Problem 43 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The equation is an identity.

Step 1: Recall definitions of hyperbolic functions

We start by recalling the definitions of the hyperbolic cosine and hyperbolic sine functions. The hyperbolic cosine, \(\cosh x\), is defined as \(\cosh x = \frac{e^x + e^{-x}}{2}\), and the hyperbolic sine, \(\sinh x\), is defined as \(\sinh x = \frac{e^x - e^{-x}}{2}\).

Step 2: Substitute hyperbolic identities

Next, substitute the definitions of \(\cosh x\) and \(\sinh x\) into the right side of the equation \(e^{-x} = \cosh x - \sinh x\). Substitute to get: \[\cosh x - \sinh x = \left(\frac{e^x + e^{-x}}{2}\right) - \left(\frac{e^x - e^{-x}}{2}\right)\].

Step 3: Simplify the equation

Simplify the expression we have obtained: \[\cosh x - \sinh x = \frac{e^x + e^{-x} - e^x + e^{-x}}{2} = \frac{2e^{-x}}{2} = e^{-x}\].

Step 4: Verifying the identity

After simplification, we find that the right side of the equation simplifies to \(e^{-x}\), which matches the left side. Therefore, we have verified that \(e^{-x} = \cosh x - \sinh x\) is indeed an identity.

Key Concepts

Identity Verification

Identity verification involves confirming that two expressions are entirely equivalent for all allowed values of the involved variables. In mathematics, checking such equivalence often requires algebraic manipulation. This process also heavily involves understanding the foundational identities.

In the exercise, we are tasked with showing that an expression involving hyperbolic functions is an identity. The equation to be verified is \( e^{-x} = \cosh x - \sinh x \).

• First, recall the definitions of the functions involved to rewrite them in known terms.
• Use algebraic manipulation to simplify either or both sides of the equation to see if they become identical.
• After simplification, if both sides match for all permissible values of \( x \), the identity is verified.

This technique ensures that the given equation holds in all scenarios defined by its parameters.

Hyperbolic Sine

The hyperbolic sine function (\( \sinh x \)) is a special function used in calculus and hyperbolic geometry, analogous to the sine function but defined using exponential functions. Its formula is \(\sinh x = \frac{e^x - e^{-x}}{2}\). This formula stems from considering properties of exponential growth and decay.

Some interesting properties include:

• \(\sinh x\) is an odd function, meaning \(\sinh(-x) = -\sinh x\).
• It grows exponentially, as its components are exponential functions.
• It is used in various applications including defining the shape of a hanging cable, known as a catenary curve in physics.

Understanding \(\sinh x\) is essential for solving equations that involve hyperbolic identities.

Hyperbolic Cosine

Hyperbolic cosine (\( \cosh x \)) shares similarities with the regular cosine function but diverges in definition due to its use of exponential expressions. The formula for \(\cosh x\) is \(\cosh x = \frac{e^x + e^{-x}}{2}\). Unlike its trigonometric counterpart, \(\cosh x\) does not oscillate but rather increases due to its exponential nature.

Key characteristics include:

• \(\cosh x\) is an even function, fulfilling \(\cosh(-x) = \cosh x\).
• Its minimum value is 1, occurring specifically at \(x = 0\).
• It frequently appears in problems concerning hyperbolas and the description of physical phenomena like the shape of a suspension bridge.

The relevance of \(\cosh x\) extends widely across multiple mathematical disciplines and practical applications.

Algebraic Manipulation

Algebraic manipulation is a fundamental skill in mathematics that involves rearranging and simplifying expressions to reveal underlying truths or verify identities. This process consists of transforming equations while maintaining equivalence.

In the context of the exercise, we perform algebraic manipulation by substituting known expressions for the hyperbolic functions: \(\cosh x\) and \(\sinh x\). After substitution, the task is to simplify the resulting expression:

• Distribute and combine like terms: \(\cosh x - \sinh x = \left(\frac{e^x + e^{-x}}{2}\right) - \left(\frac{e^x - e^{-x}}{2}\right)\)
• Combine terms: \(\frac{e^x + e^{-x} - e^x + e^{-x}}{2} = \frac{2e^{-x}}{2}\)
• Final simplification: \(e^{-x}\)

The goal is to achieve the expression on the other side of the equation, thus proving the original statement is indeed an identity.