The trigonometric functions deal with circles, whereas hyperbolic functions deal with hyperbolae. They are analogues of each trigonometric function, given the same names but with an h on the end: sinh, cosh and tanh, usually pronounced 'shine', 'cosh', and 'tanch' or 'than'.
The hyperbolic functions can be expressed in terms of exponentials.
\begin{align} \sinh \theta = \dfrac{ e^{\theta} - e^{-\theta} }{2} \\ \cosh \theta = \dfrac{ e^{\theta} + e^{-\theta} }{2} \\ \tanh \theta = \dfrac{ e^{\theta} - e^{-\theta} }{ e^{\theta} + e^{-\theta} } \end{align}
The hyperbolic identities are similar to the trigonometric identities.
\begin{align} \cosh^2 \theta - \sinh^2 \theta &= 1 \\ \\ 1-\tanh^2 \theta &= \mathrel{\text{sech}}^2 \theta \\ \\ \sinh(\theta \pm \phi) &= \cosh\theta\sinh\phi \pm \sinh\theta\cosh\phi \\ \\ \cosh(\theta \pm \phi) &= \sinh\theta\sinh\phi \pm \cosh\theta\cosh\phi \\ \\ \tanh(\theta \pm \phi) &= \dfrac{\tanh \theta \pm \tanh \theta}{1\pm\tanh\theta\tanh\phi} \\ \\ \sinh 2\theta &= 2\sinh\theta\cosh\theta \\ \\ \cosh 2\theta &= 2\cosh^2\theta -1 \\ \\ \tanh 2\theta &= \dfrac{2\tanh\theta}{1+\tanh^2\theta} \end{align}
Hyperbolic functions can be differentiated and integrated.
$f(x)$ |
$f'(x)$ |
|---|---|
$\sinh x$ |
$\cosh x$ |
$\cosh x$ |
$\sinh x$ |
$\tanh x$ |
$\mathrel{\text{sech}}^2 x$ |
Note: The derivative of $\cosh x$ does not change sign, like the derivative of $\cos x$ is.
As with trig functions, hyperbolic functions can be used in equations. To solve, convert the hyperbolic function into its exponential equivalent and solve as an exponential equation.
Solve $\cosh x = 1$.
Using the formula $\cosh x = \dfrac{ e^x+ e^{-x} }{2}$,
\begin{align} \frac{ e^x+ e^{-x} }{2} &= 1\\ e^x + e^{-x} &=2. \end{align}
Set $y= e^x$, which means $e^{-x} = \dfrac{1}{y}$.
\begin{align} e^x + e^{-x} &= 2\\ y + \frac{1}{y} &= 2\\ y^2 + 1 &= 2y\\ y^2 - 2y + 1 &= 0\\ (y-1)(y-1)&=0 \end{align}
Therefore $y=1$, but $y= e^x$, so:
\begin{align} e^x &= 1 \\ x &= \ln 1=0. \end{align}
Prof. Robin Johnson shows that $\cosh^2x = \dfrac{1}{2}(1+\cosh 2x)$ and derives the formulas for $\cosh 2x$, $\sinh 2x$ and $\tanh 2x$.
Prof. Robin Johnson solves the hyperbolic equation $\sinh 4x = -1$.
This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.