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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?

  • 3
    Use that $|\sin(x)| \leq 1$.2012-12-11
  • 0
    So, instead of writing $\frac{|x+y|}{2}|x-y|$ I could've written $\leq |-2*1*\frac{x-y}{2}|=|x-y|$ and then I can just set delta equal to epsilon?2012-12-11
  • 0
    @user1242967 Only drop the $|x+y|/2$ part. Turn that into an answer to your own question and you have my upvote. You have shown it in the best way yet.2012-12-11
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    @DavidMitra I think you have to be slightly more involved than that, $|x-y| < \delta$ for $\delta < 4\pi$ doesn't necessarily mean that we can we can "shift" both $x$ and $y$ to be in $[-2\pi, 2\pi]$. I may be missing something, but I think you have to be slightly more fiddly, as in my hint below.2012-12-11
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    @TomOldfield You'd of course demand that $\delta$ be less than $\min\{1,\delta_\epsilon\}$, say. I think this would work... Regardless, I deleted my comment as the OP's solution together with WimC's hint provides the best solution, in my opinion.2012-12-11
  • 0
    @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

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