Shape of Cotangent Function

Theorem
The nature of the cotangent function on the set of real numbers $\R$ is as follows:


 * $\cot x$ is continuous and strictly decreasing on the interval $\left({0 \,.\,.\, \pi}\right)$
 * $\cot x \to + \infty$ as $x \to 0^+$
 * $\cot x \to - \infty$ as $x \to \pi^-$
 * $\cot x$ is not defined on $\forall n \in \Z: x = n \pi$, at which points it is discontinuous
 * $\forall n \in \Z: \cot \left({n + \dfrac 1 2}\right) \pi = 0$.

Proof

 * $\cot x$ is continuous and strictly decreasing on $\left({0 \,.\,.\, \pi}\right)$:

Continuity follows from the Quotient Rule for Continuous Functions:


 * $(1): \quad$ Both $\sin x$ and $\cos x$ are continuous on $\left({0 \,.\,.\, \pi}\right)$ from Sine Function is Continuous and Cosine Function is Continuous
 * $(2): \quad \sin x > 0$ on this interval.

The fact of $\cot x$ being strictly decreasing on this interval has been demonstrated in the discussion on Cotangent Function is Periodic on Reals.


 * $\cot x \to + \infty$ as $x \to 0^+$:

From Sine and Cosine are Periodic on Reals, we have that both $\sin x > 0$ and $\cos x > 0$ on $\left({0 \,.\,.\, \dfrac \pi 2}\right)$.

We have that:
 * $(1): \quad \cos x \to 1$ as $x \to 0^+$
 * $(2): \quad \sin x \to 0$ as $x \to 0^+$

Thus it follows that $\cot x = \dfrac {\cos x} {\sin x} \to + \infty$ as $x \to 0^+$.


 * $\tan x \to - \infty$ as $x \to \pi^-$:

From Sine and Cosine are Periodic on Reals, we have that $\sin x > 0$ and $\cos x < 0$ on $\left({\dfrac \pi 2 \,.\,.\, \pi}\right)$.

We have that:
 * $(1): \quad \cos x \to -1$ as $x \to \pi^-$
 * $(2): \quad \sin x \to 0$ as $x \to \pi^-$

Thus it follows that $\cot x = \dfrac {\cos x} {\sin x} \to - \infty$ as $x \to \pi^-$.


 * $\cot x$ is not defined and discontinuous at $x = n \pi$:

From the discussion of Sine and Cosine are Periodic on Reals, it was established that $\forall n \in \Z: x = n \pi \implies \sin x = 0$.

As division by zero is not defined, it follows that at these points $\cot x$ is not defined either.

Now, from the above, we have:
 * $(1): \quad \cot x \to + \infty$ as $x \to 0^+$
 * $(2): \quad \cot x \to - \infty$ as $x \to \pi^-$

As $\cot \left({x + \pi}\right) = \cot x$ from Cotangent Function is Periodic on Reals, it follows that $\cot x \to + \infty$ as $x \to \pi^+$.

Hence the left hand limit and right hand limit at $x = \pi$ are not the same.

From the periodic nature of $\cot x$, it follows that the same applies $\forall n \in \Z: x = n \pi$.

The fact of its discontinuity at these points follows from the definition of discontinuity.


 * $\cot \left({n + \dfrac 1 2}\right) \pi = 0$:

Follows directly from Sine and Cosine are Periodic on Reals: $\forall n \in \Z: \cos \left({n + \dfrac 1 2}\right) \pi = 0$.

Also see

 * Shape of Sine Function
 * Shape of Cosine Function
 * Shape of Tangent Function
 * Shape of Secant Function
 * Shape of Cosecant Function