Prove that:

Question:

Prove that: $\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{\pi}{9}+\sin ^{2} \frac{7 \pi}{18}+\sin ^{2} \frac{4 \pi}{9}=2$

Solution:

$\mathrm{LHS}=\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{\pi}{9}+\sin ^{2} \frac{7 \pi}{18}+\sin ^{2} \frac{4 \pi}{9}$

$=\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{2 \pi}{18}+\sin ^{2} \frac{7 \pi}{18}+\sin ^{2} \frac{8 \pi}{18}$

$=\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{2 \pi}{18}+\sin ^{2}\left(\frac{7 \pi}{18}\right)+\sin ^{2}\left(\frac{8 \pi}{18}\right)$

$=\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{2 \pi}{18}+\sin ^{2}\left(\frac{\pi}{2}-\frac{2 \pi}{18}\right)+\sin ^{2}\left(\frac{\pi}{2}-\frac{\pi}{18}\right)$

$=\sin ^{2} \frac{\pi}{18}+\sin ^{2} \frac{2 \pi}{18}+\cos ^{2} \frac{2 \pi}{18}+\cos ^{2} \frac{\pi}{18}$

$=\sin ^{2} \frac{\pi}{18}+\cos ^{2} \frac{\pi}{18}+\sin ^{2} \frac{2 \pi}{18}+\cos ^{2} \frac{2 \pi}{18}$

$=1+1$

$=2$

= RHS

Hence, proved.

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