StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

Q29E

Expert-verifiedFound in: Page 216

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

**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.

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

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

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.

94% of StudySmarter users get better grades.

Sign up for free