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

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Solution

$L H S = cos^{2} 2 x - cos^{2} 6 x$
$= (cos 2 x + cos 6 x) (cos 2 x - cos 6 x)$
$= (2 cos (\frac{2 x + 6 x}{2}) cos (\frac{2 x - 6 x}{2}) (- 2 sin (\frac{2 x + 6 x}{2}) sin (\frac{2 x - 6 x}{2})$
[Since, $cos C + cos D = 2 cos (\frac{C + D}{2}) cos (\frac{C - D}{2}),$
[ $cos C - cos D = - 2 sin (\frac{C + D}{2}) sin (\frac{C - D}{2}$ ]
$= (2 cos 4 x cos (- 2 x)) (- sin 4 x sin (- 2 x))$
$= (2 cos 4 x cos 2 x) (sin 4 x sin (2 x))$
$= (2 sin 4 x cos 4 x) (2 sin 2 x cos 2 x)$
$= sin 2 (4 x) sin 2 (2 x)$ [Since, $sin 2 θ = 2 sin θ cos θ$ ]
$= sin 8 x sin 4 x$
$∴ {cos}^{2} 2 x - {cos}^{2} 6 x = sin 4 x sin 8 x$ .

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