Question
If $A = B = 45^\circ ,$ verify that $\cos (A - B) = cosA.\cos B + \sin A.\sin B$
✓
Answer
$A=B=45^{\circ}$
$ \text { L.H.S. }$
$ =\cos (A-B)$
$ =\cos \left(45^{\circ}-45^{\circ}\right)$
$ =\cos 0^{\circ}$
$ =0$
$\text{R.H.S.}$
$=\cos A \cos B+\sin A \sin B$
$=\cos 45^{\circ} \times \cos 45^{\circ}+\sin 45^{\circ} \times \sin 45^{\circ}$
$=\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}$
$=\frac{1}{2}+\frac{1}{2}$
$=1$
$\Rightarrow \sin (A-B)=\cos A \cos B+\sin A \sin B$.
$ \text { L.H.S. }$
$ =\cos (A-B)$
$ =\cos \left(45^{\circ}-45^{\circ}\right)$
$ =\cos 0^{\circ}$
$ =0$
$\text{R.H.S.}$
$=\cos A \cos B+\sin A \sin B$
$=\cos 45^{\circ} \times \cos 45^{\circ}+\sin 45^{\circ} \times \sin 45^{\circ}$
$=\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}$
$=\frac{1}{2}+\frac{1}{2}$
$=1$
$\Rightarrow \sin (A-B)=\cos A \cos B+\sin A \sin B$.
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