Hey simon, just a word: don't delete your threads after you've made them, simply because somebody answered your question. It's better if they are left for others to read, as they may help them, too.
Now, onto your question:
You are aware of how to derive the following formula, right?
2sin(A)cos(B) = sin(A-B)+sin(A+B)
It's done by:
sin(A+B)=sinAcosB+cosAsinB
sin(A-B)=sinAcosB-cosAsinB
Add these:
sin(A+B)+sin(A-B) = 2sinAcosB
Ok, that's the easy bit. Now you should make a substitution:
Let A=(a+b)/2, B=(a-b)/2
You can see that A+B=a, A-B=b.
Thus we have:
sin(A+B)+sin(A-B) = 2sinAcosB
sin(a)+sin(b) = 2sin([a+b]/2)cos([a-b]/2)
That is to say: the two identities are equivalent in what they mean.
EDIT: lol found a typo a few days late