$\sin 3\alpha =\sin \left(2\alpha +\alpha \right)=\sin 2\alpha \cos \alpha +\cos 2\alpha \sin \alpha =2\sin \alpha \cos \alpha \cdot \cos \alpha +\left({\cos }^2\alpha -{\sin }^2\alpha \right)\cdot \sin \alpha =\sin \alpha \left[2{\cos }^2\alpha +{\cos }^2\alpha -{\sin }^2\alpha \right]=$  

$=\sin \alpha \left[3{\cos }^2\alpha +{\cos }^2\alpha -1\right]=4\sin \alpha {\cos }^2\alpha -\sin \alpha$