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To put it more rigorously: Does there exist constants $0 < a < 1$ and $0 < b < 1$ such that for all real $\theta$ which are non-rational multiples of $\pi$ and all natural numbers $n$ it holds that $b a^n\leq \sin(\theta n) $. My reasoning is that because $\theta$ is restricted to be a non-rational multiple of $\pi$, we can't find such constants as $\sin$ gets arbitrarily close to 0. But, I can't find a proof. Are there any related theorems that might help me?

Update: I changed the question so that $\theta$ has to be a non-rational multiple of $\pi$, to make it more interesting. I stumbled upon this problem in a different context, and made a mistake when reformulating it here.

Another alternative would be to modify the inequality as $b a^n\leq y \sin(\theta n) + x$ with the restriction that $x\geq 2y > 0$.

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  • $\begingroup$ An interesting question could develop out of this one. But if we look at it exactly as it stands, the answer is clearly no. Take for example $\theta=\pi$, or, more extremely, $\theta=0$. $\endgroup$ – André Nicolas Apr 5 '11 at 4:14
  • $\begingroup$ Even after the edit, the answer is still clearly no, because for any fixed $ab$, you can choose $\theta < ab$ (so $\sin\theta < ab$) such that $\theta/\pi$ is irrational (cf. @Ross Millikan's answer). To make the question interesting, you might instead ask: for what values of $\theta$ does there exist $a>0$ such that $\sin(n\theta) > a^n$ for $n \in \mathbb{N}$? If $(\theta/\pi) \notin \mathbb{Q}$, then the behavior for large $n$ is all that matters, and you can replace $\sin(n\theta)$ with $n\theta$ there. I think the result will involve the irrationality measure of $\theta$. $\endgroup$ – mjqxxxx Apr 5 '11 at 19:14
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Since $a,b \gt 0$, if $\theta=0$ (or $\pi$) you will fail. Even if you restrict $\theta$ to be not a rational multiple of $\pi$, the way you have the quantifiers, you have to choose $a$ and $b$ before I choose $\theta$, so I can choose $\theta \lt ba$ and since $\sin x \lt x$ for $x \gt 0$ you lose for $n=1$

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