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Prove that:  cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A - Mathematics

Sum

Prove that: 
cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A

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Solution

Consider LHS: 
\[ \cos A + \cos 3A + \cos 5A + \cos 7A\]
\[ = 2\cos \left( \frac{A + 3A}{2} \right) \cos \left( \frac{A - 3A}{2} \right) + 2\cos \left( \frac{5A + 7A}{2} \right) \cos \left( \frac{5A - 7A}{2} \right) \left\{ \because \cos A + \cos B = 2\cos\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right) \right\}\]
\[ = 2\cos 2A \cos\left( - A \right) + 2\cos 6A \cos\left( - A \right)\]
\[= 2\cos 2A \cos A + 2\cos 6A \cos A\]
\[ = 2\cos A(\cos 2A + \cos 6A)\]
\[ = 2\cos A \times 2\cos \left( \frac{2A + 6A}{2} \right) \cos \left( \frac{2A - 6A}{2} \right)\]
\[ = 4\cos A \cos 4A \cos\left( - 2A \right)\]
\[ = 4\cos A \cos 2A \cos 4A\]
 = 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 6.2 | Page 18
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