Sign of Cosecant

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Theorem

Let $x$ be a real number.

Then:

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

where $\csc$ is the real cosecant function.


Proof

For the first part:

\(\ds \sin x\) \(>\) \(\ds 0\) if there exists an integer $n$ such that $2 n \pi < x < \paren {2 n + 1} \pi$ \(\quad\) Sign of Sine
\(\ds \leadsto \ \ \) \(\ds \frac 1 {\sin x}\) \(>\) \(\ds 0\) if there exists an integer $n$ such that $2 n \pi < x < \paren {2 n + 1} \pi$ \(\quad\) Reciprocal of Strictly Positive Real Number is Strictly Positive
\(\ds \leadsto \ \ \) \(\ds \csc x\) \(>\) \(\ds 0\) if there exists an integer $n$ such that $2 n \pi < x < \paren {2 n + 1} \pi$ \(\quad\) Cosecant is Reciprocal of Sine


For the second part:

\(\ds \sin x\) \(<\) \(\ds 0\) if there exists an integer $n$ such that $\paren {2 n + 1} \pi < x < \paren {2 n + 2} \pi$ \(\quad\) Sign of Sine
\(\ds \leadsto \ \ \) \(\ds \frac 1 {\sin x}\) \(<\) \(\ds 0\) if there exists an integer $n$ such that $\paren {2 n + 1} \pi < x < \paren {2 n + 2} \pi$ \(\quad\) Reciprocal of Strictly Negative Real Number is Strictly Negative
\(\ds \leadsto \ \ \) \(\ds \csc x\) \(<\) \(\ds 0\) if there exists an integer $n$ such that $\paren {2 n + 1} \pi < x < \paren {2 n + 2} \pi$ \(\quad\) Cosecant is Reciprocal of Sine

$\blacksquare$


Also see