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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.

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    I think you mean $(\tan (\frac\pi6))^2$. – 2017-02-24
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    @flytothesurface that's what the OP wrote. – 2017-02-24
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    I must confess I never really understood why only for trigonometric functions do people use $f^2$ to mean $f\cdot f$ rather than $f\circ f$. – 2017-02-24
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    It's also customary to use it with logarithms, as in $\log_2^3 32 = (\log_2 32 )^3=5^3$. On the other hand, I've only seen $f^2(x)=(f\circ f)(x)$ with letters or symbols like $f$ or $\phi$ representing generic functions, never with actual function names like $\sin$, $\log$ or $\Gamma$. It is (it always is) a matter of context. – 2018-03-18

4 Answers 4

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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.

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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.

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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}

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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 $\approx 0.3217505544$.