Sign of Tangent
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Theorem
Let $x$ be a real number.
Then:
| \(\ds \tan x\) | \(>\) | \(\ds 0\) | if there exists an integer $n$ such that $n \pi < x < \paren {n + \dfrac 1 2} \pi$ | |||||||||||
| \(\ds \tan x\) | \(<\) | \(\ds 0\) | if there exists an integer $n$ such that $\paren {n + \dfrac 1 2} \pi < x < \paren {n + 1} \pi$ |
where $\tan$ denotes the tangent function.
Proof
From Tangent is Sine divided by Cosine:
- $\tan x = \dfrac {\sin x} {\cos x}$
Since $n$ is an integer, $n$ is either odd or even.
Case 1
Let $n$ be odd.
Hence let $m$ be an integer such that $n = 2 m + 1$.
Case 1.1
- $n \pi < x < \paren {n + \dfrac 1 2} \pi \implies \paren {2 m + 1} \pi < x < \paren {2 m + \dfrac 3 2} \pi$
From Sign of Sine, $\sin x$ is negative.
From Sign of Cosine, $\cos x$ is negative.
Therefore:
- $\tan x = \dfrac {\sin x} {\cos x} > 0$
Case 1.2
- $\paren {n + \dfrac 1 2} \pi < x < \paren {n + 1} \pi \implies \paren {2 m + \dfrac 3 2} \pi < x < \paren {2 m + 2} \pi$
From Sign of Sine, $\sin x$ is negative.
From Sign of Cosine, $\cos x$ is positive.
Therefore:
- $\tan x = \dfrac {\sin x} {\cos x} < 0$
Case 2
Let $n$ be even.
Hence let $m$ be an integer such that $n = 2 m$.
Case 2.1
- $n \pi < x < \paren {n + \dfrac 1 2} \pi \implies 2 m \pi < x < \paren {2 m + \dfrac 1 2} \pi$
From Sign of Sine, $\sin x$ is positive.
From Sign of Cosine, $\cos x$ is positive.
Therefore:
- $\tan x = \dfrac {\sin x} {\cos x} > 0$
Case 2.2
- $\paren {n + \dfrac 1 2} \pi < x < \paren {n + 1} \pi \implies \paren {2 m + \dfrac 1 2} \pi < x < \paren {2 m + 1} \pi$
From Sign of Sine, $\sin x$ is positive.
From Sign of Cosine, $\cos x$ is negative.
Therefore:
- $\tan x = \dfrac {\sin x} {\cos x} < 0$
Therefore:
| \(\ds \tan x\) | \(>\) | \(\ds 0\) | if there exists integer $n$ such that $n \pi < x < \paren {n + \dfrac 1 2} \pi$ | |||||||||||
| \(\ds \tan x\) | \(<\) | \(\ds 0\) | if there exists integer $n$ such that $\paren {n + \dfrac 1 2} \pi < x < \paren {n + 1} \pi$ |
$\blacksquare$