Question

prove the following indentities

a) sin(x+y)sin(x-y)=sin^2x-sin^2y
a) sin2x+sin4x+sin6x=4cosxcos2xsin3x

Answer 1

Prove the following identities

a) $\sin(x + y)\sin(x - y) = \sin^2 x - \sin^2 y$

b) $\sin(2x) + \sin(4x) + \sin(6x) = 4\cos x\cos(2x)\sin(3x)$

Solution:

a) We'll use next identities

$$\sin(\alpha \pm \beta) = \sin \alpha \cdot \cos \beta \pm \sin \beta \cdot \cos \alpha,$$ $$\cos^2 \alpha = 1 - \sin^2 \alpha.$$

Thus we have

$$\sin(x + y)\sin(x - y) = (\sin x \cdot \cos y + \sin y \cdot \cos x) \cdot (\sin x \cdot \cos y - \sin y \cdot \cos x) =$$ $$= \sin^2 x \cdot \cos^2 y - \sin^2 y \cdot \cos^2 x = \sin^2 x \cdot (1 - \sin^2 y) - \sin^2 y \cdot (1 - \sin^2 x) =$$ $$= \sin^2 x - \sin^2 x \cdot \sin^2 y - \sin^2 y + \sin^2 y \cdot \sin^2 x = \sin^2 x - \sin^2 y.$$

b) We'll use next identities

$$\sin(2\alpha) = 2 \sin \alpha \cdot \cos \alpha,$$ $$\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \cdot \cos \frac{\alpha - \beta}{2}.$$

Thus we have

$$\sin(2x) + \sin(4x) + \sin(6x) = (\sin(2x) + \sin(6x)) + \sin(4x) =$$ $$= 2 \sin \frac{2x + 6x}{2} \cdot \cos \frac{2x - 6x}{2} + \sin(4x) = 2 \sin(4x) \cdot \cos(2x) + \sin(4x) =$$ $$= 2 \sin(4x) \cdot \cos(2x) + 2 \sin(2x) \cdot \cos(2x) = 2 \cos(2x) (\sin(4x) + \sin(2x)) =$$ $$= 2 \cos(2x) \cdot 2 \sin \frac{2x + 4x}{2} \cdot \cos \frac{2x - 4x}{2} = 4 \cos(2x) \cdot \sin(3x) \cdot \cos x =$$ $$= 4 \cos x \cos(2x) \sin(3x).$$