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

It is known that

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

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

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

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

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

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

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

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

${ = \sin 8x\sin 4x = R.H.S}$