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Question
Prove that:
Solution
Consider LHS:
\[ \frac{\sin 3A \cos 4A - \sin A \cos 2A}{\sin 4A sin A + \cos 6A \cos A}\]
Multiplying numerator and denominator by 2, we get
\[ = \frac{2\sin 3A \cos 4A - 2\sin A \cos 2A}{2\sin 4A \sin A + 2\cos 6A \cos A}\]
\[ = \frac{\sin \left( 3A + 4A \right) + \sin \left( 3A - 4A \right) - \sin \left( A + 2A \right) - \sin \left( A - 2A \right)}{\cos \left( 4A - A \right) - \cos \left( 4A + A \right) + \cos \left( 6A + A \right) + \cos \left( 6A - A \right)}\]
\[ = \frac{\sin 7A + \sin \left( - A \right) - \sin 3A - \sin \left( - A \right)}{\cos 3A - \cos 5A + \cos 7A + \cos 5A}\]
\[ = \frac{\sin 7A - \sin A - \sin 3A + \sin A}{\cos 3A - \cos 5A + \cos 7A + \cos 5A}\]
\[ = \frac{\sin 7A - \sin 3A}{\cos 3A + \cos 7A}\]
\[ = \frac{2\sin \left( \frac{7A - 3A}{2} \right) \cos \left( \frac{7A + 3A}{2} \right)}{2\cos \left( \frac{3A + 7A}{2} \right) \cos \left( \frac{3A - 7A}{2} \right)}\]
\[ = \frac{\sin 2A \cos 5A}{\cos 5A \cos \left( - 2A \right)}\]
\[ = \frac{\sin 2A \cos 5A}{\cos 5A \cos 2A}\]
\[ = \tan 2A\]
= RHS
Hence, LHS = RHS.
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