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I want to show the following. Assume $n>2$. Then, for each $b=1,\ldots,n$, there exists some $a\in\{2,3,\ldots,n\}$ such that

$$\left|\cos\left(\frac{\pi}{2n}\left(a-1\right)(2b-1)\right) \right| \geqslant \frac{1}{\sqrt{2}}.$$

I’ve checked this inequality holds for $b=1$ (it is very simple), but don’t know how to proceed to show the above systematically for each $b=2,\ldots,n$ and $n>2$.

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First suppose $2b-1$ is relatively prime to $2n$. Then there exists some $c\in\{1,\dots,2n-1\}$ such that $c(2b-1)$ is $1$ mod $2n$. Moreover, we have $c\neq n$ since $n$ is not relatively prime to $2n$. If $c<n$, let $a=c+1$, and if $c>n$, let $a=(2n-c)+1$. In either case, we have $a\in\{2,\dots,n\}$ and $(a-1)(2b-1)$ is $\pm1$ mod $2n$. For some integer $k$ we then have $$\cos\left(\frac{\pi}{2n}(a-1)(2b-1)\right)=\cos\left(\pm\frac{\pi}{2n}+k\pi\right)=(-1)^k\cos\left(\frac{\pi}{2n}\right).$$ Since $n>2$, $0<\frac{\pi}{2n}\leq\frac{\pi}{4}$ and it follows that $$\cos\left(\frac{\pi}{2n}\right)\geq\frac{1}{\sqrt{2}}.$$

Now suppose $2b-1$ is not relatively prime to $2n$. Since $2b-1$ is odd, $\gcd(2b-1,2n)>2$ and $c=\frac{2n}{\gcd(2b-1,2n)}<n$ satisfies $c(2b-1)=2kn$ for some integer $k$. Again letting $a=c+1$, we have $$\cos\left(\frac{\pi}{2n}(a-1)(2b-1)\right)=\cos(k\pi)=(-1)^k.$$

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  • $\begingroup$ Thanks for your reply, Eric. In the first case ($2b-1$ and $2n$ are relatively prime), how do you conclude that the set to which $c$ may belong is exactly $\{1,\ldots,n-1\}$? $\endgroup$ – Rene Jun 13 '18 at 14:14
  • $\begingroup$ Oops, I've fixed that. If $c$ is too big, you can just negate it mod $2n$. $\endgroup$ – Eric Wofsey Jun 13 '18 at 14:57
  • $\begingroup$ (1) Yes, but I don't know what you're trying to accomplish. Why would you want $c(2b-1)$ to be of the form $d+k2n$? If $d$ is large, that may not make the cosine large. (Specifically, it will fail to work if $d=2n/3$.) $\endgroup$ – Eric Wofsey Jun 14 '18 at 15:09
  • $\begingroup$ (2) My goal is to make the cosine large (in absolute value), so I want $\frac{\pi}{2n}c(2b-1)$ to be close to a multiple of $\pi$, so I want $c(2b-1)$ to be close to a multiple of $2n$. Choosing $c=2n/d$ makes $c(2b-1)$ not just close to a multiple of $2n$ but equal to a multiple of $2n$. $\endgroup$ – Eric Wofsey Jun 14 '18 at 15:11

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