Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 5

Do the following: a. Write \((\cosh x-\sinh x)^{6}\) in terms of exponentials. b. Prove \(\cosh (x-y)=\cosh x \cosh y-\sinh x \sinh y\) using the exponential forms of the hyperbolic functions. c. Prove \(\cosh 2 x=\cosh ^{2} x+\sinh ^{2} x\) d. If \(\cosh x=\frac{13}{12}\) and \(x<0\), find \(\sinh x\) and \(\tanh x\). e. Find the exact value of \(\sinh (\operatorname{arccosh} 3)\).

Short Answer

Expert verified
a. \(e^{-6x}\) b. \(cosh (x-y) = cosh x cosh y - sinh x sinh y \) is true. c. \(cosh 2 x = cosh ^{2} x + sinh ^{2} x\) is true. d. \(sinh x = -\frac{5}{12}\), \(tanh x = -\frac{5}{13}\). e. \(sinh (arccosh 3) = \sqrt{8}\).

Step by step solution

01

a. Writing hyperbolic function in terms of exponentials

Recall that \( \cosh x = \frac{e^x + e^{-x}}{2} \) and \( \sinh x = \frac{e^x - e^{-x}}{2} \). Substituting these into \((\cosh x - \sinh x)^6\) gives: \( (\frac{e^x + e^{-x}-e^x + e^{-x}}{2})^6 = (\frac{2 e^{-x}}{2})^6 = e^{-6x}\).
02

b. Proving an identity using exponential forms

Using the definition of \( \cosh x \) and \( \sinh x \) and the exponentiation rule, we get: \( \cosh (x-y)=\frac{e^{x-y} + e^{-(x-y)}}{2} =\frac{e^x}{e^y}+\frac{e^y}{e^x} =\frac{e^{2x} + 1}{2 e^x}=\cosh x \cosh y - \sinh x \sinh y \).
03

c. Proving an identity involving squares of hyperbolic functions

Applying the definitions from above, we derive \( \cosh^{2} x - \sinh^{2} x = (\frac{e^x + e^{-x}}{2})^2 - (\frac{e^x - e^{-x}}{2})^2 = 1 = \cosh 2x \).
04

d. Calculating other hyperbolic functions given a value for \( \cosh x \)

To find \( \sinh x \), we know that \( \cosh^2 x - \sinh^2 x = 1 \), so \( \sinh x = \sqrt{\cosh^2 x - 1} = \sqrt{(\frac{13}{12})^2 - 1 } = -\frac{5}{12} \). For \( \tanh x \), we know that \( \tanh x = \frac{\sinh x}{\cosh x} = \frac{-5/12}{13/12} = -5/13. \)
05

e. Calculating hyperbolic function of inverse hyperbolic function

Recall that \( \sinh(\operatorname{arccosh} x) = \sqrt{x^2 - 1} \). Thus, \( \sinh(\operatorname{arccosh} 3) = \sqrt{3^2 - 1} = \sqrt{8}.\)

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Use dimensional analysis to derive a possible expression for the drag force \(F_{D}\) on a soccer ball of diameter \(D\) moving at speed \(v\) through air of density \(\rho\) and viscosity \(\mu .\) [Hint: Assuming viscosity has units \(\frac{[M]}{[L][T]}\), there are two possible dimensionless combinations: \(\pi_{1}=\mu D^{\alpha} \rho^{\beta} v^{\gamma}\) and \(\pi_{2}=F_{D} D^{a} \rho^{\beta} v^{\gamma} .\) Determine \(\alpha, \beta\), and \(\gamma\) for each case, and interpret your results.]

Find the Taylor series centered at \(x=a\) and its corresponding radius of convergence for the given function. In most cases, you need not employ the direct method of computation of the Taylor coefficients. a. \(f(x)=\sinh x, a=0\) b. \(f(x)=\sqrt{1+x}, a=0\). c. \(f(x)=\ln \frac{1+x}{1-x}, a=0\) d. \(f(x)=x e^{x}, a=1\). e. \(f(x)=\frac{1}{\sqrt{x}}, a=1\) f. \(f(x)=x^{4}+x-2, a=2\) g. \(f(x)=\frac{x-1}{2+x}, a=1\)

Determine the exact values of a. \(\sin \frac{\pi}{8}\). b. \(\tan 15^{\circ}\). c. \(\cos 105^{\circ}\).

Evaluate the following expressions at the given point. Use your calculator or your computer (such as Maple). Then use series expansions to find an approximation of the value of the expression to as many places as you trust. a. \(\frac{1}{\sqrt{1+x^{3}}}-\cos x^{2}\) at \(x=0.015\). b. \(\ln \sqrt{\frac{1+x}{1-x}}-\tan x\) at \(x=0.0015\). c. \(f(x)=\frac{1}{\sqrt{1+2 x^{2}}}-1+x^{2}\) at \(x=5.00 \times 10^{-3}\). d. \(f(R, h)=R-\sqrt{R^{2}+h^{2}}\) for \(R=1.374 \times 10^{3} \mathrm{~km}\) and \(h=1.00 \mathrm{~m}\). e. \(f(x)=1-\frac{1}{\sqrt{1-x}}\) for \(x=2.5 \times 10^{-13}\).

Consider Gregory's expansion $$ \tan ^{-1} x=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\cdots=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{2 k+1} x^{2 k+1} $$ a. Derive Gregory's expansion using the definition $$ \tan ^{-1} x=\int_{0}^{x} \frac{d t}{1+t^{2}} $$ expanding the integrand in a Maclaurin series, and integrating the resulting series term by term. b. From this result, derive Gregory's series for \(\pi\) by inserting an appropriate value for \(x\) in the series expansion for \(\tan ^{-1} x\).
See all solutions

Recommended explanations on Combined Science Textbooks

Synergy

Read Explanation
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.

Sign-up for free