Editing Hyperbolic functions (section)

==Useful relations==
{{Anchor|Osborn}}
The hyperbolic functions satisfy many identities, all of them similar in form to the [[trigonometric identity|trigonometric identities]]. In fact, '''Osborn's rule'''<ref name="Osborn, 1902" /> states that one can convert any trigonometric identity (up to but not including sinhs or implied sinhs of 4th degree) for <math>\theta</math>, <math>2\theta</math>, <math>3\theta</math> or <math>\theta</math> and <math>\varphi</math> into a hyperbolic identity, by:
# expanding it completely in terms of integral powers of sines and cosines,
# changing sine to sinh and cosine to cosh, and
# switching the sign of every term containing a product of two sinhs.

Odd and even functions:
<math display="block">\begin{align}
 \sinh (-x) &= -\sinh x \\
 \cosh (-x) &=  \cosh x
\end{align}</math>

Hence:
<math display="block">\begin{align}
               \tanh (-x) &= -\tanh x \\
               \coth (-x) &= -\coth x \\
 \operatorname{sech} (-x) &=  \operatorname{sech} x \\
 \operatorname{csch} (-x) &= -\operatorname{csch} x
\end{align}</math>

Thus, {{math|cosh ''x''}} and {{math|sech ''x''}} are [[even function]]s; the others are [[odd functions]].

<math display="block">\begin{align}
 \operatorname{arsech} x &= \operatorname{arcosh} \left(\frac{1}{x}\right) \\
 \operatorname{arcsch} x &= \operatorname{arsinh} \left(\frac{1}{x}\right) \\
 \operatorname{arcoth} x &= \operatorname{artanh} \left(\frac{1}{x}\right)
\end{align}</math>

Hyperbolic sine and cosine satisfy:
<math display="block">\begin{align}
 \cosh x + \sinh x &= e^x \\
 \cosh x - \sinh x &= e^{-x} 
\end{align}</math>

which are analogous to [[Euler's formula]], and

<math display="block">
 \cosh^2 x - \sinh^2 x = 1
</math>

which is analogous to the [[Pythagorean trigonometric identity]].

One also has
<math display="block">\begin{align}
 \operatorname{sech} ^{2} x &= 1 - \tanh^{2} x \\
 \operatorname{csch} ^{2} x &= \coth^{2} x - 1
\end{align}</math>

for the other functions.

===Sums of arguments===
<math display="block">\begin{align}
 \sinh(x + y) &= \sinh x \cosh y + \cosh x \sinh y \\
 \cosh(x + y) &= \cosh x \cosh y + \sinh x \sinh y \\ 
 \tanh(x + y) &= \frac{\tanh x +\tanh y}{1+ \tanh x \tanh y } \\
\end{align}</math>
particularly
<math display="block">\begin{align}
\cosh (2x) &= \sinh^2{x} + \cosh^2{x} = 2\sinh^2 x + 1 = 2\cosh^2 x - 1 \\
\sinh (2x) &= 2\sinh x \cosh x \\
\tanh (2x) &= \frac{2\tanh x}{1+ \tanh^2 x } \\
\end{align}</math>

Also:
<math display="block">\begin{align}
 \sinh x + \sinh y &= 2 \sinh \left(\frac{x+y}{2}\right) \cosh \left(\frac{x-y}{2}\right)\\
 \cosh x + \cosh y &= 2 \cosh \left(\frac{x+y}{2}\right) \cosh \left(\frac{x-y}{2}\right)\\
\end{align}</math>

===Subtraction formulas===
<math display="block">\begin{align}
 \sinh(x - y) &= \sinh x \cosh y - \cosh x \sinh y \\
 \cosh(x - y) &= \cosh x \cosh y - \sinh x \sinh y \\
 \tanh(x - y) &= \frac{\tanh x -\tanh y}{1- \tanh x \tanh y } \\
\end{align}</math>

Also:<ref>{{cite book|last1=Martin|first1=George E.|title=The foundations of geometry and the non-Euclidean plane|date=1986 | publisher=Springer-Verlag|location=New York|isbn=3-540-90694-0|page=416|edition=1st corr.}}</ref>
<math display="block">\begin{align}
 \sinh x - \sinh y &= 2 \cosh \left(\frac{x+y}{2}\right) \sinh \left(\frac{x-y}{2}\right)\\
 \cosh x - \cosh y &= 2 \sinh \left(\frac{x+y}{2}\right) \sinh \left(\frac{x-y}{2}\right)\\
\end{align}</math>

===Half argument formulas===
<math display="block">\begin{align}
 \sinh\left(\frac{x}{2}\right) &= \frac{\sinh x}{\sqrt{2 (\cosh x + 1)} } &&= \sgn x \, \sqrt \frac{\cosh x - 1}{2} \\[6px]
 \cosh\left(\frac{x}{2}\right) &= \sqrt \frac{\cosh x + 1}{2}\\[6px]
 \tanh\left(\frac{x}{2}\right) &= \frac{\sinh x}{\cosh x + 1} &&= \sgn x \, \sqrt \frac{\cosh x-1}{\cosh x+1} = \frac{e^x - 1}{e^x + 1}
\end{align}</math>

where {{math|sgn}} is the [[sign function]].

If {{math|''x'' ≠ 0}}, then<ref>{{cite web|title=Prove the identity tanh(x/2) = (cosh(x) - 1)/sinh(x) | url=https://math.stackexchange.com/q/1565753 |website=[[StackExchange]] (mathematics) | access-date=24 January 2016}}</ref>

<math display="block"> \tanh\left(\frac{x}{2}\right) = \frac{\cosh x - 1}{\sinh x} = \coth x - \operatorname{csch} x </math>
===Square formulas===
<math display="block">\begin{align}
\sinh^2 x &= \tfrac{1}{2}(\cosh 2x - 1) \\
\cosh^2 x &= \tfrac{1}{2}(\cosh 2x + 1)
\end{align}</math>

===Inequalities===

The following inequality is useful in statistics:<ref>{{cite news |last1=Audibert |first1=Jean-Yves |date=2009 |title=Fast learning rates in statistical inference through aggregation |publisher=The Annals of Statistics |page=1627}} [https://projecteuclid.org/download/pdfview_1/euclid.aos/1245332827]</ref>
<math display="block">\operatorname{cosh}(t) \leq e^{t^2 /2}.</math>

It can be proved by comparing the Taylor series of the two functions term by term.