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

From specialfunctionswiki
Jump to: navigation, search
(Created page with "<div class="toccolours mw-collapsible mw-collapsed"> <strong>Theorem:</strong> The following formula holds: $$\cosh^2(z)-\sinh^2(z)=1,$$ where $\cosh$ denotes the cosh|hyper...")
 
 
(3 intermediate revisions by the same user not shown)
Line 1: Line 1:
<div class="toccolours mw-collapsible mw-collapsed">
+
==Theorem==
<strong>Theorem:</strong> The following formula holds:
+
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]].
<div class="mw-collapsible-content">
+
 
<strong>Proof:</strong> █
+
==Proof==
</div>
+
From the definitions
</div>
+
$$\cosh(z)=\dfrac{e^{z}+e^{-z}}{2}$$
 +
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==
 +
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Period of tanh|next=Pythagorean identity for tanh and sech}}: $4.5.16$
 +
 
 +
[[Category:Theorem]]
 +
[[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