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Question
Prove that:
\[\sin^2 \frac{\pi}{18} + \sin^2 \frac{\pi}{9} + \sin^2 \frac{7\pi}{18} + \sin^2 \frac{4\pi}{9} = 2\]
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Solution
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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