Question : If $\sin (A-B)=\sin A \cos B–\cos A\sin B$, then $\sin 15°$ will be:
Option 1: $\frac{\sqrt{3}+1}{2\sqrt{2}}$
Option 2: $\frac{\sqrt{3}}{2\sqrt{2}}$
Option 3: $\frac{\sqrt{3}–1}{–\sqrt{2}}$
Option 4: $\frac{\sqrt{3}–1}{2\sqrt{2}}$
Correct Answer: $\frac{\sqrt{3}–1}{2\sqrt{2}}$
Solution :
Given: $\sin (A–B)=\sin A\cos B–\cos A\sin B$
Let $A=45°$ and $B=30°$ respectively.
$\sin (45°–30°)=\sin 45°\cos 30°–\cos 45°\sin 30°$
⇒ $\sin 15°=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}–\frac{1}{\sqrt{2}}\times \frac{1}{2}$
$\therefore \sin 15°=\frac{\sqrt{3}–1}{2\sqrt{2}}$
Hence, the correct answer is $\frac{\sqrt{3}–1}{2\sqrt{2}}$.
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