Question
Prove that: $\sin 60^{\circ} \cdot \cos 30^{\circ}-\sin 60^{\circ} \cdot \sin 30^{\circ}=\frac{1}{2}$
✓
Answer
$ \text { L.H.S. }=\sin 60^{\circ} \cdot \cos 30^{\circ}-\cos 60^{\circ} \cdot \sin 30^{\circ}$
$ =\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}-\frac{1}{2} \times \frac{1}{2}$
$ =\frac{3}{4}-\frac{1}{4}$
$ =\frac{2}{4}$
$ =\frac{1}{2}$
$ =\text { R.H.S. }$
$ =\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}-\frac{1}{2} \times \frac{1}{2}$
$ =\frac{3}{4}-\frac{1}{4}$
$ =\frac{2}{4}$
$ =\frac{1}{2}$
$ =\text { R.H.S. }$
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