Cotangent Function is Periodic on Reals

Theorem

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

Proof

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

We have:

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

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

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

provided $\sin x \ne 0$.

From Shape of Sine Function, we have that $\sin$ is strictly positive on the interval $\left({0 \,.\,.\, \pi}\right)$.

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

Hence the result.

$\blacksquare$