Cosine in terms of Tangent
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Theorem
Let $x$ be a real number such that $\cos x \ne 0$.
Then:
| \(\ds \cos x\) | \(=\) | \(\ds +\frac 1 {\sqrt {1 + \tan^2 x} }\) | if there exists an integer $n$ such that $\paren {2 n - \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 1 2} \pi$ | |||||||||||
| \(\ds \cos x\) | \(=\) | \(\ds -\frac 1 {\sqrt {1 + \tan^2 x} }\) | if there exists an integer $n$ such that $\paren {2 n + \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 3 2} \pi$ |
where $\sin$ denotes the real sine function and $\tan$ denotes the real tangent function.
Proof
| \(\ds \sec^2 x - \tan^2 x\) | \(=\) | \(\ds 1\) | Difference of Squares of Secant and Tangent | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sec^2 x\) | \(=\) | \(\ds 1 + \tan ^2 x\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac 1 {\cos^2 x}\) | \(=\) | \(\ds 1 + \tan^2 x\) | Secant is Reciprocal of Cosine | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \cos^2 x\) | \(=\) | \(\ds \frac 1 {1 + \tan^2 x}\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \cos x\) | \(=\) | \(\ds \pm \frac 1 {\sqrt {1 + \tan^2 x} }\) |
Then from Sign of Cosine:
| \(\ds \cos x\) | \(>\) | \(\ds 0\) | if there exists an integer $n$ such that $\paren {2 n - \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 1 2} \pi$ | |||||||||||
| \(\ds \cos x\) | \(<\) | \(\ds 0\) | if there exists an integer $n$ such that $\paren {2 n + \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 3 2} \pi$ |
When $\cos x = 0$, $\tan x$ is undefined.
$\blacksquare$