2024/021) Show that $\sin 3\theta = \sin \theta + 2 \sin \theta \cos 2\theta$ and use it to prove $\sin \frac{\pi}{9} + 2 \sin \frac{\pi}{9} \cos \frac{2\pi}{9} = \frac{\sqrt{3}}{2}$

LHS

$= \sin 3 \theta = \sin ( \theta + 2\theta) = \ sin \theta \cos 2 \theta + \cos \theta \sin 2\theta$ usng $\sin(A+B)$ formula 

$= \ sin \theta \cos 2 \theta + \cos \theta (2\sin \theta \cos \theta)$ using $\sin 2\theta$ formula

$= \ sin \theta (\cos 2 \theta + 2  \cos^2  \theta)$

$= \ sin \theta ( 2 \cos 2 \theta +  \cos 2  \theta + 1 )$ using formula for $\cos 2\theta$

$= \ sin \theta ( 2 \cos 2 \theta +   1 )$

$= 2 \ sin \theta  \cos 2 \theta +  \sin   \theta$ which is RHS

In the above putting $\theta = \frac{\pi}{3}$  we get

$\sin \frac{\pi}{9} +  2 \sin \frac{\pi}{9} \cos \frac{2\pi}{9} = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$