Question

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

Answer 1

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

Then $A+B={60}^o+{30}^o={90}^o$.

Now $\cos (A+B)=\cos {90}^o=0$.

Now,

$\cos {60}^o.\cos {30}^o-\sin {60}^o.\sin {30}^o$

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

$=0$.

So,

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