Shape of Tangent Function

== Theorem ==
The nature of the tangent function on the set of real numbers $\R$ is as follows:
:$\tan x$ is continuous and strictly increasing on the interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$
:$\tan x \to +\infty$ as $x \to \dfrac \pi 2 ^-$
:$\tan x \to -\infty$ as $x \to -\dfrac \pi 2 ^+$
:$\tan x$ is not defined on $\forall n \in \Z: x = \paren {n + \dfrac 1 2} \pi$, at which points it is discontinuous
:$\forall n \in \Z: \map \tan {n \pi} = 0$.
== Proof ==
$\tan x$ is continuous and strictly increasing on $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$:
Continuity follows from the Quotient Rule for Continuous Real Functions:
:$(1): \quad$ Both $\sin x$ and $\cos x$ are continuous on $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$ from Real Sine Function is Continuous and Cosine Function is Continuous
:$(2): \quad$ $\cos x > 0$ on this interval.
The fact of $\tan x$ being strictly increasing on this interval has been demonstrated in the discussion on Tangent Function is Periodic on Reals.
$\tan x \to + \infty$ as $x \to \dfrac \pi 2^-$:
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:
:$(1): \quad \cos x \to 0$ as $x \to \dfrac \pi 2^-$
:$(2): \quad \sin x \to 1$ as $x \to \dfrac \pi 2^-$
From the Infinite Limit Theorem it follows that:
:$\tan x = \dfrac {\sin x} {\cos x} \to + \infty$ as $x \to \dfrac \pi 2 ^-$
$\tan x \to - \infty$ as $x \to -\dfrac \pi 2 ^+$:
From Sine and Cosine are Periodic on Reals, we have that $\sin x < 0$ and $\cos x > 0$ on $\openint {-\dfrac \pi 2} 0$.
We have that:
:$(1): \quad \cos x \to 0$ as $x \to -\dfrac \pi 2 ^+$
:$(2): \quad \sin x \to -1$ as $x \to -\dfrac \pi 2 ^+$
Thus it follows that $\tan x = \dfrac {\sin x} {\cos x} \to -\infty$ as $x \to -\dfrac \pi 2 ^+$.
$\tan x$ is not defined and discontinuous at $x = \paren {n + \dfrac 1 2} \pi$:
From the discussion of Sine and Cosine are Periodic on Reals, it was established that:
:$\forall n \in \Z: x = \paren {n + \dfrac 1 2} \pi \implies \cos x = 0$
As division by zero is not defined, it follows that at these points $\tan x$ is not defined either.
Now, from the above, we have:
:$(1): \quad \tan x \to + \infty$ as $x \to \dfrac \pi 2^-$
:$(2): \quad \tan x \to - \infty$ as $x \to -\dfrac \pi 2^+$
As $\map \tan {x + \pi} = \tan x$ from Tangent Function is Periodic on Reals, it follows that:
:$\tan x \to - \infty$ as $x \to \dfrac \pi 2 ^+$
Hence the left hand limit and right hand limit at $x = \dfrac \pi 2$ are not the same.
From Tangent Function is Periodic on Reals, it follows that the same applies $\forall n \in \Z: x = \paren {n + \dfrac 1 2} \pi$.
The fact of its discontinuity at these points follows from the definition of discontinuity.
$\map \tan {n \pi} = 0$:
Follows directly from Sine and Cosine are Periodic on Reals::
:$\forall n \in \Z: \map \sin {n \pi} = 0$
=== Graph of Tangent Function ===
== Also see ==
* Shape of Sine Function
* Shape of Cosine Function
* Shape of Cotangent Function
* Shape of Secant Function
* Shape of Cosecant Function
== Sources ==
* : $\S 5$: Trigonometric Functions: Signs and Variations of Trigonometric Functions
* : $\S 16.5 \ (2)$
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Category:Tangent Function