$\sin ^{2} \frac{17 \pi}{18}+\sin ^{2} \frac{5 \pi}{8}+\cos ^{2} \frac{37 \pi}{18}+\cos ^{2} \frac{3 \pi}{8}=?$

Solve:$\begin{aligned} = & \sin ^{2}\left(\pi-\frac{\pi}{18}\right)+\sin ^{2}\left(\pi-\frac{3 \pi}{8}\right)+ \\ & \cos ^{2}\left(2 \pi+\frac{\pi}{18}\right)+\cos ^{2} \frac{3 \pi}{8} \\ = & \sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{3 \pi}{8}+\cos ^{2} \frac{\pi}{18}+\cos ^{2} \frac{3 \pi}{8} \\ = & \left(\sin ^{2} \frac{\pi}{18}+\cos ^{2} \frac{\pi}{18}\right)+\left(\sin ^{2} \frac{3 \pi}{8}+\cos ^{2} \frac{3 \pi}{8}\right) \\ = & 1+1=2(\text { Ans. }) \end{aligned}$