Question

In $A,B,C$ are the interior angles of a $\Delta ABC$, then show that $\sin \left(\frac{B+C}{2}\right)=\cos \left(\frac{A}{2}\right)$

Answer 1

$\begin{array}{c}A+B+C=180\\ \frac{A+B+C}{2}=90\\ \frac{B+C}{2}=90-\frac{A}{2}\\ \sin \left(\frac{B+C}{2}\right)=\sin \left(90-\frac{A}{2}\right)\\ =\cos \frac{A}{2}=RHS\end{array}$