Problem 4 - 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. $$\cosh x=\frac{13}{5}, \quad x>0$$

Step-by-Step Solution

Answer

sinh x = 12/5, tanh x = 12/13, sech x = 5/13, csch x = 5/12, coth x = 13/12.

Step 1: Find sinh x using identity

Given that \( \cosh x = \frac{13}{5} \) and \( x > 0 \), use the identity \( \cosh^2 x - \sinh^2 x = 1 \). Substitute the given \( \cosh x \) value into the identity:\[ \left( \frac{13}{5} \right)^2 - \sinh^2 x = 1 \]Simplify to find \( \sinh x \):\[ \frac{169}{25} - \sinh^2 x = 1 \]\[ \sinh^2 x = \frac{169}{25} - \frac{25}{25} = \frac{144}{25} \]Taking the square root, \( x > 0 \), so:\[ \sinh x = \frac{12}{5} \]

Step 2: Find tanh x

Now, use the formula \( \tanh x = \frac{\sinh x}{\cosh x} \).Given \( \sinh x = \frac{12}{5} \) and \( \cosh x = \frac{13}{5} \), we have:\[ \tanh x = \frac{\frac{12}{5}}{\frac{13}{5}} = \frac{12}{13} \]

Step 3: Find sech x

To find \( \sech x \), use the formula \( \sech x = \frac{1}{\cosh x} \).\[ \sech x = \frac{1}{\frac{13}{5}} = \frac{5}{13} \]

Step 4: Find csch x

To find \( \csch x \), use the formula \( \csch x = \frac{1}{\sinh x} \).\[ \csch x = \frac{1}{\frac{12}{5}} = \frac{5}{12} \]

Step 5: Find coth x

For \( \coth x \), use the formula \( \coth x = \frac{\cosh x}{\sinh x} \).\[ \coth x = \frac{\frac{13}{5}}{\frac{12}{5}} = \frac{13}{12} \]

Key Concepts

sinh x

The hyperbolic sine function, denoted as \( \sinh x \), is one of the fundamental hyperbolic functions, much like the sine function in trigonometry. It can be defined using the exponential function as follows:

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

This formula illustrates that \( \sinh x \) is derived from the difference of two exponential functions, divided by 2. In terms of personal intuition, \( \sinh x \) can be considered a measure of an imaginary angle in hyperbolic geometry.

In our problem, we found sinh \( x \) using an identity involving \( \cosh x \). By solving \( \cosh^2 x - \sinh^2 x = 1 \), we calculated that \( \sinh x = \frac{12}{5} \). This shows how \( \sinh x \) and \( \cosh x \) are intricately linked, similar to sin and cos in circular trigonometry.

cosh x

The hyperbolic cosine function, denoted as \( \cosh x \), is similar to the cosine function in trigonometry, but defined using hyperbolic geometry. Its definition through exponential functions is:

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

This describes \( \cosh x \) as the average of two exponential functions. Unlike \( \sinh x \), the \( \cosh x \) function is always positive and does not oscillate, akin to the trigonometric cosine.

In the given exercise, \( \cosh x = \frac{13}{5} \). This value was utilized to find other hyperbolic functions using hyperbolic identities, showcasing the importance of \( \cosh x \) in computations involving hyperbolic functions.

hyperbolic identities

Hyperbolic identities are relationships involving hyperbolic functions that mirror trigonometric identities. One of the most important hyperbolic identities is:

• \( \cosh^2 x - \sinh^2 x = 1 \)

This equation is the hyperbolic equivalent of the Pythagorean identity \( \cos^2 x + \sin^2 x = 1 \) in trigonometry.

In our exercise, this identity was crucial to solve for \( \sinh x \). It demonstrates the deep connection between \( \cosh x \) and \( \sinh x \), allowing us to solve for remaining functions when one is known. Understanding how to apply this identity is essential in many areas of mathematics, including calculus and hyperbolic geometry.

tanh x

The hyperbolic tangent function, denoted as \( \tanh x \), is analogous to the tangent function in trigonometry. It is defined as:

• \( \tanh x = \frac{\sinh x}{\cosh x} \)

This function measures the steepness or slope at a particular point on the hyperbola.

For the problem at hand, we found \( \tanh x \) to be \( \frac{12}{13} \) using the calculated values of \( \sinh x \) and \( \cosh x \). This shows how hyperbolic tangent can be derived easily with a grasp on its component functions, \( \sinh x \) and \( \cosh x \). It is a useful function, particularly in the areas of hyperbolic geometry and complex analysis.

sech x

The hyperbolic secant function, \( \sech x \), is the hyperbolic counterpart to the secant function in trigonometry. It is defined as:

• \( \sech x = \frac{1}{\cosh x} \)

It represents the reciprocal of the hyperbolic cosine function. A unique property of \( \sech x \) is that it approaches zero as \( x \) increases, reflecting the non-periodic nature of hyperbolic functions.

In this exercise, \( \sech x \) was found to be \( \frac{5}{13} \) by taking the reciprocal of \( \cosh x = \frac{13}{5} \). Understanding \( \sech x \) and its properties is key in contexts where \( \cosh x \) is used, such as engineering disciplines dealing with structures and cables.