<s>[INST] Irrationality of this trigonometric function I'd like to prove the following conjecture.
Let $x = \frac{p}{q}\pi$ be a rational angle ($p,q$ integers, $q \geq 1$).
Then
$f(x) = \frac{2}{\pi} \arccos{\left(2\cos^4(2x)-1 \right)}$
is irrational if $x$ is not an integer multiple of $\frac{\pi}{8}$. Is this true? Is the other direction true too?
I've tried to use the standard manipulations, Chebyshev polynomials, etc. (see https://math.stackexchange.com/questions/398687/how-do-i-prove-that-frac1-pi-arccos1-3-is-irrational), but am otherwise quite stuck.
Some evaluations just to see a trend (we can restrict domain of $x$ to be $[0,\pi/2]$):
\begin{align}
x & = \frac{\pi}{2} \implies f(x)= 0 \text{ (Rational) }, \\
x & = \frac{\pi}{3},  \implies f(x) = 2\arccos(-7/8)/\pi, \\
x & = \frac{\pi}{4},  \implies f(x) = 2 \text{ (Rational) }, \\
x & = \frac{\pi}{5}, \implies f(x) = 2\arccos(-3(3+\sqrt{5})/16)/\pi, \\
x & = \frac{2\pi}{5}, \implies f(x) = 2\arccos(3(-3+\sqrt{5})/16)/\pi, \\
x & = \frac{\pi}{6}, \implies f(x) = 2\arccos(-7/8)/\pi, \\
x & = \frac{\pi}{7}, \implies f(x) = 2\arccos(-1+2\sin^4(3\pi/14))/\pi \\
x & = \frac{2\pi}{7}, \implies f(x) = 2\arccos(-1+2\sin^4(\pi/14))/\pi \\
x & = \frac{2\pi}{7}, \implies f(x) = 2\arccos(-1+2\sin^4(\pi/7))/\pi \\
x & = \frac{\pi}{8}, \frac{3\pi}{8}, \implies f(x) = 4/3  \text{ (Rational) }\\
\cdots
\end{align}
 [/INST] Lemma. Let $n$ be a positive integer such that each number in the open interval $(n/4,3n/4)$ is not coprime with $n$. Then $n\in \{1,4,6\}$.
Proof.

*

*If $n=2m+1$ is odd, and $m\geqslant 1$, then $m\in (n/4,3n/4)$.

*If $n=4m+2$ and $m>1$, then $2m-1\in (n/4,3n/4)$

*If $n=2$, then $1\in (n/4,3n/4)$

*If $n=4m$ and $m>1$, then $2m-1\in (n/4,3n/4)$
Now assume that $f(x)=4t$ with rational $t$. Then $2\cos^42x-1=\cos (2\pi t)$ and $2\cos^42x=2\cos^2(\pi t)$, $\cos \pi t=\pm \cos^2 2x$. Replacing $t$ to $t+1$ if necessary, we may suppose that $\cos \pi t=\cos^22x$. Let $\pi t=2\pi a/b$, $2x=2\pi c/b$ for coprime (not necessarily mutually coprime) integers $a,b,c$. In other words, let $b$ be a minimal positive integer for which $\pi tb$, $2xb$ are both divisible by $2\pi$. Denote $w=\exp(2\pi i/b)$. In these notations we get $$\frac{w^a+w^{-a}}2=\frac{(w^c+w^{-c})^2}4.$$
This is polynomial relation for $w$ with rational coefficients. Thus, all algebraic conjugates of $w$ satisfy it aswell. These algebraic conjugates are all primitive roots of unity of degree $b$ (here I use the well known fact that the cyclotomic polynomial $\Phi_b$ is irreducible). So, we are allowed to replace $w$ to $w^m$, where $\gcd (m,b)=1$. RHS remains non-negative, therefore so does LHS, and we get
$\cos (2\pi am/b)\geqslant 0$ for all $m$ coprime to $b$.
This is rare. Namely, denote $a=da_1$, $b=db_1$ where $d=\gcd(a,b)$ and $\gcd(a_1,b_1)=1$. Let $k$ be arbitrary integer coprime with $b_1$. Denote
$m=k+Nb_1$, where $N$ equals to the product of those prime divisors of $d$ which do not divide $k$. Now each prime divisor of $d$ divides exactly one of the numbers $k$ and $Nb_1$, therefore it does not divide their sum $m$. Clearly $m$ is also coprime with $b_1$, and totally $\gcd(m,b)=1$. Therefore $\cos (2\pi a_1k/b_1)=\cos(2\pi a_1m/b_1)=\cos (2\pi am/b)$ is non-negative. Next, $a_1k$ takes all residues modulo $b_1$ which are coprime to $b_1$. It follows that there are no residues coprime with $b_1$ in the interval $(b_1/4,3b_1/4)$. Thus by Lemma we get $b_1\in \{1,4,6\}$, and $\cos \pi t=\cos 2\pi a/b=\cos 2\pi a_1/b_1\in \{0,1/2,1\}$ and $\cos 2x=\pm \sqrt{\cos \pi t}\in \{0,\pm \sqrt{2}/2,\pm 1\}$. This indeed implies that $2x$ is divisible by $\pi/4$, in other words, $x$ is divisible by $\pi/8$.
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