Prove that, $\sin \left(\frac{2\pi }{3} + \theta \right) – sin\left(\frac{2\pi }{3} – \theta \right) + sin\theta$ = 0
$L\ldotp H\ldotp S\sin \left(\frac{2\pi }{3} + \theta \right) – sin\left(\frac{2\pi }{3} – \theta \right) + sin\theta$
= $2cos\left(\frac{\frac{2\pi }{3} + \theta + \frac{2\pi }{3} – \theta }{2}\right)sin\left(\frac{\frac{2\pi }{3} + \theta – \frac{2\pi }{3} + \theta }{2}\right) + sin\theta$
= $2cos\left(\frac{4\pi }{6}\right)sin\theta + sin\theta$
= $2cos\frac{2\pi }{3}sin\theta + sin\theta$
= $2cos\left(\pi – \frac{\pi }{3}\right)\sin \theta + sin\theta$
= – $2cos\frac{\pi }{3}sin\theta + sin\theta$
= – $2 \times \frac{1}{2} \times sin\theta + sin\theta$
= – $sin\theta + sin\theta$
= 0 = $R\ldotp H\ldotp S$ [Proved]