Prove : sin [(A+B)/2] = cos C/2 Please explain thank you?

A, B and C are the three interior angles of a triangle.
A + B + C = 180° (sum of int.∠s of Δ)

A + B = 180° - C
(A + B)/2 = (180° - C)/2
(A + B)/2 = [90° - (C/2)]
sin[(A + B)/2] = sin[90° - (C/2)]

But sin[90° - (C/2)] = cos(C/2)
Then, sin[(A + B)/2] = cos(C/2)

I assume these are the angles of a triangle

A + B + C = 180

A + B = 180 - C transposing C

(A + B)/2 = (180 - C)/2 .............dividing both sides by 2

sin[ (A + B)/2] = sin[ ( 180 - C)/2] taking sines on both sides

= sin[ 90 - C/2)

= sin((90)cos(C/2) - cos(90)sin(C/2).............expanding rhs

= (1) cos(C/2) - (0)sin(C/2)

= cos(C/2)

You have 3 variables and 1 equation. The variables don't have an established relationship, so there's nothing to prove.