Tangent Function is Periodic on Reals

Theorem

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

Proof

From the definition of the tangent function, we have that $\tan x = \dfrac {\sin x} {\cos x}$.

We have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \tan \left({x + \pi}\right)\)	\(=\)	\(\displaystyle \)	\(\displaystyle \frac {\sin \left({x + \pi}\right)} {\cos \left({x + \pi}\right)}\)	\(\displaystyle \)	\(\displaystyle \)	from Sine and Cosine are Periodic on Reals
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \frac {-\sin x} {-\cos x}\)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \)	\(\displaystyle \tan x\)	\(\displaystyle \)	\(\displaystyle \)

Also, from Derivative of Tangent Function, we have that:

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

provided $\cos x \ne 0$.

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

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): $\S 16.5 \ (2)$