Show that $|\sin{a}-\sin{b}| \le |a-b| $ for all $a$ and $b$

Question

I've recently been going over the mean value and intermediate value theorems, however I'm not sure where to start on this.

Answer 1

The mean value theorem will do just fine. Setting $f(x) = \sin(x)$, note that $$ |f(x) - f(y)| = |f'(c)||x - y| $$ for some value $c$. Can we find an upper bound for $|f'(c)|$?

Answer 2

As an alternative to the application of the mean value theorem, you can write $$ \sin a - \sin b = 2 \sin \frac{a-b}2 \cos \frac{a+b}2 $$ and use $$ |\sin u| \le |u| $$