Question

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

Answer 1

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