Problem 42 - Calculus Early Transcendentals: Pearson New International Edition

Question

Verify that the given equations are identities. \(e^{2 x}=\cosh 2 x+\sinh 2 x\)

Step-by-Step Solution

Answer

The equation is verified as an identity using definitions of hyperbolic functions.

Step 1: Understanding Identities

An identity is an equation that is true for all values of the variable. To prove an identity, we often manipulate one side of the equation to match the other side or show both sides equal a common expression.

Step 2: Recall Definitions of Hyperbolic Functions

Recall that the hyperbolic cosine and hyperbolic sine functions are defined as: \[ \cosh 2x = \frac{e^{2x} + e^{-2x}}{2} \quad \text{and} \quad \sinh 2x = \frac{e^{2x} - e^{-2x}}{2} \]

Step 3: Simplify Right-Hand Side

Add the expressions for \( \cosh 2x \) and \( \sinh 2x \): \[ \cosh 2x + \sinh 2x = \frac{e^{2x} + e^{-2x}}{2} + \frac{e^{2x} - e^{-2x}}{2} \]

Step 4: Combine Fractions

Combine the fractions from Step 3:\[ \cosh 2x + \sinh 2x = \frac{(e^{2x} + e^{-2x}) + (e^{2x} - e^{-2x})}{2} = \frac{2e^{2x}}{2} \]

Step 5: Simplify the Expression

Simplify the expression obtained in Step 4:\[ \cosh 2x + \sinh 2x = e^{2x} \]

Step 6: Verification of Identity

We have shown that the right-hand side \( \cosh 2x + \sinh 2x = e^{2x} \), which matches the left-hand side of the identity. Therefore, the equation \( e^{2x} = \cosh 2x + \sinh 2x \) is indeed verified as an identity.

Key Concepts

Hyperbolic Functions

Hyperbolic functions are mathematical functions that have properties similar to the traditional trigonometric functions but are based on hyperbolas rather than circles. Common hyperbolic functions include hyperbolic sine (\(\sinh x\)) and hyperbolic cosine (\(\cosh x\)). These functions are useful in many areas, such as calculus, physics, and engineering, particularly when dealing with hyperbolic geometry, complex numbers, and special relativity.

• Definitions: The hyperbolic cosine and sine functions are defined as: \[ \cosh x = \frac{e^x + e^{-x}}{2} \] \[ \sinh x = \frac{e^x - e^{-x}}{2} \]
• Properties: Unlike circular functions, hyperbolic functions involve real exponential functions. For example, \(\cosh x\) always yields positive values just like \(\cos x\) in trigonometry but grows exponentially as \(x\) increases.

These functions often arise in solutions to differential equations, and they are integral to hyperbolic geometry, where they help model phenomena in hyperbolic space. Understanding how these functions interact aids in recognizing their relationships and verifying identities like the one given in the exercise.

Simplifying Expressions in Calculus

Simplifying expressions is a critical skill in calculus and other branches of mathematics. It involves reducing expressions to their most basic form, making them easier to analyze or integrate into more complex equations. Here’s how simplification was applied in this exercise:First, we had to express the expression on the right-hand side. We started with the functions \(\cosh 2x\) and \(\sinh 2x\). By their definitions:

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

Next, we added these functions together and simplified:

• The denominators were the same, so we combined terms.
• This reduced the expression to \(\cosh 2x + \sinh 2x = \frac{2e^{2x}}{2}= e^{2x}\).

The simplification process made it clear that both expressions were equivalent, verifying the identity.

Equivalence of Expressions

When two expressions are equivalent, they are different representations of the same value or quantity. Verifying the equivalence of expressions is a fundamental aspect of mathematical proofs.In this exercise, the objective was to demonstrate that \(e^{2x}\) is equivalent to \(\cosh 2x + \sinh 2x\). Using the definitions and properties of hyperbolic functions, we rewrote and simplified the expressions to show their equality. This proof hinges on showing that both sides help represent the same outcome for any given value of \(x\).

• Initial Equation: Starting with the identity \(e^{2x} = \cosh 2x + \sinh 2x\).
• Identification: Both sides simplify to \(e^{2x}\).
• Verification: By manipulating the right-hand side, we matched it with the left-hand side, proving their equivalence.

Practicing equivalence proofs helps reinforce understanding of mathematical identities and builds skills in logical reasoning and manipulation of expressions. This is crucial not just in verifying known equations but also in deducing new ones.