<s>[INST] Why does $\sin(x) - \sin(y)=2 \cos(\frac{x+y}{2}) \sin(\frac{x-y}{2})$? Why does this equality hold?
$\sin x - \sin y = 2 \cos(\frac{x+y}{2}) \sin(\frac{x-y}{2})$.
My professor was saying that since 
(i) $\sin(A+B)=\sin A \cos B+ \sin B \cos A$ 
and 
(ii) $\sin(A-B) = \sin A \cos B - \sin B \cos A$
we just let $A=\frac{x+y}{2}$ and $B=\frac{x-y}{2}$. But I tried to write this out and could not figure it out. Any help would be appreciated 
 [/INST] Following your professor's advice, let $A=\frac{x+y}{2}$, $B=\frac{x-y}{2}$. Then $$x=A+B\\y=A-B$$So the LHS of your equation becomes $$\sin(A+B)-\sin(A-B)$$Now you just use the usual addition/subtraction trigonometric identities (i) and (ii) listed to evaluate this. It should give $2\cos A\sin B$ as required.
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