Question

Prove that $\frac{\sin x-\sin 3x}{{\sin }^{2}x-{\cos }^{2}x}=2\sin x$

Answer 1

It is known that

$\sin A-\sin B=2\cos \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right),{\cos }^{2}A-{\sin }^{2}A=\cos 2A$

$\therefore L.H.S.=\frac{\sin x-\sin 3x}{{\sin }^{2}x-{\cos }^{2}x}$

$=\frac{2\cos \left(\frac{x+3x}{2}\right)\sin \left(\frac{x-3x}{2}\right)}{-\cos 2x}$

$=\frac{2\cos 2x\sin (-x)}{-\cos 2x}$

$=-2x(-\sin x)$

$=2\sin x=R.H.S.$