Prove that sin A + B / 2 = cos c / 2. A, B, C are the corners of the triangle.

Since A, B, C are the angles of one triangle, so A + B + C = 180 degrees. So A + B = 180-C

sin ((A + B) / 2) = sin ((180-C) / 2) = sin (90-C / 2) = sin90cos (C / 2) -cos90sin (C / 2) = 1 * cos (C / 2) -0 * sin (C / 2) = cos (C / 2), as required.

Also, the problem could be solved even faster by applying the reduction formulas, from which it immediately follows that sin (90-C / 2) = cos (C / 2)