Chapter 4.2, Problem 59E

To determine

To prove: The given statement.

Explanation of Solution

Given information:

The statement is:

  $|\sin b-\sin a|\le |b-a|$

Calculation:

Let, $f(x)=\sin x$

Hence $f(x)$ 'is differentiable on the interval $\text{(a,b)}$ and continuous on the interval $\text{(a,b)}$ for all $\text{a}$ and $\text{b}$ , by the MVT, there must exist a $\text{c}$ in $\text{(a,b)}$ such that

  $\begin{array}{c}{f}^{\prime }(c)=\frac{f(b)-f(a)}{b-a}\\ =\frac{\sin b-\sin a}{b-a}\end{array}$

The derivative of $f(x)$ is $f^{\prime }(x)=\cos x$ , so at

  $\begin{array}{c}x=c,\\ f^{\prime }(c)=\cos c\end{array}$

Set the two expressions for $f^{\prime }(c)$ equal then gives

  $\frac{\sin b-\sin a}{b-a}=\cos c$

Hence $-1\le \cos c\le 1$ for all $\text{c}$ then $\left|\cos c\right|\le 1$

Take the absolute vale of both sides of $\frac{\sin b-\sin a}{b-a}=\cos c$ gives

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

Hence $\left|\cos c\right|\le 1$ then

  $\left|\frac{\sin b-\sin a}{b-a}\right|\le 1$

Solve this for $\left|\sin b-\sin a\right|$ then gives,

  $\begin{array}{c}\left|\frac{\sin b-\sin a}{b-a}\right|\le 1\\ \Rightarrow \frac{|\sin b-\sin a|}{|b-a|}\le 1\\ \Rightarrow |\sin b-\sin a|\le |b-a|\end{array}$

Therefore the required given statement $|\sin b-\sin a|\le |b-a|$ (Proved).