Question

if A,B and c are interior angle of a triangle ABC then show that
sin (B+C) =cos
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

Answer:

I hope it is helpful.................