4sinαsin(π/3+α)sin(π/3-α)=sin3α

Решите задачу:

$4sin \alpha*sin(\frac{\pi}{3}+\alpha)*sin(\frac{\pi}{3}-\alpha)=\\\\ 4sin \alpha*\frac{cos((\frac{\pi}{3}+\alpha)-(\frac{\pi}{3}-\alpha))-cos((\frac{\pi}{3}+\alpha)+(\frac{\pi}{3}-\alpha))}{2}}=\\\\ 2sin \alpha*(cos(\frac{\pi}{3}+\alpha-\frac{\pi}{3}+\alpha)-cos(\frac{\pi}{3}+\alpha+\frac{\pi}{3}-\alpha))=\\\\ 2sin \alpha*(cos(2\alpha)-cos(\frac{2\pi}{3}))=\\\\ 2sin \alpha*(cos(2\alpha)-cos(\pi-\frac{\pi}{3}))=\\\\$

$2sin \alpha *(cos(2\alpha)+cos(\frac{\pi}{3}))=\\\\ 2sin \alpha *(1-2sin^2 \alpha+\frac{1}{2})=\\\\ sin \alpha *(2-4sin^2 \alpha+1)=\\\\ sin \alpha *(3-4sin^2 \alpha)=\\\\ 3sin \alpha-4sin^3 \alpha=sin (3\alpha)$