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Prove That: Cos 3 a + 2 Cos 5 a + Cos 7 a Cos a + 2 Cos 3 a + Cos 5 a = Cos 5 a Cos 3 a

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Question

Prove that:

\[\frac{\cos 3A + 2 \cos 5A + \cos 7A}{\cos A + 2 \cos 3A + \cos 5A} = \frac{\cos 5A}{\cos 3A}\]
Sum
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Solution

Consider LHS: 
\[ \frac{\cos 3A + 2\cos 5A + \cos 7A}{\cos A + 2\cos 3A + \cos 5A}\]
\[ = \frac{\cos 3A + \cos 7A + 2\cos 5A}{\cos A + \cos 5A + 2\cos 3A}\]
\[ = \frac{2\cos \left( \frac{3A + 7A}{2} \right) \cos \left( \frac{3A - 7A}{2} \right) + 2\cos 5A}{2\cos \left( \frac{A + 5A}{2} \right) \cos \left( \frac{A - 5A}{2} \right) + 2\cos 3A}\]
\[ \]
\[ = \frac{2\cos 5A \cos \left( - 2A \right) + 2\cos 5A}{2\cos 3A \cos \left( - 2A \right) + 2\cos 3A}\]
\[ = \frac{2\cos 5A \cos 2A + 2\cos 5A}{2\cos 3A \cos 2A + 2\cos 3A}\]
\[ = \frac{2\cos 5A \left[ \cos 2A + 1 \right]}{2\cos 3A \left[ \cos 2A + 1 \right]}\]
\[ = \frac{\cos 5A}{\cos 3A}\]
 = RHS
Hence, RHS = LHS. 

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Transformation Formulae
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Chapter 8: Transformation formulae - Exercise 8.2 [Page 18]

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R.D. Sharma Mathematics [English] Class 11
Chapter 8 Transformation formulae
Exercise 8.2 | Q 8.02 | Page 18

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