Difference between revisions of "Relationship between cot and coth"
From specialfunctionswiki
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| − | + | ==Theorem== | |
| − | + | The following formula holds: | |
$$\cot(z)=i\coth(iz),$$ | $$\cot(z)=i\coth(iz),$$ | ||
where $\cot$ denotes the [[cotangent]] and $\coth$ denotes the [[coth|hyperbolic cotangent]]. | where $\cot$ denotes the [[cotangent]] and $\coth$ denotes the [[coth|hyperbolic cotangent]]. | ||
| − | + | ||
| − | + | ==Proof== | |
| − | + | ||
| − | + | ==References== | |
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Unproven]] | ||
Latest revision as of 07:48, 8 June 2016
Theorem
The following formula holds: $$\cot(z)=i\coth(iz),$$ where $\cot$ denotes the cotangent and $\coth$ denotes the hyperbolic cotangent.
