Difference between revisions of "Relationship between cos and cosh"

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==Theorem==
<strong>[[Relationship between cos and cosh|Theorem]]:</strong> The following formula holds:
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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]].
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<strong>Proof:</strong> █
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==Proof==
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From the definition of $\cosh$ and the definition of $\cos$,
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$$\cosh(iz)=\dfrac{e^{iz}+e^{-iz}}{2}=\cos(z),$$
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as was to be shown.
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==References==
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[[Category:Theorem]]
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[[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.

References