Question

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

Answer 1

We know that sum of all the angles of triangle is ${180}^{\circ }$

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

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

$\Rightarrow \sin \left(\frac{B+C}{2}\right)=\sin ({90}^{\circ }-\frac{A}{2})$

We know that, $\sin ({90}^{\circ }-\theta )=\cos \theta$

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

Hence, proved.