Mean value theorem(mvt) to prove the inequality

[Khayyam89]

Homework Statement

Essentially, the question asks to use the mean value theorem(mvt) to prove the inequality: $$\left|$$sin a -sin b $$\right|$$ $$\leq$$ $$\left|$$ a - b$$\right|$$ for all a and b

The Attempt at a Solution

I do not have a graphing calculator nor can I use one for this problem, so I need to prove that the inequality basically by proof. What I did was to look at the mvt hypotheses: if the function is continuous and differetiable on closed and open on interval a,b, respectively. However, the problem I am having is that I am getting thrown off by the absolute values and the fact that I've never used mvt on inequalities. I know the absolute value of the sin will look like a sequence of upside-down cups with vertical tangents between them. Hints most appreciated.

 

[dirk_mec1]

The mean value theorem states that in an interval [a,b]:

$$f'(c) = \frac{ \sin b - \sin a}{b-a} = \cos(c)$$

Now put absolute value signs there and make use of $$|\cos(x)| \leq 1$$