Sine, &c. of an angle greater than 180°.

65. Problem. To find the sine, &c. of an angle which exceeds 180°.

Solution. Make M = 180° in (33) and (35). They become, by means of (72) and (73),

that is, the tangent and cotangent of an angle, which exceeds 180°, are equal to those of its excess above 180°; and the sine, cosine, secant, and cosecant of this angle are the negative of those of its excess.

66. Corollary. If the excess of the angle above 180° is less than 90°, the angle is contained between 180° and 270°; so that the tangent and cotangent of an angle which exceeds 180°, and is less than 270°, are positive; while its sine, cosine, secant, and cosecant are negative.

67. Corollary. If the excess of the angle above 180° is greater than 90° and less than 180°, the angle is contained between 270° and 360°; so that, by § 65 and 62, the cosine and secant of an angle, which ex

Sine, &c. of an angle greater than 360o.

ceeds 270° and is less than 360°, is positive; while its sine, tangent, cotangent, and cosecant are negative.

68. Corollary. The results of the two preceding corollaries might have been obtained from (34) and (36). For by making M 360°, we have, by $58,

that is, the cosine and secant of an angle are the same with those of the remainder after subtracting the angle from 360°; while its sine, tangent, cotangent, and cosecant are the negative of those of this remainder.

69. Problem. To find the sine, &c. of an angle which exceeds 360°.

Solution. Make M = 360° in (33) and (35). They become, by means of (84) and (85),

sin. (360°N) = sin. N

cos. (360° N) =cos. N;

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that is, the sine, &c. of an angle which exceeds 360° are equal to those of its excess above 360°.

Increase of sine, &c. of an acute angle.

70. Theorem. The sine, tangent, and secant, of an acute angle increase with the increase of the angle; the cosine, cotangent, and cosecant decrease.

Proof. I. The excess of the sine of M+m over the sine of M is, by (17), equal to sin. m cos. M, which is a positive quantity when M is acute; and, therefore, the sine of the acute angle increases with the increase of the angle.

II. The excess of cos. M over cos. (M+m) is, by (19), equal to sin. m sin. M, which is a positive quantity; and, therefore, the cosine of the acute angle decreases with the increase of the angle.

III. The tangent of an angle is, by (7), the quotient of its sine divided by its cosine. It is, therefore, a fraction whose numerator increases with the increase of the angle, while its denominator decreases. Either of these changes in the terms of the fraction would increase its value; and, therefore, the tangent of an acute angle increases with the increase of the angle.

IV. The cosecant, secant, and cotangent of an angle are, by (6), the respective reciprocals of the sine, cosine, and tangent. But the reciprocal of a quantity increases with the decrease of the quantity, and the reverse. It follows, then, from the preceding demonstrations, that its secant increases with the increase of the acute angle, while its cosecant and cotangent decrease.

71. Theorem. The absolute values (neglecting their signs) of the sine, tangent, and secant of an obtuse

Increase of sine, &c. of obtuse angle.

angle decrease with the increase of the angle; while those of the cosine, cotangent, and cosecant increase.

Proof. The supplement of an obtuse angle is an acute angle, of which the absolute values of the sine, &c. are, by 61, the same as those of the angle itself. But this acute angle decreases with the increase of the obtuse angle, and at the same time its sine, tangent, and secant decrease, while its cosine, cotangent, and cosecant increase.

Sines proportional to sines of opposite angles.

CHAPTER VI.

OBLIQUE TRIANGLES.

72. Theorem. The sides of a triangle are directly proportional to the sines of the opposite angles. [B. p. 13.]

Proof. In the triangle ABC (figs. 2 and 3), denote the sides opposite the angles A, B, C, respectively, by the letters a, b, c. We are to prove that

sin. A: sin. B : sin. C = a :b:c.

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perpen

From the vertex B, let fall on the opposite side the dicular BP, which we will denote by the letter p. Then, in the triangle BAP, we have by (1)

Also, in the triangle BPC, we have, by (1) and (93), and from the consideration that BCP is the angle C (fig. 2), and its supplement (fig. 3),