### 3.476 Find the maximum value of the expression

$A=\sin ^{2}\left(\frac{15 \pi}{8}-4 \alpha\right)-\sin ^{2}\left(\frac{17 \pi}{8}-4 \alpha\right)$ for $0 \leq \alpha \leq \frac{\pi}{8}$.
Solution.

$A=\frac{1-\cos \left(\frac{15 \pi}{4}-8 \alpha\right)}{2}-\frac{1-\cos \left(\frac{17 \pi}{4}-8 \alpha\right)}{2}=\frac{1}{2}\left(\cos \left(\frac{\pi}{4}-8 \alpha\right)-\cos \left(\frac{\pi}{4}+8 \alpha\right)\right)=$

$=\sin \frac{\pi}{4} \cdot \sin 8 \alpha=\frac{\sin 8 \alpha}{\sqrt{2}}$. Since $0 \leq \alpha \leq \frac{\pi}{8}$, the maximum value of $A$ is achieved when $\sin 8 \alpha=1$, i.e., when $\alpha=\frac{\pi}{16}$.

Answer: $\frac{1}{\sqrt{2}}$ when $\alpha=\frac{\pi}{16}$.