Problem 5 - A Primer of Real Analysis

Question

Show that for any real number \(x,\) $$ \cosh ^{2}(x)-\sinh ^{2}(x)=1 $$

Step-by-Step Solution

Answer

The identity \( \cosh^{2}(x) - \sinh^{2}(x) = 1 \) holds for any real number \( x \).

Step 1: Review Hyperbolic Function Definitions

Recall that the hyperbolic cosine and hyperbolic sine functions are defined as follows:\[\cosh(x) = \frac{e^{x} + e^{-x}}{2}\]\[\sinh(x) = \frac{e^{x} - e^{-x}}{2}\]These definitions will be critical to both substituting into the equation and simplifying it further.

Step 2: Substitute Definitions into the Equation

Substitute the definitions of \(\cosh(x)\) and \(\sinh(x)\) into the given identity.\[\cosh^{2}(x) - \sinh^{2}(x) = \left(\frac{e^{x} + e^{-x}}{2}\right)^{2} - \left(\frac{e^{x} - e^{-x}}{2}\right)^{2}\]Now, we will simplify this expression.

Step 3: Expand Both Squares

Apply the squaring formula \((a \, \pm \, b)^2 = a^2 \, \pm \, 2ab \, + \, b^2\) to each of the terms:\[\left(\frac{e^{x} + e^{-x}}{2}\right)^{2} = \frac{(e^{x} + e^{-x})^2}{4} = \frac{e^{2x} + 2 + e^{-2x}}{4}\]\[\left(\frac{e^{x} - e^{-x}}{2}\right)^{2} = \frac{(e^{x} - e^{-x})^2}{4} = \frac{e^{2x} - 2 + e^{-2x}}{4}\]

Step 4: Subtract the Expressions

Subtract the expanded \(\sinh^2(x)\) from \(\cosh^2(x)\):\[\frac{e^{2x} + 2 + e^{-2x}}{4} - \frac{e^{2x} - 2 + e^{-2x}}{4} = \frac{e^{2x} + 2 + e^{-2x} - e^{2x} + 2 - e^{-2x}}{4}\]Simplifying the numerator:\[= \frac{4}{4} = 1\]So, the identity \(\cosh^{2}(x) - \sinh^{2}(x) = 1\) is verified.

Key Concepts

Hyperbolic Functions

Hyperbolic functions are similar to trigonometric functions, but they are based on hyperbolas instead of circles. The two most common hyperbolic functions are hyperbolic sine, denoted as \( \sinh(x) \), and hyperbolic cosine, denoted as \( \cosh(x) \). These functions can be thought of as extensions of the exponential function.

• Unlike circular functions, hyperbolic functions exhibit exponential growth and decay.
• They are useful in various areas of mathematics, including real analysis, calculus, and differential equations.
• Hyperbolic functions describe the shape of a hanging cable, known as a catenary, among other applications.

Understanding these functions requires appreciation of their unique properties and how they relate to their trigonometric counterparts.

Cosh and Sinh Definitions

The concept of hyperbolic cosine and hyperbolic sine starts with their definitions, which are:

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

These definitions use exponential functions for their calculations. Derived from these are key identities, similar to trigonometric identities.

For instance, while the Pythagorean identity in trigonometry is \( \cos^2(\theta) + \sin^2(\theta) = 1 \), the hyperbolic version is \( \cosh^2(x) - \sinh^2(x) = 1 \). This identity is crucial in various mathematical proofs and derivations.

By using the property of exponentials and treating them with algebraic manipulations, we easily demonstrate many such identities.

Real Analysis Problem Solving

In real analysis, problem-solving often involves rigorous justification of results. Verifying identities such as \( \cosh^2(x) - \sinh^2(x) = 1 \) showcases the analytical skills in simplifying expressions and utilizing fundamental properties.


Here's how you tackle such a problem:

• Start by understanding the problem statement. Here it asks you to prove an identity.
• Use known definitions, like those of \( \cosh(x) \) and \( \sinh(x) \), to substitute into your equation.
• Simplify following algebraic rules, including distributing terms and factoring.
• Make sure each simplification step is accurate, as errors can lead to incorrect conclusions.

These steps give insights into fostering analytical thinking and improving one's aptitude in handling complex mathematical equations.