Exponential and Logarithmic Functions: Learn It 5

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Exponential and Logarithmic Functions: Learn It 5

Hyperbolic Functions Cont.

Identities Involving Hyperbolic Functions

Just as trigonometric functions have identities that allow for the simplification and transformation of expressions, hyperbolic functions also possess their own set of identities.

hyperbolic function identities

Hyperbolic Reflection Identities:

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

Hyperbolic Pythagorean Identities:

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

Hyperbolic Squared Identities:

$1-\tanh^2 x=\text{sech}^2 x$
$\coth^2 x-1=\text{csch}^2 x$

Hyperbolic Addition Formulas:

$\sinh(x \pm y)=\sinh x \cosh y \pm \cosh x \sinh y$
$\cosh (x \pm y)=\cosh x \cosh y \pm \sinh x \sinh y$

Exponential Definitions of Hyperbolic Functions

$\cosh x+\sinh x=e^x$
$\cosh x-\sinh x=e^{−x}$

Simplify $\sinh(5 \ln x)$ .
If $\sinh x=\frac{3}{4}$ , find the values of the remaining five hyperbolic functions.

Using the definition of the $\sinh$ function, we write
$\sinh(5 \ln x)=\large \frac{e^{5 \ln x}-e^{-5 \ln x}}{2} \normalsize = \large \frac{e^{\ln(x^5)}-e^{\ln(x^{-5})}}{2} \normalsize =\large \frac{x^5-x^{-5}}{2}$ .
Using the identity $\cosh^2 x-\sinh^2 x=1$ , we see that
$\cosh^2 x=1+\big(\frac{3}{4}\big)^2=\frac{25}{16}$ .

Since $\cosh x \ge 1$ for all $x$ , we must have $\cosh x=5/4$ . Then, using the definitions for the other hyperbolic functions, we conclude that $\tanh x=3/5, \, \text{csch} \, x=4/3, \, \text{sech} \, x=4/5$ , and $\coth x=5/3$ .

Watch the following video to see the worked solution to this example.

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Inverse Hyperbolic Functions

From the graphs of the hyperbolic functions, we see that all of them are one-to-one except $\cosh x$ and $\text{sech} \, x$ . If we restrict the domains of these two functions to the interval $[0,\infty)$ , then all the hyperbolic functions are one-to-one, and we can define the inverse hyperbolic functions. Since the hyperbolic functions themselves involve exponential functions, the inverse hyperbolic functions involve logarithmic functions.

inverse hyperbolic functions

\begin{matrix} {sinh}^{- 1} x = arcsinh x = ln (x + \sqrt{x^{2} + 1}) & {cosh}^{- 1} x = arccosh x = ln (x + \sqrt{x^{2} - 1}) {tanh}^{- 1} x = arctanh x = \frac{1}{2} ln (\frac{1 + x}{1 - x}) & {coth}^{- 1} x = arccot x = \frac{1}{2} ln (\frac{x + 1}{x - 1}) {sech}^{- 1} x = arcsech x = ln (\frac{1 + \sqrt{1 - x^{2}}}{x}) & {csch}^{- 1} x = arccsch x = ln (\frac{1}{x} + \frac{\sqrt{1 + x^{2}}}{| x |}) \end{matrix}

$\begin{array}{cccc}\sinh^{-1} x=\text{arcsinh } x=\ln(x+\sqrt{x^2+1})\hfill & & & \cosh^{-1} x=\text{arccosh } x=\ln(x+\sqrt{x^2-1})\hfill \\ \tanh^{-1} x=\text{arctanh } x=\frac{1}{2}\ln\big(\frac{1+x}{1-x}\big)\hfill & & & \coth^{-1} x=\text{arccot } x=\frac{1}{2}\ln\big(\frac{x+1}{x-1}\big)\hfill \\ \text{sech}^{-1} x=\text{arcsech } x=\ln\big(\frac{1+\sqrt{1-x^2}}{x}\big)\hfill & & & \text{csch}^{-1} x=\text{arccsch } x=\ln\big(\frac{1}{x}+\frac{\sqrt{1+x^2}}{|x|}\big)\hfill \end{array}$

Let’s look at how to derive the first equation, $\sinh^{-1} x=\text{arcsinh } x=\ln(x+\sqrt{x^2+1})$ . The others follow similarly.

Suppose $y=\sinh^{-1} x$ . Then, $x=\sinh y$ and, by the definition of the hyperbolic sine function, $x=\frac{e^y-e^{−y}}{2}$ .

Multiplying both sides of this equation by $2$ to get rid of the fraction and setting that equal to zero, we get

e^{y} - 2 x - e^{- y} = 0

$e^y-2x-e^{−y}=0$

Multiplying this equation by $e^y$ , we obtain

e^{2 y} - 2 x e^{y} - 1 = 0

$e^{2y}-2xe^y-1=0$

This can be solved like a quadratic equation, with the solution

e^{y} = \frac{2 x \pm \sqrt{4 x^{2} + 4}}{2} = x \pm \sqrt{x^{2} + 1}

$e^y=\large \frac{2x \pm \sqrt{4x^2+4}}{2} \normalsize =x \pm \sqrt{x^2+1}$

Since $e^y>0$ , the only solution is the one with the positive sign.

Applying the natural logarithm to both sides of the equation, we conclude that

y = ln (x + \sqrt{x^{2} + 1})

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

Evaluate each of the following expressions

$\sinh^{-1}(2)$
$\tanh^{-1}\left(\frac{1}{4}\right)$

$\sinh^{-1}(2)=\ln(2+\sqrt{2^2+1})=\ln(2+\sqrt{5}) \approx 1.4436$

$\tanh^{-1}(\frac{1}{4})=\frac{1}{2}\ln(\frac{1+1/4}{1-1/4})=\frac{1}{2}\ln(\frac{5/4}{3/4})=\frac{1}{2}\ln(\frac{5}{3}) \approx 0.2554$