Question

If $A={60}^{\circ },B={30}^{\circ }$ then verify that $\cos (A+B)=\cos A.\cos B-\sin A.\sin B$

Answer 1

$LHS=\cos (A+B)$
$=\cos ({60}^{\circ }+{30}^{\circ })$
$=\cos {90}^{\circ }$
$=0$
$RHS=\cos A.\cos B-\sin A.\sin B$
$=cos{60}^{\circ }.\cos {30}^{\circ }-\sin {60}^{\circ }.\sin {30}^{\circ }$
$=(\frac{1}{2}\times \frac{\sqrt{3}}{2})-(\frac{\sqrt{3}}{2}\times \frac{1}{2})$
$=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}$
$=0$
$LHS=RHS$.