| If \[\sin \,(A-B)=\sin A\,\cos B-\cos A\,\sin B,\] then \[\sin 15{}^\circ \] will be [SSC (CPO) 2015] |
A) \[\frac{\sqrt{3}-1}{\sqrt{2}}\]
B) \[\frac{\sqrt{3}}{2\sqrt{2}}\]
C) \[\frac{\sqrt{3}+1}{2\sqrt{2}}\]
D) \[\frac{\sqrt{3}-1}{2\sqrt{2}}\]
Correct Answer: D
Solution :
| \[\sin 15{}^\circ =\sin \,(45{}^\circ -30{}^\circ )\] |
| \[=\sin 45{}^\circ \cdot \cos 30{}^\circ -\cos 45{}^\circ \cdot \sin 30{}^\circ \] |
| \[=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\times \frac{1}{2}\] |
| \[=\frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2\sqrt{2}}=\frac{\sqrt{3}-1}{2\sqrt{2}}\] |
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