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Let $\sin^2(x\pi)$ be a function from $\mathbb{Q}$ to $\mathbb{R}$. Let $A$ be the image of this function. Is it true that $A\subset\mathbb{Q}$?

In basic trigonometry classes we learn that $\sin{\frac{\pi}{3}}=\frac{\sqrt{3}}{2} $ and that $\sin{\frac{\pi}{4}}=\frac{\sqrt{2}}{2}$. So in this case the property holds. Is it true $\forall x\in\mathbb{Q}$?

If not, is it possible to find all $x$ rational that do obey the rule and those $x$ rational that doesn't?

DaifM
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  • Be careful!! You do not calculate/evaluate the codomain of a function. In fact, it is part of the function itself.

    Moreover, I think that you are mixing up the image of a function and the codomain of a function.

     – Fede1 Oct 30 '19 at 18:17
  • Yes, I misunderstood that word. English is not my first language. Thank you! – DaifM Oct 30 '19 at 18:22
  • As Matt's answer points out, this is not true. However, something similar but weaker is true: $\sin(\pi x)$ is always algebraic (that is, the root of a polynomial). But in most cases, the polynomial in question is more complicated than it is for your two examples. – Micah Oct 30 '19 at 18:36

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If $x=1/8$ then this is false, as the resulting value is $\frac{2-\sqrt2}4$. I think this is the simplest value to see. This is not an isolated example; I'd say it fails to be rational for most values of $x$.

Matt Samuel
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  • I see. Then is it possible to find all $x$ rational that do obey the rule and those $x$ rational that doesn't? – DaifM Oct 30 '19 at 18:38
  • @Daif Probably, but I don't know the answer off the top of my head. – Matt Samuel Oct 30 '19 at 18:40
  • Only those few obvious ones work - use the double angle to transform the problem into the rationality of $\cos \pi q$ for rational $q$ and that is a well known result following from the Bernstein polynomials ($\cos n\pi q$ as a polynomial in $\cos \pi q$ and usual rational roots results) – Conrad Oct 30 '19 at 19:29