# 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 $\openint 0 \pi$
$\cot x \to +\infty$ 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
$\forall n \in \Z: \map \cot {n + \dfrac 1 2} \pi = 0$

## Proof

$\cot x$ is continuous and strictly decreasing on $\openint 0 \pi$:

Continuity follows from the Quotient Rule for Continuous Real Functions:

$(1): \quad$ Both $\sin x$ and $\cos x$ are continuous on $\openint 0 \pi$ from Real 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 $\openint 0 {\dfrac \pi 2}$.

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 $\openint {\dfrac \pi 2} \pi$.

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 $\map \cot {x + \pi} = \cot x$ 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.

$\map \cot {n + \dfrac 1 2} \pi = 0$:

Follows directly from Sine and Cosine are Periodic on Reals:

$\forall n \in \Z: \map \cos {n + \dfrac 1 2} \pi = 0$

$\blacksquare$

### Graph of Cotangent Function 