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How would one prove $$\frac{\arccos\frac15}\pi\not\in\Bbb Q$$ Fiddling around with numbers hasn't led me anywhere:

Suppose $\frac{\arccos\frac15}\pi\in\Bbb Q$, suppose it is equal to $\frac ab$. Then $\arccos\frac15=\frac{a\pi}b$ and thus $\cos\frac{a\pi}b=\frac15$. But how would this lead to a contradiction?

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As you pointed out,we would obtain that $\cos{\frac{a\pi}{b}}=\frac{1}{5}$ for some integers $a,b,b\neq 0$. Now the problem is finished by the following lemma:

Lemma: Suppose that $a,b$ are integers such that $\cos{\frac{a\pi}{b}}\in\mathbb{Q}$. Then $\cos{\frac{a\pi}{b}}\in\{\pm1,\pm\frac{1}{2}\}$.

Proof of the lemma: Let $r = \frac{a}{b}$. With De Moivre's Formula we deduce that $\cos{r\pi}+i\sin{r\pi}$ and $\cos{r\pi}-i\sin{r\pi}$ are algebraic integers, so their sum is also an algebraic integer, which means that $2\cos{r\pi}$ is an algebraic integer. But $2\cos{r\pi}\in\mathbb{Q}$, so we must have $2\cos{r\pi}\in\mathbb{Z}$. Now from $-1\le\cos{r\pi}\le 1$ we deduce that $\cos{r\pi}\in\{\pm 1,\pm\frac{1}{2}\}$. This ends the proof of the lemma.

From the above lemma we can observe that $\cos{\frac{a\pi}{b}}\neq\frac{1}{5}$ for all $a,b\in\mathbb{Z},b\neq 0$.

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I try a different, perhaps more elementary, solution. I suppose $b\geq 1$.

1) First note that if $n\geq 1$, you have $\displaystyle \sum_{k even}{n\choose k}=2^{n-1}$ (easy computation with $(1+x)^n$ and $(1-x)^n$ for $x=1$.)

2) Note that as $\cos(a\pi/b)=1/5$, we get $\sin(a\pi/b)=2\sqrt{6}/5$. Hence by De Moivre's formula

$$(1+2i\sqrt{6})^b=(-1)^a5^b$$

Now using the binomial theorem, we get that

$$\sum_{l=0}^{[b/2]}{b \choose 2l}2^{2l}(-1)^l6^l=(-1)^a5^b$$ Modulo $5$, we have $2^{2l}(-1)^l 6^l$ congruent to $1$ for all $l$. Hence by 1) we get $2^{b-1}$ divisible by $5$, a contradiction.

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