Question

Prove that $\frac{sinx-sin3x}{cosx+cos 3x}=tan 2x.$

Answer 1

It is known that

$sinA-sinB=2cos(\frac{A+B}{2})cos(\frac{A-B}{2}),cosA+cosB=2cos(\frac{A+B}{2})cos(\frac{A-B}{2})$

$L.H.S=\frac{sinx-sin3x}{cosx+cos 3x}$

$=\frac{-2\,sin(\frac{x+3x}{2}).cos(\frac{x-3x}{2})}{2cos(\frac{x+3x}{2}).cos(\frac{5x-3x}{2})}$

$=\frac{sin2x}{cos2x}$

$=tan2x$

$=R.H.S.$