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

Given:

\( A, B, C \), are the interior angles of a triangle \( A B C \).

To do:

We have to show that \( \sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2} \).

Solution:  

We know that,

$sin\ (90^{\circ}- \theta) = cos\ \theta$

Sum of the angles in a triangle is $180^{\circ}$.

This implies,

$\angle A+\angle B+\angle C=180^{\circ}$

$\Rightarrow \frac{\angle A+\angle B+\angle C}{2}=\frac{180^{\circ}}{2}$

$\Rightarrow \frac{\angle A}{2}+ \frac{\angle B}{2}+ \frac{\angle C}{2}=90^{\circ}$

Therefore,

$\sin \left(\frac{B+C}{2}\right)=\sin (\frac{B}{2}+\frac{C}{2})$

$=\sin (90^{\circ}-\frac{A}{2})$

$=\cos \frac{A}{2}$

Hence proved.