Question

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

Answer 1

By angle sum property, $\angle A+\angle B+\angle C=\pi$

$\Rightarrow \angle B+\angle C=\pi -\angle A$

$\Rightarrow \frac{\angle B+\angle C}{2}=\frac{\pi }{2}-\frac{\angle A}{2}$

$\Rightarrow \sin \left(\frac{\angle B+\angle C}{2}\right)=\sin (\frac{\pi }{2}-\frac{\angle A}{2})$

$\therefore \sin \left(\frac{\angle B+\angle C}{2}\right)=\cos \frac{\angle A}{2}$

Hence proved.