Question

If $A,B,C$ are the angle of a $△ABC$, show that $\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}$.

Answer 1

According to the problem

In $\Delta ABC$

Sum of angles of a triangle is $180$

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

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

Hence,

Multiplying both side by $\frac{1}{2}$

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

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

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

Now taking L.H.S.

$\sin \frac{B+C}{2}$

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

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

$\cos \frac{A}{2}$=R.H.S.