Question

Given $A={60}^0$ and $B={30}^0$, then $\cos (A+B)=\cos A\cos B-\sin A\sin B$.

If above statement is true enter $1$ else $0.$

Answer 1

$A={60}^{\circ }$ and $B={30}^{\circ }$
Now, $\cos (A+B)=\cos A\cos B-\sin A\sin B$
L.H.S. $\cos (A+B)=\cos ({60}^0+{30}^0)=\cos {90}^0=0$
R.H.S $\cos A\cos B-\sin A\sin B$
= $\cos {60}^0\cos {30}^0-\sin {60}^0\sin {30}^0$

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

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

= $0$
Hence, $L.H.S=R.H.S.$

$\cos (A+B)=\cos A\cos B-\sin A\sin B$ is true, when $A={60}^{\circ }$ and $B={30}^{\circ }$.