3.138. $\sin 5 \alpha-\sin 6 \alpha-\sin 7 \alpha+\sin 8 \alpha$.
Solution.

$$
\begin{aligned}
& \sin 5 \alpha-\sin 6 \alpha-\sin 7 \alpha+\sin 8 \alpha=(\sin 8 \alpha+\sin 5 \alpha)-(\sin 7 \alpha+\sin 6 \alpha)= \\
& =2 \sin \frac{13 \alpha}{2} \cos \frac{3 \alpha}{2}-2 \sin \frac{13 \alpha}{2} \cos \frac{\alpha}{2}=2 \sin \frac{13 \alpha}{2}\left(\cos \frac{3 \alpha}{2}-\cos \frac{\alpha}{2}\right)= \\
& =2 \sin \frac{13 \alpha}{2}\left(-2 \sin \alpha \sin \frac{\alpha}{2}\right)=-4 \sin \frac{\alpha}{2} \sin \alpha \sin \frac{13 \alpha}{2}
\end{aligned}
$$

Answer: $-4 \sin \frac{\alpha}{2} \sin \alpha \sin \frac{13 \alpha}{2}$.