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.

$\blacksquare$


Also see