3.071. $\sin ^{2}\left(\frac{\alpha}{2}+2 \beta\right)-\sin ^{2}\left(\frac{\alpha}{2}-2 \beta\right)$.
## Solution.

$$
\begin{aligned}
& \sin ^{2}\left(\frac{\alpha}{2}+2 \beta\right)-\sin ^{2}\left(\frac{\alpha}{2}-2 \beta\right)=\frac{1-\cos (\alpha+4 \beta)}{2}-\frac{1-\cos (\alpha-4 \beta)}{2}= \\
& =\frac{1}{2}-\frac{\cos (\alpha+4 \beta)}{2}-\frac{1}{2}+\frac{\cos (\alpha-4 \beta)}{2}=\frac{1}{2}(\cos (\alpha-4 \beta)-\cos (\alpha+4 \beta))= \\
& =-\sin \alpha \sin (-4 \beta)=\sin \alpha \sin 4 \beta
\end{aligned}
$$

Answer: $\sin \alpha \sin 4 \beta$.