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I've been struggling with this question. I don't know how to get rid of the sine function after getting to $$|\sin x - \sin a| \leq 2 \left| \sin \dfrac{x - a}{2} \right|$$ Half angle formula isn't really useful imo, not sure what to do from here.

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    $\begingroup$ $|\sin x|\leqslant |x|$ for all $x\in\mathbb{R}$. $\endgroup$
    – RRL
    Jan 25, 2021 at 3:11

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Thinking about an identity was a good idea. However, use $\sin(\alpha)-\sin(\beta)=2\cos(\frac{\alpha+\beta}{2})\sin(\frac{\alpha-\beta}{2})$.

We want to chek that $\forall \epsilon>0$, $\exists$ $\delta>0$ such that:

$|x-c|<\delta \implies |\sin(x)-\sin(c)|<\epsilon$.

Let $\epsilon=\delta$, then $|\sin(x)-\sin(c)|=2\cos(\frac{x+c}{2})\sin(\frac{x-c}{2})\leq2|\frac{x-c}{2}|=|x-c|<\delta=\epsilon$.

We can justify this last step with the following:

  • $|\cos(x)|\leq 1 ~\forall x\in\mathbb{R}$
  • $|\sin(x)|<|x| ~\forall x\in\mathbb{R}$
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Use the mean value theorem. That is, $|\sin(x) - \sin(a)| \leq |\cos(\psi)||x-a|$ for some $\psi \in (x, a)$. I will let you finish the proof.

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    $\begingroup$ this question is essentially asking to show that sine is continuous. are you not assuming that it is continuous by assuming that it is differentiable in the mean value theorem? $\endgroup$
    – C Squared
    Jan 25, 2021 at 4:08
  • $\begingroup$ @C Squared you are correct I overlooked this. $\endgroup$ Jan 25, 2021 at 4:12

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