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Question
Prove that: \[\cos^2 \left( \frac{\pi}{4} - x \right) - \sin^2 \left( \frac{\pi}{4} - x \right) = \sin 2x\]
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Solution
\[LHS = \cos^2 \left( \frac{\pi}{4} - x \right) - \sin^2 \left( \frac{\pi}{4} - x \right)\]
\[ = \cos2\left( \frac{\pi}{4} - x \right) \left[ \because \cos^2 \alpha - \sin^2 \alpha = \cos2\alpha \right]\]
\[ = \cos\left( \frac{\pi}{2} - 2x \right)\]
\[ = \sin2x = RHS \left[ \because \cos\left( \frac{\pi}{2} - 2\alpha \right) = \sin2\alpha \right]\]
\[\text{ Hence proved } .\]
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