Prove that $\sin ^{2} 6 x-\sin ^{2} 4 x=\sin 2 x \sin 10 x$
Prove that $\sin ^{2} 6 x-\sin ^{2} 4 x=\sin 2 x \sin 10 x$
It is known that
$\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right), \sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$
$\therefore$ $L.H.S.$ $=\sin ^{2} 6 x-\sin ^{2} 4 x$
$=(\sin 6 x+\sin 4 x)(\sin 6 x-\sin 4 x)$
$=\left[2 \sin \left(\frac{6 x+4 x}{2}\right) \cos \left(\frac{6 x-4 x}{2}\right)\right]\left[2 \cos \left(\frac{6 x+4 x}{2}\right) \cdot \sin \left(\frac{6 x-4 x}{2}\right)\right]$
$=(2 \sin 5 x \cos x)(2 \cos 5 x \sin x)=(2 \sin 5 x \cos 5 x)(2 \sin x \cos x)$
$=\sin 10 x \sin 2 x$
$= R . H.S$
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