Question:
If $A, B$ and $C$ are interior angles of a triangle $A B C$, then $\sin \left(\frac{B+C}{2}\right)=$
(a) $\sin \frac{A}{2}$
(b) $\cos \frac{A}{2}$
(c) $-\sin \frac{A}{2}$
(d) $-\cos \frac{A}{2}$
Solution:
We know that in triangle $A B C$
$A+B+C=180^{\circ}$
$\Rightarrow B+C=180^{\circ}-A$
$\Rightarrow \frac{B+C}{2}=\frac{90^{\circ}}{2}-\frac{A}{2}$
$\Rightarrow \sin \left(\frac{B+C}{2}\right)=\sin \left(90^{\circ}-\frac{A}{2}\right)$
Since $\sin \left(90^{\circ}-A\right)=\cos A$
So
$\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}$
Hence the correct option is $(b)$