Prove that:
$\sin ^{2} \frac{\pi}{8}+\sin ^{2} \frac{3 \pi}{8}+\sin ^{2} \frac{5 \pi}{8}+\sin ^{2} \frac{7 \pi}{8}=2$
$\mathrm{LHS}=\sin ^{2} \frac{\pi}{8}+\sin ^{2} \frac{3 \pi}{8}+\sin ^{2} \frac{5 \pi}{8}+\sin ^{2} \frac{7 \pi}{8}$
$=\sin ^{2}\left(\frac{\pi}{2}-\frac{3 \pi}{8}\right)+\sin ^{2}\left(\frac{\pi}{2}-\frac{\pi}{8}\right)+\sin ^{2} \frac{5 \pi}{8}+\sin ^{2} \frac{7 \pi}{8}$
$=\cos ^{2} \frac{3 \pi}{8}+\sin ^{2} \frac{\pi}{8}+\sin ^{2}\left(\pi-\frac{3 \pi}{8}\right)+\sin ^{2}\left(\pi-\frac{\pi}{8}\right)$
$=\cos ^{2} \frac{3 \pi}{8}+\sin ^{2} \frac{\pi}{8}+\sin ^{2} \frac{3 \pi}{8}+\cos ^{2} \frac{\pi}{8}$
$=\left(\cos ^{2} \frac{\pi}{8}+\sin ^{2} \frac{\pi}{8}\right)+\left(\cos ^{2} \frac{3 \pi}{8}+\sin ^{2} \frac{3 \pi}{8}\right)$
$=1+1=2=$ RHS
Hence proved.
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