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'
On
"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.
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.