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Question
Prove that:
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Solution
\[\frac{\pi}{3} = 60^\circ, \frac{\pi}{6} = 30^\circ\]
\[\text{ LHS }= \sin\left( 60^\circ - x \right) \cos\left( 30^\circ + x \right) + \cos\left( 60^\circ - x \right) \sin\left( 30^\circ + x \right)\]
\[ = \sin\left[ \left( 60^\circ - x \right) + \left( 30^\circ + x \right) \right] (\text{ Using the formula }\sin A \cos B + \cos A \sin B = \sin\left( A + B \right) \]
\[\text{ and taking }A = 60^\circ - x\text{ and }B = 30^\circ + x \]
\[ = \sin90^\circ\]
\[ = 1\]
= RHS
Hence proved.
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