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प्रश्न
Prove that: \[\sin^2 \left( \frac{\pi}{8} + \frac{x}{2} \right) - \sin^2 \left( \frac{\pi}{8} - \frac{x}{2} \right) = \frac{1}{\sqrt{2}} \sin x\]
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उत्तर
\[LHS = \sin^2 \left( \frac{\pi}{8} + \frac{x}{2} \right) - \sin^2 \left( \frac{\pi}{8} - \frac{x}{2} \right)\]
\[ = \frac{1}{2}\left\{ 1 - \cos2\left( \frac{\pi}{8} + \frac{x}{2} \right) \right\} - \frac{1}{2}\left\{ 1 - \cos2\left( \frac{\pi}{8} - \frac{x}{2} \right) \right\}\]
\[ = \frac{1}{2}\left\{ \cos\left( \frac{\pi}{4} - x \right) - \cos\left( \frac{\pi}{4} + x \right) \right\}\]
Using the identit
\[\text{ cos } C - \text{ cos } D = - 2\sin\frac{C + D}{2}\sin\frac{C - D}{2}\] , we get
\[= \frac{1}{2}\left\{ - 2\sin\left( \frac{\left( \frac{\pi}{4} - x \right) + \left( \frac{\pi}{4} + x \right)}{2} \right)\sin\left( \frac{\left( \frac{\pi}{4} - x \right) - \left( \frac{\pi}{4} + x \right)}{2} \right) \right\}\]
\[ = - \sin\frac{\pi}{4}\sin\left( - x \right)\]
\[ = \sin\frac{\pi}{4}\text{ sin } x \left[ \because \sin\left( - x \right) = - \text{ sin } x \right]\]
\[ = \frac{1}{\sqrt{2}}\text{ sin } x = RHS\]
\[\text{ Hence proved } .\]
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