Proof that sin a + sin b= 2sin [(a+b)/2]cos [(a-b)/2]i need it very urgent please

Imagine that a is the sum of 2 angles, u and v, so u+v=a, and b is the difference of the same 2 angle, so b=u-v

If we'll add a+b=(u+v)+(u-v), we'll reduce the similar terms v and (-v) and the result will be a+b=2u, so u=(a+b)/2.

If we'll make the difference between the 2 terms 

a-b=(u+v)-(u-v) and we'll reduce the similar terms, the result will be  a-b=2v, so v=(a-b)/2

Let's substitute  what we have found:

sin (u+v) + sin (u-v)=2 sin u*cos v

In the right side of the equal, we'll open the paranthesis in this way:

sin (u+v)=sin u*cos v+ sin v*cos u

sin (u-v)=sin u*cos v- sin v*cos u

If we'll substitute the sin (u+v) + sin (u-v) with what we've found out, the result will be:

sinu*cosv+sin v*cosu+sin u*cosv-sinv*cosu=2 sin u*cos v

We'll reduce the similar terms (-sinv*cosu) with (sinv*cosu) and the result will be:

sinu*cosv+sinu*cosv=2 sin u*cos v

2 sin u*cos v=2 sin u*cos v

q.e.d!