3.477 Find the minimum value of the expression

$$
A=\frac{\operatorname{ctg} 2 \alpha-\operatorname{tg} 2 \alpha}{1+\sin \left(\frac{5 \pi}{2}-8 \alpha\right)} \text { for } 0<\alpha<\frac{\pi}{8}
$$
## Solution.

$\operatorname{ctg} 2 \alpha-\operatorname{tg} 2 \alpha=\frac{\cos ^{2} 2 \alpha-\sin ^{2} 2 \alpha}{\sin 2 \alpha \cdot \cos 2 \alpha}=\frac{2 \cos 4 \alpha}{\sin 4 \alpha} \Rightarrow A=\frac{2 \cos 4 \alpha}{\sin 4 \alpha \cdot(1+\cos 8 \alpha)}=$ $=\frac{2 \cos 4 \alpha}{2 \sin 4 \alpha \cdot \cos ^{2} 4 \alpha}=\frac{2}{\sin 8 \alpha}$. Since $0<8 \alpha<\pi$, the minimum value of $A$ is achieved when $\sin 8 \alpha=1$, i.e., when $\alpha=\frac{\pi}{16}$.

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