Problem 48 - Calculus Early Transcendentals: Pearson New International Edition

Question

Verify that the given equations are identities. \(\cosh (x-y)=\cosh x \cosh y-\sinh x \sinh y\)

Step-by-Step Solution

Answer

The equation is an identity, as both sides are equal after simplification.

Step 1: Recall Hyperbolic Function Definitions

Recall that the definitions for hyperbolic cosine and hyperbolic sine are given by:\[ \cosh z = \frac{e^z + e^{-z}}{2} \] \[ \sinh z = \frac{e^z - e^{-z}}{2} \] where \( z \) is any variable or expression.

Step 2: Express LHS Using Definitions

Start by expressing the left-hand side \( \cosh (x-y) \) using the definition of hyperbolic cosine:\[ \cosh (x-y) = \frac{e^{(x-y)} + e^{-(x-y)}}{2} \]

Step 3: Simplify LHS Expression

Simplify the expression \( \cosh (x-y) = \frac{e^{(x-y)} + e^{-(x-y)}}{2} \) by expanding it:\[ \frac{e^x e^{-y} + e^{-x} e^y}{2} \]

Step 4: Express RHS Using Definitions

Express the right-hand side \( \cosh x \cosh y - \sinh x \sinh y \) using definitions:\[ \cosh x \cosh y = \left(\frac{e^x + e^{-x}}{2}\right) \left(\frac{e^y + e^{-y}}{2}\right) \] \[ \sinh x \sinh y = \left(\frac{e^x - e^{-x}}{2}\right) \left(\frac{e^y - e^{-y}}{2}\right) \]

Step 5: Simplify RHS Expression

Simplify the right-hand side:\[ \cosh x \cosh y = \frac{e^{x+y} + e^{x-y} + e^{-x+y} + e^{-(x+y)}}{4} \] \[ \sinh x \sinh y = \frac{e^{x+y} - e^{x-y} - e^{-x+y} + e^{-(x+y)}}{4} \] Subtract the two expressions:\[ \cosh x \cosh y - \sinh x \sinh y = \frac{(e^{x-y} + e^{-x+y} + e^{-(x-y)} + e^{-(x-y)})}{4} \] Simplifying further results in:\[ \frac{e^{x-y} + e^{-(x-y)}}{2} \]

Step 6: Compare LHS and RHS Expressions

Notice that the simplified expressions for LHS and RHS are the same:- LHS: \( \frac{e^{x-y} + e^{-(x-y)}}{2} \)- RHS: \( \frac{e^{x-y} + e^{-(x-y)}}{2} \)Since both sides are equal, the given equation is indeed an identity.

Key Concepts

Hyperbolic Functions

Hyperbolic functions are a set of mathematical functions that parallel the trigonometric functions but are based on hyperbolas instead of circles.
These functions include hyperbolic sine (\(\sinh\)) and hyperbolic cosine (\(\cosh\)), among others.
They can be expressed using exponential functions:

• Hyperbolic cosine: \( \cosh z = \frac{e^z + e^{-z}}{2} \)
• Hyperbolic sine: \( \sinh z = \frac{e^z - e^{-z}}{2} \)

One of the remarkable properties of hyperbolic functions is their ability to represent many real-world phenomena, such as the shape of a hanging cable (the catenary) and the growth of populations.
Their mathematical behaviors also mimic those of circular functions, but with unique identities that help solve various calculus and differential equations problems.

Verification of Identities

In mathematics, verifying an identity involves proving that an equality holds true for all values of the variables involved.
It is not enough just to show an equation works for one or a few values; it has to work universally.
For instance, to verify the identity \(\cosh (x-y) = \cosh x \cosh y - \sinh x \sinh y\), we express both sides using known definitions.

• First, the left-hand side (LHS) is expressed using hyperbolic definitions as \( \cosh(x-y) = \frac{e^{(x-y)} + e^{-(x-y)}}{2} \).
• Then, the right-hand side (RHS) \( \cosh x \cosh y - \sinh x \sinh y \) is expanded and simplified to match the LHS using exponential properties.

By demonstrating that both sides simplify to the same expression, we confirm the given equation is an identity.

Mathematical Proofs

Mathematical proofs provide a logical and definitive demonstration that mathematical statements or identities are universally true.
There are several approaches to mathematical proofs, each serving different purposes based on what needs to be shown.

• Direct Proof: Proves a statement directly by using facts, definitions, and logical operations.
• Proof by Contradiction: Assumes that the opposite of a statement is true, then demonstrates a contradiction, proving the original statement must be true.
• Proof by Induction: Often used to prove statements about natural numbers, involving a base case and an inductive step.

In our example, proving the identity using direct verification ensures each step logically follows from mathematical definitions and operations.
Such rigorous methodology strengthens understanding and affirms mathematical truths in a clear and concise manner.