Question

If $\sin A\cos B=\frac{1}{\sqrt{2}}$ and $\sin B\cos A=-\frac{1}{2\sqrt{3}}$, then $\sin \left(A-B\right)=$

• A. $-\frac{2\sqrt{3}+\sqrt{2}}{2\sqrt{6}}$
• B. $\frac{2\sqrt{3}-\sqrt{2}}{2\sqrt{6}}$
• C. $\frac{2\sqrt{3}+\sqrt{2}}{2\sqrt{6}}$
• D. $-\frac{2\sqrt{3}-\sqrt{2}}{2\sqrt{6}}$

Answer 1

The correct option is C $\frac{2\sqrt{3}+\sqrt{2}}{2\sqrt{6}}$

We have,

$\sin A\cos B=\frac{1}{\sqrt{2}}$

And $\sin B\cos A=-\frac{1}{2\sqrt{3}}$

We know that

$\sin (A-B)=\sin A\cos B-\sin B\cos A$

So,

$\sin (A-B)=\frac{1}{\sqrt{2}}-(-\frac{1}{2\sqrt{3}})$

$\sin (A-B)=\frac{1}{\sqrt{2}}+\frac{1}{2\sqrt{3}}$

$\sin (A-B)=\frac{2\sqrt{3}+\sqrt{2}}{2\sqrt{2}\sqrt{3}}$

$\sin (A-B)=\frac{2\sqrt{3}+\sqrt{2}}{2\sqrt{6}}$

Hence, this is the answer.