3.247. $\left(1-\operatorname{ctg}^{2}\left(\frac{3}{2} \pi-2 \alpha\right)\right) \sin ^{2}\left(\frac{\pi}{2}+2 \alpha\right) \operatorname{tg}\left(\frac{5}{4} \pi-2 \alpha\right)+\cos \left(4 \alpha-\frac{\pi}{2}\right)$.
## Solution.

$$
\begin{aligned}
& \left(1-\operatorname{ctg}^{2}\left(\frac{3}{2} \pi-2 \alpha\right)\right) \sin ^{2}\left(\frac{\pi}{2}+2 \alpha\right) \operatorname{tg}\left(\frac{5}{4} \pi-2 \alpha\right)+\cos \left(4 \alpha-\frac{\pi}{2}\right)= \\
& =\left(1-\left(\operatorname{ctg}\left(\frac{3}{2} \pi-2 \alpha\right)\right)^{2}\right)\left(\sin \left(\frac{\pi}{2}+2 \alpha\right)\right)^{2} \times \\
& \times \operatorname{tg}\left(\pi+\left(\frac{\pi}{4}-2 \alpha\right)\right)+\cos \left(\frac{\pi}{2}-4 \alpha\right)= \\
& =\left(1-\operatorname{tg}^{2} 2 \alpha\right) \cos ^{2} 2 \alpha \operatorname{tg}\left(\frac{\pi}{4}-2 \alpha\right)+\sin 4 \alpha= \\
& =\left(1-\frac{\sin ^{2} 2 \alpha}{\cos ^{2} 2 \alpha}\right) \cos ^{2} 2 \alpha \operatorname{tg}\left(\frac{\pi}{4}-2 \alpha\right)+\sin 4 \alpha=
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0189.jpg?height=125&width=778&top_left_y=63&top_left_x=162)

$=\cos 4 \alpha \operatorname{tg}\left(\frac{\pi}{4}-2 \alpha\right)+\sin 4 \alpha=\frac{\cos 4 \alpha\left(1-\cos \left(\frac{\pi}{2}-4 \alpha\right)\right)}{\sin \left(\frac{\pi}{2}-4 \alpha\right)}+\sin 4 \alpha=$

$=\frac{\cos 4 \alpha(1-\sin 4 \alpha)}{\cos 4 \alpha}+\sin 4 \alpha=1-\sin 4 \alpha+\sin 4 \alpha=1$.

Answer: 1.