If A, B, C are the interior angles of a ∆ABC, show that :

Solution:

(i) We have to prove: $\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}$

Since we know that in triangle $A B C$

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

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

Dividing by 2 on both sides, we get

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

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

$\Rightarrow \sin \frac{B+C}{2}=\cos \frac{A}{2}$

Proved

(ii) We have to prove: $\cos \left(\frac{B+C}{2}\right)=\sin \frac{A}{2}$

Since we know that in triangle $A B C$

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

 

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

Dividing by 2 on both sides, we get

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

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

$\Rightarrow \cos \frac{B+C}{2}=\sin \frac{A}{2}$

Proved