Question

Prove that $\frac{sinx-sin3x}{sin^2x+cos^2x}=2\,sin\,x$.

Answer 1

It is known that

$sinA-sinB=2cos(\frac{A+B}{2})sin(\frac{A-B}{2}),cos^2A-sin^2A=cos2A$

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

$=\frac{2cos(\frac{x+3x}{2})sin(\frac{x-3x}{2})}{-cos2x}$

$\frac{2cos\,2x\,sin(-x)}{-cos2x}$

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

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