Difference between revisions of "Pythagorean identity for sinh and cosh"

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==Theorem==
<strong>Theorem:</strong> The following formula holds:
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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]].
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<strong>Proof:</strong> From the definitions
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==Proof==
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From the definitions
 
$$\cosh(z)=\dfrac{e^{z}+e^{-z}}{2}$$
 
$$\cosh(z)=\dfrac{e^{z}+e^{-z}}{2}$$
 
and
 
and
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\end{array}$$  
 
\end{array}$$  
 
as was to be shown. █  
 
as was to be shown. █  
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==References==
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* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Period of tanh|next=Pythagorean identity for tanh and sech}}: $4.5.16$
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[[Category:Theorem]]
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[[Category:Proven]]

Latest revision as of 22:25, 21 October 2017

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. █

References