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