is one solution. Why? Show the same by sin -4. = -. Solve for the remaining parts and check the work. X c^ N SOLUTION OF TRIANGLES. 88 ((?) Given a = 0.236, b = 0.189, B = 36° 28' 20". Solve. „. J a sin 5 ^ bsinO smjS log a = 9.37107 log 6 = 9.27646 9.27646 logsin5=: 9.77411 log sin (7= 9.99772 or 9.28774 colog b = 0.72354 colog sin B = 0.22589 0.22589 log sin A = 9.86872 log c = 9.50007 or 8.79009 A =: 47° 89' 25" c = 0.31628 or 0.06167 or 132° 20' 85". .-. (7= 95° 52' 15" or 11° 11' 5". Solve for A, and check. Show the same by sin 5 < — a Solve : (1) Given 6 = 216.4, (? = 593.2, 5= 98° 15'. (2) Given a = 22, 6 = 75, -B = 82° 20'. (8) Given a = 0.353, c = 0.295, A = 46° 15' 20".. (4) Given a = 293.445, b = 460, A = 40° 42'. (6) Given b = 631.03, (? = 629.20, jB = 84° 28' 16". Solve completely, given : a .ft \ c 1. 50 60 2. 10 11 3. 4 6 6 4. 10 5. 40 51 l^ 6. 852.25 513.27 482.68 ^ ^ 7. 0.573 0.394 ^ 8. 107.087 9. v^ 10. 197.63 246.35 11. 4090 8850 3811 3795 234.7 185.4 14. 26.234 22.6925 15. 273 136 I c / ^ ^ ^13. A B C 78° 27' 47'* 03° 35' 109° 28' 16" 38° 56' 54" 49° 28' 32" 112° 4' 56° 15' 48° SS' 117° 45° 34° 27' 73° 15' 15" 42°18'30' 84° 36' 49° 8' 24" 72° 25' 13" 84 PLANE TRIGONOMETRY. APPLICATION& 70. Measarements of heights and distances often lead to the solution of oblique triangles. With this exception, the methods of Chapter V apply, as will be illustrated in the following problems. The bearing of a line is the angle it makes with a north and south line, as determined by the magnetic needle of the mariner's compass. If the bearing does not correspond to any of the points of the compass, it is usual to express it thus: N. 40** W., meaning that the line bears from N. 40® toward W. EXAMPLES. 1. When the altitude of the sun is 48^, a pole standing on a slope inclined to the horizon at an angle of 15° casts a shadow directly down the slope 44.3 ft. How high is the pole? 2. A tree standing on a mountain side rising at an angle of 18° 3(K breaks 82 ft. from the foot. The top strikes down the slope of the moun- tain 28 ft. from the foot of the tree. Find the height of the tree. 3. From one corner of a triangular lot the other corners are found to be 120 ft. £* by N., and 150 ft. S. by W. Find the area of the lot, and the length of the fence required to enclose it. 4. A surveyor observed two inaccessible headlands, A and B. A was W. by N. and B, N.E. He went 20 miles N., when they were.S.W. and S. by E. How far was A from B ? 5. The bearings of two objects from a ship were N. by W. and N.E. by N. After sailing E. 11 miles, they were in the same line W.N. W. Find the distance between them. 6. From the top and bottom of a vertical column the elevation angles of the summit of a tower 225 ft. high and standing on the same hori- zontal plane are 45° and 55°. Find the height of the column. 7. An observer in a balloon 1 mile high observes the depression angle of an object on the ground to be 35° 20'. After ascending vertically and uniformly for 10 mins., he observes the depression angle of the same object to be 55° 40'. Find the rate of ascent of the balloon in miles per hour. 8. A statue 10 ft. high standing on a column subtends, at a point 100 ft. from the base of the column and in the same horizontal plane, the same angle as that subtended by a man 6 ft. high, standing at the foot of the column. Find the height of the column. 9. From a balloon at an elevation of 4 miles the dip of the horizon is 2° 83' 40". Required the earth's radius. TRIANGLES — APPLICATIONS. 85 10. Two ships sail from Boston, one S.E. 50 miles, the other N.E. by £. 60 miles. Find the bearing and distance of the second ship from the first. 11. The sides of a valley are two parallel ridges sloping at an angle of dO°. A man walks 200 yds. up one slope and observes the angle of eleva- tion of the other ridge to be 15°. Show that the height of the observed ridge is 273.2 yds. 12. To determine the height of a mountain, a north and south base line 1000 yds. long is measured ; from one end of the base line the sum- mit bears E. 10° N., and is at an altitude of 13° 14'. From the other end it bears E. 46° SO' N. Find the height of the mountain. 13. The shadow of a cloud at noon is cast on a spot 1600 ft. due west of an observer. At the same instant he finds that the cloud is at an ele- vation of 23° in a direction W. 14° S. Find the height of the cloud and the altitude of the sun. 14. From the base of a mountain the elevation of its summit is 54° 20'. From a point 3000 ft. toward the summit up a plane rising at an angle of 25° 30' the elevation angle is 68° 42'. Find the height of the mountain. 15. From two observations on the same meridian, and 92° 14' apart, the zenith angles of the moon are observed to be 44° 54' 21" and 48° 42' 57". CaUing the earth's radius 3956.2 miles, find the dis- tance to the moon. 1 ^^-^C^'ZenUh angle 16. The distances from a point to three objects are 1130, 1850, 1456, and the angles subtended by the distances between the three objects are respectively 102° 10', 142°, and 115° 50'. Find the distances between the three objects. 17. From a ship A running N.E. 6 mi. an hour direct to a port dis- tant 35 miles, another ship B is seen steering toward the same port, its bearing from A being E.S.E., and distance 12 miles. After keeping on their courses IJ hrs., B is seen to bear from A due E. Find B's course and rate of sailing. 18. From the mast of a ship 64 ft. high the light of a lighthouse is just visible when 30 miles distant. Find the height of the lighthouse, the earth's radius being 3956.2 miles. 19. From a ship two lighthouses are observed due N.E. After sailing 20 miles E. by S., the lighthouses bear N.N.W. and N. by E. Find the distance between the lighthouses. 20. A lighthouse is seen N. 20° E. from a vessel sailing S. 25° E. A mile further on it appears due N. Determine its distance at the last observation. EXAMPLES FOR REVIEW. In connection with each problem the student should review all principles involved. The following list of problems will then furnish a thorough review of the book. In solving equations, find all values of the unknown angle less than SCO"* that satisfy the equation. 1. If tan a = ^, tan j3 = }, show that tan ()3 ~ 2 a) = |^. 2. Prove tan a + cot a = 2 esc 2 a. 3. From the identities sin^4 + cos^— = 1, and 2 sin •— cos-r- = sin il, 2 2 2 2 prove 2 sin -- = ± V 1 + siu -4 db V 1 — sin -4, and 2 cos — = db VI + sin A qp VI — sin^. 4. Remove the ambiguous signs in £x. 3 when il is in tnm an angle of each quadrant. 5. A wall 20 feet high bears S. SQ"" 5' E. ; find the width of its shadow on a horizontal plane when the sun is due S. and at an altitude of 60^ 6. Solve sin X + sin 2 X + sin 3 X = 1 + cos X + cos2 «• 7. Prove tan-i i + tan-i i = |. 8. K il = 60% B = 45% C = SO^*, evaluate tan A + tan B + tan C tan A tan B + tan B tan C + tan C tan A 9 Prove ^^^ (^ + B) cos C _ 1 — tan A tan B cos (il + C) cos B 1 — tan A tan C 10. Solve completely the triangle whose known parts are h = 2.35, c = 1.96, C = 38^ 45'.4. 11. Fmd the functions of 18°, 36^ 54°, 72°. Let«-18°. Then 2x = 36°, 3z = 54°, and 2x + 3x-90°, 12. If cot a = -, find the value of sin a + cos a + tan a + cot a + sec a + esc a. 86 EXAMPLES FOR REVIEW. 87 13. Prove "°.^«"».2g-«!°^^''!°^«=l + 4cog«co8i8. Sin 2 a Sin )3 — sm 2 p Bin a 14. From a ship sailing due N., two lighthouses bear N.E. and N.N.E., respectively; after sailing 20 miles they are observed to bear due £. Find the distance between the lighthouses. 15. Solve 1 — 2 sin X = sin 3 X. 16. Prove sin-K/-^ = tan-^V?- ^a + 6 ^b 17. If cos tf — sin tf = V2 sin ^, then cos tf + sin tf = "\/2 cos A 18. Solve completely the triangle ABC, given a = 0.256, b = 0.387, C = 102° 20'.5. 2 cos 2 a — 1 19. Prove tan (30° + a) tan (B0°- a) = 2 cos 2 a + 1 20. Solve tan (45° - 6) + tan (45° + ^) = 4. , 21. Prove sin^ a cos* P — cos* a sin* j8 = sin* a — sin* j8. 22. Prove cos* a cos* j8 - sin* a sin* j8 = cos* a — sin* j8. 23. A man standing due S. of a water tower 150 feet high finds its elevation to be 72° 30' ; he walks due W. to A street, where the elevation is 44° 50' ; proceeding in the same direction one block to B street, he finds the elevation to be 22° 30'. What is the length of the block between A and B streets ? 24. Prove tan-* i + tan-i 1 + tan-i i + tan-i g = |. 25. U P = 60°, Q - 45°, R = 30°, evaluate sin P cos Q + tan P cos Q sin P cos P + cot P cot 72 26. If cos (90° + a) = - i evaluate 3cos 2 a + 4 sin 2 a. 27. If sin B + sin C = m, cos B + cos C = n, show that tan — 5— = — 2 n 28. Show that sin 2 p can never be greater than 2 sin j3. 29. Prove sin-* J + sin-^ ^ = tan-* JJ. 30. Solve cot-^x + sin-»| V^ = ^. 5 4 31. Solve sin-* x + sin-* (1 — a;) = cos-* x. 32. A man standing between two towers, 200 feet from the base of the higher, which is 90 feet high, observes their elevations to be the same ; 70 feet nearer the shorter tower he finds the elevation of one is twice that of the other. Find the height of the shorter tower, and his original distance from it. 88 PLANE TRIGONOMETRY. 33. Solve cos 3 j3 + 8 cos* j3 = 0. 34. Solve cot m - tan (ISO** + m) = sec m + sec (OO** - m). 35. Solve i-=^-5?:2J = 2 cos 2 ^ 1 + tan < 36. Prove cotil + cot^ = "°(^ t ^^ ' sin A sin 2^ 37. Prove cot P - cot Q = - ^^P ^^ 7 9 ' sin P sin Q 38. In the triangle ABC prove a = 6 cos C + c cos 5, 6 = c cos il + a cos C, c = a cos 5 + ft cos -4. 39. Solve completely the triangle, given a = 027.56, b = 648.25, c = 738.42. 40. Prove cos^ a - sin (30° + a) sin (30° - a) = |. ^1 "D- ^ A-. Q i. cos 2 a: — cos 4 x 41. Prove tan3xtanx = — • cos 2 X + cos 4 X 42. Simplify cos (270° + a) + sin (180° + «) + cos (90° + a), 43. Simplify tan (270° -$)- tan (90° + ^ + tan (270° + 0). 44. Solve cos 3 ^ - cos 2 ^ + cos ^ = 0. 45. Solve cos il + cos 3 -4 + cos 5 il + cos 7 -4 =0. 46. The topmast of a yacht from a point on the deck subtends the same angle a, that the part below it does. Show that if the topmast be a feet high, the length of the part below it is a cos 2 a. 47. A horizontal line ^4^ is measured 400 yards long. From a point ID. AB SL balloon ascends vertically till its elevation angles at A and B are 64° 15' and 48° 20', respectively. Find the height of the balloon. sin a a n 48. If cos ^ = n sin a, and cot ^ = - — ^, prove cos P = . tan^ Vl+n^cos^a 49. Find cos 3 a, when tan 2 a = — }. 50. Solve completely the triangle, given a = 0.296, B = 28°47'.3, C = 84° 25'. 51. Evaluate sin 300° + cos 240° + tan 2250. 52. Evaluate sec ^ - esc ^ + tan Is:. O o u EXAMPLES FOR REVIEW. 89 53. If ^^^_ f&nacmy-n\nPtmy cos a cos y — cos fi siu y and t^^^8in«siDY-8i"^co8Y, COS a sin y — cos p cos y show that tan(0 + ^) = tan (a + p), 54. If tan iee** 16' 38" = - ^, find the sine and cosine of 233° T 49". CSC a — cot a sec ce — tan a 55. Prove sec a + tan a esc a + cot a 56. Prove co8(« - 3/3)-co8(f« -/?) = 2 8in(« - /9). sm2a + sin2p 57. Prove sin 80° = sin 40° + sin 20°. 58. Prove cos 20° = cos 40° + cos 80°. 59. Prove 4 tan-i - - tan'i — = E. 5 239 4 60. From the deck of a ship a rock bears N.N.W. After the ship has sailed 10 miles E.N.E., the rock bears due W. Find its distance* from the ship at each observation. 61. Find the length of an arc of 80° in a circle of 4 feet radius. 62. Given tan ^ = |, tan ^ = A, evaluate sin(^ + ^) + cos(^ — ^). 63. If tan ^ = 2 tan ^, show that sin(^ -f ^) = 3 sin(^ — ^). 64. Prove cos(a+)8)cos(a-j8)+8in(a+i8)sin(a-i8)=^^I^5^., l4-tan*/S 65. Solve 4 cos 2 ^ + 3 cos ^ = 1. 66. Solve 3 sin a = 2 sin (60° - a). 67. Prove (sin a — esc a)' — (tan a — cot ay +(cos a — sec a)*='l. 68. Prove 2 (sin® a + cos* a) + 1 = 3(sin* a + cos* a). 69. Prove esc 2 )8 + cot 4 j8 = cot j8 — esc 4 j8. 70. K tanj3=A, cos2g=5|Z, thencsc-2_=L2 = 5vlg. 71 Solve completely the triangle, given a = 0.0654, 5 = 0.092, 5 = 38°40'.4. 72. Solve completely the triangle, given 6 = 10, c = 26, 5 = 22°37'. 73. A railway train is travelling along a curve of \ mile radius at the rate of 25 miles per hour. Through what angle (in circular measure) will it turn in half a minute ? 90 PLANE TRIGONOMETRY. 74. Express the following angles in circular measure : 630, 4°3(y, 6° 12' 36". 75. Express the following angles in sexagesimal measure: «" 3jr 17 IT 6' 8* 64 76. 11 Af By C are angles of a triangle, prove cos il + cos 5 + cos C = 1 + 4 sin — sin — sin ^ 77. Prove sin 2 x + sin 2 y + sin 2 2 = 4 sin :c sin y sin 2, when x, y^ z are the angles of a triangle. 78. Prove sec a = 1 + tan a tan ^* 79. Prove sin* (a + jS) - sin2(a - )3) = sin 2 a sin 2 j8. 80. Prove cos«(a + /3)- sin2(a - j8) = cos 2 a cos 2 j8. 81. Prove ^? ^^? + ^\" ^JP = 2cos9;). sm 10 j3 + 8m8/> 82. Consider with reference to their ambiguity the triangles whose known parts are : (a) a = 2743, h = 6452, B = 43° 15' (6) a = 0.3854, c = 0.2942, C = 38°20r (c) 6= 5, c = 53, 5 = 15^22' ((/) a = 20, 6 = 00, il = 63° 28'.5. 83. From a ship at sea a lighthouse is observed to bear S.E. After the ship sailed N.E. 6 miles the bearing of the lighthouse is S. 27° 30' E. Find the distance of the lighthouse at each time of observation. 84. Prove «ai^)i«aM±^.2co8((? + «. sm 2 u -\- sin J ^p 85. Prove cos 15° - sin 15° = — • y/2 86. Show that cos (a + j8) cos (a - )8) = cos« a - sin* p = cos*)8-sin*a. 87. Show that tan (a + 45°) tan (a - 45°) = 2 8in*tt-l , ^ ^ ^ ^ 2co82«-l 88. Solve sin (x + y) sin (x — y) = }, cos (x + y) cos (x - y) = 0. 89. Prove ^ + T" " '^" = tang. 1 + sm a + COS a 2 EXAMPLES FOR REVIEW. 91 90. Prove tan2tf + sees tf = 52S4±«n|. co$ ^ — sin 9 91. If tan ^ = -9 then a cos 2 ^ + 6 sin 2 ^ = a. a 92. Prove sin-*— + cot-i 3 = ?• 93. Solve cos A + cos 7 -4 = cos 4 -4. 94. Two sides of a triangle, including an acute angle, are & and 7 the area is 14 ; find the other side. 95. Show that ^.^^"^^"^.'"''o^'r.^^ = *an 2 ft sin5^-3sin30 + 4sin^ 96. A regular pyramid stands on a square base one side of which is 173.6 feet. This side makes an angle of 67° with one edge. What is the height of the pyramid? 97. From points directly opposite on the banks of a river 600 yards wide the mast of a ship lying between theqi is observed to be at an eleva- tion of 10** 28'.4 and 12° 14'.5, respectively. Find the height of the mast. 98. Show that (sin 60° - sin 45°) (cos 30° + cos 45°) = sin^SO^ 99. Find ar if sin-i X + sin-i ^ = ?. 2 4 100. Trace the changes in sign and value of sin a + cos a as a changes from 0° to 360°. CHAPTER VIII. MZBCSLZiANBOnS PROPOSITIONS. 71. The circle inscribed in a given triangle is often called the ineircle of the triangle, its centre the incentre^ and its radius is denoted by r. The incentre is the point of inter- section of the three bisectors of the angles of the triangle (geometry). The circle circumscribed about a triangle is called the circumdrcle^ its centre the circumcentre^ and its radius M, The circumcentre is the point of intersection of perpendicu- lars erected at the middle points of the three sides of the triangle (geometry). Incirde. Cirenmcircle. Escribed circle opposite A. PlO. 37. The circle which touches any side of a triangle and the other two sides produced is called the escribed circle; its radius is denoted by r^, r^, or r^, according as the escribed circle is opposite angle -4, 5, or C. Again, the altitudes from the vertices of a triangle meet in a point called the orthoeentre of the triangle. Finally, the medians of a triangle meet in a point called the centroid^ which is two-thirds of the length of the median from the vertex of the angle from which that median is drawn (geometry). Certain properties of the above will now be considered. 02 c MISCELLANEOUS PEOPOSITIONS. 7Z To find the radius of the incircle. Let A, A', A," A'" represent the areas of triangles ABO, COB, A 00, BOA, respectively. Then A = A' + A" + A'" 93 And since A = V«(« — a)(« — 6)(« — (?), (Art. 63) 8 Cob. To express the angles in terms of r and the sides, divide each member of the above equation by « — a. Then s -I,=J (^"/)(^ Z£)=.tan^A (Art. 62) In like manner tan i -B = ^ , ; tan i (7 = — ^« '^ « — 6 " 8 — 73. To find the radius of the circumcirde. Fio. 39. In the figure ABO is the given triangle, and A'O sl diam- eter of the circumcirde. Then, angle A = A', or 180®— A'. .•. sin J. = sin -4.'. Since A* BO is a right angle, BO sin A^ = a AIO 2R • . ./I/ — f r- 94 PLANE TRIGONOMETRY. CoE, 1. As above, 25 = -: — 7 = -: — =r = -: — :^ which is » sin A sill B sin (7* another proof of the ^^ law of sines. a Cob. 2. From 5= , we have E 2sin^ abe 2 be sin ^ 4 A = --—, where A= area ABO. 74i To find the radii of the escribed circles. Now Represent areas ABC^ BOA^ AOO, BOO, by A, A', A'', A"', respectively. Then r^ is the altitude of each of the triangles ^ BOA, AOO, BOO. A = A' + A" — A'" = J^a(<? + 6 - a)= r„(« - a> A •*• * « ^ In like manner, r^ = « — a A 1'» ^^ = « — c 75. The ortfaocentre. Denote the perpendiculars on the sides a, 6, {?, by AP^ BPf^ OP^ and let it be p^ required to find the distances from their intersection to the sides of the tri- angle, and also to the vertices. ^ OPf, = APf, tan OA 0. But APt^'CcosA, and (740 = 90^-0: ••. 0PA = <?cosilcot(7=-: — -cosAcosO. sin (7 = 2Rco8A cos 0. (Art. 73, Cor. 1) MISCELLANEOUS PROPOSITIONS. 95 In like manner, OPc = 2Bco8 £ cos A, OPa^2Eco8Cco8£. Again, the distances from the orthocentre to the vertices are, cosCAO sin (7 = 2RcobA. Also, 0B = 2R €08 B, and 00 =^ 2 R 608 0. 76. Centroid and medians. The lengths of the medians may be computed as follows : In the figure the medians to the ^ sides a, J, <?, are AMa^ BM^^ OM^^ meeting in the centroid 0. Now, by the law of cosines, from ^ /^^-""^ ^\ ^""^ ^^^ g the triangle -B-Mi (7, BM^^ =^ ^2 + jf^ (72 _ 2 a . Jf, (7 . c OS (7 = a^ + — ■ — a6 cos (7. 4 But, cos O = — ~— -r ^ zoo . »if2_^2 . J' a2^.62_^ _ 2aa+2cg-6g whence, BMf, = ^■V2a^-\-2(^-b^ = ^^a^ + €^+ 2aeco8B a^ + ^-V^ J, since — — = cos-o. 2ae In like manner, OM, = ^^/2b^ + 2a^-(^=^i^/b^ + a^ + 2baco8 0, and AM^=lW2€^ + 2V^-a^=^Vc^+b^ + 2cbco8A. 96 PLANE TRIGONOMETRY, EXAMPLES. t 1. In the triangle, a = 25, ( = 35, c = 45, find 22, r, r.. 2. Given a = 0.354, h = 0.548, C = 28"" 34' 20V, find the distances to C and By from the circumcentre, the incentre, the oentroid, and the orthooentre. 3. In the ambiguous triangle show that the circumcircles of the two triangles, when there are two solutions, are equal. 4. Prove that 1 + 1 + 1 = -. r. r^ r« r 5. In any triangle prove A ^y/rr^ri^r^ 6. Prove that the product of the distances of the incentre from the vertices of the triangle is 4 r^R. 7. Prove that the area of all triangles of given perimeter that can be circumscribed about a given circle is constant. 8. Prove that the area of the triangle ABC is Jf2r(8in il + sin £ + sin C). CHAPTER IX. SERIES— DE MOIVRE'S THEOREM— HYPEBBOLIC FUNCTIONS. 77. First consider some series by means of which loga- rithms of numbers and the natural functions of angles may be computed. For this purpose the following series is important : 6 = 1 H 1 1- • ••■ ( It may be derived as follows : By the binomial theorem, 1 , 1 , nxCnx—l') 1 , nxCiix—l')(nx-'2) 1 , n \2 n^ [3 rfi X (^-D 4-S(^-!) . • a. = 1+^ + [2-+ 2 ^ and if n increase without limit, This is called the exponential series^ and is represented by 6*, so that e* = 1 + » + — +—+ •.. + ^+ .... It is shown in higher algebra that this equation holds for all values of x ; whence, if a; = 1, 97 98 PLANE TRIGONOMETRY. This value of e is taken as the base of the natural or Naperian system of logarithms. Thifl yalue «, howeyer, is not the base of the system of logarithms computed by Napier, but its reciprocal instead. The natural logarithm is used in the theoretical treatment of logarithms, and, as will presently appear, it is customary to compute the common logarithm by first finding the natural, and then multiplying it by a constant multiplier called the modulus. Art. 82 ; t.6. in the Naperian system the modulus is taken as 1, and the base is computed. In the common system the base 10 is chosen and the modulus computed. 78l From the exponential series the value of e may be computed to any required degree of accuracy. 1+1 + 1-2.5 i = 0.1666666666 i = 0.0416666666 ,4 = 0.0083333333 i = 0.0013888888 i = 0.0001984126 1 = 0.000Q248016 1 = 0.0000027557 ,47 = 0.0000002765 |10 Adding, e = 2.7182818, correct to 7 decimal places. SERIES. 99 79. To expand a* in ascending powers of jr. Let «* = 6', then z = log^ a* = a; • log^ a. (Arts. 35, 40) Substituting Now put 1 + a for a, and (l + a)' = l + a;.log.(l + a) [2 ^ [3 ^ V But by the binomial theorem, [2 [3 Equating coefficients of a; in the second members of the above equations, log.(l + a)=a-| + |-f + ...; or writing « f or a, %.(l + a:) = a:-| + |-| + .... In this form the series is of little practical use, since it converges very slowly, and only when x is between + 1 and — 1 (higher algebra). Put — X for a;, and log,(l-a:)=-a;-----~ ; .-. loge(l + a;) - log,(l - a;) ..og.l±|-2(. + f + f+...) SS1730A 100 PLANE TRIGONOMETRY. Finally, put 2n + l for Xy and W 4- 1 loge — ^ = log, (w + 1) - log, n n .Mog.(»+l)=lo,.»+2{^-Kl(^)V|(^)*+...j, a series which is rapidly convergent. > 80. From this series a table of logarithms to the base e may be computed. To find log, 2 put w = 1. Then, since log, 1 = 0, the series becomes log,2 = log.l + 2{| + ^ + ^ + ^ + ^_ 11-3" 13. 3» }-«• 693147. The computations may be arranged thus : 3 9 9 9 9 9 9 2.00000000 .66666667 = .66666667 .07407407-*- 3 = .02469136 .00823046 -»- 5 = .00164609 .00091449 -f- 7 = .00013064 .00010161+ 9 = .00001129 .00001129 -i- 11 = .00000103 .00000125 -f- 13 = .00000009 .69314717 whence log; 2 = 0.693147, correct to 6 decimal places. To find log, 3, put » = 2, and log.3 = log,2 + 2(l + ^ + ^ + ^ + ^ + ...). SERIEa 6 2.00000000 25 .40000000 .40000000 25 .01600000 + 3 = .00633338 25 .00640000 + 5 = .00012800 25 .00025600 -J- 7 = .00000366 .00000102 + 9 = .00000011 .40546510 log. 2 = .69314717 101 .% log, 3 = 1.098612, correct to 6 decimal places. log, 4 = 2 X log, 2, log, 6 = log, 3 + log^ 2, etc. (Why ?) The logarithms of prime numbers may be computed as above by giving proper values to w. 8L Having computed the logarithms of numbers to base «, the logarithms to any other base may be computed by means of the following relation : Let loga n=:x; then a^=n. Also, logft w = y ; then b^ = w. Hence, log^ (a^ = log« (J^i and .'. x^i/logah. It follows that loga n = log^ n • log^ h ; whence log^ n = logfa n • - • logo ft This factor ; is called the modulvs of the system of logarithms to base 5. Using it as a multiplier, logarithms of numbers to base b are computed at once from the loga- rithms of the same numbers to any other base a. 102 PLANE TRIGONOMETRY. 82. To compute the common logarithms. Common logarithms are computed from the Naperian by use of the modulus log. 10 9 ».^. logio n = log, n loge 10 By Art. 80, log^lO can be, found, and •- — — - as .434294, the moduliis of the common system. loffe 10 Ex. Compute the common logarithms of : 2, 3, 4, 6, 6, 10, 15, 216, 3375. COMPLEX NUMBERS. 83. In algebra it is shown that the general expression for complex numbers is a + W, where a represents all the real terms of the expression, b the coefficients of all the imagi- nary terms, and % is so defined that i^ = — 1 ; whence t = V— 1, i* = — 1, t^ = — 2, f* = 1, etc. The laws of operation in algebra are found to apply to complex numbers. Moreover, it is further shown that if two complex numbers are equal, the real terms are equal, and the imaginary terms are equal ; i,e, if a + bi = c + di^ then a=sb and c = d. Finally, the complex niunber may be graphically represented as follows : '0 r Flo. 43. The real number is measured along OX^ a units ; the imaginary parallel to OF, 6 units. The line r is a graphic representation of a + DE MOIVBE'S THEOREM. 103 Since a = r cos 6 and J = r sin 6^ .". a + if = r (cos 6 + i sin ^). The properties of complex numbers are best developed by using this trigonometric form. If r be taken as unity, then cos 6 + % sin 6 represents any complex number. 84. De Home's Theorem. To prove that, for any value of w, (608 + i sin 0)** = cos nO + i sinnO. I. When w is a positive integer. By multiplication, (cos a -♦- i sin a) (cos /8 + i sin ff) = cos a cos fi — sin a sin, /8 + f (sin a cos /8 + cos a sin /8) = cos(a + )8) + tsin(a + /8). In like manner, (cos a + i sin a) (cos /8 + f sin )8)(cos 7 + 1 sin 7) = cos(a + )8 + 7) + i sin (a + /3 + 7) 5 and finally, (cos a+i sin a) (cos P+i sin /8)(cos 7+i sin 7) ••• to n factors = cos(a + )8 + 7 + •••) + « sin(« + /8 + 7+ ••*)• Now let a = /8 = 7 = •••, and the above becomes (cos a + i sin «)*» = cos na + i sin na. II. When w is a negative integer. Let w = — m ; then (cos a + f sin «)*» = (cos a + i sin «)""• (cos a + i sin a)*" cos ma + i sin wa cos ma — f sin ma (cos Twa + i sin m«)(cos ma — t sin ma) cos ?w« — i sin ?w« cos^ ???« + sin^ ma = cos ma — f sin ma = cos (— m) a + 1 sin (— wi) a. 104 PLANE TRIGONOMETRY. Substituting n for — m, the equation becomes (cos a + i sin a)" = cos na + i sin na. III. When w is a fraction, positive or negative. p Let n = — , ^ and q being any integers. Now cos - + 1 sin - = cos a • - + i sin 7 • - = cos a + 1 sm a (dv I > Then (COS - + f sin - J = (cos a + i sin «)?• Raising each member to the power p^ Tcos a + % sm «)« = cos - -♦- i sm - = cos ~ a + 1 sin ^ a. \ 9 qJ q q COMPUTATIONS OF NATURAL FUNCTIONS. 85. The radian measure of an acute angle is greater than its sine and less than its tangent, i.e. sin a<a< tan a. Let a be the circular, or radian, measure of any acute angle A OP. Then, in the figure, area of sector OAF < area of triangle OAT^ i.e. iOA*2LTcAP<^OA'AT. .-. arc.4P<^T: Now, since N^P < arc AP, NP ^ SLTcAP AT OP OP OP But whence bxgAP OP = circular measure of AOP = a ; 8ina<a<tana. NATURAL FUNCTIONS. 105 86. Since sina<a<tan€e, i<-A-< ^ Sin a cos a Hence, however small a may be, -: — lies between 1 and ^ sm a When a approaches 0, cos a approaches unity. cos a Therefore, by diminishing a sufficiently, we may make -, — differ from unity by an amount less than any assign- sma able quantity. This we express by saying that when a approaches 0, -: — approaches unity as a limit, i.e. — — = 1, sin a sm a approximately. Multiplying by co«a(= 1, nearly), we have ^ = 1, approximately. Whence, if a approaches 0, tan a tan az=sina=za^ approximately. 87. Sine and cosine series, cos na + { sin na = (cos a + i sin «)•, (De Moivre's Theorem). Expanding the second member by the binomial formula, it becomes, , r^Cw— l)Cw — 2) - . •9*3 ^ — ^^ ^Av ^cos" ' a . p sm' a li Substituting the values of »*, t*, «*, etc., we have . • • « w (n — 1 ) -_» • * cos na + 1 sin »i« = cos" «6 i-— — ^-cos" 'asin'a If ^ n(n-lXn-2Xn-3) ^^^,_ „ ^^, „ _ ... li + ifn cos*-' g sin « - " (" ~ |X» - ^) cos*-* a sin* «+...) lOa FLANE TKIGONOMETRY. Equating the real and imaginary parts in tlie two members, ^ nTw - V)(n - 2)(w - 3) -.4 ^ . 4 ^ [4 and m na ss n co*"*"^ a ^ "" ^^ """"^ cofi""* a mV a + —. Ex.1. Find cos 8a; sin 3 a. In the above put n = 8| and cos 3 a = cos' a — 3 cos a sin^a = 4 cos* a — 3 cos a; = 3 sin a — 48in'a. also "Mvv ^ '\- sinr a = 3 cos^ a sin a — sin' a 2. Find sin 4a; cos 4 a; sin 5 a; cos 5 a. It will be noticed that in the series for cosna and Wnna the terms are alternately positive and negative, and that the series continues till there is a zero factor in the numerator. 88l If now in the above series we let na == d, then -(--1) a\a J cos 6 = cos" a r^ 008""^ a sin^ a a\a J\a J\a J . . . ^ — ^— 1-; ^ cos* * a sin* a— ••• 0(0^ a) -_2 /sinaV = cos" a ^^— - — ^cos*. ^ al 2 V a / 11 (4 \ « y If now remain constant, and a decrease without limit, then will n become indefinitely great, and -. — - and every a NATURAL FUNCTIONS. 107 power thereof, and cos a and every powet of cos« will approach unity as a limit, so that Similarly, **^* = *"^"*'^""^"^':- By algebra it is shown that these series are convergent for all values of 0. By their use we can compute values of sin and C09 to any required degree of accuracy. Show from the above that tan 0=^0 + -^ + -— - + -.. Ex. 1. Compute the value of sin V, correct to 5 places. In sin $ = $ ^ — + _ — — -f •.., make $ the radian [3+16 [7 + measmp of 1» = -|L = 0,01745 +. 180 Then, 6 = 0.01745 + .? = 0.0000008. Li .-. sin tf = 0.01745 +. The terms of the series after the first do not affect the fifth place, so that the value is given by the first term, an illustration of the fact that, if a is small, sin a=: a, approximately. Compare the value of tan 1°. 2. Show that sin 10° = 0.17365 ; cos 10° = 0.98481 ; sin 15° = 0.25882 ; cos 60° = 0.50000. 3. Find the sine and cosme of 18° 30' ; 22° 15' ; 67° 45'. It is unnecessary to compute the functions beyond 30°, for since sin (30° + tf) + pin (30° - tf) = cos tf (why ?), .-. sin (30° + ^) = cos tf - sin (30° - $}. So,also, cos (30° + ^) = cos (30° - ^) - sill tf . Giving proper values the functions of any angle from 30° to 45° are determined at once from the functions of angles less than 30°. Thus, sin 31° = cos 1° - sin 29° ; cos 31° = cos 29° - sin 1°. 4. Find sine and cosine of 40° ; of ^50°. 108 PLANE TRIGONOMETRY. I The following are sometimes useful in applied mathematics : Ex. 1. To find the Bum of a series of sines of angles in A. F^ such as sin a + sin (a + )8)+ sin (a + 2)8)+ ... + sin (a-^-ln- l]/3). 2sinasin^= cosf a — ^J — cosf a + q)» 2sin(a + )8)sin| = cos(a + |)-co8^a+^), 2sin(a + 2)8)sin| = co8fa+^)-co8fa+^], 28m (a + [n - l]/3) Bm^ = ooB(a +^Jt^p\ - cos f « +?iLli^y Adding 2{sin « + sin (a + )3)+ sin (a + 2/3)+ ... + sin (a-\-[n- l])3)}sin| = co8(«-|)-ooe(« + 2iL^V) = 2 sin (a +2^/3) sin ^^. . .•. sin a + sin (« + fi}+ sin (« + 2^)+ ... + sin (« + [n - 1]^) sin (« + r^p) sin5^ Similariy it can he shown that cosa + co8(a + /?)+ co8(a + 2)3) + — + cos(a + [n - 1])8) cos (a + 2^)3) sin|i3 sin^ 2 HYPERBOLIC FUNCTIONS. 109 90. Th^ series e^==l-|-a;-|--~ + ~H |-±l + ...i8 proved in higher algebra to be true for all values of x^ real or imaginary. Then if a; = iff^ c** = 1 + t^+—- + -— + -. + -— + ... [2 |3 [r .-. e«* = <?(?« ^ + i «m e (Art. 87). In like manner, e*'* ==co80 — i sin tf . Adding, co« ^ = e<« -I- e'^ subtracting, sin = 2» HYPERBOLIC FUNCTIONS. 9L Since sin ^ = — tt-, , and cos == ^ are 2t 2 true for all values of ^, let 6 = i^. Then, sin (i0) = ^^^^^-^ = i ^——^ = i sinh 0, and cos (i0) « ^ "^^ =a cosh ^, so that tan (f ^) = ^Hllg} » \l^ = ; tanh ^, COS {%0) cosh ^ where tinh 0^ co%h 0^ tanh 0^ are called the hyperbolic sine^ cosine^ and tangent of 0. The hyperbolic cotangent, secant, and cosecant of are obtained from the hjrperbolio sine, cosine, and tangent, just as the corresponding circular func- tions, cotangent, secant, and cosecant, are obtained from tangent, cosine, and sin6. The hyperbolic functions have the same geometric relations to the rectangular hyper- 110 PLANE TRIGONOMETRY. bola that the circular functions have to the circle, hence the napie hyperbolic functions. stnh6 = ^ J , .'. C8ch6= ^ ° A eogh6 = — \^ ■, .-. gech0 = -T- -A tanh6 = ^^-^"^ . ... cothe = 4-±-^. 92. From the relations of Art. 91 it appears that to any relation between the circular functions there corresponds a relation between the hyperbolic functions. Since cos? (id^ + sin^ (i^) = 1, cosh^ e + i^ sinh^ 5 = 1, or cosh' d — sinh' 5 = 1. This may also be derived thus : cosy - sinh* e =(^'s^'-(^lifrj c» + 2 + e-» - e» + 2 - e-» . 4 ^' Also since sin (ia + i)8) = sin (ia) cos (i)8) + cos (ia) sin (t)8), .•. i sinh (a + ^) = i sinh a cosh y9 + cosh a • i sinh )8, and sinh (a + y9) = sinh a cosh )8 + cosh a sinh )8. Let the student verify this relation from the exponential values of sinh and cosh. EXAMPLES. Prove 1. cosh (a + )3) = cosh a cosh + sinh a sinh p. 2. cosh ( «+ )3) — cosh (a — )8) = 2 sinh a sinh j8. 3. co8h2tf=l + 28inh2tf = 2cosh»tf-l. 4. sinh 2 a = 2 sinh a cosh a. EXAMPLES. Ill ^ « ^ ^ ^ I 6. sinli 3 = 3 sinh tf + 4 sinh* $. 7. sinhtf + sinh ff>= 2 sinh "^ ^ cosh ~ r , ' ..... 8. sinh a+sinh (a + )8) + sinh (a + 2 )3) + — + sinh (« + [n - 1])3) sinh ia + ^^/S) sinh|)3 sinh? 2 ^ ^ 1 + tanh tanh 6 10. sinh^* X = cosh"! VT+^ = tanh-^— - 11. cosh (a + P) cosh (a - /8) = cosh' a + sinh^ /J = cosh* ft + sinh* a. 12. 2 cosh na cosh a = cosh (n + 1) <% + cosh (n — 1) a. 13. cosh a = J (c« + e-*) = 1 + |L + ^ + .... 14. -sinh a = J (g* - e-*) = a + J^ + ^ + «.. ^^ ^"^ 15. tanh-ia + tanh-ii = tanh-i-^-tA. 1 + cw SPHERICAL TRIGONOMETRY. -•>oj»:o«- CHAPTER X. SPHERICAL TRIAITOLE8. I 93. Spherical trigonometry is concerned chiefly with the solution of spherical triangles. Its applications are for the most part in geodesy and astronomy. The following definitions and theorems of geometry are for convenience of reference stated here. A great circle is a plane section of a sphere passing through the centre. Other plane sections are %mall circles. The shortest distance between two points on a sphere is measured on the arc of a great circle, less than 180^, which joins them. A spherical triangle is any portion of the surface of a sphere bounded by three arcs of great circles. We shall consider only triangles whose sides are arcs not greater than 180^ in length. The polar triangle of any spherical triangle is the triangle whose sides are drawn with the vertices of the first triangle as poles. If ABC is the polar of ^'5'(7', then A'B^O^ is the polar of ABC, In any spherical triangle, The sum of two sides > the third side. The greatest side is opposite the greatest angle, and conversely. .Each angle < 180^ , the sum of the angles > 180°, and < 540^ JEach side < 180° ; the sum of the sides < 360°. 112 SPHERICAL TRIANGLES. 113 The sides of a spherical triangle are the supplements of the angles opposite in the polar triangle^ and conversely. If two angles are equal the sides opposite are equals and conversely. The sides of a spherical triangle subtend angles at the centre of the sphere which contain the same number of angle degrees as the arc does of arc degrees ; i.e. an angle at the centre and its arc have the same measure numerically. The arc does not measure the angle for they have not the satne unit of measurement, but we say they have the same numerical measure ; ue. the arc contains the unit arc as many times as the angle contains the unit angle. * The angles of a spherical triangle are said to be measured by the plane angle included by tangents to the sides of the angle at their intersection. They have therefore the same numerical measure as the dihe- dral angle between the planes of the arcs. In the figure the following have the same numerical meas- ure : arc a and angle a ; arc ( and angle fi ; arc c and angle 7 ; plane angle A'BC; spherical angle B and dihedral angle A-BO-0; spherical angle O and dihedral angle B^CO^A ; spherical angle A and dihedral angle C-A O^B. A^C'B and C'A'B have not the same measure as spherical angles C and A, for BA^, A'C'f C'B are not perpendicular to OA or OC. 94. In plane trigonometry the trigonometric functions were treated as fimctions of the angles. But since an angle and its subtending arc vary together and have the same 114 SPHERICAL TRIGONOMETRY. numerical measure, it is clear that the trigonometric ratios are functions of the arcs, and may be so considered. ^11 the relations between the functions are the same whether we consider them with reference to the angle or the arc, so that all the identities of plane trigonometry are true for the func- tions of the arcs. Thus in the figure we may write, y . y sin a ss - or sin a = - ; r r sin^a + cos*a = 1, or sin^a + cos^a = 1 ; cos 2 a = 2 cos^ a — 1, or cos 2 a = 2 cos^ a — 1. GENERAL FORMULiE FOR SPHERICAL TRIANGLES. 95. The solutions of spherical triangles may be effected by formulae now to be developed: First it will be shown that in any spherical triangle cos a = cos 6 cos c + sin 6 sin c cos ^^ cos 6 = cos c COS a + sin c sin a cos By COS c = cos a cos 6 + sin a sin 6 cos C* The following cases must be considered : I. Both 6 and <? < 90^ III. Both 6 and (? > 90^ II. h > 90% c < 90^ IV. Either lore = 90^. V. J = <? = 90^ The figure applies to Case I. Let ABO be a spherical tri- angle, a, 5, c its sides, and the centre of the sphere. Draw AC^ and AS tangent to the sides 6, c at A. (The same result would be obtained by drawing AB\ AO perpen- dicular to OA at any point to GENERAL FORMULiE. 115 meet OjB, 0(7.) Since these tangents lie in tbe planes of the circles to which they are drawn, they will meet 00 and OB in (7' and B\ and the angle C^AB^ will be the measure of the angle A of the spherical triangle ABC. Since OAB\ OAC* are right angles, AOB\ AOO^ must- be acute, and hence sides ^, h are each < 90^. In the triangles 0* AS and COB, (7'5'2 ^AC'^ + AB'^ --2AC^ ^ Aff cos C^AB\ and B' 0"^ = 0(7'a + p5'2 ^2 00' ^ OB' cos (7'05'. Subtracting and noting that cos CAB' = cos A and cos COB' — cos a, we have = pC'a - ^ C"2 + 05'2 - ^IjB'a . +2AC 'AB> cos A-^OC -OB' co6a. But OfT'a - ^ (T's = OA" and 0B>^ - AB'^ = OAK Hence, = O^" ^. ^(7/ . ^^f cos A-OC • OB' cosa; cosa = — .— + _._co8A ••• cos a =s cos h cos <? + sin h sin c cos -4.. Similarly, cos h = cos a cos (? + sin a sin e? cos J?, and cos (? = cos a cos 6 + sin a sin 6 cos C. These formulae are important, and should be carefully memorized. II. 6 > 90^; (?<90°. In the triangle ABO, let h >90* and c < 90°. Complete the lune BACA'. Then in the triangle AVB the sides a and J.'(7are both less than 90®, and by (I) cos A'B S3 cos A'O cos a + sin J.'(7 sin a cos A' OB. 116 SPHERICAL TRIGONOMETRY. But ^'^ = 180^-c, ^'(7= 180^- J, and ^'C!B=180*'-.a .-. cos (180^ - c)= cos (180^ - 6) cos a + sin (180^ - J) sin a cos (180*^ - C) ; or — cos (? =a (— COS h) COS a + sin 5 sin a (— cos (7), and COS c = cos a cos i + sin a sin I cos (7. A similar proof will apply in case e > 90°, I < 90*^. III. Both J and (? > 90^ In the triangle AB(7, let both I and (?>90®. Complete the lune ABA'C. Then since A^O and ^'5 are both < 90'', cos a = cos A^O cos ^'5 + sin A^C sin ^'jB cos J.'. But A' = ^, ^'(7= 180° -ft, ^'5=180°-c. .-. cos a = cos (180° - 6) cos (180° - c) + sin (180° - 6) sin (180° - <?) cos A ; or cos a ^ cos 5 cos c? + sin I sin ^ cos A, Cases IV and Y are left to the student as exercises. 96. Since the angles of the polar triangle are the supple- ments of the sides opposite in the first triangle, we have -4' a' = 180°-^, 6' = 180° -5, c'=180°-(7, -4' = 180° -a. Substituting in cos a' = cos V cos <?' + sin V sin c' cos A\ we have cos (180° - ^)= cos (180° - jB) cos (180° - (7) + sin (180° - 5) sin (180° - (7) cos (180° - a); or — cos-4.=s( — cos£)(— cos(7)+8in£ sin (7(— cos a). GENERAL FORMULA 117 Changing signs, eos ^ = - eo8 B cos C + sin B sin C cos a. Similarly, CMB = -co8^co8C + 8ln^8iiiCGOs&9 and eos C = - cos^ cosB + sin ^ sin B cos c# tvn T u • 1 4. • 1 i. sin A sin B sin C 97. In any spherical triangle to prove — : =« — : — r= — r Since cos A = sin a sin 6 &mc cos a — cos I cos c sin I sin c ,*. sin^-4. _ ^ __ /cos a — cos 6 cos gV \ sin 6 sin c J __ sin^ 6 sin^ c — (cos a — cos b cos g)^ sin^ b sin^ c _ (1 — cos^ 6) (1 — cos^ c) — (cos g -^ cos b cos g)^ sin^ b sin^ g 1— cos^a— cos^J — cos^g+2cosacos6 cose? sin^ 6 sin^ c Hence, ^ Vl — cos^a— cos^J— cos^g— 2cosa cosJ cose sin -A = : — ;— : sin 6 sm c J sin^ __ VI — cos^g — cos^ft—cos^g— 2cosa cos 6 cose sin a sin a sin 6 sin c By a similar process, 51 — _ and 51 — will be found equal sm sin c to the same expression. • sin ^ _, sin B _, sin C^ " sina sin 6 sine 118 SPHERICAL TRIGONOMETRY. 9B. Expressions for sine, cosine, and tangent of half an angle in terms of functions of the sides. We have 2 sin^— = 1 — cos A _ 1 cosg — cost cosg sin b sin c _ cos h cos € + sinb sin c — cos a sin b sin c _ cos (6 — g) — cos a Bin b sin c Then 2 sin'^ = ^Binl(^a+b-c)Bin^(a-b+e) ^^^ ^^^ 2 sin b sm g ^ _ 2 sin (g — t) sin (g — e) sin ( sin g when 2s = a + b + c. 2 "-^ sin^sinc SimUarly, «in f = Jii*Hii^^EZ, '' 8 ^ gin a slue and gin C^J^^iiLz ^Tan^s - a) , 2 ^ Bin a sin 6 ; Also from the relation 2cos2^=l + cos^ -I , cos a — cos b cos ^ = IH : — i— r- — ' sm sin g we have em 4= >fa*8i» (*-«) . Also, cogf=A^pi^i^^, 2 ^ sine sin a and cwig=J»M£MiL=^, 2 ^ sin a sin 6 GENEKAL FORMULA. 119 From the above, . a 8in -:r- 1^;^^;; 2 ^ J8iii(« - 6) sin (« - c) 2 ^g:4 ^ 8in«8lii(«-a) 2 Also, tan f = v/gtoIJbi- «) "l" (* - «) , 2 ' 8ill«8lll(« — 6} and t^gJglnl^^ a) gin (*-&) . 2 ^ 8iii«8iii(tf-c) Compare the formulaB thus far derived with the torresponding for- mulae for solving plane triangles. The similarity in forms will assist in memorizing the formulae for solving spherical triangles. 99. From the formulae of Art. 96, the student can easily prove the following relations : sin ^ - J - C08 S cos (;S - A) 2 ^ sin £ sin C wherjB 2Sf = J. + J5+C* i Bln| = l, rin|=f. 2 ^ sinBBinC C08| = f, eos| = t. 2 tan — -^ / - cos H cos (S — ^) 2 ^cos(S-B)co8(S-C) tonf^l. tan| = ?. 120 ' SPHERICAL TRIGONOMETRY. 100. Napier's Analogies. Since 2 _ ^ sin g sin (^ — a) tan:? J sin(g- g)8in(g~ a) 2 ^ sin « sin (« — 6) • _ /sin^(8 — 6) _ sin (g — S) ^ ^sin^ (« — a) sin (« — a) * by composition and division, A B tan — + tan— . , , 2 2 __ sm (« — 6) + sin (« — a) tan^-tan^"®^^(*-*)-«^^C^-^)' 2 2 sm — • sm -rr- COS — COS — 2 2 ^ sin j^(2 g - g - 6)cos ^(g - V) sin- sin-"^^®^<^^*'"^'"*)®^^i<^^~*)' 2 2 (Art. 61) A B cos- cos- sin ^jA + ^) _ tan ^(2 g -- g - 6) sin ^(^ - J5) "" tan J(g - 6) tan- tan ^ (g — i) , since 2« — g — 6 = (?. .-. tan I (a - 6) = — ^ tan|. To find an expression for tan J (J. — jB) we have only to consider the polar triangle, and by substituting 180° — A for g, etc., 180° — a for A, etc., we have the following relations : i(g - J)= K180° - ^- 180° + 5) = - JC^ - 5); GENERAL FORMULA 121 also, J(^-5) = -i(a-6); J(-4 + 5) = i(180''-a + 180°-i)=180»-J(a + J); and £. = 90°-^. 2 2 The formula then becomes, applying Art. 29, ton^(^-B) = — ? coti. 8liil(a+6) * Formulae for tan J (a + 6), tan J (A + B) are derived as follows : Since tan^ ^ tan^ ^ J 8in(g~6)sin(i?-c) ^ / 8in(i?.^g)8in(j^~a)^ 2 2 ^ sin « • sin (« — a) ^ sin « • sin (« — J) sm — sin — 2 2 ^ 8in(^-c) cos4cos^ si^« By composition and division, A B^ . A . B cos— cos— + sm— sm— 2 2 2 2 _ sing + sin(g-g) ^^A^^B '^A'^B sin « — sin r« — {?") cos— cos— — sm— sm-— oxx^^o ^/^ 2 2 2 2 whence ^"^f^^--g) = ^^^K^ + ^), (Art. 61) cosi(A + 5) ^^^_c since 2« — <?=a + J» or, • taii|(a + &) = — ? tan^. C08|(^ + B) * 122 SPHERICAL TRIGONOMETRY. The value of tan ^- QA + 5) is derived by substituting in terms of the corresponding elements of the polar triangle. cosK^-^) ^ -tan^(^ + ^) -ooH^ia + b) cot- 2 .-. tsakhA + B)= — f-^ i^cot^. cwjCa + ft) * Similar relations among the other elements of the triangle may be derived, or they may be written from the above by proper changes of J., B^ (7, a, 6, c in the formulae. The stu- dent should write them out as exercises. 101. Delambre's Analogies. Since sin J (-4 + -B) = sin -5- cos — + cos — sin •-, then «a^UA+B\ = 8in (»-&)+ sin Co -g) ^ /sin « ■ sin (« -_£2. sin c ^ sin a • sin 6 (Art. 98) Hence s^p^(^ + -S) ^ sin (« - i) + sin (« - a) cos^' "«'•' 2 2 sin ^ cos J (« — 5) 2» c c sin-cos- cog I (a - 6) f, and aaUA + B)= — ?— - — co8|j C08| 2 In like manner derive 8ini * 2 ^ (Art. 61) BIGHT SPHERICAL TRIANGLES. 123 C08| 2 C08|(-4-B)=— S— Sillf. 8ill| * These formulae are often called Gauss's Formulse, but they were first discovered by Delambre in 1807. Afterwards Gauss, independently, dis* covered them, and published them in his Theoria Motus, 102. Formulae for solving right spherical triangles are derived from the foregoing by putting = 90°, whence sin (7=1, cos (7=0. cos e = cos a cos 6 + sin a sin b cos (Art. 96) becomes cos c = cos a cos 6. (1) Substituting the value of cos a from (1), and simplifying, .._ . cos a -cos 6 cos <? (Art. 96) becomes cos ^ = ^^. (2) in the right triangle is 8in I sine? COS^ tanc* sin J. sin a sin (7 sine? \ is gin /I _glna sine* Again, Siii^ = ^^^ (Art. 97) (3) Dividing (3) by (2), sin a cos h sin a cos a cos h sin a tan J. = cos c sin h cos c cos a sin h cos a sin h since cos a cos h = cos <?. .*. tan^ gin 6 = tana. (4) 124 SPHERICAL TRIGONOMETRY. From (4) tan a = tan J. sin 5, also, tan b s= tan B sin a. Multipljring, tan a tan b =: tan A tan B sin a sin b^ or, oot^cotB = oasaoas& = oosc. (5) From (2) and (3), by division, tan 5 cos A tan c cos c — : — - = -: — r = 7 = cos a. sm ^ sm cos 6 sin c .*. cos^ = cosa sinB. (6) Let the student write formulae (2), (3), (4), (6) for B. It will be noticed that (1) and (5) give values for c only, while (2), (3), (4), (6) apply only to A and B. 103. Formulae (l)-(6) are sufficient for the solution of right spherical triangles if any two parts besides the right angle are given. They are easily remembered by comparison with corresponding formulae in plane trigonometry. Two rules, invented by Napier, and called Napier^s Mules of Cir- cular Parts, include all the formulae of Art. 102. Omitting (7, and taking the comple- ^*^-S ments of A, <?, and B, the parts of thie triangle taken in order are a, 5, 90*^—^1, 90^ - (?, 90^ - B. These are called the circular parts of the triangle. Any one of the five parts may be selected as the middle part, the two parts next to it are called the adjacent parts, and the remaining two the opposite parts. Thus, if a be taken as the middle part, 90® — B and b are the adjacent parts, and 90® — <?, 90 — A the opposite parts. NAPIER'S RULES. 125 Napier's Two Rules are as follows : The sine of the middle part equals the product of the tangents of the adjacent parts. The sine of the middle part equals the product qf the. cosines of the opposite parts. It will aid the memory somewhat to notice that i occurs in sine and middle, a in tangent and adjacent, and o in cosine and opposite, these words being associated in the rules. The value of the above rules is frequently questioned, most computers preferring to associate the formulsd with the corresponding formulae of plane trigonometry. These rules may be proved by taking each of the parts as the middle part, and showing that the formulsB derived from the rules reduce to one of the six formulae of Art. 102. Then, if 5 is the middle part, by the rules, sinJ = tanatan(90*' — A)=tanacotA, or tanJ.=-r^, sin sin I = cos (90^ - <?) cos (90^ - B) =sin c sin B, • i> sinJ • or sm -B = -: — , sin<? results which agree with (4) and (8), Art. 102. If any other part be taken as the middle part, the rules will be found to hold. 104. Area of the spherical triangle. If r = radius of the sphere, • U^ spherical excess of the triangle = A+-B+ (7— 180% A = area of spherical triangle, then by geometry A = Er^ X -?-. ISO If the three angles are not known, U may be computed by one of the following methods, and A found as above. 126 SPHERICAL TRIGONOMETRY. Cagnoli's Method. sinf = sin J(A + B^C- 180^) Si = sin J (^ + 5) sin^- COS ^ (^ + ^) COS § . sin— cos— = [cob J (a - J) - cos J (« + ^)] (^Ajij. 101) cos^ 2sin-sm- . __ 2 2 ^ V sin 8 sin (« — a) sin (« — J) sin (« — c) cos - sin a sin 6 * 2 (Arts. 51, 98) ^■E_. Vglng gin (8 - a)gln(8 - 6)rin(8 — e) ^ 2GOg|cog|GOg| 8 8 2 Lhttllier's Method. 4 008^(^ + 5+ (7- 180") Now, multiply each term of the fraction by 2coa{CA + B- + 180°), and by Art.. 51, (1) and (3), the equation becomes fi ^ sinJ(^ + 5)-co8- tan- = \ , * cosJ(^ + -B)+8in-^ [cos J (« - *) - cos|1cos - -•- ^-^ ^ (Art. 101) cos J(a + J)4- co8| sin- sin 1^(8 — })sin \(^8 — cos -cos J ( in J(« — a) / sin<sin(g — g) ,(s-e) ^sin(» - a)sin(« - 6)' (Art. 51) AREA OF SPHERICAL TRIANGLES. 127 By Art. 52, introducing the coe£Bcient under the radical, tan^= Vtan ftaii^ (« - a)taiik8 - &)taii^ (« - e). If two sides and the included angle are given, U may be determined as follows : cos:f = cos iCA + B + 0- 180^) = cos ^(A + £) 8in-^+ sin ^(A + B) cos-^ = COS K« + *) sin^-^ 4- cos J (^ - *) ^os^-^ (Art. 101) 2 2 cos - COS ;- 4- sm - sm - cos C 2 2 2 2 cos- 2 « sm - sin - • 2 sm-- cos— But sin— = . (Cagnoli's Method) cos- 2 Dividing this equation by the above, mn - «w - 8in U , U 2 2 2 « 6 , . a . 6 /v cos - co« - 4- ««w :r 8in-co8 U 2 2 2 2 .a, This formula is not suitable for logarithmic computations. Usually it is better to compute the angles by Napier's Analo- gies, and solve by A = JEr^ x 180 128 SPHERICAL TRIGONOMETRY. EXAMPLES. 1. Show that cos a = cos ( cos c + sin b sin c cos A becomes sec ^ = 1 + sec a, when a = h = c. 2. Ifa + ft + c = T, prove B C (a) cos a = tan -- tan ■—. /1.x ^^„2 A cos a {p) cos^ •— = -:- 2 sin 5 sin c (c) sin^ — = cot 6 cot c. (d) cos -4 + cos 5 + cos C = 1. (e) 8in« I + sina | + sin^ |^ = 1. 8in:=^cos i (^ — ^ ) sin^sini (« — a) 3. Prove ^ \ =- = (Art. 104) Sin -— cos - 2 2 4. Show that cos a sin & :± sin a cos 6 cos C + sin c cos A. CHAPTER XL SOLUTION OF SPHERICAL TRIANGLES. 105. According to the principles of Bfpherical geometry any three parts are sufficient to determine a spherical tri- angle ; the other parts are computed^ if any three are given, by the formulae of trigonometry. The known parts may be : I. Three sides, or three angles. II. Two sides and the included angle, or two angles and the included side. III. Two sides and an angle opposite one, or two angles and a side opposite one. It will appear that, as in plane geometry. III may be ambiguous. The signs of the functions in the formulae are important since the cosines and tangents of arcs and angles greater than 90° are negative; whether the part sought is greater or less than 90*^ is therefore determined by the sign of the function in terms of which it is found unless this function be sine. In this case the result is ambiguous, since sin a and sin (180® — a) have the same sign and value. Thus if the solution gives log sin a = 9.56604, we may have either a = 21'* 33', or 168^*27'. The conditions of the problem must determine which values apply to the triangle in question. The negative signs, when they occur, will be indicated *^^® • log cos 115® 20' = 9. 63185-, indicating, not that the logarithm is negative, but that in the final result account must be made of the fact that cos 115® 20' is negative. 129 180 SPHERICAL TRIGONOMETR¥w 106. FormuloB for the solution of triangles. I. sin ^ _ gin B _ gin C sin a sinfr ginc* II. tiiii^=\^^2^ZE35lZES. 2 ^ sin 8 gin (« — a) III. tan^-V"^^^^^^"^^, 2 ^cog(iSf-B)cos(iSf-C) glnJC^-B) IV. tani(a-6) = -^ tanf. * gln|(^ + B) 2 c©gl(^-B) * cogJ(-4 + B) 2 ginl(a-6) ^ VI. tan|(^-B) = — f; --cot|. VII. tan|(^ + B) = — I- -cot|. * co9|(a+6) * VIIL A = Er^ 180' where H is determined by tan^=A/*«n5tani(«-a)tain|(«-6)tan|(«-c). 4 « 8 8 8 8 Right triangles may be solved as special cases of oblique triangles, or by the following : (1) cogc =co8acos&. (4) tan ^ sin & = tan a. (2) C08^=^. (5) cotAcotB = c©9C. (3) 8*n-^=f^* (6) cog^ = co8a8inB. The formula to be used in any case may be determined by applying Napier's Rule of Circular Parts. 107. In solving a triangle the student should select formulae MODEL SOLUTIONS. 181 in which all parts save one are known, and solve for that one (see page 77). Referring to Arts. 105 and 106, it will appear that solutions are effected as follows : Case I by formulaB II, or III, check by I. Case II by formulsB VI, VII, I, or IV, V, I, check by IV orVL Case III by formulaB I, IV, or I, VI, check by VI or IV. MODEL SOLUTIONS. lOa 1. Given a = 46<> 24', b = 67^ 14', c = 81^ 12'. Solve. tan^=\^HiLE^}ililIiZ£i, tan ^ -J^^ (' " ^) "^ (' ~ ^, 2 ^ sin 5 sin (« — a) 2 ^ sin « sin (s — b) ^^ C Jsm(s^a)sin(8--b)^ ^^^^^ . iBina ^ jm6 . 2 ^ 8in5sin(5— c) sin il sin^ Arrange and solve as in Example 1, page 80. Ans. A = 46** 13'.5, B= , C = Solve: (1) ^=96° 45', 5 = 108° 30', C = 116*»15'. (Use formal» III in the same manner as in Example 1.) (2) a = 108^14', 6 = 75° 29', c = 56°37'. (3) ^ = 57*50', 5 = 98^20', C = 63°40'. 2. Given b = 113° 3', c = 82° 39', A = 138° 50'. Solve, "^ cosl(6 + c) 2 ^ ^ sinj(6 + c) 2 i(B + Cr)±i(B--C) = B,OTC, 8ina=515LiiJ^. ^ sin J5 Check: tan2 = tanKft~c)8in}(^ + C) 2 sin J (5 - C) 6=113° 3' log cos 1 (6-c) =9.98453 logsm J (&-c) =9.41861 c = 82° 39' colog cos J (6 + c) = 0.86461" oolog sin i (6 + c) = 0.00409 J (&+<?)= 97° 51' log cot- =9.57466 log cot -=9.57466 J(ft-.c)= 15° 12' ^ 2 ^ 2 J -4 = 69° 25' log tan J (5 + C) = 0.42380" log tan J (5 - C) = 8.99736 } (5+ 0=110° 39' i (5-C)=5°40'.6 i(B-C) = 5°40'.6 .-. 5 = 116° 19'.6 and C=104°68'.4 132 SPHERICAL TRIGONOMETRY. Check: log Bin A = 9.81839 log tan i (5 - c) = 9.43408 log sin h = 9.96387 log sin J (5 + C) = 9.97116 cologs in 5 = 0.04756 colog sin J (5 - C) = 1.00474 log sin a = 9.82982 log tan - = 0.40998 a = 137^ 29' 2 a = 137^29' Notice that tan I (A + C) is -. Hence, HB-h C)\b greater than 90^, i.e. IIQP 39^ Solve: (1) ^ = 68° 40', B= 56° 20', c= 84° 30'. (Use formuliB IV, V, I. Compare Example 2.) (2) a = 102° 22', ,b= 78° 17', C = 125° 28'. (3) ^=130° 5', B= 32° 26', c= 51° 6'. 109. Ambiguotts cases. By the principles of geometry the spherical triangle is not necessarily determined by two sides and an angle opposite, nor by two angles and a side opposite. The triangle may be ambiguous. By geometrical principles it is shown that the marks of the ambiguous spherical tri- angle are: 1. The parts given are two angles and the side opposite one, or two sides and the angle opposite one. 2. The side, or angle, opposite differs from 90® more than , the other given side, or angle. 3. Both sides, or angles, given are either greater than 90®, or less than 90®. In the right triangle ABO^^ sin a = sin A sin c, (formula (3)) Therefore there will be no solution, one solution, or two solutions, according as sin a = sin J. sin c, i.e. according as a = the perpendicular j3. (See Art. 65.) But the most expeditious means of determining the am- biguity is found in the solution of the triangle. The use of formula I gives the solution in terms of sine, so that it is to be expected that two values of the part sought may be possible ; and whether the triangle be ambiguous or not, there must be some means of determining which of the two \ AMBIGUOUS SPHERICAL TRIANGLES. 133 angles, a and 180° — a, that have the same sine is to be used. If there are two solutions, both values are used. This is determined in the further solution of the triangle by formula V, which may be written , J _ cos ^ (A -h 0) tan ^ (a + c) ^2 cos J (A -C) Now -<90% whence tan - is 4-. Then if for both values of (7, found by the sine formula, the second member is + , there are two solutions; if the second member is — for either value of (7, there is but one solution ; while if both values of make the second member — , there is no solution. The various cases will be illustrated by problems. 3. Given a = 62° 15'.4, h = 103o 13* g^ ^ = 63° 42'.6. Solve. gj^^^smJsin^ ^^^^c_ co8iM+^)tanH« + ft), sin a 2 qo8\{A — B) sin C = S2L^-51Ld. Check: cot^ = ^^^ K^ + ^) 8^^ J (« + ^). sin a 2 sin ^ (a — 6) Solving the first formula gives log sin B = 9.94766, whence B^ = 62° 24'.4, 5j = 117° 36'.6. For each of the values B^ and B^ cos \(A -{-B) tan \(a + h) cos \{A -^B) is + and therefore equal to tan ^* Hence there are two solutions. Find c = 163° 9'.6, or 70° 26'.4 and C = 155° 43'.2, or 69° 6'.2 4. Given a = 46° 46'.5, ^=73°ir.3, 5 = 61°18'.2. Solve. gin ^- sing sin jB ^ ^q^ C _ tan j (^4 - B) cos ^(a + h) sin ^ 2 cos i (a — &) 8inc=?lBJ«8inC^ ^^^^^. ^^^t^i^u\{a ^h)^m\{A + Bl sin il 2 sin \iA-B) 134 SPHERICAL TRIGONOMETRY. Solving for h gives log sin 6 = 9.82446, whence 6i = 4P62'.6, and 6j = 138° 7'.5. For the value h^ the fraction tan H-^ + B) cos j (o + 5) cos i(a — b) is +, but for 62 cos i (a + 6) is — , making the fraction — , and hence it can not equal cot — , which is +• There is then but one solution. Find C = 60* 42'.7, c = 41° 35M. 5. Given a = 162* 30', A = 49* 50', B = 57* 52'. Solve. Solving gives log sin 6 = 9.52274, whence b^= 19*27'.9, ft, = 160* 32'.1. For both values, h^ and h^, cos ) (a + ft) is — . Therefore, tan ^ (^ + jB) cos^ (a + 6) cos 1 (a — ft) is — and not equal to cot — • Hence the triangle is impossible. « Solve, testing for the number of solutions : (1) ft = 106*24'.5, c= 40*20', C= 38*45'.6. (2) a = 80* 50 , ^ = 131* 40', B = 65* 25'. (3) a= 60*31'.4, ft = 147*32'.l, 5 = 143*50'. (4) a = 55* 30', c = 139* 5', A = 43* 25'. RIGHT TRIANGLES. 110. Right triangles are a special case of oblique triangles, but are usually solved by formulae (1) to (6), Art. 106. Students should have no difficulty in applying these. Computers generally question the utility of Napier's Rules of Circular Parts. For those who prefer the rules a problem will be solved by their use. SPECIES. 135 6. Given c = 86*» 51', B = 18° 3'.5, C = 90«. The parts sought are a,h^ A, and it is immaterial which is computed first, a and A are adjacent to c and B^ while h is the middle part of c and £. Then by Napier's first rule sin (90^ - 5) = tan (90* - c) tan a ; ^""^ or tan a = ?2?j5 = cos B tan c, cote which is formula (2). By the same rule go—A sin (90* - c) = tan (90* - ^) tan (90° - B), or cot A = £2i£ = cos c tan 5, formula (5). cot 5 . ^ ^ Finally by the second rule sin ft = cos (90* - c) cos (90* - 5) = sin c sin 5, formula (3). The solutions give a = 86* 41'.2, 6 = 18* 1'.8, A = §8* 68'.4. Verify. 111. Species. Two angles or sides of a spherical triangle are said to be of the same species if they are both less, or both greater, than 90°. They are of opposite species when one is greater and the other less than 90°. Since the sides and angles of a spherical triangle may, any or all, be less or greater than 90°, it is necessary in solutions to determine whether each part is more or less than 90°. The directions already given are sufficient in oblique triangles. In right triangles the sign of the function will determine if the solu- tion gives the result in terms of cosine or tangent, but not if the result is found in terms of sine. Thus in Example 6, above, we have log sin 5 = 9.49068, whence 5 = 18° 1'.8, or 161° 58'. 2. By formula (4) sin 5 ==^5:54. Now sin b is •^ ^ "^ tan ^ always +? therefore, tan a and tan A must be of the same sign, whence in any right spherical triangle an ollique angle and its opposite side must he of the same species. Again by formula (1) cos c = cos a cos J. Now cos (? is + or — according as c is less or greater than 90°. If then c<90°, cos a and cos h are of the same sign, but if (?>90°, cos a and cosi are of opposite sign. Therefore, if the Bin a 8inc = 186 SPHERICAL TRIGONOMETRY. hypotenu9e of a right spherical triangle is less than 90^, the other sides, and hence the angles opposite, are of the same species; Ivt if the hypotenuse be greater than 90*^, the other sides, and the angles opposite, are of opposite species. 112. Ambiguous right triangles. When the parts given are a side adjacent to the right angle, and the angle opposite this side, the triangle is ambiguous, for solving for the hypot- 4si, enuse by formula (8) gives jj\^^^ ^y/ Sin -4. Fio. 54. from which there result two values of c. By the last rule of species it follows that to the values of c, one <90°, the other >90®, there will cor- respond two values for i, one of the same species as a, the other of opposite species. Clearly sin c ^ 1, according as sin a = sin A, and hence there will be no solution, one solution, or two solutions, according as sin a ^ sin A, Solve the spherical triangles, right angled at (7, given : (1) 5 = 73° 21'.4, <?= 84° 48'.7. (2) (? = 54°28', 5 = 128°12'.6. (3) I = 45° 42', B = 185° 42'. (4) a = 108° 22'.8, I = 120° 14'.5. (5) a = 70° 50', A = 170° 40'. (6) h = 32° 8'.4, B = 46° 2'.8. (7) I = 34° 28', c = 62° 50'. (8) c = 102° 35', B = 17° 45'. (9) a = 92° 16', c = 57° 35'. ■N X. EXAMPLES 137 EXAMPLES. Solve, given : a be ABC 1. 97° 35' 27° 8'.4 119° 8'.4 2. 67°33'.4 94° 5' 99°57'.6 3. 40° 20' 70° 40' 40° 4. 82°39'.5 116° 20' . 70° 7 5. 155°47'.l 110°46'.4 90° 6. 49°44'.3 121°10'.4 26° 6'.3 7. 144° 10' 41°44'.2 130° 8. 127° 30' 132° 16' 139° 44' 9. 155° 5'.3 110° 10' 70°20'.8 10. 62° 42' 50° 12' 58° 8' 11. 120° 30' 70°20'.3 69° 35' 12. 50° 15' 75° 30' 90° 13. 116° 20' 104°59'.l 138°50'.2 14. 84°14'.5 32°26'.l 36°45'.4 15. 100° 50° 60° 16. 87° 12' 88° 12' 90° 17. 63° 50' 80° 19' 50° 30' 18. 34° 15' 42°15'.2 121°36'.2 19. 50° 63° 15' 90° 20. 159° 50' 159° 43' 123° 40' 21. 124°12'.5 54° 18' 97°12'.5 22. 48°31'.3 62°55'.7 125°18'.9 23. 76° 36' 40° 20' 42°15'.2 24. 28°45'.l 44°22'.2 122°25'.l 25. 44° 53' 53° 52' 90° 26. 98°21'.7 109°50'.4 115°13'.5 27. 99°40'.8 64°23'.2 95° 38'.] y" 188 SPHERICAL TRIGONOMETRY. For APPLICATIONS TO GEODESY AND ASTRONOMY. 113. Geodesy is concerned in measuring portions of the earth's surface, considering the eartli as a sphere. To find the distance on the earth's surface between two points whose latitudes and longitudes are known. If A and B are two places on the earth, P the north pole, ECDW the equator, and PUP' the principal merid- ian, e.(j. the meridian of Greenwich, and if the latitude and longitude of A and B are known, then AB can be computed. AP = 90° - latitude A, BP = 90° - latitude B, angle APB = longitude A — longitude B. .". two sides and the included angle of the triangle APB are known, and AB can be computed. Ex. 1. Find the distance between Ann Arbor, 42^ 19' N., 83°43'.8 W., and San Juan, 18° 29' N., 66° 7' W. 2. How far is Manila, 14° 36' N., 120° 58' E., from Honolulu, 21° 18' N., 157^55' W.? Honolulu from San Francisco, 37°47'.9 N., 122° 24'.5 W. ? San Francisco from Manila ? • 114. The celestial sphere. The heavenly bodies appear to be situated on a sphere of indefinitely great radius with the centre at the point of observation. This is called the celes- tial sphere. A tangent plane to the earth at the point of observation cuts the celestial sphere in a great circle called the horizon. The points of the horizon directly south, west, north, east are called the south, west, north, east points. A vertical line through the point of observation cuts the celestial sphere above in the zenith, and below in the nadir, the zenith and nadir being poles of the horizon. APPLICATIONS. 139 The earth's axis produced is the axis of the celestial sphere, cutting it in the north and south poles of the equator. The altitude of a star is its distance from the horizon measured on an are of a great circle drawn through the star and the zenith. The azimuth, or hearing, of a star, is the are o£ the horizon measured from some fixed point to the foot of the great circle through the star and the zenith. The fixed point is usually the south point. The declination of a star is its distance from the celestial equator. The circle d rawn through the pole and the star is the hour circle, and the angle at the pole between the prime meridian and the hour circle is the hour angle oi the star. Let an observer be at on the surface of the earth, and let P be the position of a star. Then Z is the zenith, Z' the nadir, EQE' the celestial equator, iV its north pole, S its south pole, BRH' the horizon, NFS the meridian, or hour circle, of P, and ZNP the hour angle. The deelina- ' '"" ""' tion of the star is PQ, its altitude PR, and its azimuth, or bearing, NZP. The astronomical triangle NZP can be solved if any three of its parts are known. EXAMPLES. 1. What will be the altitude of the hqii ai, 9 a.m. in Betroit, lat. 42= 20' N., its declination being 17° 30'.5? 2. At what time will the sun rise at San Francisco, lat. 37° 47'.9, if its declination is 12°46'.2? 3. Find the azimuth and altitude of a star to an observer in lat. 42° SO" N., when the hour angle of the star ia 3 h. 42.3 m. E., and the declination is 42° 31' N. 4. The latitude of Sayre Observatory is 40°3e'.4 N.; the sun's alti- tude is 47''15'.3, its azimuth 80°23'.l. Find its declination and hour angle. 5. At Ann Arbor, March 13, 18fll, the altitude of Regulus ia 32° 10'.3, and the azimuth is 283° o'.l. Find the declination and hour angle.