# Can both $x$ and $\sin(x)$ be rational at the same time?

Can both $x$ and $\sin(x)$ be rational at the same time?

Except, of course, trivial $x=0$ case ($\sin0=0$); $x$ is measured in radians.

The question turned out to be more complicated than it seemed to me at the first sight.

All I came up with, that posed question is equivalent to the question of chord and corresponding arc being together of rational length.

• This might be of interest to you: mathworld.wolfram.com/NivensTheorem.html – Gahawar Jan 8 '15 at 13:33
• @Gahawar That doesn't really help, since it only relates $\sin(x)$ with $x/\pi$. – Arthur Jan 8 '15 at 13:49
• @Gahawar interesting – nicks Jan 8 '15 at 13:51
• @Arthur I believe the Wikipedia article is more descriptive: en.wikipedia.org/wiki/Niven%27s_theorem – Gahawar Jan 8 '15 at 13:53
• @Gahawar if you claim it to be the answer - sadly it'd not, since it concerns only angles measured in degrees, not in radians. – nicks Jan 8 '15 at 13:56

## 2 Answers

If $\sin x$ is rational then $$e^{ix} = \pm\sqrt{1-\sin^2 x}+i\sin x$$ is an algebraic number over $\mathbb{Q}$ of degree at most $4$. However, if $x\in\mathbb{Q}^+$ then $e^{ix}$ is a trascendental number, since $e^{i}$ is a trascendental number.

• so, if $sin$ is algebraic, then $cos$ is algebraic too and algebraic + algebraic $i$ = algebraic ? – nicks Jan 8 '15 at 15:25
• @NikaGamkrelidze: yep. On the other hand, $e^{ix}$ cannot be algebraic because that would imply that $e^i$ is algebraic, contradiction. – Jack D'Aurizio Jan 8 '15 at 15:46

The Lindemann–Weierstrass theorem shows that if $\alpha$ is a non-zero algebraic number, then $e^{\alpha}$ is transcendental.

If $\alpha=ix$ when $x$ is rational, then $\alpha$ is algebraic, so $e^{ix}=\cos x + i\sin x$ is transcendental.

If $z$ is transcendental, then so is $w=z-\frac{1}{z}$. Otherwise then $z^2-w z -1 =0$ and thus $z$ is algebraic.

But if $z=e^{ix}$, with $x$ rational, then, $z$ is trascendental and hence so is $z-\frac{1}{z} = 2i\sin x$. So if $2i\sin x$ is transcendental, then so is $\sin x$.