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