Problem 104 - Calculus Early Transcendentals

Question

Verify the following identities. \(\sinh (x+y)=\sinh x \cosh y+\cosh x \sinh y\)

Step-by-Step Solution

Answer

Question: Prove the identity \(\sinh (x+y)=\sinh x \cosh y+\cosh x \sinh y\). Answer: By using the definitions of hyperbolic sine and hyperbolic cosine, as well as simplifying both sides of the equation, we are able to verify that \(\sinh (x+y)=\sinh x \cosh y+\cosh x \sinh y\).

Step 1: Recall the definitions of hyperbolic sine and cosine

The definitions of hyperbolic sine and cosine are: \(\sinh x =\frac{e^x - e^{-x}}{2}\) \(\cosh x =\frac{e^x + e^{-x}}{2}\)

Step 2: Evaluate the left side of the identity (\(\sinh (x+y)\))

Using the definition of hyperbolic sine, we can rewrite \(\sinh (x+y)\) as: \(\sinh (x+y) =\frac{e^{x+y} - e^{-(x+y)}}{2}\)

Step 3: Simplify the left side of the identity

By applying the properties of exponents, we can further simplify \(\sinh (x+y)\) as follows: \(\sinh (x+y) =\frac{e^{x}e^y - e^{-x}e^{-y}}{2}\)

Step 4: Evaluate the right side of the identity (\(\sinh x \cosh y+\cosh x \sinh y\))

Using the definitions of hyperbolic sine and cosine, we can rewrite the right side of the identity as: \(\sinh x \cosh y+\cosh x \sinh y = \left(\frac{e^{x} - e^{-x}}{2}\right) \left(\frac{e^{y} + e^{-y}}{2}\right) + \left(\frac{e^{x} + e^{-x}}{2}\right) \left(\frac{e^{y} - e^{-y}}{2}\right)\)

Step 5: Simplify the right side of the identity

Expand and simplify the expression on the right side of the identity: \(\sinh x \cosh y+\cosh x \sinh y = \frac{e^{x}e^y - e^{-x}e^{-y}}{4} + \frac{-e^{x}e^{-y} + e^{-x}e^y}{4}\) Notice the right side simplifies to: \(\sinh x \cosh y+\cosh x \sinh y = \frac{e^{x}e^y - e^{-x}e^{-y}}{2}\)

Step 6: Compare left side and right side of the identity

Upon comparing the left side: \(\sinh (x+y) =\frac{e^{x}e^y - e^{-x}e^{-y}}{2}\) and the simplified right side: \(\sinh x \cosh y+\cosh x \sinh y = \frac{e^{x}e^y - e^{-x}e^{-y}}{2}\) We find that the expressions are equal, thus proving the identity: \(\sinh (x+y)=\sinh x \cosh y+\cosh x \sinh y\)

Key Concepts

Identity Verification

Verifying mathematical identities is like solving a puzzle where every piece must fit perfectly for the whole picture to make sense. In the context of hyperbolic functions, identity verification involves showing that two mathematical expressions are indeed equivalent. Here, the goal is to prove the identity \(\sinh(x+y) =\sinh x \cosh y + \cosh x \sinh y\).
You begin by recalling or looking up the definitions of these hyperbolic functions.

• \(\sinh x\) is defined as \(\frac{e^x - e^{-x}}{2}\)
• \(\cosh x\) is defined as \(\frac{e^x + e^{-x}}{2}\)

These functions use exponential expressions to define values that mimic the behavior of hyperbolic sine and cosine on a geometric plane.
By applying these definitions, both sides of the given identity must be simplified into the same expression to confirm their equivalence, thus verifying the identity.

Algebraic Manipulation

Algebraic manipulation involves rewriting equations or expressions in different forms, using mathematical rules to achieve simplification or alteration of their appearance without changing their value. This is a key tool when working to prove an identity.
In our hyperbolic identity, algebraic manipulation is used to break down and simplify both sides of the equation. It involves several steps:

• Expanding products: In the right side of the identity, each term is expanded to its full using basic distribution, (e.g., \((a + b)(c + d) = ac + ad + bc + bd\)). This is essential to match both sides of the identity.
• Combination of like terms: After expansion, terms are combined to create a more organized form.
• Simplification: Both sides simplify down to the form \(\frac{e^x e^y - e^{-x} e^{-y}}{2}\), which proves the identity.

Through careful algebraic manipulation, we ensure that each step logically leads to the establishment of the identity as true.

Exponential Functions

Exponential functions are integral to understanding hyperbolic functions. They provide the backbone for hyperbolic definitions and identity proofs.
The exponential function is expressed as \(e^x\), where \(e\) is the base and \(x\) is the exponent. This function is continuous and differentiable, and it serves as an underpinning for expressions like \(\sinh\) and \(\cosh\).

• Exponents: In both the left and the right sides of our identity, the properties of exponents like \(e^{a+b}=e^a e^b\) are used to manipulate and simplify the expressions.
• Negative exponents: \(e^{-x}\) demonstrates how exponents can handle reciprocal values, which is critical in calculating hyperbolic functions.

Without understanding and using exponential functions, verifying identities involving hyperbolic functions would be challenging. These mathematical building blocks are crucial for managing and simplifying the complex relationships in hyperbolic identities.