Question

Prove $\sin ({60}^0+\theta )-\sin ({60}^0-\theta )=\sin \theta .$

Answer 1

$\sin (60+\theta )-\sin (60-\theta )=\sin \theta$

$60+\theta =x,60-\theta =y$

$\sin x-\sin y$

$2\cos \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)$

$=2\cos \left(\frac{60+\theta +60-\theta }{2}\right)\sin \left(\frac{60+\theta -60+\theta }{2}\right)$

$=2\left(\frac{1}{2}\right)\sin \theta =\sin \theta$