2
$\begingroup$

Consider rational numbers $\frac{m}{n}$ and $\frac{m'}{n'}$, where $0<\frac{m}{n}, \frac{m'}{n'} <1$.

Then $$\sin^2 (\tfrac{m}{n} \pi) = 2 \sin^2 (\tfrac{m'}{n'} \pi)$$

When $\frac{m}{n} = \frac{1}{4}, \frac{3}{4}$ and $\frac{m'}{n'} = \frac{1}{6}, \frac{5}{6}$. (Both sides equal $\frac{1}{2}$.)

Is it possible that other $\frac{m}{n}$ and $\frac{m'}{n'}$ satisfy these conditions? Is it possible to prove that there are or are not other solutions?

$\endgroup$
  • 2
    $\begingroup$ Some MathJax advice: Named math operators should appear upright, and the common ones have their own MathJax code for this purpose (e.g. \sin, \log - see entry 11 in our MathJax guide). $\endgroup$ – Zev Chonoles Jul 9 '13 at 3:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.