sin ( 1 A) sin A Consequently the sine and cosecant of an angle are the same as those of its supplement; the cosine, tangent, cotangent, and secant, are equal in magnitude with contrary signs. = : The following are left as exercises for the student :cot (+A)= tan A.....(20), sec († τ+A) covers (+A) = 1 − cos A. . . . (22), vers (Ț — A) sec (π + A) = − sec A.....(24), tan (☛ + A) cot A.......(26), vers (π + A) = -sin A.....(28), cos (T+A) cot (+A) sin (+A) sec (2-A) = = = 1 + cos A. (23), sec A.......(30), vers (2π- A) = vers A. If, moreover, the angle AO B be positive when measured towards C, it will be negative when measured in the contrary direction (Art. 12). Hence the angle A O b will be denoted by A, and consequently by Arts. 7, 8, 11, 12, Sin (-A)-bF-BF-sin A. (32), cos(-A) OF= cos A..(33). And similarly for other functions of The numerical values of the trigonometrical functions of 30°, 60°, 45° may now be found. The values of other functions of 30, 45, and 60' may be found in a similar way. EXERCISES. 1. The tangent of an angle is ; find the surd expressions for the sine, cosine, secant, and cosecant of the same. Ans. If a be the angle, sin α = √√5, cos α = √√5, sec α = √5, cosec α = 5. 2. Find the complements of the angles 26 7' 8" 21, 98 16' 30", and 218 5' 6" 67. 3. Find the supplements of 155' 7' 8" 9, and 224 5′ 8′′. 4. Find the circular measures of the angles 61°, 72° 5' 20", and 88° 19′30′′. Ans. 1.0646, 1.2582, 1.5415. 5. Find the angles whose circular measures are and †. Ans. 43 nearly, and 22·9. 6. Required the arc whose supplement is to its complement as four Ans. 0 60. to one. 7. Required the arc the sum of whose supplement and complement is to their difference as two to one. Prove the following relations :- Ans. 0 == 45%. 8. Sin a sin 28+ cos a cos 28+ sin a cos 2B + sin 2B cos a 10. Vers (+ 0) vers († π − 0) + vers 0 vers (π - = 1. = 1. 11. Seca tana (sec2ß — 1) (tan2ß + 1) — sec 2ß tan 3ß (sec3a — 1) x (tana + 1) = 0. 12. (Sin'a-sin'a sin B) (cosa - cosa cos 2B)-(sin 28 sin a sin ẞ) (cos B cos'a cos 2B) = 0. 13. Tan (+ 0) — cot 2(π + 0) = cosec 20 14. Cot'a cosa cota - cos 2a, sec 3a cosec 2a = sec2a + cosec2x. = FUNCTIONS OF TWO OR MORE ANGLES. 16. Expressions for the sine and cosine of the sum and difference of any two angles, and formula deducible from these. GE(= HD): GO:: BF: BO, or HD = BF.GO = = GC.FO sin B cos A, = FO.GO = cos A cos B, FB.GC sin A sin B. = ...... (1),* ..... (2), ...... ...... ...... . (3), . (4). HC : CG:: FO: BO, or HC ΕΟ :OG:: FO: BO, or EO HG(= DE): GC:: FB: BO, or DE Hence, taking fig. 1 and fig. 2 alternately, sin (A+B)=CD = HD+HC = sin A cos B+sin B cos A. sin (A-B) = CD=HD-HC=siu A cos B-sin B cos A cos (A+B) = OD OE-DE=cos A cos B-sin A sin B cos (A-B) = OD=OE+DE = cos A cos B+sin A sin B It is assumed in this investigation that A + B < π, and A > B ; it therefore remains to be proved that the formulæ (1, 2, 3, 4) hold for every positive or negative value of A and B. For this purpose assume A π- x, B= π - y; and let these values be such that A + B =T- (x + y), may be positive, and less than then it will be obvious that x + y > $T. sin (A+B) = sin {(x) + ( − y) } = sin{ - (x + y)}; since A+B is equal to each of these expressions within the brackets. But by (1), sin {(† π − x) + († π − y)} = sin (} π − x) cos († τ − y) +sin(- y) cos (π x) = cos x sin y+cos y sin x; sin {(x+y)} = sin (x + y). And in general, if we put (m and n being any positive or negative integers) The following is another very simple method of proving the formulæ (1, 2, 3, 4). Let A OB, BOC be two angles of the circle A CG whose radius is the linear unit. From B draw BD, BF perpendicular respectively to the radii O A, O C, and from C draw CE perpendicular to O A, meeting the circle again in G. Bisect O B in H; join F H, H D, O G; and denote the angles A OB, BOC by A and B. Then because each of the angles O D B, OFB is a right angle, a circle described from H, with OB as radius will pass through the points O, D, B, F. Hence the angle FHD 2 COA = COG; the triangles FHD, CO G, are consequently similar, and therefore FD=GC EC, since FH = CO. Also the perpendicular from H on F DOE; whence the difference of the perpendiculars from O and B on FD, B H E is equal to EO, as each of these quantities (Euc. i. Ex.) is double of the perpendicular from H on DF. Hence (Euc. vi. D.C.), OB.ECBD.OF + BF.O D, and O B.O E OD.OF-BD.BF; = or sin (A+B)=sin A cos B+sin B cos A, cos (A+B) = =cos A cos B-sin A sin B. If we interchange the perpendiculars B D, C E, so that CE sin (A -- B), we shall get in a similar way, sin (A-B) = sin A cos B-sin B cos A, cos (A-B) = cos A cos B + sin A sin B. In this case, O E is equal to the sum of the perpendiculars from O and B on F D. COSA 1 + cos A By (5) also and the formula sin 2A + cos 2A = 1, we have which give the values (10)*. sin A={√(1 + sin 2 A) + (1 sin 2 A)}. . . (9), √ cos A=√(1 + sin 2 A) (1 sin 2 A)}.. 7 √ Again, by addition and subtraction of the expressions (1, 2, 3, 4), sin (A + B) + sin (A — B) = 2 sin A cos B... sin (A+B) sin (A — B) =2 cos A sin B cos (A + B) + cos (A - B) In (11) and (13), respectively, let An B; then we have the general expressions, sin (n + 1) B= 2 sin n B cos B — sin (n − 1) B....(15), Next, to find the tangents of the sum and difference of two angles in terms of the tangents of the angles, we have by (1) and (3) of this Art., and (2) of Art. 15, The last form is got by dividing both numerator and denominator of the preceding, by cos A cos B. *The signs of these formulæ will be best interpreted by special cases. For values of A between 0 and 45°, sin A and cos A will both be positive, and cos Asin A; sin A + cos A, or its equal (1 + sin 2 A), will therefore be positive, and sin A -cos A, or its equal(1 sin 2 A), will be negative. Hence in this case (9) and (10) become, by means of the preceding expressions from which they are derived, sin A = {√(1 + sin 2A) — √ (1 − sin 2A)}, Again, for values of A between 45° and 90° (the values of the trigonometrical functions of 45° have already been determined in Art. 15), sin A + cos A and sin A-cos A will both be positive, hence sin A (1-sin 2A)}, (1 + sin 2A) - And in a similar way for other cases. By different combinations of the preceding, a great number of other formule might be deduced; but in an elementary work like the present, those only can be given that are of frequent occurrence. The following are left as exercises for the student : 1. Sin A = sin B cos (A – B) + cos B sin (A – B). 2. Cos A = sin B sin (A + B) + cos B cos (A + B). |