Problem 45 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 41-52, verify that the given equations are identities. \(\sinh (x+y)=\sinh x \cosh y+\cosh x \sinh y\)

Step-by-Step Solution

Answer

The given equation is an identity.

Step 1: Understand the identity to verify

The identity given is \(\sinh(x+y) = \sinh x \cosh y + \cosh x \sinh y\). We need to verify this equation is true for any real numbers \(x\) and \(y\).

Step 2: Recall the definitions of hyperbolic functions

Recall the definitions: \(\sinh x = \frac{e^x - e^{-x}}{2}\) and \(\cosh x = \frac{e^x + e^{-x}}{2}\). These definitions will help in proving the identity.

Step 3: Expand \(\sinh(x+y)\) using its definition

Using the definition of \(\sinh\), we have \(\sinh(x+y) = \frac{e^{x+y} - e^{-(x+y)}}{2}\).

Step 4: Simplify \(\sinh(x+y)\)

Simplify \(\frac{e^{x+y} - e^{-(x+y)}}{2} = \frac{e^x e^y - e^{-x} e^{-y}}{2}\). Factoring this result helps to see patterns that match the terms on the right side of the identity.

Step 5: Express \(\sinh x \cosh y + \cosh x \sinh y\) in terms of exponentials

Calculate \(\sinh x \cosh y\) and \(\cosh x \sinh y\) using definitions: - \(\sinh x \cosh y = \left(\frac{e^x - e^{-x}}{2}\right)\left(\frac{e^y + e^{-y}}{2}\right) = \frac{e^{x+y} + e^{x-y} - e^{-x+y} - e^{-x-y}}{4}\) - \(\cosh x \sinh y = \left(\frac{e^x + e^{-x}}{2}\right)\left(\frac{e^y - e^{-y}}{2}\right) = \frac{e^{x+y} - e^{x-y} + e^{-x+y} - e^{-x-y}}{4}\)

Step 6: Combine expressions from Step 5

Add the two expressions: \(\frac{e^{x+y} + e^{x-y} - e^{-x+y} - e^{-x-y}}{4} + \frac{e^{x+y} - e^{x-y} + e^{-x+y} - e^{-x-y}}{4}\) results in \(\frac{2e^{x+y} - 2e^{-x-y}}{4} = \frac{e^{x+y} - e^{-x-y}}{2}\).

Step 7: Compare simplified expressions

Both expressions \(\sinh(x+y)\) and \(\sinh x \cosh y + \cosh x \sinh y\) simplify to \(\frac{e^{x+y} - e^{-(x+y)}}{2}\). Hence, the original identity is verified to be true.

Key Concepts

Identity Verification

Identity verification in mathematics helps demonstrate that two expressions are equivalent. When we verify an identity, we confirm it holds true for all values of the variables involved. In our exercise, the identity given is: \[\sinh(x+y) = \sinh x \cosh y + \cosh x \sinh y\]To verify, we manipulate both sides of the equation using known definitions and properties of hyperbolic functions. The goal is to show that both sides simplify to the same expression. This means using algebraic manipulations to prove that each step logically follows the previous one. This logical procession helps confirm that the equation is valid for any real values of \(x\) and \(y\).
Understanding identity verification broadens your comprehension of how algebraic expressions can transform, providing a solid foundation for more complex problems.

Hyperbolic Sine

The hyperbolic sine function, denoted as \(\sinh\), is a fundamental hyperbolic function. It's defined using exponential functions as follows:\[\sinh x = \frac{e^x - e^{-x}}{2}\]This definition can be compared to the circular sine function \(\sin\), but in the context of the hyperbola instead of a circle. The function describes a smooth curve reflecting across the origin, unlike its trigonometric counterpart.
Each element in the equation contributes uniquely:

• \(e^x\) and \(e^{-x}\) represent exponential growth and decay at the same rate.
• The subtraction captures the net difference between these two opposing exponential trends.

Learning \(\sinh\) is essential as it simplifies and solves identities involving hyperbolic expressions, such as the one in our exercise.

Hyperbolic Cosine

Similar to \(\sinh\), the hyperbolic cosine function \(\cosh\) is important for working with hyperbolic identities. It's defined as:\[\cosh x = \frac{e^x + e^{-x}}{2}\]This definition uses both the sum and average of exponential growth and decay functions. Here's how it operates:

• The addition of \(e^x\) and \(e^{-x}\) emphasizes their symmetry around the vertical axis.
• Dividing by 2 creates an average, reflecting that \(\cosh\) always produces positive values, mirroring a hyperbolic curve above the \(x\)-axis.

Understanding \(\cosh\) helps distinguish between the different behaviors of exponential functions and solves identities requiring hyperbolic trig functions like the one in our exercise.

Exponential Functions

Exponential functions are crucial in the formulation and manipulation of hyperbolic functions. They are represented by \(e^x\), where \(e\) is the base of the natural logarithm, approximately equal to 2.718. Here are some vital points about exponential functions:

• Exponential growth is represented by \(e^x\), depicting rapid increase as \(x\) increases.
• Exponential decay is shown by \(e^{-x}\), illustrating rapid decrease as \(x\) decreases.

In hyperbolic functions, exponentials help construct expressions like \(\sinh x\) and \(\cosh x\). For the identity we are verifying, exponentials allow the manipulation of these hyperbolic functions into equivalent expressions for simpler calculation. The properties of exponential growth and decay are pivotal for understanding how hyperbolic sine and cosine functions interact.