2. Find the value of the expression $\cos ^{4} \frac{7 \pi}{24}+\sin ^{4} \frac{11 \pi}{24}+\sin ^{4} \frac{17 \pi}{24}+\cos ^{4} \frac{13 \pi}{24}$.
Answer: $\frac{3}{2}$.

Solution. Using reduction formulas ( $\sin (\pi-\alpha)=\sin \alpha$ ), the given expression can be rewritten as $\cos ^{4} \frac{7 \pi}{24}+\sin ^{4} \frac{13 \pi}{24}+\sin ^{4} \frac{7 \pi}{24}+\cos ^{4} \frac{13 \pi}{24}$. Next, we notice that

$$
\begin{gathered}
\cos ^{4} \gamma+\sin ^{4} \gamma=\left(\cos ^{4} \gamma+2 \sin ^{2} \gamma \cos ^{2} \gamma+\sin ^{4} \gamma\right)-2 \sin ^{2} \gamma \cos ^{2} \gamma=\left(\cos ^{2} \gamma+\sin ^{2} \gamma\right)^{2}-\frac{1}{2} \cdot(2 \sin \gamma \cos \gamma)^{2}= \\
=1-\frac{1}{2} \sin ^{2} 2 \gamma=1-\frac{1}{4}(1-\cos 4 \gamma)=\frac{3}{4}+\frac{1}{4} \cos 4 \gamma
\end{gathered}
$$

Then we get $\frac{3}{4}+\frac{1}{4} \cos \frac{7 \pi}{6}+\frac{3}{4}+\frac{1}{4} \cos \frac{13 \pi}{6}=\frac{3}{2}$.