Problem 12 - Thomas Calculus in SI Units

Question

Use the definitions of cosh \(x\) and \(\sinh x\) to show that $$\cosh ^{2} x-\sinh ^{2} x=1$$

Step-by-Step Solution

Answer

\(\cosh^2(x) - \sinh^2(x) = 1\) is proven true using definitions.

Step 1: Recall the Definitions of Hyperbolic Functions

The hyperbolic cosine function is defined as \( \cosh(x) = \frac{e^x + e^{-x}}{2} \) and the hyperbolic sine function is defined as \( \sinh(x) = \frac{e^x - e^{-x}}{2} \). These definitions will be used to prove the required identity.

Step 2: Evaluate the Square of \(\cosh(x)\)

Compute \(\cosh^2(x)\) based on its definition:\[\cosh^2(x) = \left(\frac{e^x + e^{-x}}{2}\right)^2 = \frac{(e^x + e^{-x})^2}{4}\]Expanding the square gives:\[= \frac{e^{2x} + 2 + e^{-2x}}{4}\]

Step 3: Evaluate the Square of \(\sinh(x)\)

Compute \(\sinh^2(x)\) based on its definition:\[\sinh^2(x) = \left(\frac{e^x - e^{-x}}{2}\right)^2 = \frac{(e^x - e^{-x})^2}{4}\]Expanding the square gives:\[= \frac{e^{2x} - 2 + e^{-2x}}{4}\]

Step 4: Subtract \(\sinh^2(x)\) from \(\cosh^2(x)\)

Subtract the results from step 3 from step 2:\[\cosh^2(x) - \sinh^2(x) = \frac{e^{2x} + 2 + e^{-2x}}{4} - \frac{e^{2x} - 2 + e^{-2x}}{4}\]Simplify the expression:\[= \frac{(e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x})}{4} = \frac{4}{4} = 1\]

Step 5: Conclusion of Proof

Conclude by stating that the subtraction yields 1 as required by the identity:\[\cosh^2(x) - \sinh^2(x) = 1\]This verifies the given identity using the definitions of \(\cosh(x)\) and \(\sinh(x)\).

Key Concepts

Understanding cosh(x)

The hyperbolic cosine function, denoted as \( \cosh(x) \), is a key component in the family of hyperbolic functions. It bears some similarities to the traditional trigonometric cosine function, but it arises in the context of hyperbolas instead of circles. Here's how it is defined and calculated:

• The mathematical definition of \( \cosh(x) \) is \( \cosh(x) = \frac{e^x + e^{-x}}{2} \).
• This involves taking the average of the exponential functions \( e^x \) and \( e^{-x} \).
• It's like finding the midpoint between these two exponentials on a graph.

The addition of the exponential functions shows the symmetric property of hyperbolic cosine around the y-axis.
Unlike trigonometric functions, \( \cosh(x) \) is never negative and is only equal to 1 at \( x = 0 \).
As \( x \) approaches infinity, \( \cosh(x) \) increases rapidly, mimicking exponential growth.

Understanding sinh(x)

The hyperbolic sine function, represented as \( \sinh(x) \), is another member of the hyperbolic family. Just like \( \cosh(x) \), \( \sinh(x) \) also has a connection to the exponential functions but differs in its unique structure.

• \( \sinh(x) \) is defined as \( \sinh(x) = \frac{e^x - e^{-x}}{2} \).
• Here, you subtract \( e^{-x} \) from \( e^x \) and then divide by 2.This emphasizes the asymmetry in its form, opposite to the symmetry in \( \cosh(x) \).

\( \sinh(x) \) can be visualized as a curve that passes through the origin, rising steeply as \( |x| \) increases.
It's a function that produces positive values for positive x, and negative values for negative x.
Unlike \( \cosh(x) \), \( \sinh(x) \) reflects the odd function property, meaning \( \sinh(-x) = -\sinh(x) \).

Identity Proof with cosh(x) and sinh(x)

One of the most intriguing aspects of hyperbolic functions is the identity shared by \( \cosh(x) \) and \( \sinh(x) \):\[ \cosh^2(x) - \sinh^2(x) = 1 \]This identity bears resemblance to the familiar Pythagorean identity found in circular trigonometry. Here's why it holds true:

• Start by computing \( \cosh^2(x) \) and \( \sinh^2(x) \) using their respective definitions.
• \( \cosh^2(x) = \left(\frac{e^x + e^{-x}}{2}\right)^2 = \frac{e^{2x} + 2 + e^{-2x}}{4} \)
• \( \sinh^2(x) = \left(\frac{e^x - e^{-x}}{2}\right)^2 = \frac{e^{2x} - 2 + e^{-2x}}{4} \)
• Subtract \( \sinh^2(x) \) from \( \cosh^2(x) \):\[ \cosh^2(x) - \sinh^2(x) = \frac{e^{2x} + 2 + e^{-2x}}{4} - \frac{e^{2x} - 2 + e^{-2x}}{4} \]
• When simplified, this results in \( \frac{4}{4} = 1 \), establishing the identity true.

This simple yet elegant identity is a cornerstone in understanding how hyperbolic functions relate to each other.
It showcases the interplay between \( \cosh(x) \) and \( \sinh(x) \), linking their squares to a constant value of 1.