Prove that: ${sin}^{2} 6 x - {sin}^{2} 4 x = sin 2 x sin 10 x$

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Solution

$L H S = sin^{2} 6 x - sin^{2} 4 x$
$= (sin 6 x + sin 4 x) (sin 6 x - sin 4 x)$
Using, $sin C + sin D = sin (\frac{C + D}{2}) cos (\frac{C - D}{2})$ and $sin C - sin D = sin (\frac{C - D}{2}) cos (\frac{C + D}{2})$
$= 2 sin (\frac{10 x}{2}) cos (\frac{2 x}{2}) \times 2 cos (\frac{10 x}{2}) sin (\frac{2 x}{2})$
Using $sin 2 x = 2 sin x cos x$
$= (2 sin 5 x cos 5 x) (2 sin x cos x) = sin 10 x sin 2 x$
$= R H S$

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