Prove Inequality using Mean Value Theorem

[Khayyam89]

Homework Statement

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

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.

 

[boombaby]

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

 

[Khayyam89]

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

 

[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.