Prove that $cos ^{2} 2 x-cos ^{2} 6 x=sin 4 x sin 8 x$

It is known that $\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right), \cos A-\cos B=-2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$

$\therefore$ L.H.S. $=\cos ^{2} 2 x-\cos ^{2} 6 x$

$=(\cos 2 x+\cos 6 x)(\cos 2 x-6 x)$

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

$=[2 \cos 4 x \cos (-2 x)][-2 \sin 4 x \sin (-2 x)]$

$=[2 \cos 4 x \cos 2 x][-2 \sin 4 x(-\sin 2 x)]$

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

$=\sin 8 x \sin 4 x$

$=$ R.H.S.