Chapter 6: Problem 1
Approximate the expression to four decimal places. $$ \begin{array}{ll}{\text { (a) } \sinh 3} & {\text { (b) } \cosh (-2) \quad \text { (c) } \tanh (\ln 4)} \\ {\text { (d) } \sinh ^{-1}(-2)} & {\text { (c) } \cosh ^{-1} 3 \quad \text { (f) } \tanh ^{-1} \frac{3}{4}}\end{array} $$
Short Answer
Expert verified
a) 10.0179
b) 3.7622
c) 0.8820
d) -1.4436
e) 1.7627
f) 0.9729
Step by step solution
01
Calculate \( \sinh(3) \)
Recall the definition of the hyperbolic sine function: \[ \sinh(x) = \frac{e^x - e^{-x}}{2} \]Substitute \( x = 3 \) into the formula:\[ \sinh(3) = \frac{e^3 - e^{-3}}{2} \]Using a calculator, approximate \( e^3 \approx 20.0855 \) and \( e^{-3} \approx 0.0498 \). Substitute these values in:\[ \sinh(3) \approx \frac{20.0855 - 0.0498}{2} \approx \frac{20.0357}{2} \approx 10.0179 \]
02
Calculate \( \cosh(-2) \)
Recall the definition of the hyperbolic cosine function:\[ \cosh(x) = \frac{e^x + e^{-x}}{2} \]Note that \( \cosh(-x) = \cosh(x) \), so \( \cosh(-2) = \cosh(2) \).Substitute \( x = 2 \) into the formula:\[ \cosh(2) = \frac{e^2 + e^{-2}}{2} \]Using a calculator, approximate \( e^2 \approx 7.3891 \) and \( e^{-2} \approx 0.1353 \). Substitute these values in:\[ \cosh(2) \approx \frac{7.3891 + 0.1353}{2} \approx \frac{7.5244}{2} \approx 3.7622 \]
03
Calculate \( \tanh(\ln 4) \)
Recall the definition of the hyperbolic tangent function:\[ \tanh(x) = \frac{\sinh(x)}{\cosh(x)} \]First, find \( \ln 4 \approx 1.3863 \).Substitute \( x = 1.3863 \) into the definitions for \( \sinh \) and \( \cosh \):\[ \sinh(1.3863) = \frac{e^{1.3863} - e^{-1.3863}}{2} \] \[ \cosh(1.3863) = \frac{e^{1.3863} + e^{-1.3863}}{2} \] Using a calculator, approximate \( e^{1.3863} \approx 4 \) and \( e^{-1.3863} \approx 0.25 \):\[ \sinh(1.3863) \approx \frac{4 - 0.25}{2} = 1.875 \] \[ \cosh(1.3863) \approx \frac{4 + 0.25}{2} = 2.125 \] Finally, find \( \tanh(x) \):\[ \tanh(1.3863) \approx \frac{1.875}{2.125} \approx 0.8820 \]
04
Calculate \( \sinh^{-1}(-2) \)
Recall the definition of the inverse hyperbolic sine function: \[ \sinh^{-1}(x) = \ln(x + \sqrt{x^2 + 1}) \]Substitute \( x = -2 \) into the formula:\[ \sinh^{-1}(-2) = \ln(-2 + \sqrt{(-2)^2 + 1}) \]Calculate inside the square root:\[ \sqrt{4 + 1} = \sqrt{5} \approx 2.2361 \]Thus:\[ \sinh^{-1}(-2) \approx \ln(-2 + 2.2361) \approx \ln(0.2361) \approx -1.4436 \]
05
Calculate \( \cosh^{-1}(3) \)
Recall the definition of the inverse hyperbolic cosine function:\[ \cosh^{-1}(x) = \ln(x + \sqrt{x^2 - 1}) \]Substitute \( x = 3 \) into the formula:\[ \cosh^{-1}(3) = \ln(3 + \sqrt{3^2 - 1}) \]Calculate inside the square root:\[ \sqrt{9 - 1} = \sqrt{8} \approx 2.8284 \]Thus:\[ \cosh^{-1}(3) \approx \ln(3 + 2.8284) = \ln(5.8284) \approx 1.7627 \]
06
Calculate \( \tanh^{-1}\left(\frac{3}{4}\right) \)
Recall the definition of the inverse hyperbolic tangent function:\[ \tanh^{-1}(x) = \frac{1}{2} \ln\left(\frac{1+x}{1-x}\right) \]Substitute \( x = \frac{3}{4} \) into the formula:\[ \tanh^{-1}\left(\frac{3}{4}\right) = \frac{1}{2} \ln\left(\frac{1+\frac{3}{4}}{1-\frac{3}{4}}\right) \] Calculate the expression inside the logarithm:\[ \frac{1+\frac{3}{4}}{1-\frac{3}{4}} = \frac{\frac{7}{4}}{\frac{1}{4}} = 7 \]Thus:\[ \tanh^{-1}\left(\frac{3}{4}\right) \approx \frac{1}{2} \ln(7) \approx \frac{1}{2} \times 1.9459 \approx 0.9729 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
sinh function
The hyperbolic sine function, denoted as \( \sinh(x) \), is a fundamental hyperbolic function often used in calculus and engineering. It's defined by the formula:
The calculation of \( \sinh(3) \) involves computing \( e^3 \) and \( e^{-3} \), subtracting these results, and dividing by 2. With \( e^3 \approx 20.0855 \) and \( e^{-3} \approx 0.0498 \), the computation gives:
- \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)
The calculation of \( \sinh(3) \) involves computing \( e^3 \) and \( e^{-3} \), subtracting these results, and dividing by 2. With \( e^3 \approx 20.0855 \) and \( e^{-3} \approx 0.0498 \), the computation gives:
- \( \sinh(3) \approx \frac{20.0855 - 0.0498}{2} \approx 10.0179 \)
cosh function
The hyperbolic cosine function, \( \cosh(x) \), is defined similarly to \( \sinh \) but instead involves the sum of exponential functions:
To compute \( \cosh(-2) \), we note that \( \cosh(-2) = \cosh(2) \). Substituting \( x = 2 \) results in:
- \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)
To compute \( \cosh(-2) \), we note that \( \cosh(-2) = \cosh(2) \). Substituting \( x = 2 \) results in:
- \( \cosh(2) \approx \frac{7.3891 + 0.1353}{2} \approx 3.7622 \)
tanh function
The hyperbolic tangent function, \( \tanh(x) \), describes the relationship between the hyperbolic sine and cosine functions:
For practical purposes, to find \( \tanh(\ln 4) \), you first compute the \( \ln 4 \approx 1.3863 \). Using this, find \( \sinh(1.3863) \) and \( \cosh(1.3863) \) and divide the two:
- \( \tanh(x) = \frac{\sinh(x)}{\cosh(x)} \)
For practical purposes, to find \( \tanh(\ln 4) \), you first compute the \( \ln 4 \approx 1.3863 \). Using this, find \( \sinh(1.3863) \) and \( \cosh(1.3863) \) and divide the two:
- \( \tanh(1.3863) \approx \frac{1.875}{2.125} \approx 0.8820 \)
inverse hyperbolic functions
Inverse hyperbolic functions allow us to retrieve the input used in a hyperbolic function from its output. These functions include \( \sinh^{-1}(x) \), \( \cosh^{-1}(x) \), and \( \tanh^{-1}(x) \).
For example, calculating \( \sinh^{-1}(-2) \) involves:
- \( \sinh^{-1}(x) = \ln(x + \sqrt{x^2 + 1}) \)
- \( \cosh^{-1}(x) = \ln(x + \sqrt{x^2 - 1}) \)
- \( \tanh^{-1}(x) = \frac{1}{2} \ln \left(\frac{1+x}{1-x}\right) \)
For example, calculating \( \sinh^{-1}(-2) \) involves:
- \( \sinh^{-1}(-2) \approx \ln(-2 + \sqrt{5}) \approx -1.4436 \)
- \( \cosh^{-1}(3) \approx \ln(3 + \sqrt{8}) \approx 1.7627 \)