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

We know that, $$sin(x+y) = sinxcosy + cosxsiny$$ --------(i)

and $$sin(x-y) = sinxcosy-cosxsiny$$ ------------(ii)

Adding equations (i) and (ii), we get : 

$$sin(x+y) + sin(x-y) = 2sinxcosy$$ ------------(iii)

Let $$x + y = C$$ and $$x - y = D$$

=> $$x = \frac{C+D}{2}$$

and $$y = \frac{C-D}{2}$$

Substituting above values in equation (iii)

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

=> Ans - (D)