Problem 2 - Thomas Calculus

Question

Each of Exercises \(1-4\) gives a value of sinh \(x\) or cosh \(x .\) Use the definitions and the identity \(\cosh ^{2} x-\sinh ^{2} x=1\) to find the values of the remaining five hyperbolic functions. $$\sinh x=\frac{4}{3}$$

Step-by-Step Solution

Answer

cosh x = 5/3; tanh x = 4/5; coth x = 5/4; sech x = 3/5; csch x = 3/4.

Step 1: Use the Identity to Find cosh x

We know from the identity that \( \cosh^{2} x - \sinh^{2} x = 1 \). Given \( \sinh x = \frac{4}{3} \), plug this into the equation: \[\cosh^{2} x - \left(\frac{4}{3}\right)^{2} = 1\] This simplifies to: \[\cosh^{2} x - \frac{16}{9} = 1 \] which means \( \cosh^{2} x = 1 + \frac{16}{9} = \frac{25}{9} \).

Step 2: Solve for cosh x

Taking the square root of both sides gives:\[ \cosh x = \sqrt{\frac{25}{9}} = \pm \frac{5}{3} \]Since \( \cosh x \) is always positive:\[ \cosh x = \frac{5}{3}\]

Step 3: Calculate the Remaining Hyperbolic Functions

Using the definitions of the hyperbolic functions, we can find the others:1. \( \tanh x = \frac{\sinh x}{\cosh x} = \frac{4/3}{5/3} = \frac{4}{5} \)2. \( \coth x = \frac{\cosh x}{\sinh x} = \frac{5/3}{4/3} = \frac{5}{4} \)3. \( \text{sech} \, x = \frac{1}{\cosh x} = \frac{3}{5} \)4. \( \text{csch} \, x = \frac{1}{\sinh x} = \frac{3}{4} \)

Key Concepts

sinh x

The hyperbolic sine function, known as \( \sinh x \), is one of the core hyperbolic functions. Understanding what \( \sinh x \) means in terms of exponential functions is a good start. When we say \( \sinh x \), we are talking about the function defined as:

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

This formula connects hyperbolic functions with exponential functions, offering a different perspective compared to regular trigonometric functions like sine and cosine.When you're given a value like \( \sinh x = \frac{4}{3} \), you know the output of this exponential combination.The task is then to use this information, along with other hyperbolic function definitions and identities, to solve for other functions. By doing this, we can unlock more information about \( \cosh x \) and other derived hyperbolic functions.

cosh x

The hyperbolic cosine function, represented as \( \cosh x \), plays a crucial role in the realm of hyperbolic functions. Similar to \( \sinh x \), \( \cosh x \) has its own exponential representation:

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

What's interesting about \( \cosh x \) is how it interacts with \( \sinh x \). These two are linked by a fundamental identity used extensively in calculus and algebra:

• \( \cosh^2 x - \sinh^2 x = 1 \)

This hyperbolic identity resembles the Pythagorean identity used in trigonometry — \( \cos^2 x + \sin^2 x = 1 \) — but with a notable difference in sign.In practical applications, starting with a given \( \sinh x \) value allows you to solve for \( \cosh x \) using this identity.For example, if \( \sinh x = \frac{4}{3} \), solving the identity reveals that \( \cosh x = \frac{5}{3} \) after recognizing that \( \cosh x \) must be positive.

hyperbolic identity

The hyperbolic identity \( \cosh^2 x - \sinh^2 x = 1 \) is a cornerstone of hyperbolic function analysis. This identity is key in converting known values of \( \sinh x \) or \( \cosh x \) into insights about other hyperbolic functions. Because \( \sinh x \) and \( \cosh x \) are defined through exponential relationships:

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

They naturally satisfy this identity, providing a reliable tool for calculation. Whenever you're given one of \( \sinh x \) or \( \cosh x \), use this identity to find the other.In cases where you have \( \sinh x = \frac{4}{3} \), substitute in the identity to discover \( \cosh x \). Using \( \cosh^2 x - (\frac{4}{3})^2 = 1 \), you'll arrive at \( \cosh x = \frac{5}{3} \) after taking the positive root.With both values, you can also determine other hyperbolic functions, making this identity an essential tool in your mathematical toolkit.