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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 $p,q\in \mathbb Z$.


marked as duplicate by Thomas Andrews, Simply Beautiful Art, Robert Israel, JonMark Perry, Xander Henderson Sep 13 '17 at 3:39

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  • $\begingroup$ Looking at the continued fraction of $\frac{\arctan(2)}{\pi}$, I would guess that it is not possible. Proving it is something else. I would even guess that it is transcendental. $\endgroup$ – Peter Sep 12 '17 at 23:09
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    $\begingroup$ I don't think it is a rational multiple of $\pi$ at all. $\endgroup$ – Ian Sep 12 '17 at 23:09
  • $\begingroup$ @Peter Any ideas about how I might show this? $\endgroup$ – Frpzzd Sep 12 '17 at 23:10
  • $\begingroup$ @Nilknarf No, irrationality proofs are usually very difficult. For instance, we do not know whether $e+\pi$ is rational. $\endgroup$ – Peter Sep 12 '17 at 23:13
  • $\begingroup$ @Peter Does it help that it is also equal to $\arccos \frac{1}{\sqrt 5}$ or $\arcsin \frac{2}{\sqrt 5}$? $\endgroup$ – Frpzzd Sep 12 '17 at 23:18

$\arctan 2$ 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 $n \ge 1$.

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    $\begingroup$ How does "$(1+2i)^n$ is real" follow from "$\arctan 2$ is a rational multiple of $\pi$"? $\endgroup$ – Frpzzd Sep 12 '17 at 23:21
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    $\begingroup$ @Nilknarf $$1+2i=\sqrt5e^{i\arctan(2)}$$ $\endgroup$ – Simply Beautiful Art Sep 12 '17 at 23:23
  • $\begingroup$ $1+2i = \sqrt{5} e^{i \arctan(2)}$. $\endgroup$ – Daniel Schepler Sep 12 '17 at 23:23
  • $\begingroup$ Ingenious solution! (+1) $\endgroup$ – Simply Beautiful Art Sep 12 '17 at 23:27
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    $\begingroup$ @Peter Why indeed, the induction step is also quite obvious, and everything seems to check out. $\endgroup$ – Simply Beautiful Art Sep 12 '17 at 23:30

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 $$ \frac{\pi}{2}-\frac{3/2}{1+2\sqrt{1+1/4}}=\color{blue}{\frac{1}{8} \left(3-3 \sqrt{5}+4 \pi \right)}.$$

  • $\begingroup$ >.< Darned, you're late on this one. $\endgroup$ – Simply Beautiful Art Sep 12 '17 at 23:31
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    $\begingroup$ More general solution than mine, +1. (In this case, you could also observe that the minimal polynomial implies $\alpha$ is not integral over $\mathbb{Z}$, while any $m$th root of unity is of course integral.) $\endgroup$ – Daniel Schepler Sep 12 '17 at 23:38
  • $\begingroup$ Lol, how many PDF's of your do I not know about? $\endgroup$ – Simply Beautiful Art Sep 13 '17 at 14:23

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