3.238. $1+\sin \left(3\left(\alpha+\frac{\pi}{2}\right)\right) \cos 2 \alpha+2 \sin 3 \alpha \cos (3 \pi-\alpha) \sin (\alpha-\pi)=$ $=2 \sin ^{2} \frac{5 \alpha}{2}$.

3.238. $1+\sin \left(3\left(\alpha+\frac{\pi}{2}\right)\right) \cos 2 \alpha+2 \sin 3 \alpha \cos (3 \pi-\alpha) \sin (\alpha-\pi)=$ $=2 \sin ^{2} \frac{5 \alpha}{2}$.
## Solution.

$$
\begin{aligned}
& 1+\sin \left(3\left(\alpha+\frac{\pi}{2}\right)\right) \cos 2 \alpha+2 \sin 3 \alpha \cos (3 \pi-\alpha) \sin (\alpha-\pi)= \\
& =1+\sin \left(\frac{3 \pi}{2}+\alpha\right) \cos 2 \alpha+2 \sin 3 \alpha \cos (3 \pi-\alpha)(-\sin (\pi-\alpha))= \\
& =1+\sin \left(\frac{3 \pi}{2}+\alpha\right) \cos 2 \alpha-2 \sin 3 \alpha \cos (3 \pi-\alpha) \sin (\pi-\alpha)= \\
& =1-\cos 3 \alpha \cos 2 \alpha-2 \sin 3 \alpha(-\cos \alpha) \sin \alpha= \\
& =1-\cos 3 \alpha \cos 2 \alpha+\sin 3 \alpha(2 \sin \alpha \cos \alpha)= \\
& =1-\cos 3 \alpha \cos 2 \alpha+\sin 3 \alpha \sin 2 \alpha=1-(\cos 3 \alpha \cos 2 \alpha-\sin 3 \alpha \sin 2 \alpha)= \\
& =1-\cos 5 \alpha=1-\cos 2\left(\frac{5 \alpha}{2}\right)= \\
& =1-\left(1-2 \sin ^{2} \frac{5 \alpha}{2}\right)=1-1+2 \sin ^{2} \frac{5 \alpha}{2}=2 \sin ^{2} \frac{5 \alpha}{2}
\end{aligned}
$$

The identity is proven.