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

We know that $$2\sin A\sin B=\cos\left(A-B\right)-\cos\left(A+B\right)$$

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

$$=\cos\left(\frac{C+D-D+C}{2}\right)-\cos\left(\ \frac{C+D+D-C\ }{2}\right)$$

$$=\cos C-\cos D$$