<s>[INST] Is there a simpler/better proof of this simple trigonometric property? The sine function has the following nice property : for any
$x,y$, we have $\sin(x)+\sin(y)=\sin(x+y)$ iff at least one of
$x,y,x+y$ is $0$ modulo $2\pi$.
I sketch below my current proof of it, which I find somewhat 
unsatisfying. Does anyone know a better proof ?
Let $\xi=\sin(x)$ and $\eta=\sin(y)$. Then 
$\sin(x+y)=\xi\cos(y)+\eta\cos(x)$, so 
$$
\begin{array}{lcl}
\sin^2(x+y)&=&\xi^2(1-\eta^2)+\eta^2(1-\xi^2)+2\xi\eta\cos(x)\cos(y) \\
&=& \xi^2+\eta^2-2\eta^2\xi^2+2\xi\eta\cos(x)\cos(y)
\end{array} \tag{1}$$
Whence
$$
\big(\sin^2(x+y)-(\xi^2+\eta^2-2\eta^2\xi^2)\big)^2=
4(\xi\eta)^2(1-\xi^2)(1-\eta^2) \tag{2}
$$ 
If $\sin(x)+\sin(y)=\sin(x+y)$, it follows then that $A(\xi,\eta)=0$
where 
$$
\begin{array}{lcl}
A(\xi,\eta)&=&\big((\xi+\eta)^2-(\xi^2+\eta^2-2\eta^2\xi^2)\big)^2-
4(\xi\eta)^2(1-\xi^2)(1-\eta^2) \\
&=& \big(2\xi\eta-2\eta^2\xi^2\big)^2-
4(\xi\eta)^2(1-\xi^2)(1-\eta^2) \\
&=& \big(2\xi\eta\big(1-\eta\xi\big)\big)^2-
4(\xi\eta)^2(1-\xi^2)(1-\eta^2) \\
&=& 4(\xi\eta)^2 \Bigg(\big(1-\eta\xi\big)^2-(1-\xi^2)(1-\eta^2) \Bigg) \\
&=& 4(\xi\eta)^2 \Bigg(\big(1-2\eta\xi+\eta^2\xi^2\big)-\big(1-\xi^2-\eta^2+\eta^2\xi^2\big) \Bigg) \\
&=& 4(\xi\eta)^2 \big(\xi-\eta\big)^2 \\
\end{array} \tag{3}$$
So one of $\xi,\eta,\xi-\eta$ must be zero. A little more case analysis then shows
that in the end one of $x,y,x+y$ must be
zero modulo $2\pi$ as wished. 
 [/INST] Here is a different derivation of Daniel Fischer's beautiful formula,
We have the well known,
$$\sin x+\sin y=2\sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)$$
And 
$$\sin (x+y)=2\sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x+y}{2}\right)$$
Now 
$$\cos \left(\frac{x-y}{2}\right)-\cos \left(\frac{x+y}{2}\right)= 
2\sin \left(\frac{x}{2}\right) \sin \left(\frac{y}{2}\right)$$
so
$$\sin x+\sin y-\sin (x+y)=4\sin \left(\frac{x+y}{2}\right)\sin \left(\frac{x}{2}\right) \sin \left(\frac{y}{2}\right)$$
</s>