# Trigonometric Ratios of Multiple and Substitute Angles

## Trigonometric Ratios of Multiple and Substitute Angles

1. Sin2A = 2sinA cosA

2. Cos2A = cos²A – sin²A

3. Cos2A = 2cos²A – 1 (or) 1 + cos2A = cos²A

4. Cos2A = 1 – 2sin²A (or) 1 – cos2A = 2cos²A

5. $$\,\tan 2A=\frac{2\tan A}{1-{{\tan }^{2}}A}$$

6. $$\sin 2A=\frac{2\tan A}{1+{{\tan }^{2}}A}$$

7. $$\,\cos 2A=\frac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}$$

The above relation are true for all values of angle A. replacing A by A/2 in the above relation, we obtain the following formulae.

1. sin A = 2 sin A/2 cos A/2

2. cos A = cos² A/2 – sin² A/2

3. cos A = 2 cos² A/2 – 1 or 1 + cos A = 2 cos² A/2

4. cos A = 1 – 2 sin² A/2 or 1 – cos A = 2 sin² A/2

5. $$\tan A=\frac{2\tan \frac{A}{2}}{1-{{\tan }^{2}}\frac{A}{2}}$$

6. $$\sin A=\frac{2\tan \frac{A}{2}}{1+{{\tan }^{2}}\frac{A}{2}}$$

7. $$\cos A=\frac{1-{{\tan }^{2\grave{\ }}}\frac{A}{2}}{1+{{\tan }^{2}}\frac{A}{2}}$$

Example: Find the $$\sqrt{2+\sqrt{2+\sqrt{2+2\cos \theta }}}$$

Solution: We have,

=$$\sqrt{2+\sqrt{2+\sqrt{2\left( 1+\cos \theta \right)}}}$$

= $$\sqrt{2+\sqrt{2+\sqrt{4{{\cos }^{2}}4\theta }}}$$

= $$\sqrt{2+\sqrt{2\left( 1+\cos 4\theta \right)}}$$

= $$\sqrt{2+\sqrt{4{{\cos }^{2}}2\theta }}$$

= $$\sqrt{2+2\cos 2\theta }$$

= $$\sqrt{2\left( 1+\cos 2\theta \right)}$$

Example: Prove that cos²A + cos² (A + 120) + cos² (A – 120) = 3/2

Solution: ½ [2cos²A + cos² (A + 120) + cos² (A – 120)]

= ½ [1 + cos2A + (1 + cos2 (A + 120)) + (1 + cos2 (A – 120))]

= ½ [3 + cos2A + cos2 (A + 120) + cos2 (A – 120))]

= ½ [3 + cos2A + 2cos2A cos240]

= ½ [3 + cos2A + 2 cos 2A (- ½)]

= 3/2