Question

Prove that $\sin {60}^{\circ }.\cos {30}^{\circ }-\cos {60}^{\circ }.\sin {30}^{\circ }=\sin {30}^{\circ }$

Answer 1

Apply below formula to proof

$\sin A\cos B-\cos A\sin B=\sin (A-B)\sin {60}^{\circ }\cos {30}^{\circ }-\cos {60}^{\circ }\sin {30}^{\circ }=\sin {(60-30)}^{\circ }=\sin {30}^{\circ }$

Proved