Sine in terms of Cosine
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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 $\paren {2 n + 1} \pi < x < \paren {2 n + 2} \pi$ |
$\blacksquare$