Sine in terms of Cosine

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $x$ be an real number.

Then:

\(\ds \sin x\) \(=\) \(\ds +\sqrt {1 - \cos ^2 x}\) if there exists an integer $n$ such that $2 n \pi < x < \paren {2 n + 1} \pi$
\(\ds \sin x\) \(=\) \(\ds -\sqrt {1 - \cos ^2 x}\) if there exists an integer $n$ such that $\paren {2 n + 1} \pi < x < \paren {2 n + 2} \pi$

where $\sin$ denotes the sine function and $\cos$ denotes the cosine function.


Proof

\(\ds \cos^2 x + \sin^2 x\) \(=\) \(\ds 1\) Sum of Squares of Sine and Cosine
\(\ds \leadsto \ \ \) \(\ds \sin^2 x\) \(=\) \(\ds 1 - \cos^2 x\)
\(\ds \leadsto \ \ \) \(\ds \sin x\) \(=\) \(\ds \pm \sqrt {1 - \cos^2 x}\)


Then from Sign of Sine:

\(\ds \sin x\) \(>\) \(\ds 0\) if there exists an integer $n$ such that $2 n \pi < x < \paren {2 n + 1} \pi$
\(\ds \sin x\) \(<\) \(\ds 0\) if there exists an integer $n$ such that $\paren {2 n + 1} \pi < x < \paren {2 n + 2} \pi$

$\blacksquare$


Also see