Problem 377 - Calculus Volume 1

Question

Find expressions for \(\cosh x+\sinh x\) and \(\cosh x-\sinh x\) . Use a calculator to graph these functions and ensure your expression is correct.

Step-by-Step Solution

Answer

\(\cosh x + \sinh x = e^x\) and \(\cosh x - \sinh x = e^{-x}\).

Step 1: Understanding Hyperbolic Functions

To solve the problem, we need to recall the definitions of the hyperbolic functions: \(\cosh x = \frac{e^x + e^{-x}}{2}\) and \(\sinh x = \frac{e^x - e^{-x}}{2}\).

Step 2: Simplifying \(\cosh x + \sinh x\)

Using the definitions, we add: \[ \cosh x + \sinh x = \frac{e^x + e^{-x}}{2} + \frac{e^x - e^{-x}}{2} \]Combine the fractions: \[ \cosh x + \sinh x = \frac{2e^x}{2} = e^x \].

Step 3: Simplifying \(\cosh x - \sinh x\)

Using the definitions, we subtract:\[ \cosh x - \sinh x = \frac{e^x + e^{-x}}{2} - \frac{e^x - e^{-x}}{2} \]Combine the fractions:\[ \cosh x - \sinh x = \frac{2e^{-x}}{2} = e^{-x} \].

Step 4: Plotting the Functions

Graph both functions: \(y = e^x\) and \(y = e^{-x}\) using a calculator or graphing software. Both should reflect the derived expressions from hyperbolic identities.

Key Concepts

Hyperbolic Identities

Hyperbolic identities are similar to trigonometric identities, but they are based on hyperbolic functions such as sinh and cosh. These identities help us simplify expressions involving hyperbolic functions. A key identity is the relationship between cosh and sinh: \( \cosh^2 x - \sinh^2 x = 1 \). This is analogous to the trigonometric identity \( \cos^2 x + \sin^2 x = 1 \).
Another important feature of hyperbolic identities is how they simplify expressions. In the given exercise, we used these identities to break down \(\cosh x\) and \(\sinh x\) using their definitions, making it easier to solve the problem. This showcases the power and utility of these identities in algebraic manipulation.

Cosh and Sinh Definitions

Understanding the definitions of the hyperbolic functions cosh and sinh is fundamental to solving problems involving these functions. Hyperbolic cosine, denoted as \(\cosh x\), is defined as \(\cosh x = \frac{e^x + e^{-x}}{2}\).
Similarly, hyperbolic sine, denoted as \(\sinh x\), is defined as \(\sinh x = \frac{e^x - e^{-x}}{2}\). These definitions are not only crucial for theoretical purposes, but they also help in practical applications, like simplifying and calculating expressions. By remembering these formulas, one can easily compute hyperbolic functions just as they would with regular trigonometric functions.
These expressions suggest how closely hyperbolic functions are related to exponential functions, using them as building blocks simplifying equations like \(\cosh x + \sinh x\) and \(\cosh x - \sinh x\), as demonstrated in the original problem.

Graphing e^x and e^-x

Graphing exponential functions helps us visually understand the behavior of hyperbolic functions. For the exercise, we are required to plot \(y = e^x\) and \(y = e^{-x}\). These graphs will help verify that our algebraic simplifications of \(\cosh x + \sinh x\) and \(\cosh x - \sinh x\) are accurate.
\(y = e^x\) is an exponential growth function, starting from a value of 1 when \(x = 0\) and increasing rapidly as \(x\) increases. On the other hand, \(y = e^{-x}\) is an exponential decay function, starting from a value of 1 at \(x = 0\) and decreasing as \(x\) increases.
By plotting these graphs, students can see that the results of \(\cosh x + \sinh x = e^x\) and \(\cosh x - \sinh x = e^{-x}\) align perfectly with the nature of these exponential graphs, providing a clear visual confirmation of the solution.

Function Simplification Step-by-Step

Simplifying functions involves breaking down complex expressions into simpler ones using known definitions and identities. Let's see how to simplify \(\cosh x + \sinh x\) and \(\cosh x - \sinh x\) step-by-step.

• Start with the definitions: \(\cosh x = \frac{e^x + e^{-x}}{2}\) and \(\sinh x = \frac{e^x - e^{-x}}{2}\).
• Add the two expressions for \(\cosh x + \sinh x\):
  \[\cosh x + \sinh x = \frac{e^x + e^{-x}}{2} + \frac{e^x - e^{-x}}{2}\]
• Combine the terms to get \(\frac{2e^x}{2} = e^x\).

Similarly, to simplify \(\cosh x - \sinh x\):

• Subtract the expression for \(\sinh x\) from \(\cosh x\):
  \[\cosh x - \sinh x = \frac{e^x + e^{-x}}{2} - \frac{e^x - e^{-x}}{2}\]
• Combine the terms to get \(\frac{2e^{-x}}{2} = e^{-x}\).

This step-by-step process reveals the power of hyperbolic function definitions in simplifying expressions and confirms the expressions \(e^x\) and \(e^{-x}\) are indeed correct for the problems.