Difference between revisions of "Relationship between cos and cosh"
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| − | + | ==Theorem== | |
| − | + | The following formula holds: | |
$$\cos(z)=\cosh(iz),$$ | $$\cos(z)=\cosh(iz),$$ | ||
where $\cos$ is the [[cosine]] and $\cosh$ is the [[cosh|hyperbolic cosine]]. | where $\cos$ is the [[cosine]] and $\cosh$ is the [[cosh|hyperbolic cosine]]. | ||
| − | + | ||
| − | + | ==Proof== | |
| − | + | From the definition of $\cosh$ and the definition of $\cos$, | |
| − | + | $$\cosh(iz)=\dfrac{e^{iz}+e^{-iz}}{2}=\cos(z),$$ | |
| + | as was to be shown. | ||
| + | |||
| + | ==References== | ||
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Proven]] | ||
Latest revision as of 03:48, 8 December 2016
Theorem
The following formula holds: $$\cos(z)=\cosh(iz),$$ where $\cos$ is the cosine and $\cosh$ is the hyperbolic cosine.
Proof
From the definition of $\cosh$ and the definition of $\cos$, $$\cosh(iz)=\dfrac{e^{iz}+e^{-iz}}{2}=\cos(z),$$ as was to be shown.
