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Question
Prove that:
Solution
Consider LHS:
\[ \frac{\sin 5A - \sin 7A + \sin 8A - \sin 4A}{\cos 4A + \cos 7A - \cos 5A - \cos 8A}\]
\[ = \frac{\sin 5A - \sin 7A + \sin 8A - \sin 4A}{\cos 4A - \cos 8A + \cos 7A - \sin 5A}\]
\[ = \frac{2\sin \left( \frac{5A - 7A}{2} \right) \cos \left( \frac{5A + 7A}{2} \right) + 2\sin \left( \frac{8A - 4A}{2} \right) \cos \left( \frac{8A + 4A}{2} \right)}{- 2\sin \left( \frac{4A + 8A}{2} \right) \sin \left( \frac{4A - 8A}{2} \right) - 2\sin \left( \frac{7A + 5A}{2} \right) \sin \left( \frac{7A - 5A}{2} \right)}\]
\[ \]
\[ = \frac{2\sin \left( - A \right) \cos 6A + 2\sin 2A \cos 6A}{- 2\sin 6A \sin \left( - 2A \right) - 2\sin 6A \sin A}\]
\[ = \frac{- 2\sin A \cos 6A + 2\sin 2A cos 6A}{2\sin 6A \sin 2A - 2\sin 6A \sin A}\]
\[ = \frac{2\cos 6A\left[ \sin 2A - \sin A \right]}{2\sin 6A\left[ \sin 2A - \sin A \right]}\]
\[ = \cot 6A\]
= RHS
Hence, LHS = RHS.
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