Question

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

If true or false:

Answer 1

Given, $A={60}^{\circ }$ and $B={30}^{\circ }$
Now, $\cos (A-B)=\cos A\cos B+\sin A\sin B$
L.H.S : $\cos (A-B)=\cos (60-30)=\cos 30=\frac{\sqrt{3}}{2}$
R.H.S: $\cos A\cos B+\sin A\sin B=\cos 60\cos 30+\sin 60\sin 30$
= $\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)$
= $\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}$
= $\frac{\sqrt{3}}{2}$