# How do you evaluate Tan^-1 (1)?

Mar 3, 2018

${\tan}^{- 1} \left(1\right) = \frac{\left(2 n - 1\right) \pi}{4}$, $n \in \mathbb{Z}$.

#### Explanation:

Let ${\tan}^{- 1} \left(1\right) = \theta$. That means that $\tan \left(\theta\right) = 1$.

$\tan \left(\theta\right) = 1$

$\sin \frac{\theta}{\cos} \left(\theta\right) = 1$

$\sin \left(\theta\right) = \cos \left(\theta\right)$

One of the basic identities of trigonometry is ${\sin}^{\textcolor{red}{2}} \left(x\right) + {\cos}^{\textcolor{red}{2}} \left(x\right) = 1$.

If $\sin \left(\theta\right) = \cos \left(\theta\right)$, then ${\sin}^{\textcolor{red}{2}} \left(\theta\right) = {\cos}^{\textcolor{red}{2}} \left(\theta\right)$.

${\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) = 1$
${\sin}^{2} \left(\theta\right) + {\sin}^{2} \left(\theta\right) = 1$
$2 {\sin}^{2} \left(\theta\right) = 1$
${\sin}^{2} \left(\theta\right) = \frac{1}{2}$
$\sin \left(\theta\right) = \pm \frac{1}{\sqrt{2}}$

Rationalize the denominator :

$\sin \left(\theta\right) = \pm \frac{\sqrt{2}}{2}$

We know that $\theta = \frac{\pi}{4}$ is a solution to this, so we have to deduce the other ones from it.

There are 4 basic identities which you can prove using the unit circle :

$\sin \left(x\right) = \sin \left(\pi - x\right)$
$\sin \left(x\right) = - \sin \left(x - \pi\right)$
$\sin \left(x\right) = - \sin \left(2 \pi - x\right)$

From these, you can get some more solutions for $\sin \left(\theta\right) = \pm \frac{\sqrt{2}}{2}$ :

$\theta \in \left\{\frac{\textcolor{red}{1} \pi}{4} , \frac{\textcolor{red}{3} \pi}{4} , \frac{\textcolor{red}{5} \pi}{4} , \frac{\textcolor{red}{7} \pi}{4} , \ldots\right\}$

All the highlighted numbers are $\textcolor{red}{\text{odd}}$ numbers, so we get the set of solutions to be

$\theta \in \left\{\frac{\left(\textcolor{red}{2 n - 1}\right) \pi}{4} , n \in \mathbb{Z}\right\}$ are all the values of $\theta$ for which $\sin \left(\theta\right) = \pm \frac{\sqrt{2}}{2}$ and, consequently, $\tan \left(\theta\right) = 1$. Finally, we get

${\tan}^{- 1} \left(1\right) = \theta$ , $\forall \theta \in \left\{\frac{\left(\textcolor{red}{2 n - 1}\right) \pi}{4} , n \in \mathbb{Z}\right\}$.