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Question
Prove that:
`sin A + sin 2A + sin 4A + sin 5A = 4 cos (A/2) cos((3A)/2)sin3A`
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Solution
Consider LHS:
\[ \sin A + \sin 2A + \sin 4A + \sin 5A\]
\[ = 2\sin \left( \frac{A + 2A}{2} \right) \cos \left( \frac{A - 2A}{2} \right) + 2\sin \left( \frac{4A + 5A}{2} \right) \cos \left( \frac{4A - 5A}{2} \right) \left\{ \because \sin A + \sin B = 2\sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \right\}\]
\[ = 2\sin \left( \frac{3}{2}A \right) \cos \left( - \frac{A}{2} \right) + 2\sin \left( \frac{9}{2}A \right) \cos \left( - \frac{A}{2} \right)\]
\[= 2\sin \left( \frac{3}{2}A \right) \cos \left( \frac{A}{2} \right) + 2\sin \left( \frac{9}{2}A \right) \cos \left( \frac{A}{2} \right)\]
\[ = 2\cos \left( \frac{A}{2} \right)\left\{ \sin \frac{3}{2}A + \sin \frac{9}{2}A \right\}\]
\[ = 2\cos \left( \frac{A}{2} \right) \times 2\sin \left( \frac{\frac{3}{2}A + \frac{9}{2}A}{2} \right) \cos \left( \frac{\frac{3}{2}A - \frac{9}{2}A}{2} \right)\]
\[ = 4\cos \left( \frac{A}{2} \right) \sin 3A \cos \left( - \frac{3}{2}A \right)\]
\[ = 4\cos \frac{A}{2} \cos \left( \frac{3A}{2} \right) \sin 3A\]
= RHS
Hence, LHS = RHS
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