• :00Days
  • :00Hours
  • :00Mins
  • 00Seconds
A new era for learning is coming soonSign up for free
Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration


Essential Calculus: Early Transcendentals
Found in: Page 216
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

Answers without the blur.

Just sign up for free and you're in.


Short Answer

Use the Mean Value Theorem to prove the inequality \(|\sin a - \sin b|\; \le \;|a - b|\) for all \(a\) and \(b\).

The inequality is \(|\sin a - \sin b|\; \le \;|a - b|\) is proved.

See the step by step solution

Step by Step Solution

Step 1: Given data

The given inequality is \(|\sin a - \sin b|\; \le \;|a - b|\).

Step 2: Concept of Mean value theorem

“Let \(f\) be a function that satisfies the following hypothesis:

1. \(f\) is continuous on the closed interval \(\left( {a\;,{\rm{ }}b} \right)\).

2. \(f\) is differentiable on the open interval \(\left( {a,\;b} \right)\).

Then, there is a number \(c\) in \(\left( {a\;,{\rm{ }}b} \right)\) such that \({f^\prime }(c) = \frac{{f(b) - f(a)}}{{b - a}}\).

Or, equivalently, \(f(b) - f(a) = {f^\prime }(c)(b - a)\).”

Step 3: Calculation to prove the given inequality

Consider the function, \(f(x) = \sin x\).

Obtain the derivative of, \(f(x)\).

\(\begin{aligned}{c}{f^\prime }(x) &= \frac{d}{{dx}}(\sin x)\\{f^\prime }(x) &= \cos x\end{aligned}\)

Replace \(x\) by \(c\).

Then, \({f^\prime }(c) = \cos c\).

Since \(\sin x\) is continuous and differentiable everywhere, by Mean Value Theorem mentioned above, then there exists a number c such that:

\(\cos (c) = \frac{{\sin (b) - \sin (a)}}{{b - a}}\) …… (1)

Take modulus on both sides of the equation.

\(|\cos (c)| = \left| {\frac{{\sin (b) - \sin (a)}}{{b - a}}} \right|\)

Since \( - 1 \le \cos (c) \le 1\), it implies that \(|\cos (c)| \le 1\).

Therefore, \(\left| {\frac{{\sin (b) - \sin (a)}}{{b - a}}} \right| \le 1\).

Since, by the property, \(\left| {\frac{a}{b}} \right| = \frac{{|a|}}{{|b|}}\).

\(\begin{aligned}|\sin (b) &- \sin (a)|\; \le 1 \cdot |b - a|\\|\sin (b) &- \sin (a)|\; \le \;| - (a - b)|\\|\sin (b) &- \sin (a)|\; \le \;|a - b|\end{aligned}\)

Hence, the inequality is proved.

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.