<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 notation, let $A=\dfrac{x+y}{2}$ and $B=\dfrac{x-y}{2}$.
Note that $A+B=x$ and $A-B=y$.
Now, $\sin x=\sin(A+B)=\sin A\cos B+\cos A\sin B$ and $\sin y=\sin(A-B)=\sin A\cos B - \cos A\sin B$ from your professor's advice.
To get the LHS, $\sin x-\sin y = 2\cos A\sin B$. And that's it. Replace $A,B$ in terms of $x$ and $y$.
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