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Prove That: Sin a + Sin 2a + Sin 4a + Sin 5a = 4 Cos a 2 3 a 2 - Mathematics

Sum

Prove that:
 sin A + sin 2A + sin 4A + sin 5A = 4 cos \[\frac{A}{2}\]\[\frac{3A}{2}\]

 

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Solution

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

Concept: Transformation Formulae
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 8 Transformation formulae
Exercise 8.2 | Q 6.3 | Page 18
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