Sine in terms of Tangent
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Theorem
Let $x$ be a real number such that $\cos x \ne 0$.
Then:
\(\ds \sin x\) | \(=\) | \(\ds +\frac {\tan x} {\sqrt {1 + \tan^2 x} }\) | if there exists an integer $n$ such that $\paren {2 n - \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 1 2} \pi$ | |||||||||||
\(\ds \sin x\) | \(=\) | \(\ds -\frac {\tan x} {\sqrt {1 + \tan^2 x} }\) | 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
For the first part, if there exists an integer $n$ such that $\paren {2 n - \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 1 2} \pi$:
\(\ds \cos x\) | \(=\) | \(\ds +\frac 1 {\sqrt {1 + \tan^2 x} }\) | Cosine in terms of Tangent | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac 1 {\paren {\frac 1 {\cos x} } }\) | \(=\) | \(\ds +\frac 1 {\sqrt {1 + \tan^2 x} }\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\sin x} {\paren {\frac {\sin x} {\cos x} } }\) | \(=\) | \(\ds +\frac 1 {\sqrt {1 + \tan^2 x} }\) | multiplying denominator and numerator by $\sin x$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\sin x} {\tan x}\) | \(=\) | \(\ds + \frac 1 {\sqrt {1 + \tan^2 x} }\) | Tangent is Sine divided by Cosine | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sin x\) | \(=\) | \(\ds + \frac {\tan x} {\sqrt {1 + \tan^2 x} }\) |
For the second part, if there exists an integer $n$ such that $\paren {2 n + \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 3 2} \pi$:
\(\ds \cos x\) | \(=\) | \(\ds -\frac 1 {\sqrt {1 + \tan^2 x} }\) | Cosine in terms of Tangent | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac 1 {\paren {\frac 1 {\cos x} } }\) | \(=\) | \(\ds -\frac 1 {\sqrt {1 + \tan^2 x} }\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\sin x} {\paren {\frac {\sin x} {\cos x} } }\) | \(=\) | \(\ds -\frac 1 {\sqrt {1 + \tan^2 x} }\) | multiplying denominator and numerator by $\sin x$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\sin x} {\tan x}\) | \(=\) | \(\ds -\frac 1 {\sqrt {1 + \tan^2 x} }\) | Tangent is Sine divided by Cosine | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sin x\) | \(=\) | \(\ds -\frac {\tan x} {\sqrt {1 + \tan^2 x} }\) |
When $\cos x = 0$, $\tan x$ is undefined.
$\blacksquare$