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