Quantitative Aptitude Ques 173

Question: If $ \sin ,(A-B)=\sin A,\cos B-\cos A,\sin B, $ then $ \sin 15{}^\circ $ will be

Options:

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}} $

Show Answer

Answer:

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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