I have the following problem:
Show that if $144\leq a\leq b\leq 169$ is true, then
$\frac{a-b}{64}\leq \sin\sqrt{a}-\sin\sqrt{b}\leq\frac{a-b}{36}$.
That $3<\pi<\frac{22}{7}$ can be used.
I figure this problem should be solved using the Mean Value Theorem and that you should divide by $a-b$.
$\frac{a-b}{64}\leq \sin\sqrt{a}-\sin\sqrt{b}\leq\frac{a-b}{36}$ $\Leftrightarrow$ $\frac{1}{64}\leq \frac{\sin\sqrt{a}-\sin\sqrt{b}}{a-b}\leq\frac{1}{36}$
MVT gives that there is some $c$ $\epsilon$ $[b,a]$ where $\frac{1}{64}\leq \frac{\sin\sqrt{a}-\sin\sqrt{b}}{a-b}=f'(c)\leq\frac{1}{36}$ where $f(x)=\sin\sqrt{x}$ so that $f'(c)=\frac{1}{2\sqrt{c}} \cos\sqrt{c}$
which gives $\frac{1}{64}\leq \frac{\cos\sqrt{c}}{2\sqrt{c}} \leq\frac{1}{36}$ $\Leftrightarrow$ $\frac{1}{32}\leq \frac{\cos\sqrt{c}}{\sqrt{c}} \leq\frac{1}{18}$.
Not sure where to go from here, how to get from here to the initial inequality. Also not sure how the pi inequality above factors into the solution.
Any help would be appreciated!
$$\left(4 - \frac{1}{3}\right) \pi < \frac{7 \cdot 12}{22} \pi < 12 \le \sqrt{\xi} \le 13 < \frac{13}{3} \pi = \left(4 + \frac{1}{3}\right) \pi$$
You can check that $\cos(\pi x)$ is $\ge 1/2$ for $4 - \frac{1}{3} \le x \le 4 + \frac{1}{3}$.
Combining this fact with $\frac{1}{13} \le \frac{1}{\sqrt{\xi}} \le \frac{1}{12}$ yields
$$\frac{1}{2 \cdot 13} \le \frac{\cos \sqrt{\xi}}{\sqrt{\xi}} \le \frac{1}{2 \cdot 12}$$