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Can we prove that for all $x$ in $(0,2\pi)$ $\sin(x)$ is an algebraic number?

I have seen people express various values of $\sin(x)$ like $\sin(3)$ and $\sin(30)$ using radicals so I suspect that all values of $\sin(x)$ must be algebraic. Is that correct? Can we prove it?

Ömer
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Adam
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    No. The set of algebraic numbers is countable, but sin(x) spans [-1,1], which is uncountable. – Espen Nielsen Oct 07 '13 at 16:14
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    I'd actually be interested in the question of which values of $\sin$ are algebraic. I know all values of the form $\sin(2\pi r)$ where $r$ is a rational number are. – Timotej Oct 07 '13 at 16:15
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    All values $\sin(x)$ are algebraic, where $x$ is an integer number of degrees. Maybe that's what Adam asks. Of course Timotej's answer is more general than this. – GEdgar Oct 07 '13 at 16:21
  • Yep, I was mostly thinking about integer degrees, when I asked the question, but the answers pretty much hit the spot for me anyway. – Adam Oct 12 '13 at 18:02

2 Answers2

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Quick answer: no. Consider that $\sin$ is continuous over the range $[-1,1]$, and thus can take on values such as $e^{-1}, \frac 2\pi$, etc.

abiessu
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  • From here, we could also argue that any nonconstant polynomial has nonalgebraic values. I could also dare to say that any function that is continuous somewhere takes nonalgebraic values. – chubakueno Oct 07 '13 at 16:19
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    @chubakueno: that is correct, and the answer is that they will. The missing piece is that all non-constant polynomials having algebraic coefficients will have algebraic or "Gaussian algebraic" zeroes. – abiessu Oct 07 '13 at 16:21
  • @chubakueno: I would not be quite so bold with "continuous somewhere" functions, for instance $1\over [x]$... – abiessu Oct 07 '13 at 17:52
  • True. A function that is continuous and not constant in an interval $[a,b]$ where $a<b$ should do the work, then. – chubakueno Oct 07 '13 at 20:19
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    @chubakueno: that follows from the fact that continuous images of connected sets are connected – Kris Oct 07 '13 at 21:06
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Let $\frac{m}{n}\in\mathbb{Q}$, then $\sin\left(\frac{m}{n}\pi\right)=\sin\left(\frac{m}{n}180^\circ\right)$ is always algebraic: put $$ \alpha=e^{\frac{i\pi}{n}}=\cos\frac{\pi}{n}+i\sin\frac{\pi}{n}. $$ Then $\alpha^n+1=0$, i.e. $\alpha$ is an algebraic number (it is a root of the polynomial $X^n+1$) and hence both $$ \cos\frac{m\pi}{n}=\frac{\alpha^m+\alpha^{-m}}{2} \qquad\text{and}\qquad \sin\frac{m\pi}{n}=\frac{\alpha^m-\alpha^{-m}}{2i}, $$ are algebraic numbers. This shows that for a countable number of transcendental values of $x$, the value of $\sin(x)$ is an algebraic number.

Conversely, it can be shown that if $x$ is an algebraic non-zero number, $\sin(x)$ is transcendental! This follows from the famous Lindemann-Weierstrass Theorem.

Nicky Hekster
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    A corollary to this is that there must be an uncountable set of transcendental numbers $X$ such that $x\in X\implies \sin x$ is transcendental... – abiessu Dec 07 '14 at 02:58