$4.30 \sin 10 \alpha \sin 8 \alpha + \sin 8 \alpha \sin 6 \alpha - \sin 4 \alpha \sin 2 \alpha$.
### 4.30 Method I. We have

$\sin 10 \alpha \sin 8 \alpha + \sin 8 \alpha \sin 6 \alpha - \sin 4 \alpha \sin 2 \alpha = \sin 8 \alpha \times$ $\times (\sin 10 \alpha + \sin 6 \alpha) - 2 \sin^2 2 \alpha \cos 2 \alpha = \sin 8 \alpha \cdot 2 \sin 8 \alpha \cdot$ $\cdot \cos 2 \alpha - 2 \sin^2 2 \alpha \cos 2 \alpha = 2 \cos 2 \alpha (\sin^2 8 \alpha - \sin^2 2 \alpha) =$ $= 2 \cos 2 \alpha \cdot \frac{1}{2} (1 - \cos 16 \alpha - 1 + \cos 4 \alpha) =$

Method II. We find

$\sin 10 \alpha \sin 8 \alpha + \sin 8 \alpha \sin 6 \alpha - \sin 4 \alpha \sin 2 \alpha =$

$= \sin 10 \alpha \sin 8 \alpha + \frac{1}{2} (\cos 2 \alpha - \cos 14 \alpha - \cos 2 \alpha + \cos 6 \alpha) =$

$= \sin 10 \alpha \sin 8 \alpha + \sin 10 \alpha \sin 4 \alpha =$

$= \sin 10 \alpha (\sin 8 \alpha + \sin 4 \alpha) = 2 \cos 2 \alpha \sin 6 \alpha \sin 10 \alpha$.