3.281. $\sin \left(\frac{5}{2} \pi+4 \alpha\right)-\sin ^{6}\left(\frac{5}{2} \pi+2 \alpha\right)+\cos ^{6}\left(\frac{7}{2} \pi-2 \alpha\right)$.
## Solution.

$$
\begin{aligned}
& \sin \left(\frac{5}{2} \pi+4 \alpha\right)-\sin ^{6}\left(\frac{5}{2} \pi+2 \alpha\right)+\cos ^{6}\left(\frac{7}{2} \pi-2 \alpha\right)= \\
& =\sin \left(\frac{5}{2} \pi+4 \alpha\right)-\left(\sin \left(\frac{5}{2} \pi+2 \alpha\right)\right)^{6}+\left(\cos \left(\frac{7}{2} \pi-2 \alpha\right)\right)^{6}= \\
& =\cos 4 \alpha-\cos ^{6} 2 \alpha+\sin ^{6} 2 \alpha=\cos 4 \alpha-\left(\cos ^{6} 2 \alpha-\sin ^{6} 2 \alpha\right)= \\
& =\cos 4 \alpha-\left(\cos ^{2} 2 \alpha-\sin ^{2} 2 \alpha\right)\left(\cos ^{4} 2 \alpha+\sin ^{2} 2 \alpha \cos ^{2} 2 \alpha+\sin ^{4} 2 \alpha\right)= \\
& =\cos 4 \alpha-\cos 4 \alpha\left(\left(\cos ^{4} 2 \alpha+\sin ^{4} 2 \alpha\right)+\sin ^{2} 2 \alpha \cos ^{2} 2 \alpha\right)=
\end{aligned}
$$

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

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

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

$=\cos 4 \alpha-\cos 4 \alpha\left(1-\frac{1}{4} \sin ^{2} 4 \alpha\right)=\cos 4 \alpha-\cos 4 \alpha+\frac{1}{4} \cos 4 \alpha \sin ^{2} 4 \alpha=$ $=\frac{1}{4} \cos 4 \alpha \sin ^{2} 4 \alpha=\frac{1}{4} \cos 4 \alpha \sin 4 \alpha \sin 4 \alpha=\frac{1}{8}(2 \sin 4 \alpha \cos 4 \alpha) \sin 4 \alpha=$ $=\frac{1}{8} \sin 8 \alpha \sin 4 \alpha$.

Answer: $\frac{1}{8} \sin 8 \alpha \sin 4 \alpha$.