Question

доказать: sin3A=3sinA-4sin^3A

Answer 1

$\sin 3 \alpha = \sin (2 \alpha + \alpha )=\sin 2 \alpha \cdot \cos \alpha +\cos 2 \alpha \cdot \sin \alpha =\\\\=2\sin \alpha \cdot \cos \alpha\cdot\cos \alpha +(\cos^2 \alpha -\sin^2 \alpha )\cdot\sin \alpha =\\\\=2\sin \alpha\cdot (1-\sin^2 \alpha )+(1-2\sin^2 \alpha )\cdot \sin\alpha =\\\\=2\sin \alpha -2\sin^3 \alpha +\sin \alpha -2\sin^3 \alpha =3\sin \alpha -4\sin^3 \alpha$