Hyperbolic Functions:
CONTENTS
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.
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.
\(\cosh^2(x) - \sinh^2(x) = 1\)
\(1 - \tanh^2(x) = \sech^2(x)\)
\(\coth^2(x) - 1 = \cosech^2(x)\)
\(\begin{align}\cosh(2x) & = \cosh^2(x) + \sinh^2(x) \\ & = 2\cosh^2(x) - 1 \\ & = 1 + 2\sinh^2(x)\end{align}\)
\(\sinh(2x) = 2\sinh(x)\cosh(x)\)
\(\tanh(2x) = \frac{2\tanh(x)}{1 + \tanh^2(x)}\)
Solving equations uses similar ideas to trigonometric equations.
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).`
\[∴ \cosh^{-1}(x) = \ln(x + \sqrt{x^2 - 1})\]
\[\begin{align}
y & = \sinh^{-1}(x) \\
\sinh(y) & = x \\
\frac{e^y - e^{-y}}{2} & = x \\
e^{2y} - 2xe^y - 1 & = 0 \\
e^y & = \frac{2x \pm \sqrt{4x^2 + 4}}{2} \\
& = x \pm \sqrt{x^2 + 1}
\end{align}\]
\[\sqrt{x^2 + 1} \gt x \\
∴ x - \sqrt{x^2 + 1} \lt 0\]
\[e^y \gt 0 \text{ for all y} \\
∴ e^y = x + \sqrt{x^2 + 1} \\
∴ y = \ln(x + \sqrt{x^2 + 1}) \\
∴ \sinh^{-1}(x) = \ln(x + \sqrt{x^2 + 1})\]
\[\begin{align}
y & = \tanh^{-1}(x) \\
\tanh(y) & = x \\
\frac{e^y - e^{-y}}{e^y + e^{-y}} & = x \\
e^y - e^{-y} & = x(e^y + e^{-y}) \\
e^y(1 - x) & = e^{-y}(1 + x) \\
e^{2y}(1 - x) & = 1 + x \\
e^{2y} & = \frac{1 + x}{1 - x} \\
y & = \frac{1}{2}\ln\left(\frac{1 + x}{1 - x}\right)
\end{align}\]
\[∴ \tanh^{-1}(x) = \frac{1}{2}\ln\left(\frac{1 + x}{1 - x}\right)\]
\(\cosh^{-1}(x) = \ln(x + \sqrt{x^2 - 1})\)
\(\sinh^{-1}(x) = \ln(x + \sqrt{x^2 + 1})\)
\(\tanh^{-1}(x) = \frac{1}{2}\ln\left(\frac{1 + x}{1 - x}\right)\)
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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