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This is what I've already done. Can't think of how to proceed further

$|\cos(x)-\cos(y)|=\left|-2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)\right|\leq\left|\frac{x+y}{2}\right||x-y|$

What should I do next?

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    @DavidMitra You're right, I think that works, thank you! I agree with you about their solution being better, I just wrote down the first thing that came to mind!2012-12-11

2 Answers 2

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Hint: Any continuous function is uniformly continuous on a closed, bounded interval, so $\cos$ is uniformly continuous on $[-2\pi,0]$ and $[0,2\pi]$.

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The cosine function is Lipschitz-continuous because its derivative is bounded. That follows from the mean value theorem. Every Lipschitz-continuous function is uniformly continuous.

postscript prompted by Tom Oldfield's comment: A function $f$ is Lipschitz-continuous iff there is some non-negative number $m$ such that for all $x,y$ in the domain of $f$ we have $|f(x)-f(y)|\le m|x-y|$. This is stronger than uniform continuity, in that every Lipschitz-continuous function is uniformly continuous, but not every uniformly continuous function is Lipschitz continuous. An example of a uniformly continuous function that is not Lipschitz-continuous is $x\mapsto\sqrt{1-x^2}$ on the interval $[-1,1]$. It's really easy to prove that Lipschitz continuity entails uniform continuity.

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    You can prove that it's uniformly continuous without mentioning Lipschitz continuity using exactly the same proof, in case people haven't heard of, or don't use much, Lipschitz continuity.2012-12-11