Prove that $sin 2 x+2 sin 4 x+sin 6 x=4 cos ^{2} x sin 4 x$

L.H.S. $=\sin 2 x+2 \sin 4 x+\sin 6 x$

$=[\sin 2 x+\sin 6 x]+2 \sin 4 x$

$=\left[2 \sin \left(\frac{2 x+6 x}{2}\right) \cos \left(\frac{2 x-6 x}{2}\right)\right]+2 \sin 4 x$

$\left[\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right]$

$=2 \sin 4 x \cos (-2 x)+2 \sin 4 x$

$=2 \sin 4 x \cos 2 x+2 \sin 4 x$

$=2 \sin 4 x(\cos 2 x+1)$

$=2 \sin 4 x\left(2 \cos ^{2} x-1+1\right)$

$=2 \sin 4 x\left(2 \cos ^{2} x\right)$

$=4 \cos ^{2} x \sin 4 x$

$=$ R.H.S.