Cotangent Function is Periodic on Reals

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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.

$\blacksquare$