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Suppose $p$ and $q$ are relatively prime positive integers, and that $x$ is a positive rational number. Given that $x \in [-\frac{1}{2}, \frac{1}{2}]$ and $$q\sin{\pi x} = p$$ how can we compute $p, q$ and $x$?

I'm not really sure how to approach this. Of course we must have $q > p$; however to find $x$ it would seem that one must pick a standard reference angle, for example by setting $x = 0.1$. This leads to surds and the like, which won't work. The constraints in the problem also prevent us from selecting something like $x = 1/2, p = 1, q = 2$.

Ayesha
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  • I think the key is that $x$ must be rational as well. Though I'm still not sure why there is only one solution. – Fengyang Wang Apr 26 '14 at 02:45
  • Note that if $p$ and $q$ are assumed to be positive and $q\sin{\pi x} = p$, then it follows that $\sin{\pi x}$ is also positive. Hence, $x\in (0,\frac12]$. – David H Apr 26 '14 at 02:50
  • Hint: Use Niven's Theorem as mentioned. Check http://math.stackexchange.com/questions/87756/when-is-sinx-rational – Macavity Apr 26 '14 at 02:59

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You will find a complete answer in the Wikipedia article on Niven's Theorem.

The only rational $x$ between $0$ and $1/2$ such that $\sin(\pi x)$ is rational are $x=0$, $1/6$, and $1/2$. You can deduce the answer for negative $x$.

André Nicolas
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