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Question
Prove that:
\[\sin\frac{5\pi}{18} - \cos\frac{4\pi}{9} = \sqrt{3} \sin\frac{\pi}{9}\]
Solution
\[LHS = \sin\left( \frac{5\pi}{18} \right) - \cos\frac{4\pi}{9}\]
\[ = \sin\left( \frac{5\pi}{18} \right) - \cos\left( \frac{\pi}{2} - \frac{\pi}{18} \right)\]
\[ = \sin\left( \frac{5\pi}{18} \right) - \sin\left( \frac{\pi}{18} \right)\]
\[ = 2\sin\left( \frac{\frac{5\pi}{18} - \frac{\pi}{18}}{2} \right)\cos\left( \frac{\frac{5\pi}{18} + \frac{\pi}{18}}{2} \right) \left[ \because \sin A - \sin B = 2\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right) \right]\]
\[ = 2\sin\left( \frac{\pi}{9} \right)\cos\frac{\pi}{6}\]
\[ = 2\sin\left( \frac{\pi}{9} \right)\cos\frac{\pi}{6}\]
\[ = 2 \times \frac{\sqrt{3}}{2}\sin\left( \frac{\pi}{9} \right)\]
\[ = \sqrt{3}\sin\left( \frac{\pi}{9} \right) = RHS\]
Hence, LHS = RHS.
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