Is $\arctan2$ irrational? Is $\tan^{-1}2$ an irrational number or a rational number? How to show that? 
Or generally how to show $\tan^{-1}3, \tan^{-1}4, \tan^{-1}5...$ is irrational or rational?
 A: Trasform this into $\cos x  = \frac{1}{\sqrt{5}}$. Now, you have an equation
$$e^{ix}+e^{-ix}=\frac{2}{\sqrt{5}}$$
or, with $z=e^{ix}$,
$$z^2-\frac{2}{\sqrt{5}}z+1=0$$
Now, above is an algebraic equation, so $z$, the solution of this equation, must be algebraic. By Lindemann-Weirstrass theorem, if $e^{ix}$ is algebraic, then $ix$ must be transcendental, except if $x=0$.
Of course, $\tan^{-1}1 = \frac{\pi}{4}$ is also transcendental (and therefore irrational). If you want to prove that it's an irrational multiple of $\pi$, you have to proceed a bit differently.

Consider an equation $z+z^{-1}=2a$, $|a|<1$, which has solutions
$$z=a\pm i\sqrt{1-a^2}$$
Now we require $z=e^{i\pi p/q}$. Raise this to the power of $q$:
$$e^{i\pi p}=\pm 1=(a\pm i\sqrt{1-a^2})^q$$
There must be such $q$, so that the right hand side is an integer ($\pm 1$). If you are given $a$, then you can just check that particular case. In general, you are basically looking for roots of unity in terms of their cartesian components. For example, you can set $a=\sqrt{n}/2$ and check for which $n$ this has a solution for $q$.
