58. To find the sines and cosines of 18°, 72°, 36o, and 54°.

Sin 360 = cos(90° - 36o) = cos 54°,

or, (if 18o = A), sin 2 A = cos 3 A;

... 2 sin A . cos A = 2 cos 2 A . cos A - cos 4,

Art. 55;

.. 2 sin A = 2 cos 2 A-1 = 2.{1 - 2 (sin 4)3} -
4
- 1 ;

... 4 (sin A)2 + 2 sin A = 1.

Where the positive sign is to be taken, because sin 180 is a positive quantity.

[There are other angles, satisfying the equation

sin 2 A

√5+1

Some of these

4

= cos 3 A, whose sines are angles will be determined in the next Article.]

Again; sin 540

= COS 36° 18°=

= cos 2 × 180 = (cos 1802 - (sin 1802

10+2/5 6-2√5

59. To find some of the values of A which satisfy the = cos 3 A.

equation, sin 2 A

cos 3 A = sin 2 A = cos (90o — 2 A)

cos 3 A = sin 2 A = cos (90° – 2 A)

= cos {(4n+2) 90° ± (90° - 2 A)}. Art. 25.

= COS · { 180° ± [(4n + 2) 90o± (90°— 2 A)]}. Art. 23.

3A

Let 34 180° {(4n + 2) 90o± (90° — 2A)};

.. 3A ± 2 A = 180o± (4n + 2) 90° ± 90o.

(2).

And by solving this equation, we obtain other values of A which satisfy the proposed equation.]

For n writing 0, 1, 2, 3......successively, we have

from (1); A=

180;

= 5 × 18°; = 9 × 18°; = 13 x 18°;...

from (2); A - 90°; = 3 x 90°; = 7 x 90°; = 11 x 90°;...

=

COR. The angle 13.18°, or 234o, is one of the angles al

60. If an angle receive any increment, to find the corresponding increment of the sine of the angle.

Let A receive an increment SA, and let the corresponding increment of the sine of A be represented by Asin A.

Then, ▲ sin A = sin (4 + 4) – sin A

this case, if tan A be not exceedingly large, (that is, if A be not nearly equal to 2n +1.90°, n being an integer),

SA
2

tan A. tan is a very small quantity, and may be ne

glected in comparison with unity. We have therefore, when these two conditions are fulfilled, and not otherwise,—

▲ sin A = cos A. sin d A.

NOTE. Hence it appears that if A be any angle of a triangle, this result cannot be applied in any particular case to determine the corresponding increment of the sine of the angle, when the angle itself has received a given small increment, if the angle be nearly equal to a right angle.

61. If an angle receive any increment, to find the corresponding decrement of the cosine of the angle.

COR. 1. As in the last Article, if dA be very small, and cot A be not at the at the same time exceedingly large,

(i. e. if A be not = 2n. 90° nearly), cot A. tan

SA

2

neglected with respect to unity; and we then have,

may be

NOTE. It must be carefully borne in mind, that unless, (1), A be a very small angle, and, (2), A be, besides, an angle which is not nearly equal to 0° or 180°, this result cannot be applied to any particular case where A is an angle of a triangle.

COR. 2. If A be less than 90°, cos A is positive, and sin A being also positive, in this case Acos A is necessarily negative. That is, in angles less than a right angle, as the angle increases, its cosine decreases.

If A be greater than one right angle, but less than two, cos A is negative, and sin A being positive, A cos A is negative. That is, when the angle is greater than one right angle, but less than two, as the angle increases, the cosine, being negative, also increases in magnitude.

62. If an angle receive any increment, to find the corresponding increment of the secant of the angle.

COR. If A be very small, and neither tan A nor cot A be very large, (that is, if ▲ be not nearly equal to n.90o),

and tan A. tan ASA will both be very small, and

may be neglected when compared with unity. case therefore we have

In this

NOTE. It is to be remarked, that before this result can be applied in any case where A is an angle of a triangle, we must have dA a very small angle, and moreover, A must not be nearly equal to 0, 90o, or 180o.