3.174. $\sin 2 \alpha+\sin 4 \alpha-\sin 6 \alpha$.
Solution.

$\sin 2 \alpha+\sin 4 \alpha-\sin 6 \alpha=\sin 2 \alpha+\sin 4 \alpha-\sin 2(3 \alpha)=$

$=2 \sin 3 \alpha \cos \alpha-2 \sin 3 \alpha \cos 3 \alpha=2 \sin 3 \alpha(\cos \alpha-\cos 3 \alpha)=$

$=2 \sin 3 \alpha \cdot(-2 \sin 2 \alpha \sin (-\alpha))=4 \sin 3 \alpha \sin 2 \alpha \sin \alpha$.

Answer: $4 \sin 3 \alpha \sin 2 \alpha \sin \alpha$.

Prove the validity of the equalities (3.148-3.152):