Question

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

Answer 1

Given, $A={30}^{\circ }$ and $B={60}^{\circ }$

$\Rightarrow \sin {30}^{\circ }=\cos {60}^{\circ }=\frac{1}{2},\cos {30}^{\circ }=\sin {60}^{\circ }=\frac{\sqrt{3}}{2}$

To prove:

(i) $\sin (A+B)=\sin A\cos B+\cos A\sin B$

$\sin ({30}^{\circ }+{60}^{\circ })=\sin {30}^{\circ }\cos {60}^{\circ }+\cos {30}^{\circ }\sin {60}^{\circ }$

$\sin \left({90}^{\circ }\right)=(\frac{1}{2}\times \frac{1}{2})+(\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2})$

$1=\frac{1}{4}+\frac{3}{4}$

$1=\frac{4}{4}$

$1=1$

hence verified

(ii) $\cos (A+B)=\cos A\cos B-\sin A\sin B$

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

$\cos \left({90}^{\circ }\right)=(\frac{\sqrt{3}}{2}\times \frac{1}{2})-(\frac{1}{2}\times \frac{\sqrt{3}}{2})$

$0=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}$

$0=0$

hence verified