Question

Prove that $\sin (A+B)\ne \sin A+\sin B$

Answer 1

Assuming $\angle A={60}^{\circ }$ and $\angle B={30}^{\circ }$

L.H.S$=\sin (A+B)=\sin ({60}^{\circ }+{30}^{\circ })$

$=\sin \left({90}^{\circ }\right)=1$

R.H.S$=\sin A+\sin B=\sin {60}^{\circ }+\sin {30}^{\circ }$

$=\frac{\sqrt{3}}{2}+\frac{1}{2}$

$=\frac{\sqrt{3}+1}{2}$

Thus, L.H.S$\ne$R.H.S

Thus,$\sin (A+B)\ne \sin A+\sin B$