consist of the produced part of the radius, while those represented by the full lines consist of the radius together with the produced part; the latter are considered as positive and the former as negative. Hence in the first and fourth quadrants the secants are positive, and in the second and third quadrants they are negative. § 11. These different lines may be clearly exhibited in the four quadrants as follows: In the above diagrams the dotted lines are considered as negative and the full lines as positive. The letters in the four diagrams are so arranged that in each case BC counted in the positive direction (§9) is the arc, CD the sine, CH the cosine, BE the tangent, GF the cotangent, AE the secant, and AF the cosecant. 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. sin. a=sin. (180°-a); cos. a=cos. (180°—a). $ 14. The following particular values should be carefully noted by the student: In general, if n denote any integral number, including also the value of n=0, we shall have, tang. 90°cos. 270° cos. (2n+1)×90° =0 180° = cos. (4n+2)×90°-1 0° tang. 180° tang. 2 nx90° = 0 = 90°=tang. 270° tang. (2n+1)×90° = ∞ cotan. 0° cotan. 180° cotan. 2n×90° cotan. 90° = cotan. 270° = cotan. (2n+1)× 90° =0 sec. sec. 90° sec. sec. 270° sec. (2n+1)×90° : 180° sec. (4n+2)×90°-1 =∞ cosec. 0° = cosec. 180° cosec. 2n×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 KI K D C B G H Since the sides of the triangle Draw EF perpendicular to 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, we have the triangle EFK similar to the triangle AFH, and the angle KFE equal to the angle FAH, equal to α. A C F D K cos. (a−b) AE AG AH+KE AH AF KE FE Collecting these results, we have sin. (a+b)=sin. a cos. b + cos. a sin. b. .. (1.) sin. (ɑ— b)=sin. ɑ cos. b cos. a sin. b. . (2.) cos. (a+b)=cos. a cos. b (3.) cos. (a — b) = cos. a cos. b + sin. a sin. b. . (4.) If we make b=a, equations (1) and (3) will give The sum and difference of (6) and (7) give 1 + cos. 2 α = 2 cos.2 1 — cos. 2 a = 2 sin.2 a. (8.) (9.) (10.) . (11.) 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) ___ 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. ɑ cos. b. (1.) (2.) (3.) cos. (a - b) — cos. (a+b) = 2 sin. a sin. b. (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. q2 sin. (p + q) cos. (p-q). (5.) sin. p — sin. q = 2 cos. § (p+q) 1⁄2 (p + q) sin. 1⁄2 (p−q). |