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# Hyperbolic Functions:

## INTRODUCTION TO HYPERBOLIC FUNCTIONS

$\DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\cosech}{cosech}\ \DeclareMathOperator{\csch}{csch}$

Hyperbolic functions are analogues of the trigonometric functions, but defined using the unit hyperbola (x^2 - y^2 = 1) instead of the unit circle (x^2 + y^2 = 1).

The function e^x is neither odd nor even, so it is often useful to consider the following functions:

$\cosh(x) = \frac{e^x + e^{-x}}{2} \\ \sinh(x) = \frac{e^x - e^{-x}}{2}$

These are the two basic hyperbolic functions. cosh(x) is an even function and sinh(x) is an odd function.

e^x can be written as cosh(x) + sinh(x).

e^(-x) can be written as cosh(x) - sinh(x).

The other hyperbolic functions can be derived from the two basic hyperbolic functions cosh and sinh. By analogy with the trigonometric tan, sec, $$\cosec$$ and cot functions:

$\tanh(x) = \frac{\sinh(x)}{\cosh(x)} \\ \sech(x) = \frac{1}{\cosh(x)} \\ \cosech(x) \text{ or } \csch(x) = \frac{1}{\sinh(x)} \\ \coth(x) = \frac{\cosh(x)}{\sinh(x)}$

Note that there is a relationship between the hyperbolic and trigonometric functions:

$\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2} = \cosh(i\theta) \\ \sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i} = -i\sinh(i\theta)$

Equations may be solved by going back to definitions.

EXAMPLE:

e.g.1 Solve cosh(x) + 4sinh(x) = 7

\begin{align} \cosh(x) + 4\sinh(x) & = 7 \\ \frac{e^x + e^{-x}}{2} + 4\left(\frac{e^x - e^{-x}}{2}\right) & = 7 \\ \frac{5e^x}{2} - \frac{3e^{-x}}{2} & = 7 \\ 5e^x - 14 - 3e^{-x} & = 0 \\ 5e^{2x} - 14e^x - 3 & = 0 \\ (e^x - 3)(5e^x + 1) & = 0 \\ e^x & = 3, -\frac{1}{5} \\ e^x \gt 0 \text{ for all } x ∴ e^x &= 3 \\ x & = \ln 3 \end{align}

## HYPERBOLIC IDENTITIES

Hyperbolic identities are related to trigonometric identities using Osborn's rule: "Two sines change the sign".

A trigonometric identity can be converted to an analogous hyperbolic identity by replacing every instance of a trigonometric function with the corresponding hyperbolic one, and changing the sign of every term that involves the product of two sines. Note that the sines may be hidden inside tans.

FORMULAE:

$$\cosh^2(x) - \sinh^2(x) = 1$$

$$1 - \tanh^2(x) = \sech^2(x)$$

$$\coth^2(x) - 1 = \cosech^2(x)$$

\begin{align}\cosh(2x) & = \cosh^2(x) + \sinh^2(x) \\ & = 2\cosh^2(x) - 1 \\ & = 1 + 2\sinh^2(x)\end{align}

$$\sinh(2x) = 2\sinh(x)\cosh(x)$$

$$\tanh(2x) = \frac{2\tanh(x)}{1 + \tanh^2(x)}$$

Solving equations uses similar ideas to trigonometric equations.

EXAMPLE:

e.g.1 Solve sinh(2x) = 5sinh(x)

\begin{align} \sinh(2x) & = 5\sinh(x) \\ 2\sinh(x)\cosh(x) & = 5\sinh(x) \\ 2\sinh(x)\cosh(x) - 5\sinh(x) & = 0 \\ \sinh x (2\cosh x - 5) & = 0 \end{align}

 \begin{align} \sinh(x) & = 0 \\ \frac{e^x - e^{-x}}{2} & = 0 \\ e^x & = e^{-x} \\ e^{2x} & = 1 \\ x & = 0 \end{align} $\text{or}$ \begin{align} 2\cosh(x) - 5 & = 0 \\ e^x + e^{-x} - 5 & = 0 \\ e^{2x} -5e^x + 1 & = 0 \\ e^x & = \frac{5 \pm \sqrt{21}}{2} \\ x & = \ln\left(\frac{5 \pm \sqrt{21}}{2}\right) \end{align}

∴ $$x = 0, \ln\left(\frac{5 \pm \sqrt{21}}{2}\right)$$

## INVERSE HYPERBOLIC FUNCTIONS

We can sketch the graphs of cosh(x), sinh(x) and tanh(x), and use them to sketch the graphs of their inverses.

\begin{align} y & = \cosh^{-1}(x) \\ \cosh(y) & = x \\ \frac{e^y + e^{-y}}{2} & = x \\ e^{2y} - 2xe^y + 1 & = 0 \\ e^y & = \frac{2x \pm \sqrt{4x^2 - 4}}{2} \\ y & = \ln(x \pm \sqrt{x^2 - 1}) \end{align}
\begin{align} x - \sqrt{x^2 - 1} & = x - \sqrt{x^2 - 1} \times \frac{x + \sqrt{x^2 - 1}}{x + \sqrt{x^2 - 1}} \\ & = \frac{x^2 - (x^2 - 1)}{x + \sqrt{x^2 - 1}} \\ & = \frac{1}{x + \sqrt{x^2 - 1}} \end{align}
\begin{align}∴ \ln(x - \sqrt{x^2 - 1}) & = \ln(\frac{1}{x + \sqrt{x^2 - 1}}) \\ & = -\ln(x + \sqrt{x^2 - 1}) \end{align}
$∴ y = \pm\ln(x + \sqrt{x^2 - 1})$
As cosh is not a one-to-one function, to invert it we need to restrict its domain. By convention, we take the positive branch and define cosh^-1(x) as always giving the non-negative number, y, such that x = cosh(y).
$∴ \cosh^{-1}(x) = \ln(x + \sqrt{x^2 - 1})$

\begin{align} y & = \sinh^{-1}(x) \\ \sinh(y) & = x \\ \frac{e^y - e^{-y}}{2} & = x \\ e^{2y} - 2xe^y - 1 & = 0 \\ e^y & = \frac{2x \pm \sqrt{4x^2 + 4}}{2} \\ & = x \pm \sqrt{x^2 + 1} \end{align}
$\sqrt{x^2 + 1} \gt x \\ ∴ x - \sqrt{x^2 + 1} \lt 0$
$e^y \gt 0 \text{ for all y} \\ ∴ e^y = x + \sqrt{x^2 + 1} \\ ∴ y = \ln(x + \sqrt{x^2 + 1}) \\ ∴ \sinh^{-1}(x) = \ln(x + \sqrt{x^2 + 1})$

\begin{align} y & = \tanh^{-1}(x) \\ \tanh(y) & = x \\ \frac{e^y - e^{-y}}{e^y + e^{-y}} & = x \\ e^y - e^{-y} & = x(e^y + e^{-y}) \\ e^y(1 - x) & = e^{-y}(1 + x) \\ e^{2y}(1 - x) & = 1 + x \\ e^{2y} & = \frac{1 + x}{1 - x} \\ y & = \frac{1}{2}\ln\left(\frac{1 + x}{1 - x}\right) \end{align}
$∴ \tanh^{-1}(x) = \frac{1}{2}\ln\left(\frac{1 + x}{1 - x}\right)$

FORMULAE:

$$\cosh^{-1}(x) = \ln(x + \sqrt{x^2 - 1})$$

$$\sinh^{-1}(x) = \ln(x + \sqrt{x^2 + 1})$$

$$\tanh^{-1}(x) = \frac{1}{2}\ln\left(\frac{1 + x}{1 - x}\right)$$

EXAMPLE:

e.g.1 Verify that $$\sinh^{-1}(x) - \cosh^{-1}(x) = \ln(1 + \sqrt{5}) - \ln(2)$$ has solution $$x = \frac{\sqrt{5}}{2}$$. Solve the equation to show that it is the only solution.

\begin{align} \sinh^{-1}\left(\frac{\sqrt{5}}{2}\right) - \cosh^{-1}\left(\frac{\sqrt{5}}{2}\right) & = \ln\left(\frac{\sqrt{5}}{2} + \sqrt{\left(\frac{\sqrt{5}}{2}\right)^2 + 1}\right) - \ln\left(\frac{\sqrt{5}}{2} + \sqrt{\left(\frac{\sqrt{5}}{2}\right)^2 - 1}\right) \\ & = \ln\left(\frac{\sqrt{5}}{2} + \frac{3}{2}\right) - \ln\left(\frac{\sqrt{5}}{2} + \frac{1}{2}\right) \\ & = \ln\left(\frac{\sqrt{5} + 3}{\sqrt{5} + 1}\right) \\ & = \ln\left(\frac{\sqrt{5} + 3}{\sqrt{5} + 1} \times \frac{\sqrt{5} - 1}{\sqrt{5} - 1}\right) \\ & = \ln\left(\frac{2 + 2\sqrt{5}}{4}\right) \\ & = \ln\left(\frac{1 + \sqrt{5}}{2}\right) \\ & = \ln(1 + \sqrt{5}) - \ln(2) \end{align}
∴ $$x = \frac{\sqrt{5}}{2}$$ is a solution to the equation.

\begin{align} \sinh^{-1}(x) - \cosh^{-1}(x) & = \ln(1 + \sqrt{5}) - \ln(2) \\ \ln(x + \sqrt{x^2 + 1}) - \ln(x + \sqrt{x^2 - 1}) & = \ln(1 + \sqrt{5}) - ln(2) \\ \ln\left(\frac{x + \sqrt{x^2 + 1}}{x + \sqrt{x^2 - 1}}\right) & = \ln\left(\frac{1 + \sqrt{5}}{2}\right) \\ \frac{x + \sqrt{x^2 + 1}}{x + \sqrt{x^2 - 1}} & = \frac{1 + \sqrt{5}}{2} \\ 2x + 2\sqrt{x^2 + 1} & = (1 + \sqrt{5})x + (1 + \sqrt{5})\sqrt{x^2 - 1} \\ 2\sqrt{x^2 + 1} - (1 + \sqrt{5})\sqrt{x^2 - 1} & = (\sqrt{5} - 1)x \\ 4(x^2 + 1) + (6 + 2\sqrt{5})(x^2 - 1) - 4(1 + \sqrt{5})\sqrt{x^4 - 1} & = (6 - 2\sqrt{5})x^2 \\ (4 + 4\sqrt{5})x^2 + (-2 - 2\sqrt{5}) & = 4(1 + \sqrt{5})\sqrt{x^4 - 1} \\ 4(1 + \sqrt{5})x^2 - 2(1 + \sqrt{5}) & = 4(1 + \sqrt{5})\sqrt{x^4 - 1} \\ 2x^2 - 1 & = 2\sqrt{x^4 - 1} \\ 4x^4 - 4x^2 + 1 & = 4x^4 - 4 \\ 4x^2 & = 5 \\ x & = \pm\frac{\sqrt{5}}{2} \\ x \ge 1 \text{ for } \cosh^{-1} \text{ to exist } ∴ x & = \frac{\sqrt{5}}{2} \end{align}
∴ $$x = \frac{\sqrt{5}}{2}$$ is the only solution to the equation.

## DIFFERENTIATION OF HYPERBOLIC FUNCTIONS

Differentiation can be done from the definitions to find general rules

\begin{align} \frac{\mathrm{d}}{\mathrm{d}x} (\cosh x) & = \frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{e^x + e^{-x}}{2}\right) \\ & = \frac{e^x - e^{-x}}{2} \\ & = \sinh x \end{align} \begin{align} \frac{\mathrm{d}}{\mathrm{d}x} (\sinh x) & = \frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{e^x - e^{-x}}{2}\right) \\ & = \frac{e^x + e^{-x}}{2} \\ & = \cosh x \end{align} \begin{align} \frac{\mathrm{d}}{\mathrm{d}x} (\tanh x) & = \frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{\sinh x}{\cosh x}\right) \\ & = \frac{(\sinh x)'\cosh x - (\cosh x)'\sinh x}{\cosh^2 x} \quad\text{(by quotient rule)}\\ & = \frac{\cosh^2 x - \sinh^2 x}{\cosh^2 x} \\ & = \frac{1}{\cosh^2 x} \\ & = \sech^2 x \end{align} \begin{align} \frac{\mathrm{d}}{\mathrm{d}x} (\sech x) & = \frac{\mathrm{d}}{\mathrm{d}x} (\cosh x)^{-1} \\ & = -(\cosh x)^{-2}(\cosh x)' \quad\text{(by chain rule)}\\ & = -(\cosh x)^{-2} \sinh x \\ & = -\sech x \tanh x \end{align} \begin{align} \frac{\mathrm{d}}{\mathrm{d}x} (\cosech x) & = \frac{\mathrm{d}}{\mathrm{d}x} (\sinh x)^{-1} \\ & = -(\sinh x)^{-2}(\sinh x)' \quad\text{(by chain rule)}\\ & = -(\sinh x)^{-2} \cosh x \\ & = -\cosech x \coth x \end{align} \begin{align} \frac{\mathrm{d}}{\mathrm{d}x} (\coth x) & = \frac{\mathrm{d}}{\mathrm{d}x} \left(\frac{\cosh x}{\sinh x}\right) \\ & = \frac{(\cosh x)'\sinh x - (\sinh x)'\cosh x}{\sinh^2 x} \quad\text{(by quotient rule)}\\ & = \frac{\sinh^2 x - \cosh^2 x}{\sinh^2 x} \\ & = \frac{-1}{\sinh^2 x} \\ & = -\cosech^2 x \end{align}

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