Question

If ${\sin }^{3}x\cos 3x+{\cos }^{3}x\sin 3x=\frac{3}{8},$ then the value of $16\sin 4x$ is

Answer 1

${\sin }^{3}x\cos 3x+{\cos }^{3}x\sin 3x=\frac{3}{8}$
$\Rightarrow 4{\sin }^{3}x\cos 3x+4{\cos }^{3}x\sin 3x=\frac{3}{2}$
$\Rightarrow (3\sin x-\sin 3x)\cos 3x+(3\cos x+\cos 3x)\sin 3x=\frac{3}{2}(\because \begin{array}{c}\sin 3x=3\sin x-4{\sin }^{3}x\\ \cos 3x=4{\cos }^{3}x-3\cos x\end{array})$
$\Rightarrow 3(\sin x\cos 3x+\cos x\sin 3x)=\frac{3}{2}$
$\Rightarrow \sin 4x=\frac{1}{2}\therefore 16\sin 4x=8$