If A, B, C are the angles of a triangle...?

prove that cos [(B-C) / 2] - sin (A/2) = 2 sin (B/2) sin (C/2)

A+B+C = pi

so A/2 = pi-(B+C)/2

sin (A/2) = sin (pi-(B+C)/2) = cos (B+C)/2

so
cos [(B-C) / 2] - sin (A/2) =

= cos [(B-C) / 2] - cos [(B+C)/2] = 2 sin (B/2) sin (C/2)

using COS A - COS B = 2 cos sin (A+B)/2 sin (A-B)/2