Sine, &c. of 180° and 270".

56. Problem.

To find the sine, &c. of 180°.

=

Solution. Make A 90°, in (49) and (50), they become, by means of (66) and (67),

sin. 180° 2 sin. 90° cos. 90° = 0

(72)

cos. 180° - (cos. 90°)2(sin. 90°)21.

(73)

57. Problem. To find the sine, &c. of 270°.

Solution. Make M 180° and N = 90° in (33) and (35).

They become, by means of (66, 67, 72, 73),

sin. 270° sin. 180° cos. 90°+cos. 180° sin. 90° — —

cos. 270°

=

-1 (78)

cos. 180° cos. 90° — sin. 180° sin. 90° = 0. (79)

Hence, from (6) and (7),

58. Problem. To find the sine, &c. of 360°.

Solution. Make A = 180° in (49) and (50); and they

Hence the sine, &c. of 360° are the same as those of 0°.

59. Problem. To find the sine, &c. of 45°.

=

Solution. Make C 90° in (57) and (58). They become, by means of (66),

cos. 45° = √ [(1 + cos. 90°)] = √ 1⁄2 (86)

Sine, &c. of 30°, 60°, and the supplement.

60. Problem. To find the sine, &c. of 30° and 60°.

Solution. Make A = 30° in (49). It becomes, from the consideration that 30° and 60° are complements of each other,

sin. 60° cos. 30° = 2 sin, 30° cos. 30°.

Dividing by cos. 30°, we have

or

1 = 2 sin. 30°,

sin. 30°

cos. 60°

(92)

whence, from (6), (7), and (10),

cos. 30° = sin 60° = √ (1 — 4) = 1 √/ 3.

(93)

61. Problem. To find the sine, &c. of the supplement of an angle.

Solution. Make M = 180° in (34) and (36). They become, by means of (72) and (73),

sin (180°-N)

sin. 180° cos. N

-cos. 180° sin. N

= sin. N

(98)

cos. (180° — N) cos. 180° cos. N+ sin, 180° sin. N

that is, the sine and cosecant of the supplement of an angle are the same with those of the angle itself and the cosine, tangent, cotangent, and secant of the supplement are the negative of those of the angle.

62. Corollary. Since, when an angle is acute its supplement is obtuse, it follows from the preceding proposition, that the sine and cosecant of an obtuse angle are positive, while its cosine, tangent, cotangent, and secant, are negative.

This proposition must be carefully borne in mind in using the trigonometric tables, as it affords the means of discriminating between the two angles which are given in B. Table XXVII, and of deciding which of these two angles is the required one.

63. Corollary. The preceding corollary might also have been obtained from (33) and (35). For by making M=90°, we have by (66) and (67)

that is, the sine and cosecant of an angle, which exceeds 90°, are equal to the cosine and secant of its excess above 90°, while its cosine, tangent, cotangent, and secant are equal to the negative of the sine, cotangent, tangent, and cosecant of this excess.

64. Problem. To find the sine, &c. of a negative angle.

Solution. Make N 0° in (34) and (36). They become, by means of (66) and (67),

so that the cosine and secant of the negative of an angle are the same with those of the angle itself; and the sine, tangent, cotangent, and cosecant of the negative of the angle are the negative of those of the angle.

(114)

(115)