Although the results of Arts. 12, 15, 16, 17 have been proved from diagrams where A is less than a right angle, the

Bisect it by the straight line OB, so that ZAOB = and draw CD perpendicular to OA, meeting OB in E.

(3.)

A

2(1 - sin2 4)-1=1-2 sin3 — (4.)

19. To find the trigonometrical ratios of 15°, 75°, 120°, 135°, 150°.

We have, sin 75° = sin (90° 15°) = cos 15° =

√√3+ 1

2√2

=

(5.) Ratios of 135°.

= tan (90° — 15°) = cot 15°
= 2 + √3, &c.

We have, sin 135° sin (180°- 135°) = sin 45° :

=

=

= tan 45°

1. Define a negative angle, and show that tan (-A) tan A, when A lies between – 90° and – 180°.

2. Trace the changes of sign of sin A. cos A through the four quadrants.

3. Trace the changes of sign of cos A + sin A, and of cos A - sin A, as A changes from 45° to 315°.

=

sin2 A,

4. Assuming generally that cos 2 A cos2 A trace the changes of sign of cos 2 A as A changes from -45° to 315°.

=

5. Write down the sines of 210°, 165°, 240°, 120°.

6. Show that sin (90° + A) = cos A, and cos - sin A, for any value of A from 0° to 180°.

(90° + A)