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.

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.
