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I have a slightly complicated function $f(g) = 2 \arccos\left(\frac{1}{2}(-1+2\cos(4g)+ \cos^2(4g))\right)$.

How do I show that for $g$ which is a rational multiple of $\pi$ excluding the set $\pi/8\mathbb{Z} $, $f(g)$ is not a rational multiple of $\pi$?

More generally what $g$ (assuming it is a rational multiple of $\pi$) should I exclude so that $f(g)$ is not a rational multiple of $\pi$ (I believe the condition above is necessary and sufficient).

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  • $\begingroup$ Not clear if the last term is $\cos^2(4g)$ or $\cos(16g^2)$ $\endgroup$
    – Andrei
    Commented Jul 10, 2021 at 22:11
  • $\begingroup$ @Andrei Not clear notation, but clear from the problem. "For $g$ not a multiple of $\frac{\pi}{8}$, $f(g)$ is not a rational multiple of $\pi$. So, presumably, by the wording of the question, it is a multiple of $\pi$ for $g$ a multiple of $\frac{\pi}{8}$. This doesn't happen if it's $\cos(16g^2)$. $\endgroup$
    – Moni145
    Commented Jul 10, 2021 at 22:19
  • $\begingroup$ @Andrei edited for clarity. $\endgroup$
    – nervxxx
    Commented Jul 10, 2021 at 22:48
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    $\begingroup$ The expression in the arccos is $\frac12\left((\cos(4g)+1)^2\right)-1$. $\cos(4g)+1$ can range from $0$ to $2$, so its square can rage from $0$ to $4$, so 1/2 its square can range from $0$ to $2$ again, and that minus 1 can range from $-1$ to $1$ — and more importantly, since the function is continuous, it'll take on all values within that range. For instance, it'll achieve $\cos(\frac\pi5)$, so $f()$ will be $\frac{2\pi}5$ and surely $g$ is not a multiple of $\frac\pi8$ then. $\endgroup$ Commented Jul 11, 2021 at 2:08
  • $\begingroup$ @StevenStadnicki Thanks. You are right of course. I missed out something crucial in the formulation of the problem. I want to restrict $g$ to be rational multiples of $\pi$. Then the statement I'd like is that excluding $g$ to be multiples of $\pi/8$, $f(g)$ is an irrational multiple of $\pi$. The example you gave where $f(g) = 2\pi/5$ is solved by $\cos(4g) = -1+\sqrt{(5+\sqrt{5})/2}$, so that $g$ is irrational (this fact can probably easily be proven). $\endgroup$
    – nervxxx
    Commented Jul 11, 2021 at 3:39

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