Question

Prove that $4\sin x\sin \left(\frac{\pi }{3}+x\right)\sin \left(\frac{\pi }{3}-x\right)=\sin 3x$

Answer 1

$4\sin x\sin \left(\frac{\pi }{3}+x\right)\sin \left(\frac{\pi }{3}-x\right)$

$4\sin x\sin \left(\frac{\pi }{3}+x\right)\sin (\pi -\left(\frac{\pi }{3}-x\right))$

$4\sin x\sin \left(\frac{\pi }{3}+x\right)\sin \left(\frac{2\pi }{3}+x\right)$

$=4\sin x(\sin x\cos \frac{\pi }{3}+\cos x\sin \frac{\pi }{3})(\sin x\cos \frac{2\pi }{3}+\cos x\sin \frac{2\pi }{3})$

$=4\sin x(\frac{\sin x}{2}+\frac{\sqrt{3}\cos x}{2})(-\frac{\sin x}{2}+\frac{\sqrt{3}\cos x}{2})$

$=4\sin x(\frac{3}{4}{\cos }^{2}x-\frac{1}{4}{\sin }^{2}x)$

$=\sin x(3{\cos }^{2}x-{\sin }^{2}x)$

$=\sin x\left(3\right(1-{\sin }^{2}x)-{\sin }^{2}x)$

$=\sin x(3-4{\sin }^{2}x)$

$=\sin 3x$.