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How can it be shown that $\forall \varepsilon > 0, \exists k\in\mathbb{Z}$:

$$|\sin(k)|<\varepsilon$$

Eg. If I take $\varepsilon = 0.00001$ how do I get a $|\sin(k)|$ less than that? I have tried various values on the calculator but it never seems to go below $0.01$.

rtybase
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Hanzel
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1 Answers1

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$2\pi$ is an irrational number, thus, according to Dirichlet's approximation theorem,$\forall N\in\mathbb{N}, \exists p,q\in\mathbb{Z}$: $$|q\cdot 2\pi - p| <\frac{1}{q}\leq \frac{1}{N} \tag{1}$$ and $1\leq q\leq N$. Also, recall that for small $x$'s we have $$|\sin{x}|\leq|x| \tag{2}$$ Now, choose $N$ such that $\frac{1}{N}<\varepsilon$ and plug $(1)$ into $(2)$: $$|\sin{p}|=|\sin{(q\cdot 2\pi - p)}|\le |q\cdot 2\pi - p|<\frac{1}{N}<\varepsilon$$

Of course, that's all nice in theory, but I need a specific $p$, I hear you asking. Well, look at the convergents of $2\pi$ (convergents of $\pi$ should work as well). Try $2\cdot 104348$ or $2\cdot 833719$ for example. You can find more here A046947/A002485.

rtybase
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  • why is it that sin(2*104348) has a different value from sin(208696) even though the number in the parentheses is the same – Hanzel Oct 24 '20 at 11:24
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    @Harold do you mean Wolfram showing different values? Check if it treats the values in degrees of radians. Try $\sin(208696.0)$ and $\sin(2*104348.0)$ instead ($.0$ at the end!). – rtybase Oct 24 '20 at 11:29
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    @Harold The values are the same. But a calculator may only do computations with a certain number of decimals, and thus get different answers. Perhaps to compute $\sin(208696)$ we begin by subtracting an integer multiple of $2\pi$ to get something in the range $-\pi<x<\pi$, then proceed with that. Interestingly, my calculator says 208696-33215*(2*Pi) is $0$. (Wow, $\pi$ is rational!) – GEdgar Oct 24 '20 at 11:36
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    @rtybase yes, Wolfram was treating the number without .0 as if in degress. Thanks! – Hanzel Oct 24 '20 at 11:54