UNIVERSITY OF 'CALIFORNIA. 0; GIF^T OF" Class TEXT-BOOK LOAN LIBRARY OF THE UNIVERSITY OF CALIFORNIA GIFT O Received Accessions No. ..i*~f Shelf No. ~^ J3B.0.TT >t. A TREATISE ON TRIGONOMETRY. TREATISE ON TRIGOlSrOMETPvY BY PROFS. OLIVER, WAIT AND JONES CORNELL UNIVERSITY. NEW YORK: JOHN WILEY & SONS. 1883. Entered according to Act of Congress, in the year 1881, by GEORGE WILLIAM JONES, in the oflice of the Librarian of Congress, at Washington. PKEFACE. THIS book is one of a series of text-books to be prepared by the department of mathematics of Cornell University, in accord- ance with the scheme of instruction now in force here. It was outlined and written mainty by Prof. Jones ; but it has been carefully read by all of us, the general plan, and all difficulties, have been discussed together, the proofs have been submitted to all, and it goes out as our joint production. It is designed as a drill-book for class use ; its leading features are : The general definition of the trigonometric functions in terms applicable to all angles, without regard to sign or magnitude. The expression of the functions of all angles in terms of the functions of positive angles less than a right angle, by direct reference to the definitions. The graphical representation of functions. The general proof of the formulae for the functions of the sum and difference of two angles, of double angles, half- angles, etc. The differentiation of trigonometric functions, their develop- ment thereby into scries, and the computation of the trigono- metric canon by means of these series. The solution of oblique triangles by means of right triangles, as well as by the general properties of triangles ; and by the use of natural as well as logarithmic functions. An exhaustive discussion of the ambiguous and impossible cases of right and oblique triangles. A careful choice and arrangement of topics, according to their relations to practical work and to the higher mathematics. 1 30896 iv PREFACE. The exact statement of principles in the form of theorems and corollaries, and their rigorous demonstration. Frequent reference of collateral matter to the reader for demonstration. Copious and varied exercises. In the preparation of the book, free use has been made of the works of other authors, particularly those of Briot and Bouquet, De Morgan, Todhunter, Peirce, Wheeler, Greenleaf, Loomis, and Chauveuet; and of collections of exercises representing Senate-house examination papers. The careful reader will doubtless find many typographical and other errors in this first edition ; he will confer a great favor if he will kindly communicate them to either of the authors. Any suggestions from practical teachers, looking to the improvement of the book in either matter or form, will be welcomed and esteemed of great value. Among other such improvements now in contemplation is the addition of a chapter on the applications of spherical trigonometry to astronomy, geodesy and navigation, and one on imaginaries, and an alphabetical index to the whole. The starred portions of the book may be omitted without breaking the continuity of the subject ; and to such teachers as do not desire to take up the whole treatise the following abridgment is recommended : I. 1-23, except the note to 18, and Note 4 to 19 ; selec- tions from Ex. 1-7, 9, 21-23, and 25-28. H. 1-3. III. 2 ; 3, one method, and 1 if the second method is chosen ; Ex. 119 and 26-45. IV. 1, 2, 4, 5 and 6, one method, and 3 if the second metJiod is chosen, except Thms. 8-10 ; Ex. 1-26. O. W. J. ITHACA, N.Y., April 8, 1881. CONTENTS. PLANE TRIGONOMETRY. I. PRIMARY DEFINITIONS AND FORMULAE. Page 1. Plane angles 2. Angular notation 3 3. Complement and supplement of an angle or arc 4 4. Positive and negative lines 5 5. Position of a point in a plane by abscissa and ordinate ... 5 G. Position of a point by bearing and distance 7 7. Trigonometric functions 8. Anti-functions 9 9. Constancy of functions 10. Periodicity of functions 11. Signs of functions 10 12. Functions of negative angles 1 13. Functions of the complement of an angle 11 14. Functions of .] T + 8 12 15. Functions of the supplement of an angle 12 Functions of TT + 13 Functions of :]- ft 14 18. Functions of ^ + 8 : 19. Values of functions in terms of each other 15 20. Functions of 30, GO , ] 21. Functions of 45, 135^, ..... 19 22. Functions of 0, 90, ' 23. Graphical representation of functions ' 24. Exercises vi CONTENTS. II. GENERAL FORMULAE. Page. 1. Functions of the sum, and of the difference, of two angles . . 28 2. Functions of double angles, and of half-angles 31 3. Functions of the half-sum, and of the half-difference, of two angles 33 4. Functions of the sum of three or more angles, of triple angles, etc 34 5. Differentiation of trigonometric functions 35 6. Development of trigonometric functions 37 7. The trigonometric canon 39 8. Exercises 44 III. SOLUTION OF PLANE TRIANGLES. 1. General properties of plane triangles 40 2. Solution of right triangles 51 3. Solution of oblique triangles 54 4. The area of a triangle 59 5. Inscribed, escribed and circumscribed circles Gl G. Exercises . G2 SPHERICAL TRIGONOMETRY. IV. SOLUTION OF SPHERICAL TRIANGLES. 1. Geometrical principles G8 2. Napier's rules foe the right triangle 70 3. General properties of spherical triangles 73 4. Solution of right triangles 81 5. Solution of quaclrantal triangles and isosceles triangles . . . 8G 6. Solution of oblique triangles *7 7. Relations between spherical and plane trigonometry .... 98 8. Exercises .... 100 T ' , TRIGONOMETRY. TRIGONOMETRY is that branch of mathematics which treats of the numerical relations of angles and triangles. . It is essentially algebraic in character, but is founded on Geometry. It has two parts : Plane Trigonometry, which treats of plane angles and triangles, and Spherical Trigonometry, which treats of spherical angles and triangles. PLANE TRIGONOMETRY. I. PRIMARY DEFINITIONS AND FORMULAE. 1, PLANE ANGLES. A plane angle is the opening between two straight lines which meet at a point, and is generated b}' revolving one of the lines about the common point as a hinge. The fixed line is the initial line, and the moving line is the terminal line. The revolving line has made one revolution when, by a continuous forward motion, it has come again to the position it first occupied. An angle is not limited in Trigonometry by the geometrical words " acute" and ''obtuse" ; it may exceed a half revolution, or a whole revolution, or it may be any number of revolutions, thus : PLANE TRIGONOMETRY. [I. If the line OP first coincide with ox, and then, revolving about o, take successively -the positions opj, op 2 , OP S , OP 4 , OFj, , the angles generated are XOP I? xop 2 , XOP S . xop 4 , , and these angles, by general agreement, are called positive angles, while xop 4 , XOP S , are negative angles. That is to say, revolution from right to left, and opposite to the hands of a clock, is positive revolution; and revolution from left to right is negative revolution. The order of the letters indicates whether the angle is positive or negative, thus : xopj is positive, and Pjox is negative ; xop 1 4-PiOx = 0, 4- PI op 2 4- P 2 ox = 0, but PiOp 2 4- P2O p 3 4- P 3 o p 4 4~ P 4 p i = owe entire positive revo- lution, and PiOP 4 4- P 4 OP 3 -f- P 3 OP 2 -f- P 2 OP! = one entire negative revo- lution. So, the arcs generated by a point p, moving along a circle, are positive or negative arcs, according as the movement of the point indicates positive or negative revolution of the radius OP, and therefore, according as the angles which they subtend are posi- tive or negative angles, thus : XP 15 XP 2 , are positive arcs, but xp 4 , xp 3 , are negative arcs; and xp 1 4-PiP 2 4-P 2 P 3 + P 3 P24-P2X = 0, but Xp 1 + P 1 p 2 + P 2 p 3 4-P 3 p 4 4-P 4 x = one circumference. If Y'Y J_x'x at o ; then : _^ XOY is the first quadrant, x' YOX' is the second quadrant, X'OY' is the third quadrant, Y'OX is the fourth quadrant. 2.] PRIMARY DEFINITIONS AND FORMULAE. 3 Aii angle is said to be in the first, second, third, or fourth quadrant, according as its terminal line lies in the first, second, third, or fourth quadrant, counting from the initial line. If the revolution of the terminal line is continuous, it comes again and again to the same positions which it had during the first revolution ; such returns are periodic. 2, ANGULAR NOTATION. The right angle is divided into 90 equal parts called degrees, the degree, into GO equal parts called minutes; the minute, into 60 equal parts called seconds. Degrees, minutes and seconds are marked , ', ", respectively, thus : 6 29' 33". 3 is read G degrees, 29 minutes, 33 and 3 tenths seconds. When angles are expressed in degrees, minutes and seconds, they are said to be expressed in degree-measure. Another measure called circular measure, arcual measure, or, briefl}', -rr-measure, comes from the use of the ratio of an arc which subtends an angle, to the radius of the arc ; for, since, in the same or equal circles, angles at the centre are proportional to the arcs which subtend them, and, in different circles, like arcs are proportional to their radii, [geom. therefore the ratio, arc : radius, is proportional to the angle sub- tended, and may be used as a representative or measure of it. But the ratio, half-circle : radius, equals 3.14159 , called TT ; therefore a half-revolution, or an angle of 180, is represented by TT ; a right anglo, or 90, by ^ ; a whole revolution, or 360, by 277 ; 30% by - ; and so on. G That angle whose arc is as long as the radius, and which may > 1 ono be called the unit angle of the 7r-measure, is - , = 57 17' 44".8. 7T The 7r-measure of an angle of 1 is , == .017 4533 ; 180 that of 1' is .017 4533 -j- GO, = .000 2909 ; that of I" is .000 2909 -- GO, = .000 0048. An angle given in 7r-measure may be expressed in degree- measure by multiplying it into the unit angle of 7r-measure ; 4 PLANE TEIGONOMETKY. [I. and an angle given in degree-measure may be expressed in TT-measure by multiplying the number of degrees, minutes or seconds by the --measure of 1, 1' or 1". NOTE. The reader must carefully distinguish between the two notations, for though 90 and ^ both represent a right angle, it is hardly right to say 90 = -, for 90 is a right angle, an actual geometrical magnitude, while - is merely a ratio, i.e., a number. With this caution, however, the angle, the arc, and the ratio, arc: radius, may be used almost at pleasure, and the one nota- tion or the other may be employed at convenience. The 7r-meas- ure is generally preferable for theoretical work, but the degree- measure for computation of triangles. 3, COMPLEMENT AND SUPPLEMENT OF AN ANGLE OE ARC. The complement of an angle is its defect from a right angle ; of an arc, its defect from a quadrant; i.e., it is the remainder when, from a right angle or quadrant, the given angle or arc is subtracted. The supplement of an angle is its defect from two right angles ; of an arc, its defect from a half-circle ; i.e., it is the remainder when, from two right angles or a half-circle, the given angle or arc is subtracted. Manifestly the complement of a positive angle or arc less than 90 is a positive angle or arc less than 90 ; of a positive angle or arc greater than 90, is a negative angle or arc ; of a negative angle or arc, is a positive angle or arc greater than 90. So, the supplement of a positive angle or arc less than 180 is a positive angle or arc less tha,n 180 ; of a positive angle or arc greater than 180, is a negative angle or arc ; of a negative angle, or arc, is a positive angle or arc greater than 180. Thus : the complement of 75 is 15 ; of 100, is 10 ; of 10, is 100 ; of -, is--; of27r, is'- i o 2 the supplement of 75 is 105; of 200, is 20; of 20, is 200 ; of -, is ; of 2 TT, is -TT. 6 6 5.] PRIMARY DEFINITIONS AND FORMULAE. 4, POSITIVE AND NEGATIVE LINES. If two or more points lie on a straight line, and the position of any one of them be known, then the positions of the other points are determined b}' their distances and directions from the point first named. The first point is the oriyin; and, of the two directions from this point, along the line, if either be called posi- tive, the other is negative, and the segments of the line measured in one direction are positive, in the other negative. The order of the letters at the extremities of a segment indicates the direc- tion of the segment, thus : if A and B be two points on a line, and AB be positive, then BA is negative, and vice versa; and, in either case, G F E D So, if A, B, c be any three points on a line, then, whatever their order upon the line, AB -f BC = AC, and AB + BC-f-CA = 0. So, if A, B, c, D, K be n points on a line, then, whatever their order upon the line, AB -f- BC -f CD -f = AK, and AB -f BC + CD -{-..... -J-KA = 0. 5, POSITION OF A POINT IN A PLANE BY ABSCISSA AND ORDLNATE. Let p be an}' point in a plane, o a fixed point, and ox a fixed line in the plane ; draw AP J_ ox ; then the position of P with reference to o and ox is determined when the length and directions of OA and AP are known. In this figure : the reference-point o is the origin; x AS AlX the horizontal distance OA is the abscissa, and the perpendicular AP is the ordinate, of P. X'A 6 PLANE TEIGOXOMETRY. [I. The line x'ox is the axis of abscissas, and Y'OY is the axis of ordinates. The abscissa and ordinate of a point, when spoken of together, are its coordinates. By general agreement, when the line ox is so placed before the reader that x is its right extremit}', then ox is assumed as tlie positive direction of the axis, and ox' as its negative direction ; therefore abscissas of points in the first and fourth quadrants are positive, and abscissas of points in the second and third quadrants are negative. By general agreement, also, that side of the axis from which positive revolution is had is assumed to be positive, and that side from which negative revolution is had is negative ; therefore ordinates of points in the first and second quadrants are positive, and ordinates of points in the third and fourth quadrants are negative. The signs of abscissa and ordinate when written together are : respectively, for points in the four quadrants taken in their order. For brevity, the abscissa of a point may be denoted b.y x and the ordinate by ?/, and the position of a point is then determined from the equations x = a and y = 6, wherein a, with its sign, stands for the length and direction of the abscissa of the point, and &, with its sign, for the length and direction of its ordinate. If x = 0, the point is on the axis of ordinates ; if y = 0, it is on the axis of abscissas ; if both x and y = 0, it is at the origin. If r represents the distance between two points, P A and P 2 ; then, the difference of then* abscissas, x^ x 2 , and the difference of their ordinates, y\yi, are the projections of?* upon the two axes, respectively ; and r = (x l x. 2 ) - -f- ( ij l y. 2 ) 2 . 6.] PRIMARY DEFINITIONS AND FORMULAE. 7 6. POSITION OF A POINT BY BEAEING AND DISTANCE. When two straight lines meet, the point of intersection is the origin ; one of the lines is the initial line, the other is the termi- nal line, and the angle between them is the bearing, from the initial line, of any point on the terminal line. 11', then, the distance of any point from the origin, and its bear- ing from the initial line, be known, the position of the point is determined. The positive directions of the lines which enclose an angle, taken with reference to that angle, are from the origin, i.e., from the vertex of the angle, measured along the sides of the angle ; thus, with reference to XOP, the sides ox and OP are positive, and ox' and OP' are negative. But, with reference to X'OP', the sides ox' and OP' are positive, and ox and OP are negative. So, with reference to POX' or to X'OP, the sides OP and ox' are positive, and OP' and ox are negative. Since the terminal line may revolve to the right or to the left, and since the distance may be measured in either the positive or the negative direction, it is manifest that a point may be deter- mined by four different combinations of bear- ing and distance. Let r stand for the distance and for the '^\'Q = -iif bearing of the point p ; then either : ,?/' ,\ V* r = + OP, and 6 = + XOP ; > or r = ~OP, and = + XOP'; V' *f or r = "OP, and = ~XOP'; wherein the signs + and ~ are signs of quality, and not of opera- tion ; they show whether the line OP and the angle XOP are taken in the positive or the negative direction. For the purposes of this treatise the first two combinations are sufficient. or r = + OP, and 6 = ~XOP ; * x - \ ,-V PLANE TRIGONOMETRY. [I- 7, TRIGONOMETRIC FUNCTIONS. X AX Let XOP be any angle, positive or negative, ox the initial line, OP the terminal line, and p any point upon it. Draw AP_!_OX ; then OP is the distance of the point p, OA is its abscissa, AP is its ordinate, and the six ratios between the three lines r, x, and y are the trigonometric f auctions of the angle XOP, viz. : The ratio i.e., is the written ordinate : distance abscissa : distance yir x: r sine of XOP cosine of XOP sin XOP COS XOP ordinate : abscissa abscissa : ordinate distance : abscissa y:x x:y r: x tangent of XOP cotangent of XOP secant of XOP tan XOP cot XOP sec XOP distance : ordinate r:y cosecant of XOP CSC XOP Two subsidiary functions, sometimes used, are the versed sine and coversed sine; their definitions are : vers XOP = 1 cos XOP, covers XOP = 1 sin XOP. NOTE. From these definitions result directly the six equations : ordinate = distance . sin XOP ; distance = ordinate . esc XOP ; abscissa = distance . cos XOP ; distance = abscissa . sec XOP ; ordinate = abscissa . tan XOP ; abscissa = ordinate . cot XOP. The sine and cosine may thus be called projecting factors, since their effect as multipliers is to project the distance upon the axes of ordinates and of abscissas respectively. So, the tangent and cotangent may be called interchanging factors, since their effect is to convert abscissas into ordinates, and vice versa. 10.] PRIMARY DEFINITIONS AND FORMULAE. 9 8, ANTI-FUNCTIONS. The expressions sin" 1 ^, cos -1 a, tan~ ] a, are called anti- functions, and are read the anti-sine of a, the anti-cosine of a, They moan "the angle whose sine is a," " the angle whose cosine is a," and so on ; thus : if a = sin0, then = sin" 1 a; if & = cos0, then = cos~ 1 6; 9, CONSTANCY OF FUNCTIONS. THEOREM 1. The functions of a given angle are constant, ivhatever points P', P", are taken on the terminal line. For, from P', P", draw A'P', A"P", JLox; then the triangles OA'P', OA"P", are similar, and the ratios of their homologous sides are , / , . p/ equal, viz. : y' : r' = ?/":?", = y' : x' = y" : #", = Q.E. D. \_geom. 10, PERIODICITY OF FUNCTIONS. THM. 2. The functions of any angle and o/2n?r+0 (n being any integer, positive or negative) are identical ; sine with sine, cosine icith cosine, and so on. For, '.- 2?r, 4?7, 2 me stand for one, two, n entire revolutions, forward or backward, . . OP has the same position for the angles 0, 2 77 + 0, 477+0, 2 7177 + 0; and r, x and y are identical, each with each, for the angles thus formed. .*. the ratios arc identical ; sine with sine, cosine with cosine, and so on. Q. E. D. COROLLARY 1. The functions of any negative angle, 0, and of the positive angle, 2 77 0, which has the same bounding lines, arc identical. 10 PLANE TRIGONOMETRY. [I- NOTE. The trigonometric functions are therefore periodic functions of angles, and Trigonometrj* is sometimes defined as 4 ' that branch of Algebra which treats of periodic functions." If 6 = sin'Vt, then 6 is airy one of an infinite number of angles all of which have the same sine, a. 11, SIGNS OF FUNCTIONS. THM. 3. Tlie signs of functions of angles in the four quad- rants are : Angle in sin and esc cos and sec tan and cot vers and covers First quadrant Second quadrant Third quadrant Fourth quadrant : -f ; J For, r is measured along the side of the angle, and is therefore always posijtive ; [ 6 and x is positive in the first and fourth quadrants, and negative in the second and third ; [ 5 and y is positive in the first and second quadrants, and negative in the third and fourth ; [ 5 .*. the signs of the several ratios are as given above. 12, FUNCTIONS OF NEGATIVE ANGLES. THM. 4. For any negative angle, 0, the values of the several functions, in terms of the functions of the opposite positive angle, + 0, are: 1] sin ( 0) = sin0; esc ( 0) = - csc0 ; 2] cos ( 0) = -fcos0; sec ( <9) = -f-sec0; 3] tan( 0) = tan0; cot ( 0) = cot0 ; For, let xopj be any negative angle, 0, and about the initial line ox, as an axis of symmetry, draw OP 2 , making XOP S oppo- site to xoPi ; i.e., of equal magnitude, but of opposite sign. 13.J PRIMARY DEFINITIONS AND FORMULAE. 11 Take OPi and OP^ equal distances on the terminal lines, and join PJ p 2 ; then, since ox bisects ? at right angles, 05 [geom. therefore the abscissas, x t and x. 2 , of P! and P 2 are identical, and their ordiuates y l and y 2 are opposites ; 2/1 _ _ fa X_L_%2 i~~ 'a' ?*i~V /y Q. E. D, 13, FUNCTIONS OF $TT 6, THE COMPLEMENT OF AN ANGLE 0. TI-IM. 5. Any function of the complement of an angle, co-0 or I-TT 0, is the co-function of the angle, i.e. : 4] 5] 6] \Y sin co-0 = cos 6 ; cos co-0 = sin ; tan co-0 = cot0; csc co-0 = sec ; sec co-0 = csc ; cot co-0 = tan 0. AX A o X i 4>^0 X ! ^^ V"'.XX^ p^ \JP For, let XOP be any angle 0, and POY its complement, TT 0. Draw AP, ordinate of P with reference to ox ; take OY = OP, and draw BY, ordinate of Y with reference to OP ; then, AYBO = AOAP, OB = AP, and BY = OA ; and OY = OP ; i.e., x' =y, y' = %, and r' r. sin POY =?/':?' = x : r = cos XOP ; cos POY = x' :r' = y:r = sin XOP ; and so for the other functions, as the reader may prove. \_geom. \constr. 12 PLANE TEIGONOMETHY. [I. NOTE. The words cosine, cotangent, and cosecant are ab- breviated forms for complement-sine, complement-tangent, and complement-secant ; i.e., for sine of complement, tangent of com- plement, and secant of complement. 14, FUNCTIONS OF THM. 6. For any angle, -J-7T-J-0, the values of the several func- tions, in terms of the functions of 0, are : sin (4-7T + 0) = + cos# ; CSC^TT -f 6) = -f sec<9 ; COS(TT + 0) = sin (9 ; sec^Tr -f 6) = C sc0 ; + 0) = - cot ; cot(^7r -f 0) = tantf. B_ j ij AX I A>'0 For, let XOP be any angle 6, and POQ = \ir ; then XOQ = $7r+0. Take OQ = OP, and draw AP and BQ ordinates of p and Q with reference to ox ; then, AQBO = AOAP, \_geom. .. OB = AP, and BQ = OA ; and OQ = OP ; [constr. i.e., x' = y, y' = x, and r' = r. .'. sin XOQ = y' : r' = x:r= cos XOP ; cos XOQ = x' : r' = y : r = sin XOP ; and so on, as the reader may prove. 15, FUNCTIONS OF TT (9, THE SUPPLEMENT OF AN ANGLE 6. THM. 7. For the supplement of any angle, irQ, the values of the several functions, in terms of the functions ofO, are: 10] sin O- 0) = + sin ; CSC(TT 0) = + csctf ; 11] COS(TT (9) = cos0; sec(?r #) = sec(9; 12] tan(7r- 0) = - tan0 ; cot(ir- 0) = - cote. 16.] PRIMARY DEFINITIONS AND FORMULAE. 13 For, let XOP be any angle 6, and draw OQ so that / QOX'=XOP ; then ZxoQ=7r 0. / X'"*A Take OQ = OP, and dratv AP and BQ, ordinates of P and Q with reference to ox ; then, . AOBQ = AOAP, \_geom. .-. OB = OA, and BQ = AP ; and OQ = OP ; [constr. i.e., x' = x, y' = y'> an( l r' = r. .-. sin XOQ =?/':?'= y : r = sinxop ; COSXOQ = x' : r' = x:r = COSXOP ; and so on, as the reader may prove. 16, FUNCTIONS OF TT + B. THM. 8. For any angle, n+O, the values of the several func- tions, in terms of the functions ofO, are: 13] sin(7r + #) = sin0; csc(7r + (9) = 14] cos(7r-f-<9) = cos0; 15] For, let XOP be anj* angle B, and produce PO ; then XOQ = -rr+O. Take OQ = OP, and draw AP and BQ ordinates of p and Q with reference to ox ; then, v AOBQ = AOAP, \_geom. .-. OB = OA, and BQ = AP ; and OQ = OP ; [constr. i.e., x 1 = x, y' = y, and r' = r. .'. etc., as the reader may prove. 14 PLANE TRIGONOMETRY. [I. o 17, FUNCTIONS OF -- 0. q THM. 9. For any angle, *-j- $, the values of the several func- tions, in terms of the functions ofO, are: 16] sin (^-^-costf; cscf?f _ \ = ' 17] c \2 18] tan^-^= + cot^; cot^5f \ z ' / \ 2 The reader may prove. 18, FUNCTIONS OF THM. 10. For any angle, ^+0, ^e voZites of the several func- tions, in terms of the functions of 6, are: 19] sin^ + A^-costf; cscf + = - sec0 ; \ 2 / \ 2 20] cos T _|_ = + sill $ . sec 5 + = + CSC ; \ * J \ * / 21] tan + =-cot0; cot +0 = - tan0. The reader may prove. * NOTE. The ten theorems just proved may be summarized and generalized as follows : m 19.] PlllMARY DEFINITIONS AND FORMULAE. 15 The}' also give the following : 25] sin- 1 (sm0) = csc- 1 (csc0)=2?i7r+0 or (271+1)^-0; 26] cos" 1 (cos 0) = sec" 1 (sec 0) = 2mr0 ; 27] tim-- l (tsLn6) = CQt- l (cotO)=nff+0. 19, VALUES OF FUNCTIONS IN TEEMS OF EACH OTHER. THM. 11. For any angle 6, the values of the several functions, in terms of each other, are : [28] Binds [29] cos0 = [30] tan0 = [31] C0t0 = [32) sec0 = [33] CSC0 = sin0 VI sin 2 sin0 Vl sin-0 1 1 VI cos 2 COS0 Vl-siir0 Vl-cos 2 sin0 COS0 Vl siu-0 1 siu0 1 COS0 Vl eos^ COS0 Vl-cos 2 tan0 1 t'in A 1 \/t111-^J 1 Vtair'0-j-l Vtair'0 + 1 1 Vtair^ + 1 cot^ 1 tan0 /%/-\f /3 VcoW+I tan0 Vcot 2 0-|-l Vsecr'0 1 Vcot-^+1 1; C0t0 COtt/ 1 C0t0 VCOt U-\-L sec0 sec0 1 sec^ Vcsc'-'^ 1 " 1 Vsec' 2 0-l sect/ CSC0 Vsec 2 1 csc0 csc^ Vcscr0 1 Vcsc 2 0-l Vcsc 2 0-l CSC0 cos0 x:r, csc0 = r : ?/, sec0 = r : a;, = l, cos0-sec0 = For . sin 6 ?/ : r, and 34] 35] i.e., and cot = x : y, [ 7 tan0-cot0 = l; cos0 sin sine and cosecant are reciprocals, cosine and secant are reciprocals, tangent and cotangent are reciprocals ; 16 PLANE TRIGONOMETRY. [I. whatever value is found for sin (9, its reciprocal is a value of csc#, whatever value is found for cos (9, its reciprocal is a value of sec#, whatever value is found for tan 0, its reciprocal is a value of cot 0, and whatever values are found for sin 6 and cos 0, their ratios are values of tan and cot#. So, '. 3? -f y 2 = r 2 , [geom. 36] i.e., sin 2 = Vl cos 2 6>, and cos 0= Vl siu 2 (9; and (2) !+=, 37] i.e., 1-f tan 2 <9=sec 2 0; tan<9 = Vsec 2 # 1, andsec# = Vl + tan 2 < 38] i.e., and (3) ^ + 1=^.; y- y* 1 =Vcsc 2 6' 1, andcsc<9 = Vl+cot 2 ^. So, by combination of the formulae which have been proved, others are established ; for example : sec0 Vl+tanV and cos^ sin = tan 0. cos (9= tan ^ Vl+tan 2 (9 NOTE 1. These formulae, when taken two and two, are metrical, for example : (1) Those for sine, in terms of cosine, tangent, secant, with those for cosine, in terms of sine, cotangent, cosecant, (2) Those for tangent, in terms of sine,, cosine, secant, with those for cotangent, in terms of cosine, sine, cosecant, (3) Those for secant, in terms of sine, cosine, tangent, with those for cosecant, in terms of cosine, sine, cotangent, 19.] PRIMARY DEFINITIONS AND FORMULAE. 17 NOTE 2. For a given ^ ^ the-< has but one value ; for a given < the-! has but one valae ; for a given ^ *H has but one value ; but in every other case there are two corresponding values, the one positive, and the other negative, for every function. This appears alike from the double signs of the radicals involved, and from the relations of the abscissas and ordinates of points in the several quadrants ; thus : with a given distance, to every abscissa correspond two ordi- nates ; therefore to every cosine correspond two sines. So, to ever} 7 ordinate correspond two abscissas ; therefore to every sine correspond two cosines ; and so on. NOTE 3. Of the relations given above, the most important are those between sine and cosecant, cosine and secant, tangent and cotangent ; those between sine and cosine, -tangent and secant, cotangent and cosecant ; and the expressions for tangent and cotangent as ratios of sine and cosine, and for sine and cosine in terms of tangent. These relations should be committed to memory ; the rest may be proved as exercises. * NOTE 4. The above relations may also be expressed thus : sin~ 1 a = esc" 1 -, a _ 1 sin" 1 ( a) = cos" 1 V 1 a 2 = sec" 1 , -- _ Vl-a 2 wherein Vl a 2 is either positive or negative throughout, inde- pendently of a. The reader may express in like manner the values of the five other principal anti-functions of a. 18 PLANE TBIGONOMETBY. 20, FUNCTIONS OF 30, 60, 120, 150, . ., i.e., OF - '2 o THM. 12. The functions of - are : 2 6 Angle. Sine. Cos. Tan. Cot. Sec. Csc 30, i.e., -; or2mr + - 6 6 i W3 JV3 V 3 IV 3 2 ' *' 6 ' *)* f Jv/3 i V3 W 3 2 120, i.e., ; or (2n + 2)^ + ~ 6 6 W3 -i ~V3 -iv 3 -2 IV 150, i.e., ; or(2tt+l)7r-- 6 6 -JV 3 -W 3 -V 3 -IV 3 2 6 ' 6 -i -iV 3 JV 3 V 3 -iV 3 -2 240,le., ; or(2-i)7r-^ -ii/8 2 V _, , 3 1. , 3 2 6 G 2 " o V 300 fe 107r - orf2rc i^4- 7r t) 6 -JV 3 * -V 3 -iV 3 2 II 71 " Ti" ,t.e., ;o] -- -} M -W 3 yg IV 3 -2 For, let ABC be an equilateral triangle, ^^ and from A let fall ADJ_BC ; then AD bisects s' /. A, = 60, and side BC : \qzom. *,' = 30, and DC = AC. But sin-DAC = DC : AC = J ; .-. sin 30=^; .-. cos30 =Vr =: = The reader ma} r prove the remaining values as exercises the preceding theorems ; for example : upon tan 80 = cos 30 - V 3 sin 150 = sin (180 - 30) = sin 30 = ; sec 330 = sec (-30) = sec 30 = 21.] PRIMARY DEFINITIONS AND FORMULAE. 19 21, FUNCTIONS OF 45, 135, i.e., OF mr -. THM. 13. The functions of mr - are : 4 Angle. Sine. Cos. Tan. Cot. Sec. Csc. 45, i.e., f; or 2^ + 7 4 4 'Vi .V* + 1 + 1 V^ V2 135V.e., ; or (2?i+l),r-- 4 4 Vi -V4 j -1 -V 2 V2 225, i.e., ; or(2+l)ir+- 4 4 -VI -Vi + 1 + 1 -V- ? -V2 315, i.e., ;or2tt7r-- -VI Vi -1 -1 V2 -V2 For, let ADC be a right isosceles triangle,, right-angled at D ; then Z A =45, and AD = DC = AC . *J%. [geom. But sin A = DC : AC, and cos A = AD : AC. . . sin 45 = A/J, and cos 45 = ^/-J. The reader may prove the remaining val- ues as exercises upon the preceding the- orems ; for example : tan 45 = ^i5! = ^t=l; ^ A n /I sec cos 45 -J cos4o tan 225 = tan (180 + 45) = tan 45 = 1 ; sin 405 = sin 45= -v/i; cos 405 = cos 45= cos 45 = ^/^; tan 585 = tan 225 = tan 45 = 1 ; cot 585 = cot 135 = tan 45 = 1 ; sec G 75 = sec 45= sec 45 = V 2 5 esc- 675 = esc 45== 'V 2 - 20 PLANE TRIGONOMETRY. [I- 22, FUNCTIONS OF 0, 90, 180 , i.e., OF TT. THM. 14. The functions of n -rr are : Angle. Sine. Cos. Tan. Cot. Sec. Csc 0, i.e., OTT; or 2mr 1 GO 1 CO 90, i.e., ; or (2?i + |)7r 1 00 GO 1 180, i.e., TT; or(2?i-fl)7r -1 oo 1 00 270, i.e., ^; or(2n-i) B - -1 GO GO -1 The reader may prove these values by direct reference to the definitions of the functions. [ 7 NOTE. From the values given above, and from others com- puted by methods to be given later, tables have been constructed, giving the functions of all angles, and other tables giving the logarithms of these functions ; the} 7 are called trigonometric tables. For their use the reader is referred to the tables them- selves. 23, GRAPHICAL REPRESENTATION OF FUNCTIONS. Line-Functions. Let XOP be any angle, and with o as center, and any radius ox, describe a circle, cutting the sides in x and P ; from p let 23.] PRIMARY DEFINITIONS AND FORMULAE. 21 fall PA_Lox ; at x draw XT tangent to XP, and meeting OP in T. Draw OY_Lox, and meeting the circle at Y; from Y let fall YB_Lop, and at p draw PT' tangent to PY, and meeting OY in T' ; then : if the radius be taken as the unit of length, the ratio AP : ox is the numerical measure of AP ; i.e., the number of units in the length of the line AP is equal to sinxop, and the line AP represents sin XOP. So, the line OA represents cos XOP ; the line XT represents tan XOP ; the line OT represents sec XOP ; the line AX represents versxor. So, the line BY represents sin POY, i.e., cos XOP ; the line PT' represents tan POY, i.e., cot XOP ; the line OT' represents secpOY, i.e., CSCXOP ; the line BP represents versPOY, i.e., covers XOP. These lines ma}' be called the line-functions of the angles, as distinguished from the ratio-functions heretofore defined. They are also called functions of arcs; and the others, functions of angles. In most of the earlier treatises on trigonometry the functions were always defined as lines, thus : the sine was said to be "the perpendicular from one extremit}' of an arc upon the diameter through the other extremity"; the cosine was "the distance from the center to the foot of the sine," or " the sine of the comple- ment," and so on. NOTE 1. These line-functions must never be thought of as identical with the ratio-functions, not even when the radius is unit}* ; the ratio-functions are neither lines nor the lengths of lines, but numbers merely whose values are independent of any measuring unit of length. The reader may name the line-func- tions in the different figures, may state which of them are positive and which negative, and may show that the ratios of these line- functions to the radius are identical with the ratio-functions heretofore given. 22 PLANE TRIGONOMETRY. Curve of Sines. [I- Let ox be the radius of a circle, and divide the circumference into any convenient parts at P I? P 2 , ; draw A^, A 2 p 2 , , ordinates of P 15 P 2 , with reference to ox, and sines of the arcs XP 15 xp 2 , .....; draw XY tangent to the circle at x. Conceive the circle to roll along XT ; let B 15 B 2 , be the points on XY where PJ, P 2 , rest respectively; and at B I? B 2 , erect perpendiculars to XY, and make B^ = A^, B 2 c 2 = A 2 p 2 , Through c 1? C 2 , draw a smooth curve ; it is the curve of sines. NOTE 2. The following relations are manifest : (1) At x the sine is 0. (2) While the arc is small, the sine is nearly as long as the arc. (3) The sine increases more and more slowly. (4) It is equal to + 1, its maximum, when the arc is a quad- rant ; then (5) It decreases, at first very slowly, but faster and faster as the arc approaches a half-circle. (6) When the arc is a half-circle, the sine is 0. (7) While the arc increases from 180 to 270, the sine de- creases from to 1, its minimum. (8) While the arc increases from 270 to 360, the sine in- creases from 1 to 0. (9) At the end of the first revolution the sine is again 0. (10) During every successive revolution the same phenomena are repeated in the same order, and for negative as well as posi- tive arcs. (11) While the revolution is continuous, the values of the sines are periodic, every successive revolution indicating a new cycle, and a new wave in the curve. 24.] 1MIIMAUY DEFINITIONS AND FORMULAE. 23 (12) The four parts of each wave that correspond to the four quadrants of the angle XOP, are equal and similar to each other. (13) The sine has no value greater than -f- 1 > nor less than 1 . The reader ma}' draw curves to represent the other functions, and discuss them ; lie will find, among other things, that : (14) The tangent is at = ; increases through the first quadrant to + I at 90 changes suddenly to oo ; increases through the second quadrant to at 180 ; increases through the third quadrant to + oc at 270 ; at 270 changes to oo ; increases through the fourth quadrant to at 360 ; and so on. (15) The secant is +1 at = ; increases through the first quadrant to + ; at 90 changes to oo ; increases through the second quadrant to 1 at 180 ; decreases through the third quadrant to oo at 270 ; at 270 changes to -f- oo ; decreases through the fourth quadrant to +1 at 360 ; and so on. It has no value between + 1 and 1 . (16) The cosine, cotangent and cosecant have the same limits as the sine, tangent and secant respectively ; they go through like changes and are represented b} T like curves ; but they begin, for = 0, with different values, viz., 1, oo, and oo. 24, EXERCISES. 1 . Express in degree-measure the angles : 7T 7T 5 73 5' 7' 9' ~3~ 2. Express in 7r-measure the angles : 14, 15, 24, 120, 137 15', -4800, 13', 24", -5, 19' 37".o. 3. If the radius of a circle be one inch, what is the length of the arcs : 14, 15, 120, 5717'44".8, 1, 1', 1", ?, , 2, *r+l. 5 7 PLANE TRIGONOMETRY. [I. 4. Find the complements and the supplements of the angles : 37, 215, 325, 107 12' 15", -36 12', ^ !!5 _^5 o' 7 ' 9 ' 3 ' 5. Construct the points in a plane whose abscissas and ordi- nates are : i 3 I 2,9; 8, 7; 3,4; 5, 8. [Use any convenient unit.] 6. Construct the points in a plane whose bearings and dis- tances are : 30, 10; 30, -10; 210, 10; -150, 10; 150, -10; -GO , 10; TT, 8 ; TT, 8 ; TT, 8 ; 0, 8 ; , 4 ; , -4 - 4 ?T _/L 3' 3' 3' 3'" 7. Construct the angles whose sines are -J, -f, V^pl ? -^ Oj i + V3 t cosines are .6, -, j-ys, Lz_V?. 2 tangents are , f, 0, -1, oo , V2+1. cotangents are i, 3^, 2 ^ ~ 6 , [ o and 6, any lines. -yo * 8. Write formulae for all values of : when sin = sin a, Vsin 2 a, VcosV. when cos 6) = cos a, Vcos 2 a, Vsin 2 a. when tan 9 = tana, Vtan 2 a, Vcot 2 a. when cot = cot a, Vcot 2 a, Vtan 2 a. 9. Find the remaining functions of 6 : if sin 0=. 6, . ?, -?, J_ J^-L V 3 17 4 3 -v/5' V2 6 , , 2 ' 13' 2V2 yil z a a "3~"' 1-a 2 ' ~"b' ifcsc^=-2, |, t^ 2 , ^, 2(-l)". [m any integer. 24.] PRIMARY DEFINITIONS AND FORMULAE. 25 *]0. Find sin0 from the equation, sin 0. cos = m. *11. If tan0 + cot0 = ra, express all the functions of in terms of m. *12. If tan 0= f-Y, find the value of + _A_. \aj cos0 sin0 *13. Eliminate from the equations : sin0-f-cos0 = a, tan0 + sec0 = 2&. * 14. Eliminate < from the equations : esc (f> sin < = m, sec cos = n. *15. Eliminate and $ from the equations : asin 2 0-f&cos 2 = m, 6sin 2 4-acos 2 <^ = ?i, a tan = b tan <. *16. If tan<^-hsec<^ = a, find si *17. Show geometrically that sin 2 < < 2 sin <, if <>0, ^iV^* cos50, sin 50, sin~. 16 tangents are 1, V^> tan-, cot 40, cot ^. 8 8 cotangents are 0, ' ^3, cot , tan 20, tan-^- 8 8 secants are oo , , sec ^ x esc 35, esc -^- \fo 8 8 7 7T 7 7T cosecants are oo , 2, esc , sec 35, sec - 8 o 26 PLANE TRIGONOMETRY. [I. 21. Find the 97-w sine of 225, 585, 810, 960, STT, . 4 cosine of 315, -675, 960, -1110, ^, -877. tangent of 495, -945, 1110, -1260, , -515. 6 4 cotangent of 675, -1035, 1260, -1410, , -l^ 7 ". G 4 secant of 855, -1215, 1410, -1560, , -41?. 6 4 cosecant of 1035, -1395, 1560, -1710, ^5, -13 TT. 6 22. In terms of the functions of positive angles less than 90, express the values of the sine of 135, 335, -535, -735, <0 QQKO -orO r-QKO 27?T 29 7T 5 ' cosine of 235, 435, -635, -835, ?^5, O tangent of 335, 535, -735, -935, , - 7 QO _ cotangent of 435, 635, -835, -1035, ^, -OTT. o secant of 535, 735, -935, -1135, cosecant of 635, 835, 1035, -1235, , , 5 7 23. In terms of the functions of positive angles less than 45 express the values of the sine of 50, 150, -250, -350, ?f, -4ir. cosine of 60, 160, -260, -360, 5, -^ 12 o tangent of 70, 170, -270, -370, ^T, - 12 o cotangent of 80, 180, -280, -380, , -677. 24.] P1UMAHY DEFINITIONS AND FORMULAE. 27 secant of 90, 190, -290, 390, -~, - . cosecant of 100, 200, -300, -400, , - . 12 3 * 24. Trace the changes, when 6 increases from to 2?r, in tho sign and value of each of the expressions : sin + cos 0) sin 6 cos 0, sin + esc 0, tan cot 0, sin 2 0, cos 2 0, sin 2 cos 2 0, cos 2 $ sin 2 (9, tan 2 0-f cot 2 0. 25. From the table of natural functions, find the sin of 20, 21, 20 10', 2010'45", 8918'25", 15715'23". cos of 20, 21, 20 10', 2010'45", 8918'25", 15715'23". tan of 35, 36, 35 15', 3515'47", 8958'35", 125 0'12". cot of 35, 36, 35 15', 3515'47", 8958'35", 125 0'12". 26. From the table of natural functions find the angles whose sines are .25882, .25910, .25900, .92794, .92805, .92800. cosines are .92794, .92805, .92800, .25910, .25882, .25900. tangents are .5022, .5059, .5035, .9217, .9271, .9250. cotangents are .9217, .9271, .9250, .5022, .5059, .5035. 27. From the table of logarithmic functions find the logarithmic sin of 20, 21, 20 10', 2010'45", ' 8918'25", 15715'23". cos of 20, 21, 20 10', 2010'45", 8918'25", 15715'23". tan of 35, 36, 3515', 3515'47", 8958'35", 125 0'12". cot of 35, 36, 35 15', 3515'47", 8958'35", 125 0'12". sec of 50, 51, 5020', 5020'49", 115 0'45", 17958'55". esc of 50, 51, 50 20', 5020'49", 115-0'45", 17958'55". 28. From the tabled of logarithmic functions, find the angles whose logarithmic sines are 8.580892, 8.584193, 8.582125, 9.999683. cosines are 8.580892, 8.584193, 8.582125, 9.999683. tangents are 8.581208, 8.584514, 8.583125, 11.418790. cotangents are 8.581208, 8.584514, 8.583125, 11.418790. secants are 10.367529, 11.367514, 12.367529, 13.367514. cosecants are 10.367529, 11.367514, 12.367529, 13.367514. 28 PLANE TBIGOXOMETBY. [II. II. GENERAL FORMULAE. 1, FUNCTIONS OF THE SUM, AND OF THE DIFFERENCE, OF TWO ANGLES. THEOREM 1. If Q and 0' be any two angles, then : 39] 40] 41] 42] sin (0 + 0') = sin cos0'+ cos sin 0' ; sin (0 6') = sin cos0' cos sin 0' ; cos(0+ 0'} = COS0COS0'- sin<9 sin0' ; / cos(0- 0') = cos cos 0'+ sin sin 0'. D ?/__. _K Fx'Q For, let XOP be any angle 0, positive or negative, and POQ be an}' other angle 0' ; then XOQ = + 0'. Draw BQ ordinate of Q. with reference to OP, and AQ and CB ordi- nates of Q and B with reference to ox ; draw BK parallel to ox and meeting AQ in D ; on BQ take F so that Z KBF = TT -f 0. Then : = ordinate A Q = CB + DQ . [L7 distance OQ OQ (1) but and OQ OB OQ = sin#cos0 f , Q = S.5Q OQ BQ OQ = cos0sin0'; [I. 7 [I. 7 [7 Q. E. D. 1.] GENERAL FORMULAE. 29 (2) In [39] substitute 0' for 0', then: ... -6' =0 + (-&), .-. sin (0 6') = sin0cos(-0') + cos0sin(-0') [39 = sin cos 0' cos 6 sin 0'. Q.E.D. [1,2 (3) cos(0 + 0Q = = ; [i. 7 distance OQ OQ but 2 = ^.^ OQ OB OQ = cos0cos0', [I. 7 and BD = BD.BQ OQ BQ OQ . = cos(i7r + 0)sin0' [I. 7 = sin0sin0'; [8 = cos0cos0' sin0sin0'. Q.E.D. (4) In [41] substitute 0' for 0', then cos (0 - 0') = cos cos (0') sin0 sin( 0') [41 = cos0cos0'+sin0sin0'. Q.E.D. [2,1 NOTE. Since each of the angles and 0' may be in either of the four quadrants, there are sixteen different cases possible ; but the proof is general, for it applies to all alike. The reader may enumerate the cases in detail and draw the other twelve figures. COR. 1. IfO and 0' be any two angles, then: 43] tan (0 -}- 0') = - /A /if\ tan tan 0' 44 I tan (0 0') = -. l + tan0tan0' [35 _ sin cos 0'+ cos sin 0' p . . ~cos0cos0'~ sin sin 0'* Divide both terms of the fraction by cos cos 0', then ^ _ [35 1 tan tan 0' 30 PLANE TRIGONOMETRY. [II. So ' ""^-SgEg P _ sin cos 0' cos sin f cos 6* cos 0' H- sin sin 0' tan 0- tan 0' ~l + tan0tan0'' The reader may find like formulae for cot (0+6') and cot (00') NOTE. The six formulae [39-44] may be written as three : 39, 40] sin (0 0') = sin cos 0' cos 6 sin 6' ; 41, 42] cos (0 0') = cos cos 0' q: sin sin 0' ; 43,44] tan(00') = COR. 2. 7jf0 and 0' 6e a?iy ^wo angles, then: 45] 46] 47] cos 6 cos 0' 48] 251^1 COS COS 0' 49] 50] . =cot ^ gr sin cos ; The reader may prove these formulae by performing the di visions and reductions indicated. COR. 3. I/O and O 1 be any two angles, then: 51] sin (0 + O 1 ) + sin (0 -0')= 2 sin cos 6' ; 52] sin (0 + 0') - sin (0 - 6') = 2 cos 6 sin 6' ; 53] cos (0 + 0') + cos (0 - r ) = 2 cos cos 0' ; 54] cos (0 + 0') - cos (0 - 0') = - 2 sin sin 0'. 2.] GENERAL FORMULAE. 31 For, if to [39], [40] be added, the result is [51] ; if from [39], [40] be subtracted, the result is [52] ; if to [41], [42] be added, the result is [53] ; if from [41], [42] be subtracted, the result is [54]. COR. 4. I/O and 0' be any two angles, then: 55] sin (0 + .0') sin (0 6') = sin 2 - sin 2 0'= cos 2 <9' cos 2 ; 56] cos(0 -f- 6') cos(0 - 6') = cos 2 0- sin 2 0'= cos 2 0' sin 2 <9. The reader may prove, by performing the multiplications indi- cated ; he will make use of [36] . COR. 5. If 6 and 0' be any two angles, then: sin (0 +#') _ tan + tan f . sin (0-0') ~~ tan - tan 0' ' cos (0 +0') _ 1 - tan tan 0' cos (0 - 0') ~ 1 -f tan tan 0' sin (0 0') tan tan 0' 57] 591 /7.i-/i\ Itan0tan0' The reader may prove, by aid of [47-48]. 2. FUNCTIONS OF DOUBLE ANGLES, AND OF HALF ANGLES. THM. 2. If be any angle, then: 60] sm20 = 2sin0cos0; 61] cos 2 = cos 2 sin 2 = 2 cos 2 1 = 1 2sin 2 ; not -1/1 fl COS0 63] sm|0 = ^| - - -- .-< -, /, /1 + COS0 64] cos0 sin0 1 C030 65] 32 PLANE TRIGONOMETRY. [II. (1) In [39, 41, 43] substitute for 0', then: .- 0+0' =2(9; .-. Sm20 = sin0cos0-f-cos0sin0 = 2sin0cos0. [39 So, cos20 = cos0cos0 sin0sin0='cos 2 sin 2 [41 = 2cos 2 6> -1 = 1 - 2sin 2 ; [36 and tan20 = - tan0 2tan0 E> D> 1 tau0tan0 1 tan 2 (9 (2)-.- cos20 = l-2sin 2 0, whatever the angle (9, [61 .-. cos0 = !-2sin 2 iJ-0, [substitute for [1-COS0 ' sm-J0 = ^j -- - So, . cos 2 = 2 cos 2 1 , whatever the angle 0, [61 .-. cos0 =2cos 2 1, [substitute |0 for 1+COS0 Hence, tani0 = = A l 1 -^^. [35 cos^-0 \l+cos0 Multiply both terms of the fraction by ^/(l -f cos0) , then tan0 =, Va-eo8'g) = sin0 1+COS0 1+COS0 or, multiply both terms of the fraction by ^/(l cos0) , then NOTE. Although the radical ^/(l cos 2 0) , = sin0, has two values, opposites ; yet tan -J-0 has but one sign and one value ; for, when is in the first or second quadrant, then 4-0 is in the first or third quadrant, and sin0 and tan J-0 are both positive ; so, when is in the third or fourth quadrant, then |-0 is in the sec- ond or fourth quadrant, and sin0 and tan J0 are both negative ; i.e. , whatever the sign of sin 0, the same is that of tan 0. [I. Th. 3 COR. IfObe any angle, then: 66] sin0 = ![V(l+sin0)-V(l-sin0)]; 67] cos J0 = J [V(l + sin0) + % /(l- sin0)]. [See nt. 1 p. 103 3.J GENERAL FORMULAE. 33 For, . sin 2 . = 2 sin cos 0, whatever the angle 0, [60 . . sin = 2 sin 1 cos -j- ; [substitute for but 1 = cos 2 04 sin 2 0; [36 .-. l+sin0 = (cos ^0 + sin |0) 2 , and 1 sin0 = (cosJ0 sinJ0) 2 ; .-. V( 1 + sin 0) = cos |0 + sin0, and VC 1 sin 0) = cos i# sin0. .. etc. Q. E. D 3, FUNCTIONS OF THE HALF-SUM, AND OF THE HALF- DIFFERENCE, OF TWO ANGLES. THM. 3. IfO and 0' be any two angles, then: 68] sin + sin 0'= 2 sin |(0 +0')cosJ(0 0') ; 69] sin0 sin0'= 2cosi(0 + 0')smJ(0 0') ; 70] cos + cos 0'= 2cos^(0 + 0')cosi-(0-0') ; 71] cos0 - cos0'= - 2 sin |(0 + 0')sin J(0 - r ) . For, in [51-54] substitute |(0 + 0') for and i(0 0') for f ; then also stands for (0 + 0'), and 0' for (0 - 0') ; and [51-54] reduce to [68-71] respectively. COR. If 6 and 0' be any two angles: f "J 73] 74] 75] 76] 771 sin sin 0' sin 4- sin 0' tan (0 4- 0') ; tan i (0-0'); pot 1 f/J d'"\ cos + cos 0' sin sin 0' cos0 4- cos0' sin 4- sin f cos cos 0' sin sin 0' COl -j ^(7 " / J cot i (040'); cos cos 0' cos0 4- cos0 r The reader ma} r prove, by dividing [68-71] one by another, as indicated, and reducing the quotients. 34 PLANE TRIGONOMETRY. [II. * 4. FUNCTIONS OF THE SUM OF THREE OR MORE ANGLES, OF TRIPLE ANGLES, ETC. THM. 4. If 0, 0', 0", ..... be any angles, then: 781 sin (04-0'+ 0"^=J sill<9 cos ^' cos ^' + cos sin 0' cos 0" 1 + cos cos 0' sin 0" - sin sin 0' sin 6" 79] cos(0+0'+ 0") = ( c 08 * 008 *' 009 *" - eos 6 sin 0' sin 0" } l-sin0cos0'sin0"-sin0sin0'cos0" = cos0cos0'cos0"(l tan 0' tan 0" tan0tan0" tan tan 0'); 801 tanffl I f I g^ = tang + tang f + tang ff -tngtang f tang fy l-tan0'tan0" tan tan 0"- tan tan 0'' The reader may prove, by developing sin [0 +($' + 0")]? cos[0-h(0'+0")], and tan[0+(0'+0")] as functions of the sum of two angles and (0'+0"), and then developing the functions of (0'+0") which are involved. He may also prove the result for tan (0 4- 0' + 0") by dividing that for sin (0 -f 0' + 0") by that forcos(0 + 0' + 0"). The reader ma}* in like manner get the sine, cosine, and tan- gent of the sum of four angles, of five angles, and so on. For sine and cosine, he will find that : 2tan0-2tan0tan0 f tan0" 821 cos(0+0 f +0 ff + ..... )_f 1 - Stan tan 0' cos cos 0' cos 0" ..... !-j-2tan0tan0'tan0"tan0'"- ..... ; wherein 2tan0 stands for the sum of the tangents of 0, 0', ..... , 2 tan tan 0' stands for the sum of the products of those tangents taken two and two ; and so on. By making = 0' = 0" = ..... , he will find that : 83] sin 3 = 3 sin0 cos 2 - sin 3 = 3sin0-4sin 3 0; [36 84] cos30 = cos 3 0-3cos0sin 2 = -3cos0-f 4cos 3 0; 5.] GENERAL FORMULAE. 35 85] sin 40 = 4sin0 cos 3 4sin 3 cos0 = 4sin0cos0 8sin 3 0cos0; [36 86] cos40 = cos 4 Gcos 2 0sin 2 <9-f-sin 4 = 1 8cos 2 + 8cos 4 0; 87] sin 50 = 5 sin<9 cos 4 - 10sin 3 <9 cos 2 -f sin 5 (9 = 5sin0 - 20 sin 3 + lGsin 5 ; 88] cos50*=cos 5 0-10cos 3 0sin 2 + 5cos0sin 4 = 5cos0 - 20cos 3 + 16cos 5 ; and. in general, that : 89] sin n = c n>1 sin cos"" 1 ^ c n)3 sin 3 cos n ~ 3 + c n)5 sin 5 0cos n - 5 ; 90] cosn0 = cos n 0-c n , 2 sm 2 0cos n - 2 0+c n)4 sin 4 0cos r: - 4 0-- ; wherein c n>r denotes n(n 1) (71 2) (n r + 1) 1.2.3 r the number of combinations of n things taken r at a time. *5, DIFFERENTIATION OF TRIGONOMETRIC FUNCTIONS. LEMMA. If 6 be an infinitesimal angle, then lim (sin0 : 0) =1, and lim (tan : 0) = 1 . For, let c be the circumference of a circle, o the centre, p and j> r the perimeters of two regular polygons of the same number of sides, the first inscribed in and the other circum- scribed about the circle, and having their sides PP', TT', parallel each to each. Draw OAX_LPP' and TT', then : *.* 2 ) < c whatever the number of sides, and . p approaches indefinite^- near to p 1 when the number of sides is indefinitely increased, [geom. .*. c is the common limit of p and p', .. 1 is the common limit of the ratios p : c and p' : c. 36 PLANE TRIGONOMETRY. [II. And-.- the two polygons are similar, and AP, XP and XT are like parts of p, c and p\ [geom. AP : XP = p : c, and XT : XP = p 1 : c ; Vrp "X^T* and tan 6:0, = : = XT : XP, = p' : c ; ox ox .-. Iim(sin0 : 0) = lim (p : c) = 1, and lim (tan : 0) = lim (p 1 : c) = 1 . THM. 5. If be any angle, then: 91] D0Sin0 = cos0; D csc0 = cot0csc0 ; 92] D0Cos0 = sin 0; D sec0= tan0sec0; 93] D tan^= sec 2 ^; T> e cotO = csc 2 0. For, let be any angle, and 0' an infinitesimal angle, the incre- ment of 0, then : v sin (0 + 6') - sin0 = 2 cos (0 + -J0') sin Jtf r ; [69 0' Y Q is the above series for cot0. ciri /3 /3 3 ' So, '.'-DQlogcosO^-- - = -tan0 = -(0 + + ..... ),[96 cos0 3 101] ,.logco S e . == -g + |l 3+ -|l + i ^_ 7 + ). 7.] GENERAL FORMULAE. 39 7, THE TKIGONOMETKIC CANON. The Trigonometric Canon is a set of tables which give the sine, cosine, tangent, cotangent, secant and cosecant for every angle, from to a right angle, taken at regular intervals of say 10", or 1', or 10', , as may be chosen for the particular tables. Such tables contain either the functions themselves or their logarithms. The first are called the natural functions, aud the others the logarithmic functions. PROBLEM 1. To CONSTRUCT A TABLE OF NATURAL SINES AND COSINES, TO MINUTES OF ANGLE. FIRST METHOD: Assume sinl 9 as differing insensibly from arc 1', i.e., that sin 1' = .000 290 8882, and hence, that cos 1', = Vl sin 2 !', = .999 999 9577 ; then : (1) For angles 0-30, apply formulae: sin (0 + 6') = 2 sin cos 0' sin (0 6') , [51 cos(0-h0') = 2cos cos 0' cos (0 0') ; [53 make 0= 1', 2', 3', successively, and 6' I' constantly, thus- sin 2'= 2sinl'cosl' sinO' = 2 x .000 290 8882 x .999 999 9577 - = .000581 7764 x(l-. 000 000 0423) = .000 581 7764 ; sin 3' = 2 sin 2' cos 1' sin 1' = 2 X .000 581 7764 X (1 .000 000 0423) - .000 290 8882 = .000 872 6646. So, cos2' = 2cosl'cosl' cosO' = 2 X .999 999 9577 X (1 - .000 000 0423) - 1 = .999 999 8308 ; cos 3' = 2 cos 2' cos 1' cos 1' = 2 X .999 999 8308 X (1 - .000 000 0423) -.9999999577 = .999 999 6193*. 40 PLANE TRIGONOMETRY. [II. (2) For angles 30-45, substitute 30 for 0, and 1', 2', 3', successively for 0' in formulae : sin (0 + 0') = 2 sin0 cos0'- sin (0 - 0') [51 = cos0' - sin(0 - 0') ; [sin 30 = % cos (0 -f- 0') = cos (0 - 0') - 2 sin sin 0' [54 = cos (0-0')- sin 0'. Thus : sin 30 1' = cos 1' sin 29 59' = .999 999 - .499 75 = .500 25 ; sin 30 2' = cos 2'- sin 29 58' = .999 999 - .499 50 = .500 50. So, cos30l' = cos 29 59'- sin 1' = .866 17 -.000 29 = .865 88; cos 30 2' = cos 29 58'- sin 2' = .865 73. (3) For angles 45-90, apply formulae: sin (45 + 0') = cos (45 0') ; [I. Thm. 5 cos (45+ 0') = sin (45- 0') . Thus: sin 45 1' = cos44 59' = .707 31 ; sin 45 2' = cos44 58' = .707 52. So, cos 45 1' = sin 44 59' =.706 90; cos45 2' = sin 44 58' = .706 70. * SECOND METHOD : Substitute the value of for 0', 1', 2 f , in [94, 95]. Thus: 1'- v = 3. 141 592 653 589 793 180 x 60 10 800 = .000 290 888 208 666 ; ... sin 1' = .000 290 888 208 666 - ' 00 29 ^ 208 666 + = .000 290 888 2046 ; .000 290 888 208 666 2 , 1f 1 cosl=l- 2! 9999999577; ' 7.] GENERAL FORMULAE. 41 siii 2' = 2 X .000 290 888 208 666 03 ^.000290888208666* ~TT = .000 581 7764 ; 9 ,_ n 09 ^ .000 290 888 208 666 2 , COS & J. Z A - = .999 999 8308. NOTE 1 . The process is evidently very tedious ; but the reader will notice, first, that it would be much shorter if four or five decimal places only were sought in the functions ; and, second, that once having raised the fraction to the required powers, thereafter he has only to take simple multiples of them. At first he need use but two terms of the series ; but later, when is larger, and the series therefore converges less rapidly, more terms must be used ; thus : -.-, for 30, = - = .52360 nearly ; = .52360 .02392 + .00033 - .00000 -f ..... = .5 within less than .00001 ; i.e.-, by the use of three terms of the series, the sine is found correct to four decimal places, the same degree of accuracy as that assumed for the value of TT. NOTE 2. The results may be verified by using both methods of computation ; and for certain angles there are other and inde- pendent methods : [63,64 .-. from cos 45, = Vi, [I- Thm. 13 are found in succession the sines and cosines of 22 30', 11 15', 5 37' 30", ..... . So, from cos 30, = V 3 ' P- Thm - 12 are found in succession the sines and cosines of 15, 7 30', 3 45', ..... . 42 PLANE TRIGONOMETRY. [II. (2) . sin 2 = 2 sin cos 0, [60 and cos 30 = 4cos 3 3cos0, [84 and v sin 36 = cos 54, [I. Thm. 5 .-. 2 sin 18 cos 18 = 4 cos 3 18- 3 cos 18; whence are found in succession the sines and cosines of 9, 4 30', 2 15', ..... (3) Fromcos3G = cos 2 18- sin 2 18= i(^5 + 1), [61, above and sin36 = V(l-cos 2 36)=iV( 10 -- 2 V 5 ); [36, above are found the sine and cosine of (36 30), i.e., of 6, and thence in succession the sine and cosine of 3, 1 30', 45', ..... (4) Fromsm(36+0')-sm(36 -0 f ),= 2cos36sin0' [52 = H V5 + 1) sin 0', [above subtract sin(72+0') (sin 72 0'), = 2cos72sin0' [52 = ^(V5 I)sin0'. [above then, 102] sin(36 + 0')-sm(36 -0') = sin(72 +0')-sm(72-0')-hsin0', which is Euler's. formula of verification, and serves to test the sines of all angles from to 90, if to 0' be given the different values from to 18. PROB. 2. To COMPUTE TANGENTS, COTANGENTS, SECANTS AND COSECANTS. FIRST METHOD. For the tangents, divide the sines of the angles, in order, by the cosines, each by each; for the cotangents, divide the cosines by the sines; for the secants, divide unity by the co- sines; for the cosecants, divide unity by the sines. * SECOND METHOD. Substitute the values of for 1', 2', 3', ..... in formulae [96-99] . NOTE. These series converge less rapidly than those for sine and cosine ; but for small angles they may be used. 7.] GENEEAL FORMULAE. 43 PROB. 3. To COMPUTE TABLES OF LOGARITHMIC FUNCTIONS : FIRST METHOD. From a table of logarithms of numbers take out the logarithms of the natural sines and cosines. For the tan- gents, subtract the logarithmic cosines from the logarithmic sines; for the cotangents, subtract the logarithmic sines from the loga- rithmic cosines; for the secants and cosecants, subtract the loga- rithmic cosines and sines from 0, respectively. * SECOND METHOD. For the sines and cosines, substitute the values of for 1', 2', 3', in formulae [100, 101] ; for thetan- gents, cotangents, secants and cosecants, follow the first method. NOTE. A more rapid method, applicable also to making tables of natural functions, and many others, is this : Compute the logarithms of three, four, or more angles at regu- lar intervals, and lind their several " orders of differences " ; then, by the algebraic " method of differences," find the succes- sive terms of the series of logarithms, and interpolate for other angles lying between those of the series. Repeat this process for different parts of the table, and verify by direct computation. For safety, four-place tables must be com- puted to six places ; five-place tables to seven places, and so on. A useful modification of the above rule is this : Add the last difference of the highest order to the last differ- ence of the next lower order, and that sum to the last difference of the next lower order, and so on till a term of the series is reached. Thus, in the example which follows, the numbers below the heavy rules are got by successive addition : Angle log sine 1st difs. 2d difs. 3d difs. 18 18 10' 18 20' 18 30' 18 40' 18 50' 19 9. 489982 4 9.4938513 9.4976824 9.5014764 38689 38311 37940 -378 371 7 7 7 7 -364 -357 -350 37576 37219 36869 9.5052340 9.5089559 9.5126428 44 PLANE TRIGONOMETRY. [II. * 8, EXERCISES. If A, B, c be any three plane angles whose sum is 180 (the three angles of a triangle) , prove that : 1 . tan A 4- tan B 4- tan c = tan A tan B tan c . [80 2. tanj-Atan-j-B 4- tan^B tan^c4-tanjc tan|-A= 1. [80 3. sin2A4-sin2B 4- sin2c = 4sinAsinB sine. [68,10,71,4 4. sinA4-sinB 4-sinc = 4cos^-Acos^B cos^-c. [68,4,70 Prove that: 5. siii(2sin- 1 a;)=2#V( 1 O- [60 6. sin (3 sin- 1 a) = 3 a 4 or*. [78 7. tan (2 tan' 1 a) = 2 a: (l-x 2 ). [62 8. tan(3tan~V)= (Sx x?) : (1 3^). [80 11. tan" 1 ^ tan" 1 ?/ = tan- 1 (x y) : (xy ^ 1) . [43, 44 13. ^?r = tan -1 l = tan" 1 !- 4- tan" 1 ^ [43 = tan" 1 !- 4- tan" 1 -!- _j_ tan" 1 ^ = 2 tan- 1 ^ 4- tan" 1 -}-. [80 15. If (l+ cos(9)(l-ecos^)=l- 2 then also, tan|0 : tan|-^ = J(l + e) : ^(l e). [65 16. IfO.O' and 6" be any three angles, prove that _ ~ : *V* ,, f 9 17. If 6 and 0' be any two angles , and n any integer, prove that [sm0+sin(<9-|-<9')+sm(0-j-2<9 f )-{- = cos (0 - J0') - cos (0 +^!>6 f ) , [54 [cos0+cos(0+0')+cos(0+2<9')_f- ..... +cos(0+?i^l)0'].2sin40' = sm(0 + '^=l6')-sm(0-0'). [52 18. From the results of Ex. 17, prove that 8.] GENERAL FORMULAE. 45 \ n \ n wherein n is any positive integer. 19. In the results of Ex. 18, make n = 3, and prove that sin 9 + sin (60 - 0) - sin (60+ 0) = ; cos cos (60 - 0) - 008(60+ 0) = 0. 20. In the results of Ex. 18, make n = 5, and prove that sm0+sin(72 +0)+sin(36 -0)-sin(36 +0)-sin(72-0)==0, cos0+cos(72 +0)-cos(36 -0)-cos(36 +0)+cos(72-0)=0, and from the first of these two, get Euler's formula. [102 21. In the results of Ex. 18, make n 9 and 15, in succes- sion, and thence find other formulae of verification. Show that the formula found when n = 9 verifies the sines of all angles in the quadrant, if to be given values from to 10. 22. If 6 be an}' angle and 0' be an infinitesimal angle, the in- crement of 0, prove that inc 2 sin0 = -(2sin0') 2 sm(0 + 0'), [69 inc 2 cos0 = - (2 sin 0') 2 cos (0 + 0') , [71 inc 4 sin = (2sin|0') 4 sin (0 + 20'), inc 4 cos0 = (2sn40') 4 cos(0 + 20'), Wherein inc 2 sin stands for the increment of the increment of sin 0, i.e., for [sin (0 + 20') - sin(0 + 0')] - [sin(0 -f- 0') sin0], or sin (0 -f 2 0') 2 sin (0 + 0') + sin 0, and inc 4 stands for inc inc inc inc sin 0, i.e., for sin(0+40')-4sin(0+30') + 6sin(0+20')-4sin(0+0') + sin0 and so on. [Alg., Meth. Dif. 23. Compute the sines and cosines of 22 30' and G7 30', 11 15' and 78 45', ..... ; of 15 and 75, 7 30' and 82 30', ..... ; of 6 and 84, 3 and 87, ..... ; of 9 and 81, 4 30' and 85 30', ...... [Prob. 1, note 2 24. From the logarithmic sines of 18, 18 10', 18 20', ..... find the several orders of differences, and thence, by interpola- tion, find the logarithmic sines of 18 1', 18 2', 18 3', 18 4', ..... 46 PLANE TRIGONOMETRY. Fill. HI. SOLUTION OF PLANE TRIANGLES. 1, GENERAL PROPEETIES OF PLANE TRIANGLES. THEOREM 1. In any plane triangle the sides are proportional to the sines of the opposite angles. B A Let ABC be any triangle ; a, &, c the sides opposite the angles A, B, c respectively ; then will : 103] a : b = siiiA : sinB ; b : c = sinB : sine ; c : a = sin c : sin A. For, draw DC _L AB ; then DC = AC sin A = b sin A, and DC = BCsiiiB = asinB, [I. 7, note So, and a : b = sin A : sinB. b:c = sinB: sine, era = sin c : sin A. [Thm. prop'n Q.E. D, NOTE 1. This theorem may be stated more symmetrically thus: a : b : c = sin A : sin B : sin c ; or thus : a b c sin A sin B sin c 1.] SOLUTION OF PLANE TRIANGLES. 47 COR. In any plane triangle ABC : sinB sine , csiiiB asinB b = = ; sine sin A __asinc_ b sin c . sin A shiB 105] BinA = 5LEB = 8inc; b c bs'mc b sin A sins = sin c = c a csinA csiiiB NOTE 2. The expressions - , ..... are equal to the diameter sin A of the circumscribed circle, as appears later, PROB. 4. NOTE 3. All the angles of a triangle are positive angles, and the sides of the triangle arc positive lines when taken with refer- ence to the angles included b}" them. The ordinatc of the vertex, with reference to the base, is there- fore alwa t ys positive ; but the abscissa ma} T be either positive or negative, .according as it is measured in a positive or negative direction from the origin used. THM. 2. In any plane triangle, the sum of any two sides 'is to their difference as the tangent of half the sum of the two opposite angles is to the tangent of half their difference : 106] i.e., (a + 6) : (a~ &) = tan J(A + B) :tan(A~B) ; (c + a) : (c ~a) = tan(c -f A) : tan (c ~ A) . For, .* a : b = sin A : sinB, [Thin. 1 .-. (a + b) : (a b) = (sin A -f SIHB) : (sinA sins) ; [Thm. prop'n and *. (sin A + sin B): (sin A siiiB) = tan(A+B) : tan(A B) ; [72 .-. (a+l) : (-&) = tan J(A + B) :tan'(A-is). 48 PLANE TBIGONOMETBY. [III. But*. 'the greater side of a triangle lies opposite the greater angle, \_geom. . . when a > 6, then also A > B, and (a + &) : (a - b) = tan|-(A + B) :tan|(A B). So, when b >a, then also B > A, and (a + b) : (b a) = tan(A + B) :tanJ-(B A) ; . . (a -f- b) : (a ~ b)-= tan -(A + B) : tan (A ~ B) . So, (b -f- c) : (b ~ c) = tan i (B + c) : tan (B ~ c) , and (c +a) : (c ~a) = tanJ-(c + A) : tan|(c ~ A). Q.E. D. COR. In any plane triangle ABC : 107] tan(A~B) = ^- ? a 4- / 1A 6 ~' ( tan |( , A )-^f c 4- a THM. 3. J;i any plane triangle ABC : 108] ; COSC = 2a6 A c D B A B For, . a 2 =& 2 + c 2_ 2 and AD = b COS A ; D P a 2 = 6 2 + c 2 2 6c cos A, [I. 7, note and So, and COSA = COSB = cosc = 2 be 2ca Q.E.D. 1.] SOLUTION OF PLANE TUTAXGLES. 49 TIIM. 4. In any plane triangle ABC : 1091 sin4-A = -v/ > J \ be -c)(s-] ca (s-a)(s-b) wherein s stands for (a + b + c) , or half the perimeter. For, v 2sin 2 A = 1 cos A [63 _ a2 ^ 26c (a 6 + c) (a + & c) 2 be So, 8inJB = J^- c )(^ a >; and Blnjc = --. Q.B.D THM. 5. In any plane triangle ABC 1101 cosJ-A= l s ( s - a ). . "Xl be ls(s-b)_ ~ \ ca 50 PLANE TRIGONOMETRY. [III. For, v 2cos 2 |A = l+cosA [64 c - a 2 be 2 be 2 be _4s(s a). 2bc /*(*-&). 80, COS^B =-vf - - -1 II CCv ^c = \ and cos CCv " THM. 6. Jw an?/ plane triangle ABC 111] s(s-c) For, tan^A = sin^A: COS^-A [35 s (s - a) \(s - So, and tanc=. NOTE. Since no angle is greater than 180, therefore no half- angle is greater than 90, and the radicals of Thins. 4, 5, and 6 are all positive. [I. Thm. 3 2.] SOLUTION OF PLANE TRIANGLES. 51 2. SOLUTION OF BIGHT TRIANGLES. PROBLEM 1 . To SOLVE A RIGHT TRIANGLE : Let OAF be an} r right triangle ; o, the base- angle ; A, the right angle ; x, y, and r, the base, perpen- dicular and hypothenuse. CASE 1. Given the hypothenuse and an acute angle, for example r and o ; then : P =90-o; or = rsinp; 2/ = rsino, or = rcosp, or = #tano. [I. 7 CASE 2. Given a side and an acute angle, for example x and o ; then: p=90-o; r = x : cos o, or = x : sin p ; y = xtano, or = rsino, or = rcosp. [I. 7 CASE 3. Given the hypotlienuse and a side, for example r and x ; then : coso = x : r, whence o is found ; P = 90 o; 2/ = rsino, or =rcosp, or = #tano, [I. 7 or = Vr 2 x 2 = V (r + ic) (r x) . [geom. CASE 4, Given the two sides about the right angle, then : tan o = y : x, whence o is found ; P =90-o; r = #:coso, or = a;: ship, or =y:sino, or = y : cos p, [I. 7 or = Va^H- y 2 . \_geom. NOTE 1. To test the correctness of the work, various checks may be applied : (1) Compute the part by some other process. (2) Substitute the three computed parts in some one of the six equations which express the definitions of the trigonometric functions [I. 7]; if this equation is satisfied the parts are right. 52 PLANE TEIGONOMETKY. [III. functions [I. 7] , or of the six equations which result therefrom [I. 7, note] ; if this equation is satisfied the parts are right. (3) Take the three given parts and one computed part, or two given parts and two computed parts, or one given part and the three computed parts, and substitute them in some one of the formulae of Thms. 1-6 ; if this formula is satisfied, the work is correct, for the part or parts tested. *NOTE 2. When the solution involves angles near to 0, 90 or 180, care must be used in the selection of formulae ; for, of angles near 90, those which differ very considerably have nearly the same sine, and cannot therefore be determined with precision from the table of sines, thus : the sines of all angles from 89 50' to 90, inclusive, differ from 1 by less than .000005, and the sines of all angles from 89 49' to 89 42', inclusive, differ from .99999 by less than .000005 ; but, of tangents, that of 89 is 57.2900 ; that of 89 20' is 85.9398 ; that of 89 40' is 171.885 ; that of 90 is infinity ; whereby it appears that the tangents of angles near 90 not only increase very fast, but that they also increase faster and faster. So, of angles near and 180, the natural cosines change very slowl}', and the natural cotangents ver} 7 fast. Of angles near and 180, the logarithmic sines, tangents and cotangents change very fast, and at rapidly varying rates, and the logarithmic cosines very slowly. So, of angles near 90, the logarithmic cosines, cotangents and tangents change very fast, and at rapidly varying rates', and the logarithmic sines very slowly. To avoid these angles, use the following formulae : (1) If x be very small compared with r, and therefore o be nearly 90, use : 1 7* - />* 112] sino = --r; 113] 2.] SOLUTION OF PLANE TRIANGLES. 114] For, v coso = -> [I- 7 r .'. 2sin 2 ^o, =1 coso, = 1 , [63 /W At I /V and 2cos 2 ^o, =l+coso, =l-f- =- , [64 .'. etc. Q. E. D. (2) If x be nearly equal to r, and therefore o be nearly 0, compute ?/, = V (r + x) (r x) , and use : 115] sm^p: 116] \ 2r 117] (3) If # be very small as to y, and therefore o }}e nearly 90^ compute r, =Vit >2 +2/ 2 , then use either of the formulae [112- 114] ; if ?/ be very small as to #, use [115-117]. *NOTE 3. If o be very small, then special tables may be used which give the angle in seconds, the logarithmic sine and tan- gent, and the logarithms of the ratios of the sine and tangent to the number of seconds in the angle. These ratios are nearly constant, and their differences are very nearly proportional to the differences of the corresponding angles. The formulae are : 118] o in seconds ___ EL2 __ = I ; o in seconds r 119] and o in seconds . _ tano _ = .?. o in seconds x So, if p be ver}' small. These special tables give the functions and angles with more accuracy than the ordinary tables give functions and angles in any part of the quadrant. 54 PLANE TRIGONOMETRY. [Ill 3, SOLUTION OF OBLIQUE TRIANGLES. PROB. 2. To SOLVE AN OBLIQUE TRIANGLE. jf FIRST METHOD. By means of right triangles. Let ABC be any oblique triangle ; and a, 6, c the sides oppo- site the angles A, B, c, respectively. From c draw DC JL AB, and let y stand for DC, x' for AD, x" for DB, c' for ZACD, and c" for ZDCB. CASE 1. Given two angles and a side; for example, A, c and 6 ; then : B = 180-(A + c); [geom. In rt. A ACD, b and /. CAD are known, whence y and x 1 are found ; In rt. A BCD, y and /. CBD are known, whence a and x" are found ; c = x'+x". NOTE 1. The reader must carefully distinguish between the signs of the sides and angles of a triangle, when taken with refer- ence to the triangle itself, and when taken with "reference only to some initial direction ; thus : The parts of the triangle BCD, when taken with reference to the triangle BCD, are all positive ; but, with reference to AB, as initial direction, the line DB is posi- tive or negative according as it is measured to the right or to the left from D. So, with reference to ordinary positive rotation, the angle DCB is positive or negative according as CB swings to the right or to the left from CD as the reader looks at the diagram. 3.] SOLUTION OF PLANE TKIANGLES. 55 NOTE 2. x' and c' are positive or negative, with reference to AB and to ordinary positive rotation, according as A is acute or obtuse ; and x" and c" are positive or negative, according as B is acute or obtuse. If A + c = or > 180, there is no triangle ; if A -f- c < 180, there is always one triangle, and but one. The parts are determined without ambiguity from the formulae. CASE 2. Given two sides and the included angle; for example, &, c and A ; then : In rt. A ACD, b and ZCAD are known, whence y, x' and c' are found ; x"=c-x'; In rt. A BCD, y and x" are known, whence Z CBD and c" are found ; c = c'+c". B = ZCBD or =180 CBD. NOTE, x' and c' are positive or negative, with reference to AB and to ordinary positive rotation, according as A is acute or obtuse, and x" and c" are positive or negative, and B is acute or obtuse, according as c> or ?/, there are two triangles ; (2) if A is acute and a < 6, but a = ?/, there is one triangle ; (3) if A is acute and a < 6, but a < y, there is no triangle. (4) If A is acute and a = &, there is one triangle ; (5) if A is acute and a > 6, there is one triangle. (6) If A is right or obtuse and a > 6, there is one triangle ; (7) if A is right or obtuse and a = &, there is no triangle ; (8) if A is right or obtuse and a < 6, there is no triangle. 3.]- SOLUTION OF PLANE TRIANGLES. 57 For v c" is found from its cosine, = y : a, .'. when a <6, c" and x" may be either positive or nega- tive ; andB = CBD or =180 CBD. [2 But, when a = or > 6, the negative values of x" and c" are in- admissible ; for, then c, = x'+'x", and c, = c'+"c", are both or negative, which is absurd. SECOND METHOD. By means of the general properties. CASE 1. Given two angles and a side; for example, A, B and c, then : c = 1 80 ( A + B) , [geom. (* (* a = sin A, b= sins. [104 sin c sin c CJiecTc: See formulae [120, 121]. NOTE. The formulae give one value, and but one, for each part. CASE 2. Given two sides and the included angle; for example, a, b and c ; then : . ^^ 7, tan(A~B) = cotjc, whence -J(A~B) is found, [107 J-(A + B) -f- ^(A ~ B) = the greater of the two angles ; J(A -+- B) ^(A ~ B) = the less of the two angles. c = -^ sine. [104 sin A Check : b : sin B = a : sin A. NOTE. The formulae give one value, and but one, for each part. CASE 3. Given the three sides: a, 5, c. Apply the formulae of Thms. 3-6. Check: A + B + c = 180. NOTE 1. The formulae give one value, and but one, for each part. NOTE 2. Of these formulae, those of Thm. 3 are only useful when the computation is by natural functions ; those of Thm. 4 use nine different logarithms, those of Thin. "> use ten different logarithms, those of Thm. 6 use only seven different logarithms, for the computation of all the angles. The formulae of Thm. 6 58 PLANE TRIGONOMETRY. [III. are therefore generally to be preferred ; they may be put in the form: L- l(-a)(-6)(-c) s a\ s (s a) (s b)(s c) 5 ( a) (-&)( -J1 . c \ s wherein the second factor of the right member is the same, and may be computed but once, for all. It will appear later, PKOB. 4, that this factor is the radius of the inscribed circle. CASE 4. Given two sides and an angle opposite one of them; for example, a, b and A ; then : shin = - sin A, whence B is found. [104 Ch c = 180 - (A + B) , [geom. sine. [104 a smA Clieck: See formulae [120, 121]. NOTE 1 . The formulae leave the parts in doubt ; for the same value of SIUB belongs to two angles, which are supplements of each other $ so that, in general, B may be an acute or an obtuse angle ; whence two values each for c and c, and two triangles. But this is limited b}* the conditions, that " the greater side of a triangle lies opposite the greater angle," and that "a triangle can have but one obtuse angle, and no side or negative." If, then, a = or > 6, B cannot be obtuse ; and if A is obtuse, B is acute. Moreover, the shortest length possible for a is the perpendicular DC ; for 6sinA = asiiiB = DC, .*. if a < DC, then sins > 1, which is impossible. NOTE 2. The same care must be taken in solving oblique tri- angles as in solving right triangles, to avoid the use of angles near 0, 90 or 180, unless the special . table mentioned under right triangles is used. *.] SOLUTION OF PLANE TRIANGLES. 59 The following formulae are useful ; the reader ma}^ prove. 121] 122] 123] 124] 1251 c c s sin^c sin ^ (A ~ B) COSi(A + B)' sin|(A~B) . cos^c COS A COS B . sin ^ (A -}- B) ' c s c c s-b sin^c sin 1 c sin J A COS!B . c s a cos > c cos } A sin > B 4. THE AREA OF A TRIANGLE. PROB. 3. To FIND THE AREA OP A TRIANGLE, AND THE LENGTH OF THE PERPENDICULARS FROM THE VERTICES TO THE OPPOSITE SIDES : Let ABC be aiw triangle, and let K stand for its area, and p a , p b , p c , for the perpen- diculars on a, 6, c, respectively. CASE 1. Given two sides and the included angle; for example, 6, c, and A. (1) For the area, multiply half the product of any two sides by the sine of the included A angle. (2) For the perpendicular upon either given side, multiply the adjacent side by the sine of the included angle. For, draw DC_L AB ; then, V K = ^DC.AB = |J? O C; [geom. and v _p c =&shiA; [I. 7, note 120] .'. K = ^6csinA. So, K=Tj-casinB; and K 60 PLANE TRIGONOMETRY. [Ill CASE 2. Given the angles and one side; for example, c : (1) For the area, multiply half the square of any side by the sines of the adjacent angles, and divide the product by the sine of the opposite angle. (2) For the perpendicular , multiply the side by the sines of the adjacent angles, and divide the product by the sine of the opposite angle. For, v K = ^6csinA, [126 and v b = -, [104 sine 127] .'. K= sine 4 a 2 sin B sine 1 & 2 shi c shi ^ So, K = - - , and K = 128 sin A smB 2K a sin B sine a smA b sin c sin A c sin A sin B So, p b = -- : 5 and p c smB sine CASE 3. Given the three sides, a, b, c: (1) For the area, from half the sum of the sides subtract each side separately, multiply the continued product of these remainders by the half sum, -and take the square root of the product. (2) For tfie perpendicular, divide twice the root above found by the side on which the perpendicular falls. For, v K = ia6sinc; [126 and v sine = 2sin^ccos^c [60 \ (s ~ 5) .J* (g 7 c) [109, no ab \ ab 2 ", = 2. 129] .'. K = Vs(s a) (sb) (s c), 130 1 = = J la a 2 K 2 K So, ^^-5"' and ^ = ~ 5.] SOLUTION OF PLANE TKI ANGLES. NOTE. The reader may prove the cheek-formula 61 5, INSCRIBED, ESCRIBED AND CIRCUMSCRIBED CIRCLES. PROB. 4. To FIND THE RADII OF THE CIRCLES INSCRIBED IN, ESCRIBED AND CIRCUMSCRIBED ABOUT, ANY TRIANGLE. (1) For the radius of the inscribed, circle, divide the area by half the perimeter. (2) For the radius of an escribed circle, divide the area by half the perimeter, less the side beyond which the circle lies. (3) For the radius of the circumscribed circle, divide half of either side by the sine of the opposite angle. /' For, let ABC be any triangle, and let r stand for the radius of the inscribed circle, and r', r", r'", for the radii of the escribed circles whose centers are o', o", o'", respectively ; let R stand for the radius of the circumscribed circle ; then : 62 PLANE TRIGONOMETRY. [III. (1) . K = Jr(a + b + c) = ?* [greom. 132] .-. , _ K l( s ct) (s 6)(s c) Q. E. D. s \ s (2) ..- K = ir'(- + & + c ) = ? ''( s - a )' = J r" (a - 6 + c) = r" (-&), = ir'"(a + 6-c) = r'"(s-c); [geom. 1331 K K K Q. E. D. NOTE. The reader m&y prove the check-formulae 134] _ *. o-' /)-" 1'"' and 136] .-. R = -^ J sin A So, 135] K 2 = r . r r . r". (3) About ABC circumscribe a circle and draw CA' a diameter ; join A'B, then A = A', and Z A'BC is a rt. Z, [geom. [I. 7. CA'=^ sm A' Bin A sin B sm c 6, EXERCISES. Q.E.D. Solve the right triangles [both by use of natural functions and by use of logarithmic functions] , given : p/ I. r = 36.3, o = 50. 2. x =29.28, p = 37!2'. 3. r = 125, y=l05. 4. a;=29.275, y = 39.07. 5. r = 37.09, y=.379. 6. r = !3ll, o = 89l8 f . SOLUTION OF PLANE TRIANGLES. 63 Solve the oblique triangles [both methods], given: 7. a = 25.3, 8. A = 34, 9. a=127, 10. a =18, 11. a =10, 12. a=16, !:>. a = 20, 14. a =24, a = 24, a = 20, a=16, a=127, 15. 16. 17. 18. 19. a =2000, 6=136, B = 95, 6 = 64.9, 6 = 20, 6 = 20, 6 = 20, 6=20, 6=20, 6 = 20, 6 = 20, 6 = 20, 6 = 254, 6=1999, c = 98 15'. c = 13.89. c= 152.16. A = 55 2.4'. A = 30. A = 86 40'. A = 47 9'. A = 37 36'. A =120. A = 135. A =150. c = 380. A = 91. D 15 Find the areas of the triangles, and the three perpendiculars, jp fl , p 6 , p c , given : 20. =12.5, 6 = 25, c = 36. 21. A = 37 18', B = 9218', c = 39.5. 22. a = 29. 7, 6 = 6.238, c = 34.21 Find the radii of the inscribed, escribed and circumscribed circles, given : 23. a = 12. 7, 6 = 22.8, c = 33.9. 24. A = 64 19' 8", B = 100 2' 27", c = 51.25. 25. a =136, 6 = 95.2, c=ll37'. 26. In surveying, the bearing of a point is the angle which its direction makes with the north and south line through the point of observation ; its latitude is its distance north or south, and its departure is its distance east or west, from the datum-point. If a surveyor, starting from A, run N. 22 37' E., 3.37 chs. to B ; thence N. 80 24' E., 3.81 chs. to c ; thence s. 41 12' E., 5.29 chs. to D ; thence s. 62 45' w., 6.22chs. to E ; find the latitude and departure respectively of B, c, D, E, from A ; and find the bearing and distance of A from E. PLANE TRIGONOMETRY. [III. 27. Divide the field above given into triangles and trapezoids, by means of parallels of latitude (east and west lines) through the corners, and thence find its area. 28. At 120 feet distance from the foot of a steeple standing on a plane, the angle of elevation of the top is 60 30 f ; find the height. 29. From the top of a rock 326 feet above the sea the angle of depression of a ship's hull is 24 ; find the distance of the ship. 30. A ladder 29 feet long standing in the street just reaches a window 24 feet high on one side of the street, and 21 feet high on the other side ; how wide is the street? 31. What is the dip of the horizon from the top of a moun- tain 2f miles high, the earth's mean radius being 3956 miles ? 32. From the top of a mountain 1J miles high, the dip of the horizon is 136'52" ; find the earth's diameter. 33. Given the earth's mean radius 3956 miles, and the angle which this radius subtends at the sun 8". 81 ; find the distance of the earth from the sun. Fig. 1. Fig. 2. Fig. 3. 34. What is the distance across a river, when the base AB = 475 ft., ZA = 90, Z B = 57 13' 20"? (Fig. 1.) 35. What is the distance CD? (Fig. 2.) Given: the base AB = 131-J yds., Z. BAD = 50, Z. BAC = 85 15', ZDBC = 38 43', /. DBA = 94 13'. Prove the work by making two distinct com- putations from the data. 36. What is the distance AB? (Fig. 3.) Given : CA = 131 ft. 5 in., BC = 109 ft. 3 in., and /. c = 98 34'. Prove the work. *S/TY C.] SOLUTION OF PLANE TRIANGLES. t>5 37. From the top of a hill I observe two milestones in the plain below, and in a straight line before me, and find their angles of depression 5 and 15 ; what is the height of the 1 hill? 3 r r x* + y 2 wherein dx, dy, dr, do, dp stand for total differentials of x, y, r, o. P ; i.e., for any simultaneous infinitesimal changes in the quantities x, y, ^ o, P, that are consistent with the known rela- tions of the parts of a right triangle. *49. If, in a right triangle, only the values of x and y be given, and if these have the possible errors x' and y 1 respec- tively ; i.e., if x may possibly differ from its assumed value by either 4- x 1 or x', and y by either -f y 1 or y', these signs not being necessarily alike ; show from Ex. 48 that the resulting values of r and o will have the possible errors and yx xy , = ~(x' sino + y f coso) ; r 2 r wherein x' and y' are positive. So, if only x and r be given, with the possible errors x f and r', find the possible errors of the other sides and angles. So, if only x and o be given, or only r and o, with the pos- sible. errors x f and o f , or r' and o'. G.] SOLUTION OF PL AXE TRIANGLES*. 67 *50. From the known relations of the parts of an oblique triangle ABC, A + B -f c = 180, ashiB = b sin A, ..... , prove that a] dA-fdB-Mc = 0, b~\ b cos A . dA. a cos B . dv sin B . da -f- sin A . db = 0, c COSB . cZB b cos c . dc sine . db + sinB . dc = 0, a cos c . dc c cos A . cLv sin A . do 4- sine . da = 0. From these, b}' elimination and reduction [104, Ex. 46], derive c] b . dc -f- c cos A . dn sin A . dc -f- sin c . da = 0, c.dv + bcosA.dc s'mA.db -f-sinB.da = 0, with four similar equations, which maybe written from S3 r mmetry ; d] frsinc-. dA da + cose. db + COSB. dc = 0, [Ex.47 with two similar equations, which may be written from symmetry. *51. If in an oblique triangle only side a and angles B, c be given, and if their possible errors be - , 10", and 15", respectively, find- the possible errors of A [Ex. 50, a] ; of b [Ex. 50, c] ; of c [Ex. 50, c]. Find the values of these possible errors when ABC is very nearly equilateral, 5000 feet on each side. *52. Given the values of c, a, 6, with the respective possible errors c f , a', b\ deduce the possible errors of B and A [Ex. 50, c] ; of c [Ex. 50, d]. *53. Given A, a, &, with possible errors A', a', &', deduce the possible errors of B [Ex. 50, &] ; of c and c. *54. Given A, B, 6, with possible errors A', B', 6', find the possible errors of c, a and c ; first, when, as in all the above cases, the computation is assumed to be exact ; and secondly, when c, a, c are liable to the further possible errors c", a", c" from omitted decimal-figures, etc., in the computation. *55. Given a, 6, c, with possible errors a', 6', c', find the possible error of A ; the possible error of computation being A". NOTE. If a', 6', c', A" denoted " probable errors," the probable error of A would be In the same way, Ex. 49, 51-54 are adapted to probable errors. SPHERICAL TRIGONOMETRY. IV. SOLUTION OF SPHERICAL TRIANGLES. 1, GEOMETEICAL PEINCIPLES. If about the vertex of a triedral angle as center a sphere be described, the traces [intersections] of the three faces of the triedral upon the surface of the sphere, are arcs of great circles, and together the}' constitute a spherical triangle, whose sides measure the face-angles, and whose angles measure the diedral angles, of the triedral. SPHERICAL TRIGONOMETRY is therefore the trigonometry of the triedral, and it treats of the numerical relations of the six parts of the triedral, viz. : the three face- angles and the three diedrals. As the three faces of the triedral, when produced, are in- definite planes, and divide all space into eight solid angles, so the sides of the spherical triangle, when produced, are great circles, and divide the surface of the sphere into eight triangles. Of these eight triangles, any two that are diametricall}* opposite are symmetrical, and anj r two not opposite have one side and its opposite angle the same in each, while the remaining sides and angles of the one are supplementary to the corresponding sides and angles of the other. In this treatise only those triangles are considered whose parts are positive and each less than 180 ; but in Astronomy the general spherical triangle is used, which is free from this restric- tion. In the restricted triangle above named the following principles hold true ; the}- are proved in Geometry, and grouped together here for the convenience of the reader : The sum of the three sides lies between and 360. The sum of the three angles lies between 180 and 540. Each side is less than the sum of the other two. 1.] SOLUTION OF SPHERICAL Til TANGLES. 69 Each angle is greater than the difference between 180 and the sum of the other two. Of any two unequal sides, the greater lies opposite the greater angle ; and conversely. Each side or angle is the supplement of the corresponding angle or side of the polar triangle. If two sides of a triangle are equal, so also are the opposite angles; and conversely. The perpendicular from the vertex of an isosceles triangle to the base bisects both the vertical angle and the base. Equilateral triangles, in general, are not similar. A spherical triangle is, in general, determined when any three of its six parts are known. The area is to the hemisphere as the excess of the sum of the angles over 180 is to 360. The following principles, proved later, are added : In a right spherical triangle : An oblique angle and its opposite side are of the same species [both less than 90, both greater than 90, or both equal to 90]. [Thm. 1 Cor. 1 If the Irypothenuse is less than 90, the other two sides are of the same species, and so are the two oblique angles ; but if the liypothenuse is greater than 90 they are of different species, and so are the two oblique angles. [Thm. 1 Cor. 2 If the hypotheuuse equals 90, another side and its opposite angle are each equal to 90 ; and if another side or its opposite angle equals 90, the hypothenuse equals 90. [Thm. 1 Cor. 3 No side is nearer to 90 than its opposite angle. [Thm. 1 Cor. \- In an}' spherical triangle : A side that differs from 90 not less than another side is of the same species as its opposite angle. [Thm. 3. Cor. An angle that differs from 90 not less than another angle is of the same species as its opposite side. [Thm. 4 Cor. 1 There are at least two sides which are of the same species as their opposite angles. [Thin. 4 Cor. 2 The half-sum of any two sides and the half-sum of their oppo- site angles are of the same species. [Thm. 12 Cor. 70 SPHERICAL TRIGONOMETRY. [IV. 2, NAPIEE'S EULES FOE THE EIGHT TEIANGLE. THEOREM 1. In a right spherical triangle, if the right angle be ignored, and if, of the five remaining parts, the two sides be taken, and for the hypothenuse and the two oblique angles their complements be substituted, then the sine of any one of the five parts (called the middle part) equals : ' (1) The product of the cosines of the two opposite parts; (2) The product of the tangents of the two adjacent parts. [The figure shows four right spherical triangles : PQR, p, q and r all < 90; PQR', j>'andg'>90, r<90; P'QR, 2>'andr f >90, g'"<90. The demonstration applies to all alike.] Let PQR be anj T spherical triangle, wherein R is^a right angle ; p and Q oblique angles, either acute or obtuse ; r the hypothe- nuse ; p and q sides opposite P and Q respectivel}' ; then will : 151] 15-2] 153] 154] 155] sin p = sin r sin p and sin q sin r sin Q and cos r = cosp cos q and cos P = cosp sin Q and cos Q = cos q sin p and = tang cotQ ; = tanp cotp ; = cotp cotQ; = tan q cot r ; = tanpcotr. For, join the vertices p, Q, R to o, the center of the sphere, and through either oblique angle, as P, draw PA and PB, J_ OP, and meeting OR and OQ in A and B ; then also is AB_LOA and PA, \_geom. and A OPA, OPB, OAB and PAB are right-angled at P, P, A and A. 2.] SOLUTION OF SPHERICAL TRIANGLES. 71 But arcs p, q and r measure Z^AOB, POA and FOB, [geom. .-. sinp = , cosp = , tanp = ; OB OB OA T> A OT* PA. sing =-, cosg = -, tang = -; sn,- = , OB .OB OP sinp =^, COSP = ^, tanp = ^. - [I. 7 PB PB PA OB OB PB sing, = , = .~ = tanpcotp; OA OA AB cosr, =^, = ^.2Z = OB OB OA COSP, =, = . = tangcotr. Q.E.D. PB OP PB So, either oblique angle may be p and the other Q, .. sin q = sinr sinQ ; sin p = tan q cot Q; COSQ = tanp cot?*. Q. E. D. And,-.' cotp = 2- and cotQ = , [above tanj) tang .-. cotp cotQ = sin ^ sn q = cosp cosg ; [35 tanp tan g but cosr = cos^> cosg, [above .-. cosr = cotp cotg. Q- E. D And, . COSP = tang cot r, COSJP = -, sin Q = , [above cosg smr . . cos P = cosp sin Q. So, cos Q = cos g sin p. Q. E. D. COR. 1. In a right spherical triangle, an oblique angle and its opposite side are of the same species [both less than 90, both greater than 90, or both equal to 90]. 72 SPHEKICAL TIUGOXOMETKY. [IV. For, v cos p = cosp sin Q, [154 and v shiQ is not 0, and is always positive, [ 1, 1. Thm.3 .'. COSP and cosp are both positive, or both negative, or both zero ; .'. P andp are both < 90, or both > 90, or both = 90. So, Q and q are Q. E. D. COR. 2. In a right spherical triangle, if the liypothenuse is less than 90, the other two sides are of the same species, and so are the two oblique angles; but, if the liypothenuse is greater than 90, they are of different species, and so are the two oblique angles. For, v all parts of a triangle are positive and all less than 180, and v cos r = cosp cos q, and = cotpcotQ, [153 .*. if cos?* is positive, then cosp and cos q are both posi- tive or both negative ; and so are cotp and cotQ ; i.e., if r < 90, p and q are both < 90, or both > 90, and so are p and Q ; Q. E. D. [I. Thm. 3 but, if cosr is negative, then cosp and cosg are one positive and the other negative ; and so are cot p and cot Q ; i.e., if r>90, p and q are one of them < 90 and the other > 90, and so are p and Q. Q. E. D. [I. Thm. 3 COR. 3. In a right spherical triangle, if the hypothenuse equals 90, another side and its opposite angle are also each equal to 90; and conversely, if another side or its opposite angle equals 90, the hypothenuse equals 90. For, ifcosr=0, then cosp cos 1; which is absurd. 3.] SOLUTION OF SPHERICAL TRIANGLES. 73 3, GENERAL PROPERTIES OF SPHERICAL TRIANGLES. TIIM. 2. In any spherical triangle the sines of the sides are proportional to the sines of the opposite angles. \c [151 D Let ABC be any spherical triangle ; then will : 156] sina: sin& = sinA: sinB ; sin b : sin c = sin B : sin c ; sin c : sin a = sin c : sin A. For, draw CD, = |>, _L AB ; then : sinp = sinasinB, and sinjp = sin&shiA, sin a sin B = sin b sin A. .-. sin a : sin b = sin A : sinB. [Thm. prop'n sin b : sin c = sin B : sine ; sine : sin a = sine : sin A. Q. E. D. NOTE. This theorem may be stated more symmetrically thus : sina : sin6 : sine = sinA : sinB : sine ; or thus : sina _ sin b __ sine sinB So, and sin A sine COR. In any spherical triangle ABC : sin a = sin b sin A sine sin A smA = sin B sin a sm c sin a sin b = sin c = S111B sincsiiiB sine sin a sin c sin A sine sin a sin B sin A sin & sine sinB S111B = smc = sin b sin c sin b sine sin A sin c sina sine sin A sin b sina sin B sin c sin b 74 SPHERICAL TBIGQ^OMETBY. [IV. TIJM. 3. In any spherical triangle ABC : 157] cos a = cos b cose -f- sin b sine cos A, cos 6 = cos c cos a -f- sin c sin a cos B, cos c = cos a cos & -f- sin a sin & cos c. [The figure shows four oblique spherical triangles : ABC, a, b and c all < 90 ; AB'C, a' and c' > 90, b < 90 ; A'BC, a < 90, 6' and c" > 90 ; A'B'C, a' and 6' > 90, c'" < 90. The demonstration applies to all alike.] For, join the vertices ABC to o, the center of the sphere, and through A draw AE and AF, J_OA, and meeting OB and oc in E and F ; then A OAE and OAF are right-angled at A. But arcs a, b and c measure A EOF, AOF and AOE, and A = /. EAF ; then : cosa = OE 2 + OF 2 EF S and 2 OE . OF OE 2 = OA 2 -f AE 2 , and OF 2 = OA 2 -f- AF 2 ; OA 2 4- AE 2 + OA 2 + AF 2 EF 2 2 OE . OF [geom. [108 \_geom. cosa = OA* OE .OF AE 2 +AF 2_ EF 2 2 OE . OF AF 2 EF 2 _ OA OA , AF AE AE OF OE OF OE 2 AE . AF But COSA = AE 2 + AF 2 - EF 2 2 AE . AF [108 3.] SOLUTION OF SPHERICAL TRIANGLES. 75 . . cos a = cos b cos c -f- sm & sin c cos A. So, cos b == cose cos a -f- sine sin a cos B, and cos c = cos a cos 6 + sin a sin 6 cos c. [I. 7 Q.E. D. COR. In any spherical triangle, a side that differs from 90 not less than another side, is of the same species as its opposite angle. -r, COS COS COSC r-irrrr lor, .* COSA = 1157 sm b sin c and . the product sin b sine is always positive, [I. Thm. 3 . *. cos A has the same sign as (cos a cos b cos c) . But if a differs from 90 not less than 6 or c, then cos a is as large [great numerically] as cos b or cose, and if cos a = 0, then cos b or cos c = ; and . * cos b and cos c are both smaller [less numerically] than 1, .-. cos a is larger than cosb cose ; . . cos a gives sign to (cos a cos b cos e) ; .*. cos A has the same sign as cos a, [above or ifcosa = 0, then COSA = O; i.e., both are positive, or both negative, or both zero ; .-. A and a, both < 90, or both > 90, or both =90. Q. E. D. THM. 4. In any spherical tri- angle ABC : 158] cos A = cos B cos c H-sinB sine cos a, COS B = COS C COS A -|- sine sin A cos&, COS C = COS A COS B -j-sinA siiiB cose. For, let the triangle A'B'C' be polar to ABC ; then,-.- a'=7r A, &'=TT B, C'=TT c, and A'=TT a, [geom* and cosa'= cos 6' cos c' + sin 6' sine' cos A', [157 .-. cos A = ( COSB) (cose) -f- sin B sinc( cosa), [10,11 .-. COSA = COSB cose + sinB sine cosa. So, cos B = cos c cos A + sin c sin A cos &, and cose = COSA COSB -f- sin A sin B cose. Q.E. D. 76 SPHERICAL TRIGONOMETRY. [IV. COR. 1. In any spherical triangle, an angle that differs from 90 not less than another angle, is of the same species as its oppo- site side. For, ... cos= . [158 sm B sin c and v the product sin B sine is alwa} r s positive, [I. Tbm.3 cos a has the same sign as (cos A + COSB cose). But if A differs from 90 not less than B or c, then cos A is at least as large as COSB or cose, and if cos A = 0, then COSB or cose = ; and . COSB and cose are both smaller than 1, . . cos A is larger than cos B cos c ; .-. cos A gives sign to (cos A -f- COSB cose) ; .. cos a has the same sign as cos A, [above or if cos A = then cos a = ; i.e., both are positive, or both negative, or both zero ; .-. a and A are both < 90, or both > 90, or both = 90. Q.E.D. [I. Thms.3,14 COR. 2. In any spherical triangle there are at least two sides which are of the same species as their opposite angles. This is a direct consequence of Thm. 3 Cor. and Thm. 4 Cor. 1. THM. 5. In any spherical triangle ABC : )sin(s 6) sin (8 c) sn B For, . 2 sin 2 | A = 1 cos A [63 _ .. cos a cos b cos c sin b sine _ cos b cos c + sin b sine cos a sin b sine __ cos (b c) cos a sin b sine I sin (s c),sin(s a) = \ - . \ sine sin a sin (s a) sin(s b) r *, ., . , - ' [42 3.] SOLUTION OF SPHERICAL TRIANGLES. 77 2 sin 1(6 c + fl)sini(6 c a) ,- 71 sin b sine 2 sin i (a So, sin sin b sine 2 sin (s b) sin (s c) sin b sine sin (s 6) sin (s c) s'mb sine sin (s c) sin (s a) sine sin a 1 sin (s a) sin (s b) = V~ ' sinasinft THM. 6. In any spherical triangle ABC : Isin s sin (s a) 1601 cosi-A =^l i -i \ sin sine [sin s sin (s 6) COS4-B = .* -j \ sine sin a cos^c as.j__ L \ sin a si For,*/ 2cos 2 ^A= 1 4- COSA [64 , cos a cos 6 cose sin b sin c cos a cos b cos c -f- sin 6 sin c sin b sin c COSCT cos(6-fc) sin b sin c sin 6 sine 'a 4. 5 -|- c ) s in|( a + 5 + c) sin 6 sin c 2 sins sin (s a) sin b sin c sin s sin (s a) sin b sine ri -- 7 L 1O/ [41 78 SPHERICAL TRIGOXOMETKY. [IV. (sin .9 sin (5 b) So, COS^B =x v~ sin c sin a (sins sin(s c) = \ siuasiuft THM. 7. JTI cmy spherical triangle ABC : 161] tan| A = ./ si "( s - 6 ) sin ^- c ), \ sin s sin (s a) - Isin (g~c) sin (8-g) \ sins sin(s 6) . m (s c) The reader may prove the theorem by dividing the value of sin IA by that of COS|-A [Thins. 5, 6] ; and so for tan IB and tan^c. THM. 8. In any spherical triangle ABC : I cos s cos (s A) 162] sin|a = -J- sin B sine cos s cos (s B) sine sin A cos s cos (s c) r ., - For, let the triangle A'B'C' be polar to ABC : then v a' = 7T A, b' 7r B, C' = TT c, > f aeom. A' = 7T a, B' = 7T 0, C' = 7T C, /sins' sin (s f a') and -.. sin 6' sine' cos s cos (s A) _ V [0,16,4,10 3.J SOLUTION OF SPHERICAL TRIANGLES. 79 I cos s cos (s B) So, sin & b = sine sn A I coss cos(s c) sin ^ c = V sinAsinB <** The reader may also prove the theorem directly from the formulae of Thm. 4 : cos A = cos B cos c + sin B sin c cos a in the same manner as Thm. 5 was proved from the formulae of Thm. 3. THM. 9. In any spherical triangle ABC : cos (s B) cos (s c) 1631 sin B sine Icos (s c) cos (s A) I = \ sne sn A cos (s A) cos (s B) r i , >. - siulsmB The reader may prove the theorem directly from the formulae in Thm. 4, or prove it by aid of the polar triangle in the same manner as Thm. 8 was proved. THM. .10. In any spherical triangle ABC : 164] taniq = J \cos(s B) cos(s c) =J -cossco S (s-B) \cos(s c)cos (s A) ta J -cosscos(s-c) \cos(s A)COS(S B) The reader may prove the theorem by dividing the value of sin^-a by that of cos^a [Thms. 8, 9] ; and so for tan ^6 and tan -J c. NOTE. The radicals in [159-164] are all positive ; for they are the sines, cosines and tangents of angles A, ..... Ja, ..... , which are all in the first quadrant. 80 SPHEEICAL THIGOXOMETKY. THM. 11. In any spherical triangle ABC : 165] sinKA + B) = ^ig^)co S}C ; 1GC] sini(A~B) 1C7] C osKA + B) V. . sinc cos-i-c 1G8] co S KA~B) = , . sm-|c For, v sin (A B) = sin(iA JB) = sin-i-Acos-i-B cos|AsiniB [39,40 _siii(s b) |sinssin(g c) sin(g a) /sinssin(s c) pj.^ 1GQ sine \ sin a sin b sine \ sin a sin 6 sin(s b) sin(s a) ' sine 2cos4-csini(a 6) zsin^c cos^-c and i sm-J-c [160 [68, 60 [69,60 Q.E.D. [41,42 So, . COS|-(A B) = co _sins !sin(s a) sin(s b) sin (s c) Jsin (s a) sin (s b} sinc\ sin a sin b sine \ sin a sin b sin (s-c) sin i c [159 sns sine and cos^c sin Tr . sm-Jc, . .. sm|c. Q.E.D. 3.] SOLUTION OF SPHERICAL TRIANGLES. 81 These four formulae, with the like formulae found when the other sides and angles are employed, are called Delambre's Formulae. THM. 12. In any spherical triangle ABC : 171] 172] The reader ma}' prove the theorem by dividing the formulae of Thm. 1 1 one by another, viz. : [165] by [167] ; [166] by [168] ; [168] by [167] ; [166] by [165]. These four formulae, with the like formulae found when the other sides and angles are employed, are called Napier' s Analogies. COR. In any spherical triangle the sum of any two sides is less than, equal to, or greater than, 180, according as the sum of the opposite angles is less than, equal to, or greatef~than, 180. For, v tan (a + 6) COS^(A + B) = tan^-c COS(A~B), [171 and v c and (A ~ B) are both less than 90, and ^c is not 0, whence the product tan^-c COS(A~B) is positive, and not ; .'. the product tan|-(a -+- b) cos ^(A 4- B) is positive, not 0, .*. tan J(a + &) and cos J(A + B) are both positive or both negative, or one is GO and the other ; .'. (a + 6) and i( A + B ) are both < 90, both = 90, or both > 90; [I. Thin. 3 .'. (a-f 6) and (A + B) are both < 180, both = 180, or both > 180. Q.E.D. NOTE. Thm. 1 Cor. 1, and the Cors. to Thins. 3, 4 arid 12 are summarized as follows : The half-sum of any two sides and the half-sum of their opposite angles are of the same species ; and so are either side and its opposite angle, unless the side be nearer to 90 than an}* other side, and the angle be nearer to 90 than an}- other angle. 82 SPHERICAL TRIGONOMETRY. [IV. 4, SOLUTION OF EIGHT TRIANGLES. PROB. 1. To SOLVE A RIGHT SPHERICAL TRIANGLE. Let PQR be any spherical triangle right-angled at R ; then the formulae of Thm. 1 (Napier's rules) apply directly, and with the right angle and an}' other two parts given, the remaining parts can be found. The computer will form equations containing three parts, two known and one unknown, and then solve these equations for the unknown part. He will observe that : If the three parts are contiguous to each other, then that which lies between the other two is the middle part, and the others are adjacent parts ; but if two parts lie together and the other apart from them, then that which lies apart from the other two is the middle part, and the others are opposite parts. The following rule will solve all cases : Take each of the two given parts in turn for middle part, and apply that one of Napier's rules which brings in the other given part. Take the remaining part for middle part, and apply that one of Napier's rules which brings in both of the parts just found. For a check make the part last found the middle part, and apply that one of Napier's rules wJiich brings in both the given parts. The whole work requires but nine logarithms, or seven without the check, since two of the logarithmic functions are used twice over. The check is to be applied to the sine of the part last found. If the two values got for this sine, natural or logar- ithmic, differ ~by not more than three units in the last decimal place, the work is probably right, since the defects of the tables permit this discrepancy in the two results. If such discrepancy exist the mean of the two values may be used. 4.] SOLUTION OP SPHElilCAL TRIANGLES. 83 CASE 1. Given p and q, the two sides about the right angle, then: sing = tanpcotp, ./. cot p = cotp sing ; sinp = tang cot Q, .'. cotQ = sinp cot q ; cos r = cot P cot Q ; check cos r = cosp cos g. NOTE. One triangle is alwa}~s possible, and but one ; the parts q, P and Q are determined without ambiguity by the formulae. CASE 2. Given the hypothenuse and one side, for example p, then: cosr = cosp cosg, .'. cos# = seep* cos r\ s'mp = sin r sin p, /. sin p = s'mp esc r ; cos Q = cos q sin p ; check cos Q = tanp cot r . NOTE. A triangle is possible only when r is nearer to 90 than p is, or when r andp are both 90 ; for then only can cosr tan/? cos q, = ,<1; or COSQ, =, and r are both 90, when COSQ and cos 90, and q and Q are of the op- posite species to p and P. This ambiguity appears directly from the figure. For, let PQR be a spheri- cal triangle right-angled at R ; produce the arcs PQ and PR to meet at p r ; then : 4.] SOLUTION OF SPHERICAL TRIANGLES. 85 and . A PRQ and P'RQ are both right angles, [hypoth. .*. two. right triangles exist, PQU and P'QR, which have the same two parts given, p and p, and the remaining parts of the one triangle supplementary to those of the other. CASE 5. Given the hypothenuse and one oblique angle, for ex- ample P, then : cos P = tang cot r, .'. tan q = tan r cos p ; cosr = cotPcotQ, .'. cot Q = cosr tan P; = tang cot Q ; check smp ==. sin r sin p. NOTE. One triangle is always possible, and but one. The part p is of the same species with p, and so can have but one of two possible values ; the parts q and Q are determined without ambiguity by the formulae. CASE 6. Given the two oblique angles p and Q, then : cos P = cosjp sin Q, . . cosp = cos p esc Q ; cos Q = cos g ship, .*. cosg = CSCPCOSQ ; cos r = cosp cos q ; check cos r = cot P cot Q. NOTE 1. The parts p, q and r are determined without arnbi- guit}' by the formulae ; but the solution is possible only when COSP CSCQ, CSCP COSQ and cotPcotQ are each smaller than 1 ; i.e., when COSP is smaller than sing and COSQ is smaller than siup, and this when p is nearer to 90 than Q to or 180, and when Q is nearer to 90 than p to or 180. NOTE 2. The computer may follow a different order from that given ; he ma}', at pleasure, find all the required parts directly from the given parts, or compute an} r one of the required parts, and then use that part in the computation of other parts. If, however, he make an error in the first part, that error is repeated and perhaps magnified, in the computation of all other parts which depend upon it ; hence the importance of testing the results. NOTE 3. Unless he use the special tables noted under the right plane triangle, the computer must avoid angles near 0, 90 or 86 SPHERICAL TRIGONOMETRY. [IV. 180. For this purpoca he may prove and use the following formulae : 173] sinHr = sinHp cosHg-h cos 2 |-> sin 2 -|-g ; [153,61,36 174] tanr^p = tan(r + q) tan|(r - q) ; [65, 153, 77 175] tair-i-p = sin(r - q) : sin(r + g) ; [65, 154, 57 176] tan P = sin(r q) : smp cosg ; = SJf^f; [65, 154, 151 177] sin(p q.) = sinp tau^p singtan^Q; [176,40,153 178] tan 2 |p = tan J(Q -f P - 90) : tan|(Q - p + 90) ; [65, 154, 72 179] tan 2 !?- = cos(p + Q) : cos(p - Q) . [65, 153, 58 5. SOLUTION OF QTJADRANTAL TRIANGLES, AND ISOSCELES TRIANGLES. PROB. 2. To SOLVE A QUADRANTAL TRIANGLE : Find the triangle which is polar to the given triangle; it is a right triangle; solve it, and take the supplements of the parts thus found for the corresponding parts of the given triangle. NOTE 1. Napier's rules apply directly to the quadrantal triangle, if the quadrant is ignored, and for the five parts are taken the two angles adjacent to the quadrant and the comple- ments of the opposite angle and the two oblique sides. NOTE 2. Manifestly, the biquadrantal triangle cannot be solved unless either the base or the vertical angle is given ; for the remaining parts, two right angles and two quadrants, are quite independent of these two. PROB. 3. To SOLVE AN ISOSCELES TRIANGLE : Draw an arc from the vertex to the middle of the base, thereby dividing the given triangle into two equal right triangles; solve one of these triangles. NOTE. When only the base and the vertical angle are given, there are two triangles, one triangle or none, according as the base <, = or > the vertical angle. When only the two equal sides or the two equal angles are given, there is an infinite number of triangles. Otherwise, subject to the conditions in 1, there is one triangle, and but one. O.J SOLUTION OF SPHERICAL TRIANGLES. 87 6, SOLUTION OF OBLIQUE TRIANGLES. PROB. 4. To SOLVE AN OBLIQUE TRIANGLE. FIRST METHOD. By means of right triangles. D Let ABC be any oblique spherical triangle, and a, b, c the sides opposite the vertices A, r., (5 respectively. Let N be the pole of AB, and through N and c draw a great circle, meeting the great circle AB at D and i>', whereof D stands less than 180 in the direction AB from A ; then is DC _L AB. [geom. Let p stand for DC ; or < q'. NOTE 2. There is always one triangle, and but one. The parts arc determined without ambiguity by the formulae. 88 SPHERICAL TRIGONOMETRY. [IV. CASE 2. Given two angles and the included side, for example c, A and b ; then : In rt. A ACD, b and A are known, whence p, q' and c f are found ; c" = c c'; In rt. A BCD, p and c" are known, whence a, q" and Z CBD are found ; NOTE 1 . q" is positive or negative, and B = CBD or 180 CBD, according as c" is positive or negative with reference to positive rotation, i.e., according as c > or < c'. NOTE 2. There is alwa}~s one triangle, and but one. The parts are determined without ambiguity from the formulae. NOTE 3. If for the given triangle its polar be substituted, then two sides and the included angle are known, and Case 2 is embraced in Case 1 . The required parts are the supplements of the parts found in the polar. CASE 3. Given tiuo sides and an angle opposite one of them, for example a, b and A ; then : In rt. A ACD, b and A are known, whence jp, q' and c' are found ; In rt. A BCD, p and a are known, whence q", Z. CBD and c" are found ; G.} SOLUTION OF SPHERICAL TRIANGLES. 89 NOTE 1. q' is full}' determined from the data, but q" (found from its cosine) ma}* be positive or negative ; and there are two triangles, one triangle or none, according as q'q' ( and q'+q", both, one of them or neither, lie between and 180. The perpendicular CD falls within or without the triangle, and B = CBD or 180 CBD, according as q" is taken positive or nega- tive with reference to the direction AB. And, v A and Z CBD are always of the same species with j?, [Thm. 1 Cor. I .'. A and B are of the same species or different species according as B = CBD or 180 CBD ; i.e., according as q" is positive or negative. *NOTE 2. "Whether there are two triangles, one triangle or none, may, in general, be known by inspection of the given parts : (1) If a lies between b and its supplement, there is one tri- angle, and but one. (2) If a equals b or its supplement, then : If A is of the same species with a, and is not 90, there is one triangle ; If A is of different species from a, there is no triangle ; If A, a and b are all 90, there is an infinite number of triangles. (3) If b lies between a and its supplement, and if a and A are of the same species, there are two triangles, one triangle or none, according as a is nearer 90 than p, as near it, or more remote from it ; and there is no triangle if a and A are of different species. For *. A BCD is possible if a is as near 90 as p, and of the same species ; .. it-is possible if a is as near 90 as 6, and of the same species. [Prob. 1, Case 2, note And . cos a = cos p cos q" and cos b = cosp cos q', [153 cos a : cos b = cos q" : cos q 1 , [Thm. prop'n a proportion wherein a, b and q' are known, and q" may, in gen- eral, be either positive or negative. 90 SPHERICAL TRIGONOMETRY. [IV. (1) then and (2) and But then So, then If a is nearer 90 than 5, q" is nearer 90 than q', [I. 23, Note 2 of q'-\-q" and q'q", one always, but never both, lies between and 180 ; the first or the second of them, according as q' < or > 90. If a = b, then q"=q', [2 in either case, in general, of q' -f- q" and q' q", one lies between and 180, and the other = or 180, and there is one triangle. if a and A are not of the same species, a differs from 90 not less than &, there is no triangle. [Thm. 3 Cor. if a, 6 and A are all 90, B is also 90, and the triangle is biquadrantal and in- determinate. [Thm. 1 Cor. 3 \c (3) then, and, But, then So, then If b is nearer 90 than a, if there is a triangle ACD,' 1, which is impossible ; there is no triangle ACD, and no triangle ABC. And*.* b is nearer 90 than a, [hypoth. there is no triangle when a and A are of different species. [Thm. 3 Cor. G.] SOLUTION OF SPHERICAL TRIANGLES. 91 CA-I; -1. Given two angles and a side opposite one of them, for example A, B and a. In rt. A BCD, a and Z CBD are known, whence p, q" and c" are found ; In rt. A ACD,-p and A are known, whence 6, q' and c' are found ; c = sine' and cos CBD = cos p sine", [154 .'. cos A : cos CBD = sine' : sine", [Thm. prop'n a proportion wherein A, CBD and c" are known, and c' may, in gen- eral, have two supplementary values, c : and C 2 , both positive. [10 a) then and (2) and But, then So, then If A is nearer 90 than B, c" is nearer 90 than c l and C 2 , [ 23, Note 2 of Cj -f- c" and c 2 + c", one alwaj^s, but never both, lies between and 180. KA = B, then d = c" or c 2 = c", if A = 180- B, then d = -c" or c 2 = -c"; in either case, in general, of c x + c" and C 2 -f- c", one lies between and 180, and the other = 0or 180, and there is one triangle, if A and a are not of the same species, A differs from 90 not less than B, there is no triangle. [Thm. 4 Cor. 1 if A, B and a are all 90, 6 is also 90, and the triangle is biqnadrantal and in- determinate. [Thin. 1 Cor. 3 (3) then and, But, then So, then If B is nearer 90 than A, if there is a triangle ACD, c f is nearer 90 than c", in general, Cj-f c"and c 2 +c"both lie between and 180. c' = 90, and d + c" = c, -f-c", there is but one (a qnadrantal) triangle. if p is nearer 90 than A, sinc'> 1, which is impossible ; G.] SOLUTION OF SPHERICAL TRIANGLES. 93 .'. there is no triangle ACD, and no triangle ABC. And*/ B is nearer 90 than A, [hypoth. .'. there is no triangle when A and a are of different species. [Thm. 4 Cor. 1 NOTE 3. If, for the given triangle, its polar be substituted, then the two sides and an angle opposite one of them are known, and Case 4 is embraced in Case 3. The required parts are the supplements of the parts found in the polar. CASE 5. Given the three sides, a, fr, c ; then : / cosa = cos p cos q" and cos6 = cosp cosq f , [153 .'. cos a : cos b = cosq" : cos 7', [Thm. prop'n and cosa-f-cos6 : cosa cos 6 = cos ^"+ cos g' : cos g" cos q'. But v cos a + cos 6 : cosa cosb= cot^(a-f-&)cot-J(a &) ? [77 and cos q' + cos q" : cos q" cos q' = - coti(r/+ q") coti(f/'- q') , and .c = q'+q" ; .'. cot J- (a -}- b) cot % (a b) = cot^-c cot-J- (q" q ! ) , whence i (Y'-r/) is found. Andvr/=K-W-2') and q"=$c + ( 360, or if either side = or > the sum of the other two, there is no triangle ; otherwise there is one triangle, and but one. The parts are determined without ambiguity by the formulae. CASE 6. Given the three angles, A, B, c ; then: v cos A = cosp sine', and cos B = cos p sine", [154 . ' . cos A : cos B = sin c' : sin c" ; [Thm . prop'n and cos A+ COSB : COSA COSB =sinc'+ sine" : sine' sine". 94 SPHERICAL TRIGONOMETRY. [IV. Bllt V COSA + COSB : CObA COSB = COtJ(A+B)cotJ(A B),[77 and sinc'+sinc" : sine' sinc'^=tani(c'H-c'')cot^c'-c'0,[72 and vc = c' + c" ; .*. cot(A + B) cot^(A B)=tan^c cot J(c' c"), whence i-(c'- c") is found. And v c'= ic + (c'- c") and c"= Jc - |(c'- c") , .'.In rt. A ACD, A and c' are known, whence b and q 1 are found ; In rt. A BCD, Z CBD and c" are known, whence a and q" are found ; NOTE 1. B = CBD or 180 CBD, and q" is positive or negative with reference to AB, according as c" is positive or negative with reference to positive rotation. NOTE 2. If A + B -f c does not lie between 180 and 540, or if either angle does not exceed the difference between 180 and the sum of the other two angles, there is no triangle ; otherwise there is one triangle, and but one. The parts are determined without ambiguity by the formulae. NOTE 3. If for the given triangle its polar be substituted, then the three sides are known, and Case 6 is embraced in Case 5. The required parts are the supplements of the parts found in the polar. SECOND METHOD. By means of the general properties. CASE 1. Given two sides and the included angle , for example 6, c and A. Find i(B + c) and KB c) b J Thm - 12 > theu : B = i(B-f-c)-f !(B-C), and C = -|-(B + c) KB c), s i na = . sin A, whence a is found. [Thm. 2 sin B NOTE. There is alwaj r s one triangle, and but one. The parts B and c are determined without ambiguity Uy the formulae, and the species of a is determined by Thm. 12, Cor. 6.J SOLUTION OF SPHERICAL TRIANGLES. 95 CASE 2. Given two angles and the .included side, for example c, A and b. Find |(c + a) and (c a) by Thm. 12 ; then : c = J(c + a) + (c a) , and a = (c + a) |(c a) , sinB = ^^sin6, whence B is found. [Thm. 2 sin a NOTE 1. There is always one triangle, and but one. The parts c and a are determined without ambiguity by the formulae, and the species of b is determined by Thm. 12, Cor. NOTE 2. The solution may also be obtained by applying the methods of Case 1 to the polar triangle. CASE 3. Given two sides and an angle opposite one of them, for example a, b and A ; then : shiB = sin&, whence B is found. [Thm. 2 sin a Find c and c from the formulae of Thin. 12. NOTE. There may be two triangles, one triangle or none. If sin A sin b < sin a, in general, there are two triangles ; for then sinB< 1, and B (determined from its sine) may be either of two angles which are supplementary to each other, and the side CB may lie to the right or to the left of CD. But this is limited by the condition that the greater angle lies opposite the greater side, and that no angle or side can exceed 180, or be negative. If sin A sin b = sin a, there is one (a right) triangle ; for then sinB = 1, and B is a right angle. If sin A sin b > sin a, there is no triangle ; for then sinB > 1, which is impossible. For detail of specific conditions the reader may consult the note to First Method for solving this case. CASE 4. Given two angles and a side opposite one of them, for example A, B and a ; then : ain b = !1M. si nB , whence b is found. [Thm. 2 sinA Find c and c from the formulae of Thm. 12. 96 SPHERICAL TRIGONOMETRY. [IV. NOTE 1. There ma,y be two triangles, one triangle or none. If sin a sin B < sin A, in general, there are two triangles ; for then sin 6 < 1 , and b (determined from its sine) may be either of two arcs which are supplementary to each other. But this is limited by the condition that the greater side lies opposite the greater angle, and that no side or angle can exceed 180, or be negative. If sin a sinB = sin A, there is one (a quadrantal) triangle ; for then sin& = 1, and b is a quadrant. If sin a sinB > sin A, there is no triangle ; for then sin b > 1, which is impossible. For detail of specific conditions the reader may consult the note to First Method for solving this case. NOTE 2. The solution may also be obtained by applying the methods of Case 3 to the polar triangle. CASE 5. Given the three sides, a, &, c. Apply the formulae of Thm. 5, 6 or 7. NOTE 1 . If either side equals or exceeds the sum of the other two, or if the sum of the three sides equals or exceeds 360, there is no triangle ; otherwise there is one triangle, and but one. That there is a single triangle appears also from the formulae, since the half-angles computed must be each less than 90, and but one such half-angle can be found from a given function. NOTE 2. Of these formulae those of Thm. 5 use nine different logarithms, those of Thm. 6 use ten different logarithms, and those of Thm. 7 use only seven different logarithms for the com- putation of all the angles. Those of Thm. 7 are, therefore, gen- erally to be preferred ; they may be put in the form ! 1 / sin (s a) sin (3 6) sin(s c) tan 7T A * I ; 9 sm(s a)\ sins 1 I sin (s a) sin (s b) sin(s c) tan -s- B = - ^ I ; ? sm(s b) \ sins ^ __ 1_ | sin (s a} sin (s b) sin (s c) ~~ sin (s c) V sin s 6.] SOLUTION OF SPHERICAL TRIANGLES. 97 wherein the second factor of the right member is the same, and may be computed once for all. NOTE 3. A complete check is afforded by either of Delambre's formulae, [165-168]. CASE 6. Given the three angles A, B, c. Apply the formulae of Thm. 8, 9 or 10. NOTE 1 . If either angle does not exceed the difference between 180 and the sum of the other two, or if the sum of the angles does not lie between 180 and 540, there is no triangle ; other- wise, there is one triangle, and but one. That there is a single solution appears also from the formulae, since the half-sides computed must be less than a quadrant, and but one such half-side can be found from a given function. NOTE 2. Among the formulae the same choice may be made as in Case 1. The reader may transform those of Thm. 10 for convenient use. NOTE 3. A complete check is afforded by either of Delambre's formulae, [165-168]. NOTE 4. The solution may also be obtained by apptying the methods of Case 5 to the polar triangle. The following formulae, which the reader may deduce from [159-164], are sometimes of use : 180"1 s * n s _ cos-J-Acos^-B cos s _sin^-#sin|-& m sine sin^-c sine cos^-c sin (s c) _ sin j- A sin -^ B cos(s c) __ COS^CT sine sin^c sine cos^c sin(s d) _ cos j-A sin j-B cos(s A) _ sin j-q cos-^-6 sine cos^-c sine sin^-c 183"! sm ( g c)_tanj-A cos(s c) _ sin(s a) tan^-c cos(s A) cot-^-c - A > =- 184] sm s cos s 185] sin(s a)tan ^A=sin(s 6)tan^B=sin(s c)tanjc. [183 98 SPHERICAL TRIGONOMETRY. [IV. 7, RELATIONS BETWEEN SPHERICAL AND PLANE TRIGONOMETRY. Manifestly, when the sides of a spherical triangle subtend very small angles at the center of the sphere, the spherical triangle differs but little from a plane triangle having the same vertices ; and, if the vertices be fixed in position while the center of the sphere recedes further and further away, and the radii grow longer and longer, then the spherical triangle approaches closer and closer to the plane triangle having the same vertices. Then also the small angles at the center of the sphere, which are measured by the sides of the triangle, are very nearly equivalent to their sines or tangents, and the sum of the three angles of the triangle is very nearly 180. If, in the several formulae of spherical trigonometiy, a is sub- stituted for sin a, tana, 2 sin^-a, ; p, for ship, ; s, for sins, ; and 1, for cos a, ; then, in general, these formulae reduce to the corresponding formulae of plane trigonometry, or to mere truisms. Thus : Spherical. Plane. 151] snrp = sin?* sin P= tang cot Q ; p = r sinp = gcotQ or y = r sin o = x cotp. [I. 7 154] cos P = cosjp sin Q= tang cot r; cosp=l . sinQ=g : r or coso=sinp = o': r. [I. 7 156] sina: sin6 = sinA: sins ; a : b =shiA : sine. [103 IKO-I i I sin (8 -6) sin (g-c) . 1591 smlA=A : 7- \ sin 6 sine i l\^ ) \^ ~~ ^/ nno rfB.jA.-y- - w - f sins sin (s a) , 160] COS|A = ^ Sin6sinc ' >-') 7.] SOLUTION OF SPHERICAL TRIANGLES. 99 Spherical. Plane. 166] gtoKA~B) = " in *. (g 1 ~ 6) coio; sm^c smJ(A~B) = ^^cos|c. [121 168] OAll-R-ly f s \ a "y i r--| ^\r cir ?L( ft. r^j 7A 170] sm - a ~B)==^^cotic. [107 a -j o c . [121 * Some correspondences between the formulae of plane and of spherical trigonometry appear only when functions of the sides, of their half-sum, etc., are represented by taking two or more terms of the series [94-99] instead of using the first term alone as above. Thus : Formula cosr = cos_p cos q [153 becomes a mere truism if each cosine be represented by 1 , its limit when j>, q and r become indefinitely small ; but if the values cos?* = [95 be substituted in [153] they give 100 SPHERICAL TEIGONOMETKY. [IV. .. p 2 -}- (f = j* 2 terms of higher degree, whose ratios to j9 2 , (f and r 2 have the limit when p, q and r become indefinite!} 7 small ; i.e., p 2 + # 2 , or x 2 4- ?/ 2 , = j* 2 , whence sin 2 + cos 2 = 1 ; .-. [153] corresponds to [36]. So, writing b for sin6, c for sine, 1 Ja 2 + for cosa, , then the formula cosa = cos 6 cose -J- sin 6 sine cos A [157 gives a 2 = b 2 + c 2 - 2 be cos A. [108 So, writing two terms of [94] for sin6, , and three terms of [95] for cos a, , then the formula cos a = cos b cose 4- sin b sine cos A [157 gives &C(COSA' COSA) = 1 J2-(a 2 6 2 +& 2 c 2 +c 2 a 2 ) 2^(a 4 +6 4 +c 4 ) terms whose limiting ratios to these terms, when a, b and c become indefinitely small, are ; and so for ca(cosB f COSB), and a&(cosc' cose) . [symmetry .*. be (cos A' cos A) = ca (COSB 'COSB) =ab (cose 'cose) approximately, wherein cos A'= (6 2 +c 2 a 2 ) : 2 6c, , and A', B' and c' are the angles of the plane triangle whose sides a', 6' and c' are respectively equal to the arcs a, b and c. But '. ' cos A' COSA= 2sin|-(A A')sin-J(A-f- A') = (A A') sin A', very nearly, and so for COSB' COSB and cose' cose, [symmetry &C(A A f )sinA f = C(B B')sinB f =a&(c c')sinc f , .-. A A' = B B' = C c', very nearly ; [156 186].'. each angle of sph. AABC exceeds the corresponding angle of pi. AA'B'C' by one-third of the spherical ex- cess, A + B -f- c 180; which is Legendre's theorem. 8, EXEKCISES. Solve the spherical right triangles, given : 1. p=116, g= 16, R = 90. 2. r=140, p= 20, R = 90. 3. P= 80 10', g=155 46', R = 90. 8.] SOLUTION OF SPHERICAL TRIANGLES. 101 4. p=100, j9 = 112, R = 90. 5. r=120, p=120, R = 90. 6. p= 60 47', Q= 57 16', R=90. 7. r=140, p = 140, R=90. 8. r = 120, p= 90, R = 90. Solve the quadrantal triangles, given : 9. A = 80, a= 90, 6 = 37. 10. B = 50, 6 = 130, c = 90. Solve the isosceles triangles, given : 11. a= 70, 6= 70, A = 30. 12. a = 30, A = 70, B = 70. 13. a =119, 6 = 119, c = 85. Solve the oblique triangles [both methods] , given : 14. 6= 98 12', c= 80 35', A= 10 16'. 15. A =135 15', c= 50 30', 6= 69 34'. 16. a= 40 16', 6= 47 14', A= 52 30'. 17. a = 120, 6= 70, A =130. 18. a= 40, 6= 50, -A = 50. 19. A =132 16', B = 13944', a =127 30'. 20. A = 110, B= 60, a= 50. 21. A= 70, B = 120, a= 80. 22. a =100, 6= 50, c= 60. 23. A = 120, B = 130, c= 80. 24. In astronomy the altitude of a heavenly body is its angular elevation above the horizon, and its azimuth is its angular dis- tance west from the south point. What is the angular distance between the moon, alt. 40, az. 25 w., and Venus, alt. 24, az. 110 w. ? 25. In navigation the shortest distance from port to port is the arc of a great circle. Find the course and distance from San Francisco, lat. 37 48' N., long. 122 25' w., to Cape of Good Hope, lat. 33 56' s., long. 18 29' E., no allowance being made for intervening lands. 102 SPHERICAL TRIGONOMETRY. [IV. 26. In geodetic surveys triangles upon the earth's surface are considered as spherical triangles. Assume the earth's radius to be 3956 miles ; then, if one side of a triangle be 100 miles, and the adjacent angles be 65 and 60 respectively : Find the other two sides in degrees and in miles ; find the third angle ; find the spherical excess ; \_geom. find the area of the triangle in square miles ; find the number of square miles of area which corresponds to 1" of spherical excess. 27. In a geodetic survey there were measured Z A = 30, Z B = 48 45', Z c = 101 15' 12" and side c = 70 miles : Find the angles of the plane triangle whose sides equal a, b and c of the spherical triangle, and thence find the lengths of a and b. [Legendre's theorem 28. From the two equations cos b = cos c cos a + sin c sin a cos B cos c = cos a cos b -\- sin a sin b cose [157 eliminate cose, and prove that sin a cos & = cos a sin & cose + sine COSB ; [36 thence eliminate sine, and prove that sin a cot b = cos a cose + cot B sine. [156 So, prove that sin b cot c = cos b cos A + cot c sin A, sine cot a = cose COSB -f cot A sinB, sin a cote = cos a COSB + cote sinB, sin b cota= cos b cose +cotA sine, sine cot& = cosccosA + cotB sin A. 29. Show that [173] reduces, for a plane right triangle, to a^-f- 2 / 2 = r 2 ; [174], to the same; [175], to [114]; [176], to [114]; [177], toy a=ytano ictanj-p; [178],too-f-p=90; [179], to the same. 30. Show that [180] reduces, for a plane triangle, to [122] and s = 90 ; [181], to [123] and s = 90 ; [182], to [125] and a : c = sinA : sine ; [183], to (s c) : (s a) = tan^-A : tan Jc and a : c = as above ; [184], to what? 8.] SOLUTION OF SPHERICAL TRIANGLES. 103 NOTE 1. In this corollary ^/(l+sin#) is positive for all values of 6 from -J-TT to -f-j TT, negative for all values of from -f-f7r to +TT, and so on; and ^/(l sin#) is positive for all values of 9 from f ?r to -f JTT, negative for all values of 6 from -f-^TT to +J TT, and so on. The reader may prove this by observ- ing that when 6 lies between and ^TT, then cos^-0>smJ0>0 ; that when increases or decreases continuously, the radicals A/(l-f-sin#) and A/(l sin#) can change from positive to nega- tive or from negative to positive only when passing through ; i.e. V(l + sin0), only when = ITT, JTT, f?r, JTT, , and V(l sin0), only when = |TT, TT, ITT, ITT, ; and that, when passes through either of these values, the radical - N /(lsin0), =sin0cos!-0, which becomes 0, must change its sign, since both sin^0 and cos0 continue to increase or decrease just afterward as they did just before. UNIVERSITY OF CALIFORNIA LIBRARY BERKELEY Return to desk from which borrowed. This book is DUE on the last date stamped below. 15 1948 LD 21-100m-9,'47(A5702sl6)476 YC 2230! Oil t-e