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. 
 A: We have $\DeclareMathOperator{\Ima}{Im}$
$$\begin{align}
\sin (x+y) - \sin x - \sin y &= \Ima \left(e^{i(x+y)} - e^{ix} - e^{iy}\right)\\
&= \Ima \left(e^{i(x+y)} - e^{ix} - e^{iy} + 1\right)\\
&= \Ima \left(\left(e^{ix}-1\right)\left(e^{iy}-1\right)\right)\\
&= \Ima \left(e^{i(x+y)/2}\left(e^{ix/2} - e^{-ix/2}\right)\left(e^{iy/2} - e^{-iy/2}\right)\right)\\
&= \Ima \left(e^{i(x+y)/2} \left(2i\sin \tfrac{x}{2}\right)\left(2i\sin\tfrac{y}{2}\right) \right)\\
&= - 4 \sin \frac{x+y}{2}\sin \frac{x}{2} \sin \frac{y}{2}.
\end{align}$$
That product is $0$ if and only if at least one of the factors is $0$.
A: Here is my version. Let $u=e^{ix}$, $v=e^{iy}$. By assumption $z=uv-u-v\in\mathbb{R}$.
This is equivalent to
$$
uv-u-v=\frac{1}{uv}-\frac{1}{u}-\frac{1}{v},
$$
and again,this is equivalent to
$$
(1-u)(1-v)=\left(1-\frac{1}{u}\right)\left(1-\frac{1}{v}\right)=
\frac{(1-u)(1-v)}{uv}
$$
or
$$
(1-u)(1-v)\left(1-\frac{1}{uv}\right)=0
$$
so at least one of $u$, $v$, $uv$ is equal to one.
A: Well here is (yet another) solution, this using only real variables.
By the sum formula we have
$$\sin x(1-\cos y)+\sin y(1-\cos x)=0$$
multiplying by $(1+\cos x)(1+\cos y)$ we get 
$$\sin x \sin y [\sin y(1+\cos x)+\sin x(1+\cos y)]=0$$ so 
either $\sin x=0$ or $\sin y=0$ or 
$$\sin y(1+\cos x)+\sin x(1+\cos y))=0$$
which is equivalent to 
$$\sin x  +\sin y +\sin(x+y) =0$$
so $$\sin(x+y) =0$$
Then it is easy to derive the result.
A: HINT:
For a proof in the real numbers use the fact that for a function with period $a$, $f(x+a\cdot n) = f(x)$ for $n \in \mathbb{Z}$. It's not easy but it avoids complex numbers.
A: 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
$$\sin x+\sin y-\sin (x+y)=4\sin \left(\frac{x+y}{2}\right)\sin \left(\frac{x}{2}\right) \sin \left(\frac{y}{2}\right)$$
