Question

If A,B,C are the interior angles of a $\Delta$ ABC, show that:

(i) sin $\frac{B+C}{2}=cos\frac{A}{2}$

(ii) cos $\frac{B+C}{2}=sin\frac{A}{2}$

Answer 1

A+B+C=180

So,

B + C=180-A

$\frac{B+C}{2}$ = 90 - $\frac{A}{2}$ ( Dividing the L.H.S and R.H.S by 2)

sin($\frac{B+C}{2}$) = sin(90 - $\frac{A}{2}$) ( Substituting $\frac{B+C}{2}$ value)

= cos($\frac{A}{2}$)


Similarly for second question

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

Hence Proved