Use the mean value theorem to prove that if $\displaystyle ~|x|<\frac{\pi}2 ~,~~ |y|<\frac{\pi}2~,$ then
$$|\sin(y) – \sin(x)| \leq |y-x| \leq |\tan(y) – \tan(x)|$$
What I did was using the mean value theorem using $\sin(x)$ as the function to get
$$\frac{ \sin(x) – \sin(y) }{ x-y } = \sin'(c) = \cos(c)$$
for some $c$ inbetween $x$ and $y$, and since the value of $\cos(x)$ for every $x$ goes around $1$ and $-1$
$$\left| \frac{ \sin(x) – \sin(y) }{ x-y }\right| \leq 1$$
but I'm stuck here
Best Answer
Note that if $f(x)=\tan{x}$, then by mean value theorem $$\frac{\tan{x}-\tan{y}}{x-y}=\tan'(c)=\sec^2({c})\geq1$$