Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
This appears to have a solution: (along with this hint: – Andy Bromberg May 21 '13 at 22:57
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
up vote 9 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.

share|cite|improve this answer
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.