Question

Prove that
$\sin 4A=4\sin A{\cos }^{3}A-4\cos A{\sin }^{3}A$.

Answer 1

To prove:
$\sin 4A=4\sin A{\cos }^{3}A-4\cos A{\sin }^{3}A.$
$LHS=si{n}^{4}A$
$=2\sin 2A\cos 2A$
$(\because \sin 2\theta =2\sin \theta \cos \theta )$
$=2\left(2\sin A\cos A\right)({\cos }^{2}A-{\sin }^{2}A)$
$(\because \cos 2\theta ={\cos }^2\theta -{\sin }^2\theta )$
$=4\sin A\cos A({\cos }^{2}A-{\sin }^{2}A)$
$=4\sin A{\cos }^{3}A-4\cos A{\sin }^{3}A$
$LHS=RHS$
Hence, proved.