How do I prove |cosa - cosb| <= |a-b|?

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Homework Statement


I have no idea how to approach this question.

Homework Equations

The Attempt at a Solution


I suppose ∫ |cosa - cosb| < = |a-b|
sinb-sina <= b^2/2 - a^2/2
but now what do I do?
 
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nvm. I figured it out, it's a subtle trick with MVT
 
Since it looks like you've found a solution ... assuming ##a\leq b## gives $$\left|\int_a^b\sin x\ \mathrm{d}x\right|\leq\int_a^b|\sin x|\ \mathrm{d}x\leq\int_a^b1\ \mathrm{d}x$$.
If ##a>b##, you just need to flip the limits on the last two integrals. The desired inequality isn't too incredibly difficult to get from there.

I like the MVT proof better, though. It's, like, one step.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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