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Proof. $\forall \epsilon>0$, choose $\delta=\epsilon$. Now $\forall x$, if $0<|x-c|<\delta$, then $|\sin x -\sin c|=|2\cos\frac{x+c}{2} \sin\frac{x-c}{2}|\le 2|\sin \frac{x-c}{2}|\le 2|\frac{x-c}{2}|=|x-c|<\delta=\epsilon $, which completes the proof.

The proof above uses $|\sin x|\le |x|$, which is easy to show geometrically. However, I am afraid it is not rigorous. Can anyone tell me how to show $|\sin x|\le |x|$ without using derivative or Taylor expansion? Thanks in advance.

user10354138
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toronto hrb
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    https://math.stackexchange.com/questions/125298/how-to-strictly-prove-sin-xx-for-0x-frac-pi2 – A learner Jul 20 '20 at 03:58
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    It depends on how is $\sin$ defined. Here is huge discussion https://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1/3736280#3736280 of several possible cases. Without derivative, geometry and Taylor I brought, for example, axiomatic definition. – zkutch Jul 20 '20 at 03:58
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    Why $|sinx|≤|x|$ is not rigorous? It is well established and at least in my opinion rigour enough. In that sense you also used $cost\leq1$ without showing it, trigonometric identities and absolute values properties without showing them also...It can go on forever. – Eminem Jul 20 '20 at 05:22
  • Thanks to you all. – toronto hrb Jul 20 '20 at 14:16

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