Problem 12 - Essential Calculus Early Transcendentals

Question

Prove the identity. $$ \cosh (x+y)=\cosh x \cosh y+\sinh x \sinh y $$

Step-by-Step Solution

Answer

The identity is verified: \( \cosh(x+y) = \cosh x \cosh y + \sinh x \sinh y \).

Step 1: Recall the Definitions

To start, we need to recall the definitions of hyperbolic functions. The hyperbolic cosine function is defined as \( \cosh z = \frac{e^z + e^{-z}}{2} \) and the hyperbolic sine function is defined as \( \sinh z = \frac{e^z - e^{-z}}{2} \).

Step 2: Write Cosine of a Sum for Hyperbolic Functions

We want to verify that \( \cosh (x+y) = \cosh x \cosh y + \sinh x \sinh y \). According to the definition, \( \cosh(x+y) = \frac{e^{x+y} + e^{-(x+y)}}{2} \).

Step 3: Expand Using Exponential Properties

We can expand \( e^{x+y} \) and \( e^{-(x+y)} \) using the properties of exponents to get \( e^x e^y \) and \( e^{-x} e^{-y} \), respectively. So, \( \cosh(x+y) = \frac{e^x e^y + e^{-x} e^{-y}}{2} \).

Step 4: Expand the Right-Hand Side

For the right-hand side, substitute the definitions: \( \cosh x \cosh y + \sinh x \sinh y = \left(\frac{e^x + e^{-x}}{2}\right)\left(\frac{e^y + e^{-y}}{2}\right) + \left(\frac{e^x - e^{-x}}{2}\right)\left(\frac{e^y - e^{-y}}{2}\right) \).

Step 5: Simplify the Right-Hand Side

Multiply out the terms: 1. \( \frac{e^x + e^{-x}}{2} \cdot \frac{e^y + e^{-y}}{2} = \frac{e^{x+y} + e^{x-y} + e^{-x+y} + e^{-(x+y)}}{4} \)2. \( \frac{e^x - e^{-x}}{2} \cdot \frac{e^y - e^{-y}}{2} = \frac{e^{x+y} - e^{x-y} - e^{-x+y} + e^{-(x+y)}}{4} \).Add these:3. Combine like terms: \( \frac{2e^{x+y} + 2e^{-(x+y)}}{4} = \frac{e^{x+y} + e^{-(x+y)}}{2} \).

Step 6: Compare Both Sides

Now, compare the results from Step 3 and Step 5: Both are \( \frac{e^{x+y} + e^{-(x+y)}}{2} \). Thus, the identity \( \cosh(x+y) = \cosh x \cosh y + \sinh x \sinh y \) is verified.

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogous to trigonometric functions but are based on hyperbolas instead of circles. They are used extensively in various areas of mathematics, including hyperbolic geometry, complex analysis, and in solutions of some differential equations.
Hyperbolic functions include hyperbolic sine (\( \sinh(x) \)), cosine (\( \cosh(x) \)), tangent (\( \tanh(x) \)), and their reciprocals. These functions are defined using exponential functions which offers them unique properties:

• The function \( \sinh(x) \) is defined as: \( \sinh x = \frac{e^x - e^{-x}}{2} \). It's an odd function, meaning \( \sinh(-x) = -\sinh(x) \).
• The function \( \cosh(x) \) is defined as: \( \cosh x = \frac{e^x + e^{-x}}{2} \). It's an even function, implying: \( \cosh(-x) = \cosh(x) \).

The hyperbolic functions have notable identities, similar to trigonometric identities, which can be used to simplify functions and expressions in calculus. They have unique derivatives: for instance, the derivative of \( \sinh(x) \) is \( \cosh(x) \) and vice versa.

cosh and sinh Definitions

The definitions of hyperbolic functions, specifically \( \cosh(x) \) and \( \sinh(x) \), are central to understanding these mathematical concepts.
Their definitions in terms of exponential functions provide a foundation for proving many important identities.
To break it down further:

• The hyperbolic cosine function, \( \cosh(x) \), is defined as:\( \cosh x = \frac{e^x + e^{-x}}{2} \).
• On the other hand, the hyperbolic sine function, \( \sinh(x) \), is defined as:\( \sinh x = \frac{e^x - e^{-x}}{2} \).

These functions differ from their circular counterparts because instead of cycling through repeated values, they exhibit exponential growth and decay properties. They graphically resemble exponential functions, rising or falling in a smooth, continuous curve. Understanding their exponential definitions is crucial because it forms the basis for proving identities and understanding their properties in mathematical contexts.

Identity Proofs

Identity proofs involving hyperbolic functions often rely on their definitions in terms of exponentials. One critical identity to know is the addition formula for \( \cosh \):
\[\cosh (x+y) = \cosh x \cosh y + \sinh x \sinh y\]The proof of this identity starts by substituting the definitions of \( \cosh \) and \( \sinh \) into the equation.

• Start by expressing \( \cosh(x+y) = \frac{e^{x+y} + e^{-(x+y)}}{2} \).
• Then apply the property of exponents to split \( e^{x+y} \) and \( e^{-(x+y)} \) into \( e^x \) \( e^y \) and \( e^{-x} \) \( e^{-y} \) respectively.
• Substitute back into the expanded forms of \( \cosh x \cosh y \) and \( \sinh x \sinh y \).

The process involves simplifying the expression by combining like terms and utilizing properties of exponentials, which results \[ \frac{e^{x+y} + e^{-(x+y)}}{2} \], proving the identity. This identity is very similar to the cosine addition formulas found in trigonometry, showcasing the often parallel nature of hyperbolic and trigonometric functions, even though they originate from distinctly different geometrical concepts.