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Question
\[\sin^2 \left( \frac{\pi}{18} \right) + \sin^2 \left( \frac{\pi}{9} \right) + \sin^2 \left( \frac{7\pi}{18} \right) + \sin^2 \left( \frac{4\pi}{9} \right) =\]
Options
1
2
4
none of these.
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Solution
2
\[\text{ We have, } \]
\[ \sin^2 \left( \frac{\pi}{18} \right) + \sin^2 \left( \frac{\pi}{9} \right) + \sin^2 \left( \frac{7\pi}{18} \right) + \sin^2 \left( \frac{4\pi}{9} \right)\]
\[ = \frac{1}{2}\left[ 1 - \cos\left( \frac{\pi}{9} \right) + 1 - \cos\left( \frac{2\pi}{9} \right) + 1 - \cos\frac{7\pi}{9} + 1 - \cos\frac{8\pi}{9} \right] \left( \because \sin^2 \theta = \frac{1 - \cos2\theta}{2} \right)\]
\[ = \frac{1}{2}\left[ 4 - \cos\left( \frac{\pi}{9} \right) - \cos\left( \frac{2\pi}{9} \right) - \left\{ - \cos\left( \pi - \frac{7\pi}{9} \right) \right\} - \left\{ - \cos\left( \pi - \frac{8\pi}{9} \right) \right\} \right]\]
\[ = \frac{1}{2}\left[ 4 - \cos\left( \frac{\pi}{9} \right) - \cos\left( \frac{2\pi}{9} \right) + \cos\left( \frac{2\pi}{9} \right) + \cos\left( \frac{\pi}{9} \right) \right]\]
\[ = \frac{4}{2}\]
\[ = 2\]
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