Question
Prove that:
$\sin^2\frac{\pi}{8}+\sin^2\frac{3\pi}{8}+\sin^2\frac{5\pi}{8}+\sin^2\frac{7\pi}{8}=2$

Answer

$\text{LHS}=\sin^2\frac{\pi}{8}+\sin^2\frac{3\pi}{8}+\sin^2\frac{5\pi}{8}+\sin^2\frac{7\pi}{8}$
$=\sin^2\big(\frac{\pi}{2}-\frac{3\pi}{8}\big)+\sin^2\big(\frac{\pi}{2}-\frac{\pi}{8}\big)+\sin^2\frac{5\pi}{8}+\sin^2\frac{7\pi}{8}$
$=\cos^2\frac{3\pi}{8}+\sin^2\frac{\pi}{8}+\sin^2\big(\pi-\frac{3\pi}{8}\big)+\sin^2\big(\pi-\frac{\pi}{8}\big)$
$=\cos^2\frac{3\pi}{8}+\sin^2\frac{\pi}{8}+\sin^2\frac{3\pi}{8}+\cos^2\frac{\pi}{8}$
$=\big(\cos^2\frac{\pi}{8}+\sin^2\frac{\pi}{8}\big)+\big(\cos^2\frac{3\pi}{8}+\sin^2\frac{3\pi}{8}\big)$
$=1+1=2=\text{RHS}$
Hence proved.

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