What condition that if imposed on $\alpha$ imply that $\cos^{-1} \alpha$ is a rational multiple of $\pi$? It is well known that if $x$ is a rational multiple of $\pi$ then $\cos x$, $\sin x$, etc, are algebraic numbers. What is known about the inverse problem?
That is, is there a set of conditions that if imposed on $\alpha$ imply that $\cos^{-1} \alpha$ is a rational multiple of $\pi$?
Another way of putting it is, given the complex number $z=a+ib$, is there a way of deciding whether there exists some $n$ such that $z^n=1$?
EDIT: 'factor' -> 'multiple'
 A: "Given the complex number $z=a+bi$, is there a way of deciding whether there exists some [positive integer] $n$ such that $z^n=1$?" 
That would depend on how $z$ is given. If you are given polynomials $f,g$ with integer coefficients and you are told that $a$ is the real root of $f$ between (say) $.5$ and $.7$ and $b$ is the real root of $g$ between (say) $.7$ and $.9$, then from the degrees of $f,g$ you can derive an upper bound for $n$, and you can calculate $z^n$, exactly, for $n$ up to that upper bound, and you can determine whether you ever get $1$. 
If you are given $a,b$ as decimal numbers, rounded to (say) $17$ places, then, provided $a^2+b^2=1$ to $17$ places, it is guaranteed that there is a positive integer $n$ such that $z^n=1$, to $17$ places. 
If you have a black box that gives you any finite number of decimal places of $a$ that you ask for, and if $b=\pm\sqrt{1-a^2}$, then I don't think there's any way you can ever know for certain whether there's a positive integer $n$ such that $z^n=1$, since both roots and nonroots of unity are dense on the unit circle. 
A: Consider the case of $\theta = \arcsin(1/3)$   We have $\sin(\theta) = 1/3$ and $\cos(\theta) = 2\sqrt{2}/3;$ both of these numbers is algebraic.  However,
$\theta/\pi$ has a decimal expansion with no repetitions after a couple of thousand digits.  This is not definitive, but it seems likely to me it's not rational.  
A: 
Another way of putting it is, given the complex number $z=a+ib$, is
  there a way of deciding whether there exists some n such that $z^{n}=1$?

Calculate the minimal monic polynomial $p(x)\in\mathbb{Q}[x]$ for which $z$ is a root. That is, $p(x)$ is the lowest power possible for which $p(z)=0$ and the coefficients of $p(x)$ are all rational, and the coefficient of the highest powered term is 1. Then $n$ exists iff $p(x)$ exists and is cyclotomic.
This follows from the fact that $$z^n-1=\Pi_{d|n}\Phi_d(z)$$ where $\Phi_d(z)$ is the $d$th cyclotomic polynomial, and one other unique factorization theorem.
Comment: In practice, when I have applied this theorem to a given $z$ in quantum computing problems, $p(x)$ is readily calculated. In all cases of interest, $p(x)$ contained a fractional coefficient, which immediately proved the corresponding angle was an irrational multiple of $\pi$, since all cyclotomic polynomials have integer coefficients.
Edit: Applying this to $\theta=\arcsin(1/3)$, we find $z=e^{i\theta}=2\sqrt{2}/3+i/3$. The corresponding minimal monic polynomial in $\mathbb{Q}[x]$ is $$p(x)=x^4-14/9x^2+1.$$ Because of the $-14/9$ fractional coefficient, we can immediately conclude $p(x)$ is not cyclotomic, there is no $n\in\mathbb{N}:z^n=e^{in\theta}=1$, and thus $\theta$ is an irrational multiple of $\pi$.
