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.

  $\displaystyle e^{i x} = \cos x + i \;\sin x$    
  $\displaystyle e^{-i x} = \cos x - i \;\sin x$    
  $\displaystyle \cosh ix = \frac{e^{i x} + e^{-i x}}{2} = \cos x$    
  $\displaystyle \sinh ix = \frac{e^{i x} - e^{-i x}}{2} = i \sin x$    
  $\displaystyle \tanh ix = i \tan x \,$    
  $\displaystyle \cosh x = \cos ix \,$    
  $\displaystyle \sinh x = -i \sin ix \,$    
  $\displaystyle \tanh x = -i \tan ix \,$    

Jim Branson 2012-10-21