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