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Expert-verified Found in: Page 216 ### Essential Calculus: Early Transcendentals

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.

See the 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. ### Want to see more solutions like these? 