Exact value of $\arctan(2)$ The following problem has been on my mind for a while.
Lots of exact values of the arctangent function are known, such as
$$\arctan 0=0$$
$$\arctan 1=\frac{\pi}{4}$$
$$\arctan \frac{1}{\sqrt 3}=\frac{\pi}{6}$$
However, I can't seem to find an exact value of
$$\arctan 2$$
How can I find one? Is it possible?
NOTE: By exact, I mean that I am looking for an answer in the form
$$\frac{p\pi}{q}$$
with $p,q\in \mathbb Z$.
 A: If $\arctan(2)$ were a rational multiple of $\pi$, then $\alpha=\frac{1+2i}{\sqrt{5}}$ would be a $m$-th root of unity for some $m\in\mathbb{N}$. On the other hand the minimal polynomial of $\alpha$ over $\mathbb{Q}$ is $x^4+\frac{6}{5}x^2+1$, and this is not a cyclotomic polynomial, since cyclotomic polynomials always have integer coefficients (by Moebius inversion formula, if you like). It follows that $\alpha$ is not a $m$-th root of unity and 
$$ \frac{\arctan 2}{\pi}\color{red}{\not\in}\mathbb{Q}.$$
Small variation: there are not so many cyclotomic polynomials with degree $4$. As many as the solutions of $\varphi(n)=4$, given by $n\in\{8,10,12\}$. The minimal polynomial of $\alpha$ does not belong to the set $\{\Phi_8(x),\Phi_{10}(x),\Phi_{12}(x)\}$ and the conclusion is the same.

Anyway, by the Shafer-Fink inequality a pretty good approximation of $\arctan 2$ is provided by
$$ \frac{\pi}{2}-\frac{3/2}{1+2\sqrt{1+1/4}}=\color{blue}{\frac{1}{8} \left(3-3 \sqrt{5}+4 \pi \right)}.$$
A: $\arctan 2$ is not a rational multiple of $\pi$.  If it were, then for some integer $n > 0$, we would have $(1 + 2i)^n$ is real.  On the other hand, if we define $a_n := \operatorname{Im}((1 + 2i)^n)$, it is straightforward to show that this sequence satisfies the recurrence relation:
$$a_{n+2} = 2 a_{n+1} - 5 a_n, \, n \ge 0.$$
But now $a_0 = 0$ and $a_1 = 2$, so by induction, it is straightforward to show $a_n \equiv 2^{n+1} \pmod{5}$ for $n \ge 1$.  Since no power of 2 is divisible by 5, this implies that $a_n \ne 0$ for $n \ge 1$.
