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

(N/A) We use the identity $\cos^{2} A - \cos^{2} B = \sin(A+B) \sin(B-A)$.
Alternatively,using the formula $\cos^{2} A - \cos^{2} B = (\cos A + \cos B)(\cos A - \cos B)$:
$L.H.S. = (\cos 2x + \cos 6x)(\cos 2x - \cos 6x)$
Using sum-to-product formulas:
$\cos 2x + \cos 6x = 2 \cos \left( \frac{2x+6x}{2} \right) \cos \left( \frac{2x-6x}{2} \right) = 2 \cos 4x \cos(-2x) = 2 \cos 4x \cos 2x$
$\cos 2x - \cos 6x = -2 \sin \left( \frac{2x+6x}{2} \right) \sin \left( \frac{2x-6x}{2} \right) = -2 \sin 4x \sin(-2x) = 2 \sin 4x \sin 2x$
Multiplying these results:
$L.H.S. = (2 \cos 4x \cos 2x)(2 \sin 4x \sin 2x)$
$= (2 \sin 4x \cos 4x)(2 \sin 2x \cos 2x)$
Using the double angle identity $\sin 2\theta = 2 \sin \theta \cos \theta$:
$= \sin 8x \sin 4x = R.H.S.$