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Suppose $x,y$ are fixed real numbers. Does there always exist a real number $z$ such that $\sin(x+z)$ and $\sin(y+z)$ are rational numbers?

I know that $\sin(x) \in \mathbb{Q}$ implies that $\sin(x) \in \{ 0, \frac{1}{2}, -\frac{1}{2}, \pm1 \}$. But I don't know how to use this fact to show the problem.

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    $\begingroup$ $\sin(\arcsin{q})=q\in\mathbb{Q}$ for any $q\in \mathbb{Q}\cap[-1,1]$, so your last statement is incorrect. $\endgroup$ – Peter Woolfitt May 29 '15 at 2:14
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    $\begingroup$ I think you should rethink (or explain) what you've concluded from $\sin(x) \in \Bbb Q$. $\endgroup$ – pjs36 May 29 '15 at 2:15
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If $x=0$ and $\sin y$ is transcendental - say $\sin y = \pi/4$ and $\cos y>0$ - then $\sin(x+z)=\sin z$ being rational means $\cos z$ is algebraic, so $$\sin(y+z)=\sin y\cos z + \sqrt{1-\sin^2 y} \sin z$$ being rational means $\sin y$ is algebraic.

So there is no such $z$ in this case.

Indeed, for any $x$ there are at only countably many $y$ so that there is such a $z$. This is because $\sin(x-y)$ and $\cos(x-y)$ are necessarily algebraic.

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