Hyperbolic Function Identities

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Hyperbolic Function Identities

Identities can be easily derived from the definitions.
$\displaystyle \cosh^2 x - \sinh^2 x = 1\,$
$\displaystyle \sinh(-x) = -\sinh x\,\!$
$\displaystyle \cosh(-x) = \cosh x\,\!$
$\displaystyle \tanh(-x) = -\tanh x\,\!$
$\displaystyle \coth(-x) = -\coth x\,\!$
The derivatives of the hyperbolic functions.
$\displaystyle \frac{d}{dx}\sinh(x) = \cosh(x) \,$
$\displaystyle \frac{d}{dx}\cosh(x) = \sinh(x) \,$
$\displaystyle \frac{d}{dx}\tanh(x) = 1 - \tanh^2(x) = \hbox{sech}^2(x) = 1/\cosh^2(x) \,$
$\displaystyle \frac{d}{dx}\coth(x) = 1 - \coth^2(x) = -\hbox{csch}^2(x) = -1/\sinh^2(x) \,$
Hyperbolic functions of sums.
$\displaystyle \sinh(x+y) = \sinh x \cosh y + \cosh x \sinh y \,$
$\displaystyle \cosh(x+y) = \cosh x \cosh y + \sinh x \sinh y \,$
$\displaystyle \tanh(x+y) = \frac{\tanh x + \tanh y}{1 + \tanh x \tanh y} \,$
$\displaystyle \sinh 2x\ = 2\sinh x \cosh x \,$
$\displaystyle \cosh 2x\ = \cosh^2 x + \sinh^2 x = 2\cosh^2 x - 1 = 2\sinh^2 x + 1 \,$
$\displaystyle \cosh^2\frac{x}{2} = \frac{\cosh x + 1}{2}$
$\displaystyle \sinh^2\frac{x}{2} = \frac{\cosh x - 1}{2}$
Inverse hyperbolic functions from logs.
$\displaystyle \coth ^{-1}x=\tanh ^{-1}\left( \frac{1}{x} \right)$
$\displaystyle \sinh ^{-1}x=\ln \left( x+\sqrt{x^{2}+1} \right)$
$\displaystyle \cosh ^{-1}x=\ln \left( x+\sqrt{x^{2}-1} \right);x\ge 1$
$\displaystyle \tanh ^{-1}x=\frac{1}{2}\ln \left( \frac{1+x}{1-x} \right);\left\vert x \right\vert<1$
Hyperbolic sine and cosine are related to sine and cosine of imaginary numbers.
$\displaystyle e^{i x} = \cos x + i \;\sin x$
$\displaystyle e^{-i x} = \cos x - i \;\sin x$
$\displaystyle \cosh ix = \frac{e^{i x} + e^{-i x}}{2} = \cos x$
$\displaystyle \sinh ix = \frac{e^{i x} - e^{-i x}}{2} = i \sin x$
$\displaystyle \tanh ix = i \tan x \,$
$\displaystyle \cosh x = \cos ix \,$
$\displaystyle \sinh x = -i \sin ix \,$
$\displaystyle \tanh x = -i \tan ix \,$

next up previous contents Next: ``Rotations'' in 4 Dimensions Up: Review of the Hyperbolic Previous: Review of the Hyperbolic Contents Jim Branson 2012-10-21

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