Question
If $A = 30^\circ $ and $B = 60^\circ ,$ verify that$: \cos (A + B) = \cos A \cos B - \sin A \sin B$
✓
Answer
$A=30^{\circ}$ and $B=60^{\circ}$
$ \text { L.H.S. }$
$ =\cos (A+B)$
$ =\cos \left(30^{\circ}+60^{\circ}\right)$
$ =\cos 90^{\circ}$
$ =0$
$ \text{R.H.S.}$
$=\cos A \cos B-\sin A \sin B$
$=\cos 30^{\circ} \times \cos 60^{\circ}-\sin 30^{\circ} \times \sin 60^{\circ}$
$=\frac{\sqrt{3}}{2} \times \frac{1}{2}-\frac{1}{2} \times \frac{\sqrt{3}}{2}$
$=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}$
$=0$
$\Rightarrow \cos (A+B)=\cos A \cos B-\sin A \sin B$
$ \text { L.H.S. }$
$ =\cos (A+B)$
$ =\cos \left(30^{\circ}+60^{\circ}\right)$
$ =\cos 90^{\circ}$
$ =0$
$ \text{R.H.S.}$
$=\cos A \cos B-\sin A \sin B$
$=\cos 30^{\circ} \times \cos 60^{\circ}-\sin 30^{\circ} \times \sin 60^{\circ}$
$=\frac{\sqrt{3}}{2} \times \frac{1}{2}-\frac{1}{2} \times \frac{\sqrt{3}}{2}$
$=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}$
$=0$
$\Rightarrow \cos (A+B)=\cos A \cos B-\sin A \sin B$
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