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How do i prove that $\frac{1}{\pi} \arccos(1/3)$ is irrational?

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This appears to have a solution: math.utk.edu/~jconant/pisol.html (along with this hint: math.utk.edu/~jconant/pihint2.html) –  Andy Bromberg May 21 '13 at 22:57
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This question have been answered here: link –  Johan Asplund May 21 '13 at 23:00
    
Am I right in thinking you mean $\Big(\arccos(1/3)\Big)/\pi$ and not $\arccos\Big((1/3)/\pi\Big)$? –  Michael Hardy May 21 '13 at 23:03

2 Answers 2

up vote 5 down vote accepted

Let $\theta = \arccos\dfrac13$ so that $\cos\theta=\dfrac13$.

If $\theta$ is a rational multiple of $\pi$, say $\theta=\dfrac mn \pi$, then $\cos(n\theta)=\pm1$. Now $\cos(n\theta)=T_n(\cos\theta)$, where $T_n$ is the $n$th-degree Chebyshev polynomial. Using mathematical induction and some trigonometric identities, you can show that the leading coefficient in the $n$th-degree Chebyshev polynomial is $2^{n-1}$ for $n\ge2$. We have $$ \pm1=T_n\left(\frac 13\right) = 2^{n-1}\left(\frac13\right)^n+\text{lower-degree terms}. $$ Multiplying both sides by $3^n$, we get $$ \pm3^n = 2^{n-1} + \text{terms divisible by $3$}. $$ And that says a positive power of $2$ is a multiple of $3$, which violates uniqueness of prime factorizations.

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Nice solution....Michael Hardy... –  juantheron Nov 22 '13 at 18:07
    
@juantheron : Thank you. –  Michael Hardy Nov 22 '13 at 19:22

Let’s call this angle $\Theta$. If it were rational, then there would be an $N$ such that $\cos(N\Theta)=1$. That would say that $\cos\Theta$ was a root of $T_N-1$, where $T_N$ is the appropriate Čebyšev polynomial, which it can’t be ’cause it’s transcendental.

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This seems to assume that what was intended was $\arccos\Big((1/3)/\pi\Big)$. If that's what was intended, then this answer is right. I posted an answer that is right if $\dfrac{\arccos(1/3)}{\pi}$ was intended. The latter seems more plausible, since if the former was intended, then it could have said $(1/(3\pi))$ rather than having fractions within fractions. –  Michael Hardy May 21 '13 at 23:20
    
I meant $\frac{1}{\pi} \arccos(1/3)$. Thanks for the observation. –  Hugo C. Botós May 21 '13 at 23:27
    
Are you saying $\Theta=\dfrac1\pi\arccos(1/3)$, or $\Theta=\arccos(1/3)$. In the latter case, $\cos\Theta$ is rational. In the former case, why should $\cos\Theta$ be a root of $T_N-1$? –  Michael Hardy May 21 '13 at 23:34
    
When in doubt, use parentheses! Thanks for the corrections. –  Lubin May 22 '13 at 3:02

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