Find the value of $\arctan(1/3)$ How can I calculate $\arctan\left({1\over 3}\right)$ in terms of $\pi$ ? I know that $\tan^2(\frac{\pi}{6})= {1\over3}$ but don't know if that helps in any way.
 A: The numerical computation of $\arctan\frac{1}{3}$ is pretty simple from the Maclaurin series of $\arctan$:
$$\arctan\frac{1}{3}=\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)3^{2n+1}} \tag{1}$$
To get $\arctan\frac{1}{3}\approx 0.321751$ we just need to consider the partial sum up to $n=4$. 
By multiplying this constant by $\frac{180}{\pi}$ we get
$$ \arctan\frac{1}{3}\approx 18^\circ 26'6'' \tag{2}$$
and the Shafer-Fink inequality provides the algebraic approximation $\arctan\frac{1}{3}\approx \frac{3}{3+2\sqrt{10}}$ with four correct figures. On the other hand it is not difficult to prove that $\arctan\frac{1}{3}\not\in\pi\mathbb{Q}$. Assuming that $\arctan\frac{1}{3}$ is a rational multiple of $\pi$ we have that $\alpha=\frac{3+i}{\sqrt{10}}$ is a root of unity, but its minimal polynomial over $\mathbb{Q}$ is $5x^4-8x^2+5$, which is not a monic polynomial, leading to a contradiction.
A: There is no simple way to do this.  That is because the answer is not some nice rational fraction times $\pi$.  Of course, you can look up (or use a calculator) to determine the $\arctan(1/3)$ (which equals $0.322$ radians or $18.435^\circ$ ) and then divide by $\pi$, but I don't think that is what you are looking for!
There is a way to represent the $\arctan$ using the series:
$$\arctan(x) = \frac{\pi}2 - \frac1x + \frac1{3x^3} - \frac1{5x^5} + \frac1{7x^7} - \cdots$$
However this is only usable for $x>1$ (and the similar series for $x < 1$ doesn't include $\pi$!)
The only other way to get $\pi$ in the "answer" would be to recognize that the $\arctan(x)$ can equal $\theta + n\cdot\pi$ (where $n = 0, \pm1,2,3,...$).
So the $\arctan(1/3) = 0.322 +n\cdot\pi  = (0.102 + n)\cdot\pi$
  (where $n = 0, \pm1,2,3,...$).
Not very satisfying.
A: You can use Newton-Rapson with $\tan\left(x\right) - 1/3 = 0$  and the 'starting point' $x = \pi/6$:
\begin{align}
&\texttt{Clear[i, x];}
\\
&\texttt{x = Pi/6;}
\\ &
\\
&\texttt{For[i = 0, i < 5, ++i,}
\\
&\texttt{x -= (Tan[x] - 1/3)/Sec[x]^2; Print[N[x]]]}
\end{align}

\begin{align}
&\texttt{0.340586, 0.321873, 0.321751, 0.321751}\,,\ \color{red}{0.321751}
\end{align}
A: For the evaluation of $\tan^{-1}(x)$, you can also use Padé approximants such as
$$\frac{ x+\frac{4 x^3}{15}} { 1+\frac{3 x^2}{5}}\tag 1$$
$$\frac{x+\frac{11 x^3}{21} }  {1+\frac{6 x^2}{7}+\frac{3 x^4}{35} } \tag 2$$
$$\frac{x+\frac{7 x^3}{9}+\frac{64 x^5}{945}}  {1+\frac{10 x^2}{9}+\frac{5 x^4}{21} } \tag 3$$
$$\frac{x+\frac{34 x^3}{33}+\frac{x^5}{5} }  { 1+\frac{15 x^2}{11}+\frac{5 x^4}{11}+\frac{5 x^6}{231}} \tag 4$$
Using the above formulae, you would get $\frac{139}{432}$, $\frac{250}{777}$, $\frac{20806}{64665}$, $\frac{19593}{60895}$ which are $\approx 0.3217592593$, $\approx 0.3217503218$, $\approx 0.3217505606$,  $\approx 0.3217505542$ while the "exact" value would be $\approx 0.3217505544$.
A: Don't forget you can construct it. A geometric solution is also a solution.

