3.304. $1+\cos \left(2 \alpha-\frac{3}{2} \pi\right)+\sin \left(2 \alpha+\frac{3}{2} \pi\right)-\operatorname{ctg}\left(\frac{\pi}{2}+2 \alpha\right)$.
## Solution.

$$
1+\cos \left(2 \alpha-\frac{3}{2} \pi\right)+\sin \left(2 \alpha+\frac{3}{2} \pi\right)-\operatorname{ctg}\left(\frac{\pi}{2}+2 \alpha\right)=
$$

$$
\begin{aligned}
& =1+\cos \left(\frac{3}{2} \pi-2 \alpha\right)+\sin \left(\frac{3}{2} \pi+2 \alpha\right)-\operatorname{ctg}\left(\frac{\pi}{2}+2 \alpha\right)= \\
& =1-\sin 2 \alpha-\cos 2 \alpha+\operatorname{tg} 2 \alpha=(1+\operatorname{tg} 2 \alpha)-(\cos 2 \alpha+\sin 2 \alpha)= \\
& =\left(1+\frac{\sin 2 \alpha}{\cos 2 \alpha}\right)-(\cos 2 \alpha+\sin 2 \alpha)=\frac{\cos 2 \alpha+\sin 2 \alpha}{\cos 2 \alpha}-(\cos 2 \alpha+\sin 2 \alpha)= \\
& =(\cos 2 \alpha+\sin 2 \alpha)\left(\frac{1}{\cos 2 \alpha}-1\right)=\frac{(\cos 2 \alpha+\sin 2 \alpha)(1-\cos 2 \alpha)}{\cos 2 \alpha}= \\
& =\frac{\frac{2}{\sqrt{2}}\left(\frac{\sqrt{2}}{2} \cos 2 \alpha+\frac{\sqrt{2}}{2} \sin 2 \alpha\right)\left(1-\cos ^{2} \alpha+\sin ^{2} \alpha\right)}{\cos 2 \alpha}= \\
& =\frac{\sqrt{2}\left(\cos \frac{\pi}{4} \cos 2 \alpha+\sin \frac{\pi}{4} \sin 2 \alpha\right) \cdot 2 \sin ^{2} \alpha}{\cos 2 \alpha}= \\
& =\left[\cos x \cos y+\sin ^{2} x \sin y=\cos (x-y)\right]=\frac{2 \sqrt{2} \cos \left(\frac{\pi}{4}-2 \alpha\right) \sin ^{2} \alpha}{\cos 2 \alpha} \\
& \quad \text { Answer: } \frac{2 \sqrt{2} \sin ^{2} \alpha \cos \left(\frac{\pi}{4}-2 \alpha\right)}{\cos 2 \alpha} . \\
& \text { 3.305. } 4 \cos ^{2}\left(2 \alpha-\frac{3}{2} \pi\right)+\cos (2 \alpha-\pi)+\sin \left(\frac{5}{2} \pi-6 \alpha\right)
\end{aligned}
$$

## Solution.

$4 \cos ^{2}\left(2 \alpha-\frac{3}{2} \pi\right)+\cos (2 \alpha-\pi)+\sin \left(\frac{5}{2} \pi-6 \alpha\right)=$

$=4 \cos \left(\left(\frac{3}{2} \pi-2 \alpha\right)\right)^{2}+\cos (\pi-2 \alpha)+\sin \left(\frac{5}{2} \pi-6 \alpha\right)=$

$=4 \sin ^{2} 2 \alpha-\cos 2 \alpha+\cos 6 \alpha=2(1-\cos 4 \alpha)+(\cos 6 \alpha-\cos 2 \alpha)=$

$=2\left(1-\cos ^{2} 2 \alpha+\sin ^{2} 2 \alpha\right)+(\cos 6 \alpha-\cos 2 \alpha)=$
$=4 \sin ^{2} 2 \alpha+(\cos 6 \alpha-\cos 2 \alpha)=$

$=\left[\cos x-\cos y=-2 \sin \frac{x+y}{2} \sin \frac{x-y}{2}\right]=4 \sin ^{2} 2 \alpha-2 \sin 4 \alpha \sin 2 \alpha=$

$=2 \sin 2 \alpha(2 \sin 2 \alpha-\sin 4 \alpha)=2 \sin 2 \alpha(2 \sin 2 \alpha-2 \sin 2 \alpha \cos 2 \alpha)=$

$=4 \sin ^{2} 2 \alpha(1-\cos 2 \alpha)=4 \sin ^{2} 2 \alpha\left(1-\cos ^{2} \alpha+\sin ^{2} \alpha\right)=$

$=4 \sin ^{2} 2 \alpha \cdot 2 \sin ^{2} \alpha=8 \sin ^{2} 2 \alpha \sin ^{2} \alpha$.

Answer: $8 \sin ^{2} 2 \alpha \sin ^{2} \alpha$.