Question

Prove the following identity:
$\frac{{\sin }^{2}4\alpha }{2\cos \alpha +\cos 3\alpha +\cos 5\alpha }=2\sin \alpha \sin 2\alpha$

Answer 1

$\frac{{\sin }^{2}4\alpha }{2\cos \alpha +\cos 3\alpha +\cos 5\alpha }$

$=\frac{{\sin }^{2}4\alpha }{2\cos \alpha +2\cos 4\alpha \cos \alpha }$

$=\frac{{1-\cos }^{2}4\alpha }{2\cos \alpha (1+\cos 4\alpha )}$

$=\frac{(1+\cos 4\alpha )(1-\cos 4\alpha )}{2\cos \alpha (1+\cos 4\alpha )}$

$=\frac{1-\cos 4\alpha }{2\cos \alpha }$

$=\frac{{2\sin }^{2}2\alpha }{2\cos \alpha }$

$=\frac{2\sin \alpha \cos \alpha }{\cos \alpha }\times \sin 2\alpha$

$=2\sin \alpha \sin 2\alpha .$ $\left(proved\right)$