The algebraic sines of these lines will be as follows: § 12. By carefully inspecting the diagrams of § 11, we see that denoting any arc when considered as positive by a, it will, when counted in an opposite direction, be denoted by -a, and the following relations will hold good for negative arcs : sin. (—a)=—sin. a; cos. (-a)=cos. a. tang. (-a)=-tang. a; cotan. (—a)=—cotan. a. § 13. If the arc KC' is equal to the arc BC, then will the arc BC' be the supplement (87) of the arc BC. De- K noting the arc BC by a, we have ob.viously the following relations: sin. a=sin. (180°—a); cos. a=—cos. (180°—a). §14. The following particular values should be carefully noted by the student: 90° 180° 270° 360° same as 0° 0° In general, if n denote any integral number, including also the value of n=0, we shall have, tang. tang. cotan. 90° cos. 270° cos. (2n+1)×90° =0 180° 0° = = tang. 180° tang. 2 nx90°=0 90° tang. 270° = tang. (2n+1)×90° = ∞ = cotan. 90° cotan. 270° = cotan. (2n+1)× 90° = 0 -1 sec. 180° cosec. 0° sec. (4n+2)×90° = −1 cosec. 180° cosec. 2 n ×90° = cosec. 90° =cosec. (4n+1)×90° =1 cosec. 270° = cosec. (4n+3)×90° — — 1 =∞ § 15. To find the sine and cosine of the sum and difference of two arcs. Let the angle BAC=a; that is, let a denote the length of the arc which measures this angle; also let the angle CAD be denoted by b; then will BAD be denoted by a+b. also A 2 D K C Since the sides of the triangle Draw EF perpendicular tɔ AC, and from the points E and F draw EG and FH perpendicular to AB; draw FK perpendicular to EG. EFK are respectively perpendicular to the sides of the triangle AFH, these triangles are similar. (Geom. B. III., T. VII.) Hence, the angle FEK is equal to FAH, equal to a. For the sine and cosine of a-b, let the angle BAC = a, and the angle CAD=6; then will BAD=a-b. Then, as in the former diagram, Collecting these results, we have sin. (a+b)=sin. a cos. b + cos. a sin. b. . (1.) b: If we make b= a, equations (1) and (3) will give sin. 2 a = 2 sin. a cos. ɑ, cos. 2 a cos.2 a-sin.2 a. From (5) of (B), § 8, we have (3.) (4.) The sum and difference of (6) and (7) give Equations (5) and (6) make known the sine and cosine of a double arc in terms of the sine and cosine of the arc itself. Equations (10) and (11) give, conversely, the sine and cosine of half an arc in terms of the cosine of the arc itself. Dividing (1) by (3), we have sin. (a+b) cos. (a + b) = sin. a cos. b + cos. a sin. b cos. a cos. b- sin. a sin. b' Dividing numerator and denominator of the right-hand member each by cos. a cos. b, we have § 16. Equations (1), (2), (3), and (4) (§ 15), give as follows: sin. (a+b)+sin. (a-b) = 2 sin. a cos. b. sin. (a+b) sin. (a-b) = 2 cos. a sin. b. cos. (ab)+cos. (a+b) = 2 cos. a cos. b. (1.) (2.) (3.) (4.) If in these results we substitute p for a + b, and q for a-b, which gives a = (p + q) and b = (p−q), they will become sin. p + sin. q 2 sin. Numerical Values of Sines, Tangents, &c. (9.) § 17. Having deduced a few of the most simple relations of the trigonometrical functions of angles, we will now proceed to determine their numerical values. Sines of very small angles are obviously very nearly equal to the arcs which measure these angles. We have found (§ 6) that the length of an arc of 1' is equal to 0·0002908882, &c. If we regard this as the sine of 1', from which, as will be hereafter shown, it does not differ even in its eleventh decimal figure, we may find the cosine of 1' as follows (§ 8): cos. 1'=√1-sin. 1'=0.9999999577, &c. Having thus found the sine and cosine of 1', we may continue our work for larger angles by the aid of equations (1) and (3), (§16), which give sin. (a+b)2 sin. a cos. b-sin. (a-b). cos. (a+b)=2 cos. a cos. b-cos. (a-b). Putting b=1' and a=1', 2', 3′, 4', &c., successively, we shall obtain sin. 2'2 sin. 1' cos. 1'-sin. O'. &c. = &c. |