# Prove Inequality using Mean Value Theorem

1. Oct 28, 2008

### Khayyam89

1. The problem statement, all variables and given/known data
Essentially, the question asks to use the mean value theorem(mvt) to prove the inequality: abs(sina - sinb) $$\leq$$ abs(a - b) for all a and b

3. 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 continous 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.

Last edited: Oct 28, 2008
2. Oct 28, 2008

### boombaby

Assume a>b, then sina-sinb=(a-b)cosc, for some b<c<a, which gives the inequality with no problems.

3. Oct 28, 2008

### Khayyam89

Wait, are you considering that abs(...) = absolute value of the sum?

4. Oct 28, 2008

### Dick

The mean value theorem tells you (sin(a)-sin(b))/(a-b)=sin'(c)=cos(c) for some c between a and b, as boombaby said. Take the absolute value of both sides and use that |cos(c)|<=1.