Use the Mean Value Theorem to prove that for all a,b that belongs to the reel, we have :
$\displaystyle |sin(b)-sin(a)|$ $\displaystyle \leq$ $\displaystyle | b-a |$
well, if you let $\displaystyle f(x)=\sin x,$ then by mean value theorem there exists $\displaystyle c$ between $\displaystyle a,b$ such that $\displaystyle f(b)-f(a)=(b-a)f'(c).$ that means $\displaystyle \sin b - \sin a = (b-a)\cos c.$
so $\displaystyle |\sin b - \sin a|=|b-a||\cos c| \leq |b-a|,$ because $\displaystyle |\cos c| \leq 1.$ it wan't very difficult, was it?
There is a second part to this problem :
Assume that f : $\displaystyle \Re \mapsto \Re $ is a differentiable function and that there exists a real number M > 0 such that |f '(x)|$\displaystyle \leq M$ for all x in the reel then show that :
$\displaystyle |f(a) - f(b)|\leq M|b-a|$for all x in the reel
Choose any value $\displaystyle c$, we know there exists a corresponding interval $\displaystyle [a,b]$ such that $\displaystyle |f'(c)|=\left|\frac{f(b)-f(a)}{b-a}\right|\leqslant M$ and since $\displaystyle c$ was arbitrary this must work for every interval $\displaystyle [a,b]$.
Another almost identical method would be to assume that there exists an interval $\displaystyle [a,b]$ such that $\displaystyle \left|\frac{f(b)-f(a)}{b-a}\right|>M$. But since $\displaystyle f$ is differentiable we can find a $\displaystyle c\in(a,b)$ such that $\displaystyle |f'(c)|=\left|\frac{f(b)-f(a)}{b-a}\right|\implies |f'(c)|>M$ which is a contradiction.
This is called Lipschitz's continuity by the way.
If you have any questions at all feel free to ask
this, although an interesting claim, but is not what mean value theorem says: you choose an interval [a,b] and then you find a < c < b that satisfies the condition. what you're doing here is
reversing things, i.e. you choose c and then you claim that there exists an interval [a,b] containing c with that condition! i'm not sure if this is always possible!
here's a simple counter-example: let $\displaystyle f(x)=x^3$ and choose $\displaystyle c=0.$ geometrically it's clear that there are no $\displaystyle a \neq b$ such that $\displaystyle f'(0)=\frac{f(b)-f(a)}{b-a}.$ you can also prove it very easily:
if there exist such $\displaystyle a,b,$ then: $\displaystyle 0=\frac{b^3-a^3}{b-a}=a^2+ab+b^2,$ which is possible iff $\displaystyle a=b=0,$ because: $\displaystyle 0=2a^2 + 2ab+2b^2=(a+b)^2+a^2+b^2.$
regarding Larrioto's question:
by mean value theorem for any real numbers $\displaystyle a,b,$ there exists $\displaystyle c$ (between $\displaystyle a,b$) such that $\displaystyle f(b)-f(a)=(b-a)f'c).$ thus: $\displaystyle |f(b)-f(a)|=|b-a||f'(c)| \leq M|b-a|.$