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