Question

Evaluating hyperbolic functions Evaluate each expression without using a calculator or state that the value does not exist. Simplify answers as much as possible. a. \(\cosh 0\) b. \(\tanh 0\) c. csch 0 d. sech (sinh \(0)\) e. coth ( \(\ln 5)\) f. \(\sinh (2 \ln 3)\) g. \(\cosh ^{2} 1\) h. \(\operatorname{sech}^{-1}(\ln 3)\) i. \(\cosh ^{-1} \frac{17}{8}\) j. \(\sinh ^{-1} \frac{e^{2}-1}{2 e}\)

Short answer

Question: Evaluate the following expressions: a. \(\cosh 0\) b. \(\tanh 0\) c. csch \(0\) d. sech (sinh \(0\)) e. coth (\(\ln 5\)) f. \(\sinh (2 \ln 3)\) g. \(\cosh^2 1\) h. \(\operatorname{sech}^{-1}(\ln 3)\) i. \(\cosh^{-1} \frac{17}{8}\) j. \(\sinh^{-1} \frac{e^2-1}{2e}\) Answer: a. 1 b. 0 c. undefined d. 1 e. \(\frac{26}{24}\) f. \(\frac{80}{18}\) g. \(\frac{e^2+2+\frac{1}{e^2}}{4}\) h. not exist i. \(\ln\left(\frac{17+\sqrt{257}}{2}\right)\) j. -2

Step-by-step solution

a. Evaluating \(\cosh 0\)

Using the definition of \(\cosh x\), we can write \(\cosh 0 = \frac{e^0 + e^{-0}}{2}\). We know that \(e^0 = 1\), so we get \(\cosh 0 = \frac{1+1}{2} = 1\).

b. Evaluating \(\tanh 0\)

Using the definition of \(\tanh x\), we can write \(\tanh 0 = \frac{\sinh 0}{\cosh 0}\). From step a, we know that \(\cosh 0 = 1\), and using a similar calculation for sinh, we get \(\sinh 0 = 0\). Therefore, \(\tanh 0 = \frac{0}{1} = 0\).

c. Evaluating csch \(0\)

csch is the reciprocal of \(\sinh x\), so csch \(0 = \frac{1}{\sinh 0}\). From step b, we know that \(\sinh 0 = 0\). Hence, csch \(0 = \frac{1}{0}\) which is undefined.

d. Evaluating sech (sinh \(0)\)

sech is the reciprocal of \(\cosh x\), so sech (sinh \(0\)) = \(\frac{1}{\cosh (\sinh 0)}\). From step b, we know that \(\sinh 0 = 0\), so we get sech (sinh \(0\)) = \(\frac{1}{\cosh 0}\). From step a, we know that \(\cosh 0 = 1\), so we get sech (sinh \(0\)) = \(\frac{1}{1} = 1\).

e. Evaluating coth (\(\ln 5\))

Using the definition of coth \(x\), we can write coth (\(\ln 5\)) = \(\frac{\cosh (\ln 5)}{\sinh (\ln 5)}\). We can use the definitions of \(\cosh x\) and \(\sinh x\) to substitute expressions in terms of exponentials as follows: coth (\(\ln 5\)) = \(\frac{\frac{e^{\ln5}+e^{-\ln5}}{2}}{\frac{e^{\ln5}-e^{-\ln5}}{2}}\). Simplifying, we get coth (\(\ln 5\)) = \(\frac{e^{\ln5}+e^{-\ln5}}{e^{\ln5}-e^{-\ln5}} = \frac{5+1/5}{5-1/5} = \frac{5.2}{4.8} = \frac{26}{24}\).

f. Evaluating \(\sinh (2 \ln 3)\)

Using the definition of \(\sinh x\), we can write \(\sinh (2 \ln 3) = \frac{e^{2\ln3}-e^{-2\ln3}}{2}\). We can make use of the properties of logarithms as \(e^{2\ln3}=(e^{\ln3})^2=3^2=9\) and \(e^{-2\ln3}=\frac{1}{e^{2\ln3}}=\frac{1}{9}\) and find that \(\sinh (2 \ln 3)=\frac{9-\frac{1}{9}}{2}=\frac{80}{18}\).

g. Evaluating \(\cosh^2 1\)

Using the definition of \(\cosh x\), we can write \(\cosh^2 1 = \left(\frac{e^1 + e^{-1}}{2}\right)^2\). Therefore, \(\cosh^2 1 = \left(\frac{e + \frac{1}{e}}{2}\right)^2 = \frac{e^2+2+\frac{1}{e^2}}{4}\), which cannot be simplified further.

h. Evaluating \(\operatorname{sech}^{-1}(\ln 3)\)

If we let \(y = \operatorname{sech}^{-1}(\ln 3)\), then we have \(sech(y) = \ln 3\). However, we know that sech(\(y\)) is the reciprocal of \(\cosh y\) and always lies between \(0\) and \(1\), and ln4000 since \(\ln 3 > 1\). Therefore, the value does not exist.

i. Evaluating \(\cosh^{-1} \frac{17}{8}\)

Let \(y = \cosh^{-1} \frac{17}{8}\). Then, we have \(\cosh y = \frac{17}{8}\). Using the definition of \(\cosh x\), we can write \(\frac{e^y + e^{-y}}{2} = \frac{17}{8}\). By solving this equation for \(y\), we can find the value of \(\cosh^{-1} \frac{17}{8}\). Multiplying both sides by \(2\) and \(e^y\), we get \(e^{2y} - 17e^y + 1 = 0\). This is a quadratic equation in \(e^y\). Solving it using the quadratic formula, we get two possible results for \(e^y\): \(\frac{17\pm\sqrt{17^2-4}}{2}\). Since \(e^y > 0\), we take the positive square root, and find that \(e^y = \frac{17+\sqrt{257}}{2}\). Therefore, \(y = \ln\left(\frac{17+\sqrt{257}}{2}\right)\), which is the value of \(\cosh^{-1} \frac{17}{8}\).

j. Evaluating \(\sinh^{-1} \frac{e^2-1}{2e}\)

Let \(y = \sinh^{-1} \frac{e^2-1}{2e}\). Then, we have \(\sinh y = \frac{e^2-1}{2e}\). Using the definition of \(\sinh x\), we can write \(\frac{e^y - e^{-y}}{2} = \frac{e^2-1}{2e}\). By solving this equation for \(y\), we can find the value of \(\sinh^{-1} \frac{e^2-1}{2e}\). Multiply both sides by \(2e\), \(e^y\) and simplify to get, \(e^{2y}-2e^2e^{y}+1 =0\). This is a quadratic equation in \(e^y\) and has roots \(e^y = e^2, 1/e^2\). Since the expression is \(\sinh^{-1}(x)\), we know that \(0 < \frac{e^2-1}{2e} < 2\), which means that \(0 < e^y < 2\). This leads to us to choose the solution \(e^y = 1/e^2\), and therefore \(y = \ln\frac{1}{e^2} = -2\), which is the value of \(\sinh^{-1} \frac{e^2-1}{2e}\).

Key concepts

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the variables. They simplify the process of evaluating expressions and solving equations involving these functions.

• The basic trigonometric functions, like sine and cosine, are crucial for understanding these identities.
• Hyperbolic functions, such as \(\sinh\) and \(\cosh\), have their own set of identities similar to trigonometric identities.

Some common hyperbolic identities include:

• \(\cosh^2 x - \sinh^2 x = 1\)
• \(1 - \tanh^2 x = \text{sech}^2 x\)

These identities help in evaluating expressions like \(\cosh 0\) and \(\tanh 0\) directly without a calculator, as seen from their definitions. Understanding identities simplifies complex calculations and provides a basis for derivations and proofs.

Inverse Hyperbolic Functions

Inverse hyperbolic functions serve to "undo" the hyperbolic functions. They work like regular inverse functions, allowing us to find the original value that produced a given output in a hyperbolic function.

For instance:

• \(\sinh^{-1} x\) returns the value \(y\) such that \(\sinh y = x\).
• \(\cosh^{-1} x\) returns the value \(y\) such that \(\cosh y = x\).

These functions have specific domains and ranges:

• \(\cosh^{-1} x\) is defined for \(x \geq 1\).
• \(\sinh^{-1} x\) is defined for all real numbers \(x\).

They're useful in calculating exact values like \(\cosh^{-1} \frac{17}{8}\) and \(\sinh^{-1} \frac{e^2-1}{2e}\), converting expressions back to simpler forms involving log functions and exponentials.

Exponential Functions

Exponential functions, generally in the form of \(e^x\), are essential in both calculus and algebra. They describe growth and decay processes and are defined for all real numbers.

Key properties include:

• \(e^0 = 1\), a property used to evaluate \(\cosh 0\).
• The inverse of \(e^x\) is the natural logarithm, \(\ln x\), providing solutions to equations involving exponentials.

Exponentials are deeply connected with hyperbolic functions:

• \(\cosh x = \frac{e^x + e^{-x}}{2}\)
• \(\sinh x = \frac{e^x - e^{-x}}{2}\)

These expressions allow us to evaluate hyperbolic functions using exponential identities, as seen in solving \(\sinh (2 \ln 3)\). They also appear in results for problems involving inverse hyperbolic functions, such as \(\cosh^{-1} x\) and \(\sinh^{-1} x\).