Problem 75 - Calculus Early Transcendentals

Question

Evaluate each expression without using a calculator, or state that the value does not exist. Simplify answers to the extent possible. a. \(\mathrm{cosh 0}\) b. \(\mathrm{tanh 0}\) c. \(\mathrm{csch 0}\) d. \(\mathrm{sech}(sinh 0)\) e. \(\operatorname{coth}(\ln 5) \quad\) f. \(\sinh (2 \ln 3)\) g. \(\cosh ^{2} 1 \quad\) h. \(\operatorname{sech}^{-1}(\ln 3)\) i. \(\cosh ^{-1}(17 / 8)\) j. \(\sinh ^{-1}\left(\frac{e^{2}-1}{2 e}\right)\)

Step-by-Step Solution

Answer

a. \(\mathrm{cosh} 0 = \boldsymbol{1}\) b. \(\mathrm{tanh} 0 = \boldsymbol{0}\) c. \(\mathrm{csch} 0\) is undefined or does not exist. d. \(\mathrm{sech}(\sinh 0) = \boldsymbol{1}\) e. \(\mathrm{coth}(\ln 5) = \frac{\cosh{(\ln 5)}}{\sinh{(\ln 5)}}\) f. \(\sinh(2\ln 3) = \frac{40}{9}\) g. \(\cosh^2 1 = \left(\frac{e + \frac{1}{e}}{2}\right)^2\) h. \(\operatorname{sech}^{-1}(\ln 3)\) i. \(\cosh^{-1}(\frac{17}{8})\) j. \(\sinh^{-1}\left(\frac{e^2 - 1}{2 e}\right)\)

Step 1: a. Evaluate \(\mathrm{cosh 0}\)

By definition, \(\mathrm{cosh x} = \frac{e^x + e^{-x}}{2}\). Plug in \(x = 0\) to get \(\cosh 0 = \frac{e^0 + e^{-0}}{2} = \frac{1 + 1}{2} = 1\).

Step 2: b. Evaluate \(\mathrm{tanh 0}\)

By definition, \(\mathrm{tanh x} = \frac{\sinh x}{\cosh x}\). Since we know that \(\sinh 0 = 0\) (from the definition of \(\sinh\)) and \(\cosh 0 = 1\), we find that \(\tanh 0 = \frac{0}{1} = 0\).

Step 3: c. Evaluate \(\mathrm{csch 0}\)

By definition, \(\mathrm{csch x} = \frac{1}{\sinh x}\). Since we know that \(\sinh 0 = 0\) (from the definition of \(\sinh\)), evaluating \(\mathrm{csch 0}\) would result in an undefined expression, so the value does not exist.

Step 4: d. Evaluate \(\mathrm{sech}(\sinh 0)\)

Recall that \(\sinh 0 = 0\) and \(\mathrm{sech x} = \frac{1}{\cosh{x}}\). Plug in \(x = 0\) to get \(\mathrm{sech}(\sinh 0) = \mathrm{sech}(0) = \frac{1}{\cosh 0} = \frac{1}{1} = 1\).

Step 5: e. Evaluate \(\mathrm{coth}(\ln 5)\)

By definition, \(\mathrm{coth x} = \frac{\cosh{x}}{\sinh{x}}\). Plug in \(x = \ln 5\) to get \(\mathrm{coth}(\ln 5) = \frac{\cosh{(\ln 5)}}{\sinh{(\ln 5)}}\). There is no further simplification for this expression.

Step 6: f. Evaluate \(\sinh(2\ln 3)\)

By definition, \(\sinh x = \frac{e^x - e^{-x}}{2}\). Plug in \(x = 2\ln 3\) to get \(\sinh(2\ln 3) = \frac{e^{2\ln 3} - e^{-2\ln 3}}{2} = \frac{3^2 - 3^{-2}}{2} = \frac{9 - 1/9}{2} = \frac{80}{18} = \frac{40}{9}\).

Step 7: g. Evaluate \(\cosh^2 1\)

By definition, \(\cosh x = \frac{e^x + e^{-x}}{2}\). Plug in \(x = 1\) to get \(\cosh^2 1 = (\cosh 1)^2 = \left(\frac{e^1 + e^{-1}}{2}\right)^2 = \left(\frac{e + \frac{1}{e}}{2}\right)^2\). There is no further simplification for this expression.

Step 8: h. Evaluate \(\operatorname{sech}^{-1}(\ln 3)\)

The given expression involves taking the inverse hyperbolic secant function of a number which has no straightforward simplification. Hence, we leave the expression as it is: \(\operatorname{sech}^{-1}(\ln 3)\).

Step 9: i. Evaluate \(\cosh^{-1}(\frac{17}{8})\)

We need to find the inverse hyperbolic cosine of a given number, but there is no straightforward simplification for this expression. So, we leave the expression as it is: \(\cosh^{-1}(\frac{17}{8})\).

Step 10: j. Evaluate \(\sinh^{-1}\left(\frac{e^2 - 1}{2 e}\right)\)

The given expression involves taking the inverse hyperbolic sine function of a number, but there is no further simplification possible for this expression. Hence, we leave the expression as it is: \(\sinh^{-1}\left(\frac{e^2 - 1}{2 e}\right)\).

Key Concepts

Understanding Hyperbolic Cosine (cosh)

Hyperbolic cosine, denoted as \( \cosh x \), is a fundamental hyperbolic function. It is defined by the equation: - \( \cosh x = \frac{e^x + e^{-x}}{2} \) where \( e \) is the base of the natural logarithm. This function is similar to the ordinary cosine function but uses exponential functions instead. The hyperbolic cosine function has several important properties:

• \( \cosh x \) is always positive because both \( e^x \) and \( e^{-x} \) are positive.
• The range of \( \cosh x \) is \([1, \infty)\).
• \( \cosh 0 = 1 \), since \( \frac{1 + 1}{2} = 1 \).

The graph of \( \cosh x \) is symmetrical with respect to the \( y \)-axis and resembles a "U" shape. This function grows exponentially as \( x \) moves away from zero in both the positive and negative directions. When dealing with exercises involving \( \cosh \), such as finding \( \cosh 0 \), it helps to be familiar with its formula and properties.

Exploring Hyperbolic Tangent (tanh)

The hyperbolic tangent, represented as \( \tanh x \), provides a way to compare the relative sizes of \( \sinh x \) and \( \cosh x \). Its formula is: - \( \tanh x = \frac{\sinh x}{\cosh x} \) This function effectively compresses the output of the hyperbolic sine over the output of the hyperbolic cosine. Some key details about \( \tanh x \) include:

• The range of \( \tanh x \) is \((-1, 1)\).
• \( \tanh x \) is an odd function, meaning \( \tanh(-x) = -\tanh x \).
• When \( x = 0 \), \( \tanh 0 = 0 \) because \( \sinh 0 = 0 \) and \( \cosh 0 = 1 \).

The graph of \( \tanh x \) is sigmoid-shaped, starting near \(-1\), crossing the y-axis, and approaching 1. The function is continuous and differentiable everywhere along its domain. It is particularly useful in various areas of mathematics and applied sciences, including calculus and complex analysis.

Investigating Hyperbolic Cosecant (csch)

The hyperbolic cosecant, denoted by \( \csch x \), is the reciprocal of the hyperbolic sine function \( \sinh x \). Its formula is: - \( \csch x = \frac{1}{\sinh x} \) Just like its circular counterpart in trigonometry, \( \csch x \) poses some interesting properties and considerations. Important properties and considerations include:

• The function is undefined at \( x = 0 \) because \( \sinh 0 = 0 \), leading to division by zero.
• The range of \( \csch x \) is \((-\infty, -1] \cup [1, \infty) \).
• \( \csch x \) is an odd function; hence, \( \csch(-x) = -\csch x \).

\( \csch x \) diverges away from the origin as \( x \) moves away from zero, creating vertical asymptotes at \( x = 0 \). The function is not suitable for all situations due to its undefined nature at zero. In exercises, understanding when \( \csch \) does not exist is crucial, particularly for evaluations like \( \csch 0 \).