# Tangent Function is Periodic on Reals

From ProofWiki

## Theorem

The tangent function is periodic on the set of real numbers $\R$ with period $\pi$.

## Proof

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \tan \left({x + \pi}\right)\) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \frac {\sin \left({x + \pi}\right)} {\cos \left({x + \pi}\right)}\) | \(\displaystyle \) | \(\displaystyle \) | Definition of Tangent Function | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \frac {-\sin x} {-\cos x}\) | \(\displaystyle \) | \(\displaystyle \) | Sine and Cosine are Periodic on Reals | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\) | \(\displaystyle \) | \(\displaystyle \tan x\) | \(\displaystyle \) | \(\displaystyle \) |

From Derivative of Tangent Function, we have that:

- $D_x \left({\tan x}\right) = \dfrac 1 {\cos^2 x}$

provided $\cos x \ne 0$.

From Shape of Cosine Function, we have that $\cos > 0$ on the interval $\left({-\dfrac \pi 2 \,.\,.\, \dfrac \pi 2}\right)$.

From Derivative of Monotone Function, $\tan x$ is strictly increasing on that interval, and hence can not have a period of *less* than $\pi$.

Hence the result.

$\blacksquare$

## Sources

- K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*(1977)... (previous)... (next): $\S 16.5 \ (2)$