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