Difference between revisions of "Pythagorean identity for sinh and cosh"
From specialfunctionswiki
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| − | + | ==Theorem== | |
| − | + | The following formula holds: | |
$$\cosh^2(z)-\sinh^2(z)=1,$$ | $$\cosh^2(z)-\sinh^2(z)=1,$$ | ||
where $\cosh$ denotes the [[cosh|hyperbolic cosine]] and $\sinh$ denotes the [[sinh|hyperbolic sine]]. | where $\cosh$ denotes the [[cosh|hyperbolic cosine]] and $\sinh$ denotes the [[sinh|hyperbolic sine]]. | ||
| − | + | ||
| − | + | ==Proof== | |
| + | From the definitions | ||
$$\cosh(z)=\dfrac{e^{z}+e^{-z}}{2}$$ | $$\cosh(z)=\dfrac{e^{z}+e^{-z}}{2}$$ | ||
and | and | ||
| Line 15: | Line 16: | ||
\end{array}$$ | \end{array}$$ | ||
as was to be shown. █ | as was to be shown. █ | ||
| − | + | ||
| − | + | ==References== | |
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Proven]] | ||
Revision as of 07:52, 8 June 2016
Theorem
The following formula holds: $$\cosh^2(z)-\sinh^2(z)=1,$$ where $\cosh$ denotes the hyperbolic cosine and $\sinh$ denotes the hyperbolic sine.
Proof
From the definitions $$\cosh(z)=\dfrac{e^{z}+e^{-z}}{2}$$ and $$\sinh(z)=\dfrac{e^{z}-e^{-z}}{2},$$ we see $$\begin{array}{ll} \cosh^2(z) - \sinh^2(z) &= \left( \dfrac{e^{z}+e^{-z}}{2} \right)^2 - \left( \dfrac{e^{z}-e^{-z}}{2} \right)^2 \\ &= \dfrac{1}{4} \left( e^{2z}+2+e^{-2z}-e^{2z}+2-e^{-2z} \right) \\ &= 1, \end{array}$$ as was to be shown. █
