<s>[INST] 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!
 [/INST] 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 $x= -y= \pm \frac12$
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