Question

If A, B and C are interior angles of a triangle ABC, then show that
sin [ B + C / 2 ] = cos A/2.

Answer 1

 Sin[(B+C)/2]
Since A+B+C=180 for interior angles of triangle ABC.
then B+C=180-A.

NOW   Sin [(180-A)/2]
              =Sin[90-(A/2)]                      since Sin(90-A)=CosA
               =Cos(A/2)

Answer 2

Please refer to the attachments below. The answer is given in parts.