$4.9 \sin ^{2}\left(\frac{7 \pi}{8}-2 \alpha\right)-\sin ^{2}\left(\frac{9 \pi}{8}-2 \alpha\right)=\frac{\sin 4 \alpha}{\sqrt{2}}$.
## 4.9 We have

$$
\begin{aligned}
& \sin ^{2}\left(\frac{7 \pi}{8}-2 \alpha\right)-\sin ^{2}\left(\frac{9 \pi}{8}-2 \alpha\right)= \\
& =\frac{1-\cos \left(\frac{7 \pi}{4}-4 \alpha\right)-1+\cos \left(\frac{9 \pi}{4}-4 \alpha\right)}{2}= \\
& =-\sin (2 \pi-4 \alpha) \sin \frac{\pi}{4}=\frac{\sin 4 \alpha}{\sqrt{2}}
\end{aligned}
$$

Here we successively used formulas (4.16)

and (4.22), as well as the fact that $\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}$.