# How do you find the exact value of arctan(2)?

Mar 15, 2018

This is not a rational number of degrees, nor a rational multiple of $\pi$ radians.

We can write:

$\arctan 2 = \frac{\pi}{2} - {\sum}_{k = 0}^{\infty} {\left(- 1\right)}^{k} \frac{1}{{2}^{2 k + 1} \left(2 k + 1\right)}$

#### Explanation:

$\arctan \left(2\right)$ is an angle in a right angled triangle with sides $\text{adjacent} = 1$, $\text{opposite} = 2$ and $\text{hypotenuse} = \sqrt{5}$. It is not a rational multiple of $\pi$ radians nor a rational number of degrees.

We can represent it as the sum of an infinite series.

Note that:

$\arctan x = {\sum}_{k = 0}^{\infty} {\left(- 1\right)}^{k} {x}^{2 k + 1} / \left(2 k + 1\right) = x - {x}^{3} / 3 + {x}^{5} / 5 - {x}^{7} / 7 + {x}^{9} / 9 - {x}^{11} / 11 + \ldots$

However, this only converges for $\left\mid x \right\mid \le 1$.

To get a series that does converge, we can use:

$\tan \left(\frac{\pi}{2} - x\right) = \frac{1}{\tan} x$

So:

$\arctan \left(\frac{1}{x}\right) = \frac{\pi}{2} - \arctan x$

and hence:

$\arctan 2 = \frac{\pi}{2} - \arctan \left(\frac{1}{2}\right)$

$\textcolor{w h i t e}{\arctan 2} = \frac{\pi}{2} - {\sum}_{k = 0}^{\infty} {\left(- 1\right)}^{k} \frac{1}{{2}^{2 k + 1} \left(2 k + 1\right)}$