Irrationality of $ \frac{1}{\pi} \arccos{\frac{1}{\sqrt{n}}}$ This paper arxiv.org/pdf/0911.1933 discusses, regarding the irrationality of certain trigonometric functions. Recently, i encountered this problem which says states the given function, $$ f(n)=\frac{1}{\pi} \arccos{\frac{1}{\sqrt{n}}}$$ is irrational for every odd $n \geq 3$. But i couldn't find the proof anywhere. Can anyone provide me with the proof.
 A: You could find the proof in this book: Proofs from the Book by Martin Aigner, Günter M. Ziegler, Karl H. Hofmann, pages 40-41
A: If you want to prove that a certain angle is irrational with respect to $\pi$, most of the times it follows from the following simple result:
Lemma: If $n, k \in {\mathbb Z}$ and $n>0$ then $2 \cos( \frac{k \pi}{n})$ is an algebraic integer.
Proof: We show that there exists a $P_n \in {\mathbb Z}[x]$ monic of degree $n$ so that 
$$2 \cos(nx) = P_n(2 \cos(x)) (*) \,.$$
$P_0(X)=1, P_1(X)=X$ and using
$$\cos((n+1)x)+ \cos((n-1)x)=2 \cos(x) \cos(nx) \,,$$
we get the recurence:
$$P_{n+1}(x) = XP_n(x)-P_{n-1}(X) \,.$$  
Now $(*)$ follows by induction, and then Lemma follows immediately.

Now, back to the problem. Suppose by contradiction $\frac{1}{\pi} \arccos{\frac{1}{\sqrt{n}}}$ is rational. Then we can find $n, k \in {\mathbb Z}$ and $n>0$ so that
$$\frac{1}{\pi} \arccos{\frac{1}{\sqrt{n}}}=\frac{k}{n} \,.$$
Hence
$$\frac{1}{\sqrt{n}}= \cos(\frac{2k}{n}) \,.$$
Then, by Lemma, $2\frac{1}{\sqrt{n}}$ is an algebraic integer, and hence is its square.
But then $\frac{4}{n}$ is an algebraic integer and rational, thus integer.
This shows that $4$ divides $n$. 
