Question

Prove $\sin 3x{\sin }^{3}x+\cos 3x{\cos }^{3}x={\cos }^{3}2x.$

Answer 1

$\sin 3x=3\sin x-4{\sin }^{3}x$

$\therefore {\sin }^{3}x=\frac{1}{4}(3\sin x-\sin 3x)$

Similarly, ${\cos }^{3}x=\frac{1}{4}(3\cos x+\cos 3x)$

$L.H.S.=\frac{1}{4}\left[\sin 3x\right(3\sin x-\sin 3x)+\cos 3x(3\cos x+\cos 3x\left)\right]$

$=\frac{1}{4}\left[3\cos \right(3x-x)+({\cos }^{2}3x-{\sin }^{2}3x\left)\right]$

$=\frac{1}{4}[3\cos 2x+\cos 6x]=\frac{1}{4}[3\cos A+\cos 3A]$

$={\cos }^{3}A,by\left(2\right)={\cos }^{3}2x$ $\because A=2x$