If $A, B$ and $C$ are interior angles of a triangle $ABC$,then show that $\sin \left(\frac{B+C}{2}\right) = \cos \frac{A}{2}$.

(N/A) We know that for a triangle $ABC$,the sum of interior angles is $180^{\circ}$.
$\angle A + \angle B + \angle C = 180^{\circ}$
$\angle B + \angle C = 180^{\circ} - \angle A$
Dividing both sides by $2$,we get:
$\frac{\angle B + \angle C}{2} = \frac{180^{\circ} - \angle A}{2} = 90^{\circ} - \frac{\angle A}{2}$
Taking the sine on both sides:
$\sin \left(\frac{B+C}{2}\right) = \sin \left(90^{\circ} - \frac{A}{2}\right)$
Using the trigonometric identity $\sin(90^{\circ} - \theta) = \cos \theta$,we have:
$\sin \left(\frac{B+C}{2}\right) = \cos \left(\frac{A}{2}\right)$
Hence,it is proved.