# Prove That: Sin a + 2 Sin 3 a + Sin 5 a Sin 3 a + 2 Sin 5 a + Sin 7 a = Sin 3 a Sin 5 a - Mathematics

Sum

Prove that:

$\frac{\sin A + 2 \sin 3A + \sin 5A}{\sin 3A + 2 \sin 5A + \sin 7A} = \frac{\sin 3A}{\sin 5A}$

#### Solution

Consider LHS:
$\frac{\sin A \sin 2A + \sin 3A \sin 6A}{\sin A \cos 2A + \sin 3A \cos 6A}$
Multiplying numerator and denominator by 2, we get
$= \frac{2\sin A \sin 2A + 2\sin 3A \sin 6A}{2\sin A \cos 2A + 2\sin 3A \cos 6A}$
$= \frac{\cos \left( A - 2A \right) - \cos \left( A + 2A \right) + \cos \left( 3A - 6A \right) - \cos \left( 3A + 6A \right)}{\sin \left( A + 2A \right) + \sin \left( A - 2A \right) + \sin \left( 3A + 6A \right) + \sin \left( 3A - 6A \right)}$
$= \frac{\cos\left( - A \right) - \cos 3A + \cos \left( - 3A \right) - \cos 9A}{\sin 3A \sin\left( - A \right) + \sin 9A + \sin \left( - 3A \right)}$
$= \frac{\cos A - \cos 3A + \cos 3A - \cos 9A}{\sin 3A - \sin A + \sin 9A - \sin 3A}$
$= \frac{\cos A - \cos 9A}{\sin 9A - \sin A}$
$= \frac{- 2\sin \left( \frac{A + 9A}{2} \right) \sin \left( \frac{A - 9A}{2} \right)}{2\cos \left( \frac{A + 9A}{2} \right) \sin \left( \frac{9A - A}{2} \right)}$
$= \frac{\sin5A\cos4A}{\sin 5A \cos \left( - 4A \right)}$
$= \tan 5A$
= 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 8.1 | Page 18