/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Prove the identity. $$\frac{1+\... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Prove the identity. $$\frac{1+\tanh x}{1-\tanh x}=e^{2 x}$$

Short Answer

Expert verified
\(\frac{1+\tanh x}{1-\tanh x} = e^{2 x}\) is proven using hyperbolic identities.

Step by step solution

01

Recall Hyperbolic Tangent Identity

First, remember the definition of the hyperbolic tangent function: \( \tanh x = \frac{\sinh x}{\cosh x} \). Keep this identity in mind as it will be used in this exercise.
02

Rewrite \(1 + \tanh x\)

Substitute the definition of \(\tanh x\) in \(1 + \tanh x \) to get: \[1 + \tanh x = 1 + \frac{\sinh x}{\cosh x} = \frac{\cosh x + \sinh x}{\cosh x}. \]
03

Rewrite \(1 - \tanh x\)

Similarly, substitute the definition of \(\tanh x\) in \(1 - \tanh x \) to get: \[1 - \tanh x = 1 - \frac{\sinh x}{\cosh x} = \frac{\cosh x - \sinh x}{\cosh x}. \]
04

Form the Expression

Now substitute \(1 + \tanh x\) and \(1 - \tanh x\) into the left-hand side of our original identity:\[\frac{1 + \tanh x}{1 - \tanh x} = \frac{\frac{\cosh x + \sinh x}{\cosh x}}{\frac{\cosh x - \sinh x}{\cosh x}} = \frac{\cosh x + \sinh x}{\cosh x - \sinh x}.\]
05

Simplify Expression

Notice that you can simplify \( \frac{\cosh x + \sinh x}{\cosh x - \sinh x} \) using the identity \( {a}/{b}\) transformed to \((a\cdot (b'))/(b\cdot (b'))\) where \(b'\) is the conjugate of \(b\):\[= \frac{(\cosh x + \sinh x)(\cosh x + \sinh x)}{(\cosh x - \sinh x)(\cosh x + \sinh x)}.\]The denominator becomes a difference of squares:\[= \frac{(\cosh x + \sinh x)^2}{\cosh^2 x - \sinh^2 x}.\]
06

Use Hyperbolic Identity for Simplification

Apply the hyperbolic identity \(\cosh^2 x - \sinh^2 x = 1\) to the denominator. Thus, the expression becomes:\[= \frac{(\cosh x + \sinh x)^2}{1} = (\cosh x + \sinh x)^2.\]
07

Expand and Simplify the Expression

Expand \((\cosh x + \sinh x)^2\):\[= \cosh^2 x + 2\cosh x \cdot \sinh x + \sinh^2 x.\]Using \(\cosh^2 x + \sinh^2 x\) again, rewrite it as \( 1 + 2\cosh x \cdot \sinh x \). So the expression is:\[= 1 + 2\sinh x \cdot \cosh x.\]
08

Recognize Exponential Expression

Recall the identity for the double angle of hyperbolic sine: \(\sinh(2x) = 2\sinh x \cdot \cosh x\). Recognize that our expression \(1 + \sinh(2x) \) corresponds to the well-known identity of \(e^{2x}\):\[e^{2x} = \cosh(2x) + \sinh(2x).\]Thus, we have shown:\[1 + \sinh(2x) = e^{2x},\]proving the initial identity.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hyperbolic Tangent
The hyperbolic tangent, denoted as \( \tanh x \), is a fundamental hyperbolic function that resembles the trigonometric tangent function. Its definition is given as:
  • \( \tanh x = \frac{\sinh x}{\cosh x} \)
Here, \( \sinh x \) represents the hyperbolic sine and \( \cosh x \) represents the hyperbolic cosine. Both \( \sinh x \) and \( \cosh x \) are key elements in hyperbolic functions, similar to how sine and cosine are in trigonometry.
The value of \( \tanh x \) tells us about the ratio of the length of the opposite side to the adjacent side in the context of the unit hyperbola, much like what is done with the tangent in a circle.
Hyperbolic tangent is particularly useful in mathematical modeling and engineering because it naturally appears in calculations involving entities such as waves, signals, and other phenomena that exhibit exponential growth or decay characteristics.
Exponential Functions
Exponential functions take on the form \( f(x) = a^x \), where \( a \) is a constant base and \( x \) is the exponent. One of the most common exponential functions is \( e^x \), where \( e \approx 2.71828 \) is a mathematical constant that serves as the base for natural logarithms.
In the context of hyperbolic functions, exponential functions play a crucial role in defining them. For example, both \( \sinh x \) and \( \cosh x \) can be expressed in terms of exponential functions:
  • \( \sinh x = \frac{e^x - e^{-x}}{2} \)
  • \( \cosh x = \frac{e^x + e^{-x}}{2} \)
These expressions demonstrate that hyperbolic functions are deeply intertwined with exponential functions.
This relationship is vital in proving identities like the one given in the above exercise, where the original expression containing \( \tanh x \) is transformed to match an exponential form \( e^{2x} \).
The memorable property of exponential functions is that the rate of growth is proportional to its current value, making them invaluable in modeling natural processes such as population growth, radioactive decay, and more.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that hold true for all values of the involved variables. Similarly, hyperbolic functions have their own identities, often used in proofs and problem-solving.
One of the most significant hyperbolic identities is:
  • \( \cosh^2 x - \sinh^2 x = 1 \)
This identity parallels the Pythagorean identity in trigonometry \( \cos^2 x + \sin^2 x = 1 \) but applies to hyperbolic functions.
In the given exercise, this identity is utilized to simplify complex expressions and demonstrate equivalence with exponential expressions. Specifically, it aids in transforming a difference of squares in the denominator to a simplified form, allowing further progression in proving identities involving hyperbolic functions.
Understanding these trigonometric and hyperbolic identities is crucial for solving complex algebraic and calculus problems, providing a toolbox for simplifying and manipulating expressions efficiently.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A plane flies horizontally at an altitude of 5 \(\mathrm{km}\) and passes directly over a tracking telescope on the ground. When the angle of elevation is \(\pi / 3,\) this angle is decreasing at a rate of \(\pi / 6 \mathrm{rad} / \mathrm{min}\) . How fast is the plane traveling at that time?

The frequency of vibrations of a vibrating violin string is given by $$\mathrm{f}=\frac{1}{2 \mathrm{L}} \sqrt{\frac{\mathrm{T}}{\rho}}$$ where \(L\) is the length of the string, T is its tension, and \(\rho\) is its linear density. [See Chapter 11 in D. E. Hall, Musical Acoustics, 3 \(\mathrm{d}\) ed. (Pacific Grove, CA: Brooks/Cole, \(2002 ) . ]\) (a) Find the rate of change of the frequency with respect to (i) the length (when T and \(\rho\) are constant), (ii) the tension (when L and \(\rho\) are constant), and (iii) the linear density (when L and T are constant). (b) The pitch of a note (how high or low the note sounds) is determined by the frequency f. (The higher the frequency, the higher the pitch.) Use the signs of the derivatives in part (a) to determine what happens to the pitch of a note (i) when the effective length of a string is decreased by placing a finger on the string so a shorter portion of the string vibrates, (ii) when the tension is increased by turning a tuning peg. (iii) when the linear density is increased by switching to another string.

If $$x=\ln (\sec \theta+\tan \theta), \text { show that } \sec \theta=\cosh x$$

$$ \begin{array}{l}{\text { (a) Use differentials to find a formula for the approximate }} \\ {\text { volume of a thin cylindrical shell with height } h \text { , inner }} \\ {\text { radius } r, \text { and thickness } \Delta r \text { . }} \\ {\text { (b) What is the error involved in using the formula from }} \\ {\text { part (a)? }}\end{array} $$

Find the derivative. Simplify where possible. $$f(t)=\operatorname{csch} t(1-\ln \csc h t)$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.