Single Variable Calculus Questions and Answers

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Single Variable Calculus, Chapter 7, 7.7, Section 7.7, Problem 2

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Determine the numerical value of a.) $\tan h 0$ and b.) $\tan h 1$

a.) $\tan h 0$

Using Hyperbolic Function

$ \begin{equation} \begin{aligned} \tan h x =& \frac{\sin h x}{\cos h x} = \frac{e^x - e^{-x}}{e^x + e^{-x}} \\ \\ \tan h 0 =& \frac{e^0 - e^{-0}}{e^0 + e^{-0}} \\ \\ \tan h 0 =& \frac{\displaystyle e^0 - \frac{1}{e^0}}{\displaystyle e^0 + \frac{1}{e^0}} \\ \\ \tan h 0 =& \frac{1 - 1}{1 + 1} \\ \\ \tan h 0 =& \frac{0}{2} \\ \\ \tan h 0 =& 0 \end{aligned} \end{equation} $

b.) $\tan h 1$

Using Hyperbolic Function

$ \begin{equation} \begin{aligned} \tan h x =& \frac{\sin h x}{\cos h x} = \frac{e^x - e^{-x}}{e^x + e^{-x}} \\ \\ \tan 1 =& \frac{e^1 - e^{-1}}{e^1 + e^{-1}} \\ \\ \tan 1 =& \frac{\displaystyle e - \frac{1}{e}}{\displaystyle e + \frac{1}{e}} \\ \\ \tan 1 =& \frac{\displaystyle \frac{e^2 - 1}{\cancel{e}}}{\displaystyle \frac{e^2 + 1}{\cancel{e}}} \\ \\ \tan 1 =& \frac{e^2 - 1}{e^2 + 1} \\ \\ \tan 1 =& 0.7616 \end{aligned} \end{equation} $