/*! 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 83 Prove that \(|\cos a-\cos b| \le... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 83

Prove that \(|\cos a-\cos b| \leq|a-b|\) for all \(a\) and \(b\).

Short Answer

Expert verified
By applying the Mean Value Theorem, we find that \(|\cos a-\cos b| \leq |a-b|\). This is due to the fact that the derivative of the cosine function, which is the sine function, has an absolute value less than or equal to 1. When this is multiplied by the absolute difference of \(a\) and \(b\), it must be less than or equal to the absolute difference of \(a\) and \(b\), therefore fulfilling the inequality in question.

Step by step solution

01

Apply Mean Value Theorem

The Mean Value Theorem states that if \(f\) is continuous on the interval \([a, b]\) and differentiable on the interval \((a, b)\), then there exists at least one point \(c\) in \((a, b)\) at which the derivative of \(f\) is equal to the average rate of change over the interval. This is expressed as \(f'(c) = (f(b)-f(a)) / (b-a)\). Apply the Mean Value Theorem to \(f(x) = \cos(x)\). Note that \(f(x) = \cos(x)\) is continuous and differentiable everywhere. Find \(f'(x) = -\sin(x)\).
02

Substitution into Mean Value Theorem

Now, substitute \(f(x) = \cos(x)\) and \(f'(x) = -\sin(x)\) into the Mean Value Theorem: \(f'(c) = (f(b)-f(a)) / (b-a)\) becomes \(-\sin(c) = (\cos(b)-\cos(a)) / (b-a)\). Multiply both sides by \((b-a)\) to get \(-(b-a)\sin(c) = \cos(b)-\cos(a)\).
03

Absolute Values and Inequality

Take the absolute value of both sides of the equality: \(|(b-a)\sin(c)| = |\cos(b)-\cos(a)|\). Since \(|\sin(c)|\) is always less than or equal to 1, any value, \(|\sin(c)|\), multiplied by \(|b-a|\), should be less than or equal to \(|b-a|\). Therefore, \(|(b-a)\sin(c)| \leq |b-a|\). This implies that \(|\cos(b)-\cos(a)| \leq |b-a|\).

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 91影视!

Key Concepts

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

The Cosine Function
The cosine function is an important part of trigonometry. It's a periodic function with a wave-like structure that repeats every 360 degrees or \(2\pi\) radians. The cosine of an angle in a right triangle is the ratio of the length of the adjacent side to the hypotenuse. This can be written as:\[\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}\]
  • The graph of the cosine function starts at the maximum point when \(x = 0\) and descends to the minimum point, creating a smooth curve.
  • It has a range between -1 and 1, so the output values of a cosine function are limited to these bounds.
  • It is an even function, meaning that \(\cos(-x) = \cos(x)\).
Understanding the cosine function is essential, especially when evaluating derivatives and integrals in calculus. The smooth and continuous nature of this function, along with its differentiability, is critical when applying calculus concepts such as the Mean Value Theorem.
Continuity and Differentiability
In the context of the Mean Value Theorem, understanding continuity and differentiability is crucial. A function is continuous if you can draw it without lifting your pencil. Mathematically, continuity at a point \(c\) means:\[\lim_{{x \to c}} f(x) = f(c)\]
  • A function is continuous over an interval if it鈥檚 continuous at every point within that interval.
  • For the Mean Value Theorem, continuity over a closed interval \([a, b]\) is needed.
Differentiability is about the existence of a derivative. If a function is differentiable at a point, it means there is a definite slope at that point, defined by its derivative \(f'(x)\). A differentiable function will be smooth, without any sudden breaks or corners. The cosine function \(\cos(x)\) is both continuous and differentiable everywhere on the real number line, making it suitable for applying the Mean Value Theorem.
Trigonometric Inequalities
Trigonometric inequalities involve understanding the bounds within which trig functions operate. These inequalities help compare and limit functions. For example, the inequality \(|\sin(x)| \leq 1\) reveals how \(\sin(x)\) never exceeds 1 in magnitude.In the problem at hand, \[|\cos(a) - \cos(b)| \leq |a-b|\]was derived using properties of trigonometric inequalities through the Mean Value Theorem.
  • The crucial step was recognizing that \(|\sin(c)| \leq 1\), which confines the value of trigonometric expressions.
  • This inequality is significant because it demonstrates that the change in cosine values is limited by the change in angles.
These trigonometric inequalities serve as powerful tools in proving various other inequalities and equations within trigonometric contexts.

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

Sketch the graph of the function using extrema, intercepts, symmetry, and asymptotes. Then use a graphing utility to verify your result. $$ y=\frac{2 x}{1-x} $$

\mathrm{\\{} G r a p h i c a l ~ R e a s o n i n g ~ Consider the function \(f(x)=\frac{\cos ^{2} \pi x}{\sqrt{x^{2}+1}}, \quad 0

Find \(a, b, c\), and \(d\) such that the cubic \(f(x)=a x^{3}+b x^{2}+c x+d\) satisfies the given conditions. Relative maximum: \((3,3)\) Relative minimum: \((5,1)\) Inflection point: \((4,2)\)

Find the maximum value of \(f(x)=x^{3}-3 x\) on the set of all real numbers \(x\) satisfying \(x^{4}+36 \leq 13 x^{2}\). Explain your reasoning.

Use the definition of limits at infinity to prove the limit. $$ \lim _{x \rightarrow \infty} \frac{2}{\sqrt{x}}=0 $$
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

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.