Problem 13 - Calculus Early Transcendentals

Question

Verify each identity using the definitions of the hyperbolic functions. \(\cosh 2 x=\cosh ^{2} x+\sinh ^{2} x\) (Hint: Begin with the right side of the equation.)

Step-by-Step Solution

Answer

Question: Verify the identity \(\cosh 2x = \cosh^2 x + \sinh^2 x\). Answer: We have shown that the right side of the equation, \(\cosh^2 x + \sinh^2 x\), simplifies to \(\frac{e^{2x} + e^{-2x}}{2}\), which is equal to the hyperbolic cosine of \(2x\), or \(\cosh 2x\). This verifies the given identity: \(\cosh 2x = \cosh^2 x + \sinh^2 x\).

Step 1: Define the hyperbolic functions

We start by defining the hyperbolic cosine and sine functions as follows: \(\cosh x = \frac{e^x + e^{-x}}{2}\) and \(\sinh x = \frac{e^x - e^{-x}}{2}\)

Step 2: Expand the right side of the equation

We need to expand \(\cosh^2 x + \sinh^2 x\). Using the definitions from step 1, we can write: \(\cosh^2 x + \sinh^2 x = \left(\frac{e^x + e^{-x}}{2}\right)^2 + \left(\frac{e^x - e^{-x}}{2}\right)^2\)

Step 3: Simplify the right side of the equation

Now let's simplify the expression: \begin{align*} \cosh^2 x + \sinh^2 x &= \frac{(e^x + e^{-x})^2 + (e^x - e^{-x})^2}{2^2} \\ &= \frac{e^{2x} + 2 + e^{-2x} + e^{2x} - 2 + e^{-2x}}{4} \\ &= \frac{2e^{2x} + 2e^{-2x}}{4} \\ &= \frac{e^{2x} + e^{-2x}}{2} \end{align*}

Step 4: Compare the result to the left side of the equation

Now we can see that the simplified right side of the equation is equal to the hyperbolic cosine of \(2x\), which is the left side of the equation: \(\cosh 2x = \frac{e^{2x} + e^{-2x}}{2}\) Therefore, we have verified the given identity using the definitions of the hyperbolic functions: \(\cosh 2x = \cosh^2 x + \sinh^2 x\)

Key Concepts

Understanding Hyperbolic Cosine

The hyperbolic cosine function, denoted as \(\cosh x\), is one of the primary hyperbolic functions. It resembles the regular cosine function but relates to the shape of a standard hyperbola. Algebraically, it is defined as:

• \(\cosh x = \frac{e^x + e^{-x}}{2}\)

The function calculates the midpoint (or average) of \(e^x\) and \(e^{-x}\).

In essence, \(\cosh x\) can be visualized as half of the sum of two exponential functions. If you plot \(\cosh x\), it forms a shape similar to the catenary curve, often seen in the natural form of hanging cables. Additionally, it is always positive due to the nature of exponential growth and demonstrates even function properties since \(\cosh(-x) = \cosh(x)\). The hyperbolic cosine is instrumental in various mathematical and physics applications, particularly in describing certain types of growths and spreads.

Understanding Hyperbolic Sine

The hyperbolic sine function is denoted as \(\sinh x\) and serves as another key function within hyperbolic mechanics. It is defined as:

• \(\sinh x = \frac{e^x - e^{-x}}{2}\)

This equation takes half of the difference between \(e^x\) and \(e^{-x}\), creating a function that appears similar to a stretched out or elongated sine wave.

\(\sinh x\) captures the difference direction, essentially measuring the 'dip' between two exponential expressions, unlike its cosine counterpart. Its graphical representation shows it is an odd function, meaning \(\sinh(-x) = -\sinh(x)\). This property mirrors that of the standard sine function. You can recognize \(\sinh x\) through its symmetry about the origin. It plays a vital role in fields involving wave equations, such as engineering and signal processing.

Identifying and Verifying Identities in Hyperbolic Functions

Verifying identities in hyperbolic functions involves proving that two expressions are equivalent by using predefined hyperbolic function definitions.

In our case, we are comparing \(\cosh 2x\) to \(\cosh^2 x + \sinh^2 x\), starting from the right side:

• Expand the expressions using \(\cosh x = \frac{e^x + e^{-x}}{2}\) and \(\sinh x = \frac{e^x - e^{-x}}{2}\).
• Square each, \((\cosh x)^2\) and \((\sinh x)^2\).

Initially, it appears intricate, yet through algebraic simplifications using exponential identities, the end result simplifies back to \(\cosh 2x\):

• \(\cosh^2 x + \sinh^2 x = \frac{e^{2x} + e^{-2x}}{2}\)

By showcasing step-by-step manipulations, the intention is to consolidate understanding and confirm the equivalency of the identity. Tackling such problems enhances your algebraic manipulation skills and builds a deeper appreciation for the symmetries and behaviors in hyperbolic frameworks.