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.

$\blacksquare$


Also see