2cos[(C+D)/2]sin[(C-D)/2] is equal to

Expression : 2cos[(C+D)/2]sin[(C-D)/2]

Using the formula : $$sin x . cos y = \frac{1}{2} [sin (x + y) + sin (x - y)]$$ --------------(i)

Substituting $$(x + y) = C$$ and $$(x - y) = D$$

=> $$x = \frac{C + D}{2}$$ and $$y = \frac{C - D}{2}$$ in equation (i),

=> $$sin (\frac{C + D}{2}) . cos (\frac{C - D}{2}) = \frac{1}{2} [sin C + sin D]$$

=> $$2 . cos (\frac{C - D}{2}) . sin (\frac{C + D}{2}) = sin C + sin D$$

Replacing D by -D, we get :

=> $$2 . cos(\frac{C - (-D)}{2}) . sin(\frac{C + (-D)}{2}) = sin C + sin(-D)$$

=> $$2 . cos(\frac{C + D}{2}) . sin(\frac{C - D}{2}) = sin C - sin D$$