# Prove That: Sin 5 π 18 − Cos 4 π 9 = √ 3 Sin π 9 - Mathematics

Sum

Prove that:
$\sin\frac{5\pi}{18} - \cos\frac{4\pi}{9} = \sqrt{3} \sin\frac{\pi}{9}$

#### Solution

$LHS = \sin\left( \frac{5\pi}{18} \right) - \cos\frac{4\pi}{9}$
$= \sin\left( \frac{5\pi}{18} \right) - \cos\left( \frac{\pi}{2} - \frac{\pi}{18} \right)$
$= \sin\left( \frac{5\pi}{18} \right) - \sin\left( \frac{\pi}{18} \right)$
$= 2\sin\left( \frac{\frac{5\pi}{18} - \frac{\pi}{18}}{2} \right)\cos\left( \frac{\frac{5\pi}{18} + \frac{\pi}{18}}{2} \right) \left[ \because \sin A - \sin B = 2\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right) \right]$
$= 2\sin\left( \frac{\pi}{9} \right)\cos\frac{\pi}{6}$
$= 2\sin\left( \frac{\pi}{9} \right)\cos\frac{\pi}{6}$
$= 2 \times \frac{\sqrt{3}}{2}\sin\left( \frac{\pi}{9} \right)$
$= \sqrt{3}\sin\left( \frac{\pi}{9} \right) = RHS$
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 3.5 | Page 17