Find all $(x,y)$ such that $\sin x+\sin y=\sin(x+y)$ and $|x|+|y|=1$ 
Find all $(x,y)$ such that $$\sin x+\sin y=\sin(x+y)$$ and $|x|+|y|=1$  

My work:
We have, $\sin x+\sin y=\sin x\cos y+\cos x \sin y$
By comparing both sides, we have,$\cos y=1,\cos x=1$, but certainly that is not true, as $|x|+|y|=1$.
So,$\sin x(1-\cos y)=\sin y(\cos x-1)$
Now, I am stuck. I cannot break this loop to proceed further. Please help!
 A: Using Prosthaphaeresis Formulas, 
$$\sin x+\sin y=2\sin\frac{x+y}2\cos\frac{x-y}2$$
$$\sin(x+y)=2\sin\frac{x+y}2\cos\frac{x+y}2$$
So, $$\sin x+\sin y-\sin(x+y)=2\sin\frac{x+y}2\left(\cos\frac{x-y}2-\cos\frac{x+y}2\right)$$
$$=2\sin\frac{x+y}2\cdot2\sin\frac x2\sin\frac y2 (\text{ again Prosthaphaeresis })$$
Now we know $\displaystyle \sin A=0\implies A=n\pi$ where $n$ is an integer
As $|x|+|y|=1, |x|=1-|y|\le 1$
Case $1a)$  If $\displaystyle \sin\frac x2=0\iff x=2n\pi\implies|2n\pi|\le1\implies|n|<1\implies n=0\implies x=0$
$\displaystyle|y|=1\iff y=\pm1$
Case $1b)$ Similarly,  if $\displaystyle \sin\frac y2=0$
Case $2)$ If  $\displaystyle \sin\frac{x+y}2=0, x+y=2n\pi$
Using this, $\displaystyle |x+y|\le |x|+|y|=1\implies |2n\pi|\le1\implies |n|<1\implies n=0$
$\displaystyle\implies x+y=0$ and we have $\displaystyle|x|+|y|=1$
Can you take it from here?
A: Considering $\sin(x)+\sin(y)=\sin(x+y)$, there will be solutions when $x=0$ or $y=0$ since $\sin(0)=0$.
From $|x|+|y|=1$ there are obvious solutions at $(0,1), (0,-1), (1,0), (-1,0)$.  
You should also consider four further possibilities:


*

*$x \in (0,1), y=1-x \gt 0$

*$x \in (0,1), y=x-1 \lt 0$ 

*$x \in (-1,0), y=1+x \ge 0$

*$x \in (-1,0), y=-x-1 \le 0$


You can show that the first and fourth give no solutions. The second and third give the additional solutions $\left(\frac12,-\frac12\right)$ and $\left(-\frac12,\frac12\right)$, which are also intuitively obvious since $\sin(-x)=-\sin(x)$.
A: from your identity $\sin x(1-\cos y)=\sin y(\cos x-1)$ we have 
$$ 2\sin\frac{x}2\cos\frac{x}2 2\sin^2\frac{y}2=-2\sin\frac{y}2\cos\frac{y}2 2\sin^2\frac{x}2
$$
so as long as $\sin\frac{x}2 \ne 0, \sin\frac{y}2 \ne 0$ this gives
$$ \cot \frac{x}2 = - \cot \frac{y}2 
$$ so $$\frac{x}2 = - \frac{y}2 + n\pi$$
clearly $n=0$ if $|x|+|y|=1$ hence $x= -y= \pm \frac12$
