For $0 le theta le pi/2$, When are both $theta/pi$ and $sqrt2sintheta$ rational?












0












$begingroup$


For $0 le theta le pi/2$, when are both $theta/pi$ and $sqrt2sintheta$ rational?



I think $theta=0, pi/4$ is the only cases. This problem seems to be related to Niven's theorem, but I cannot prove this.






trigonometry irrational-numbers rational-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: sine is periodic
    $endgroup$
    – Sorfosh
    Dec 5 '18 at 14:57










  • $begingroup$
    In addition to the hint provided, you can also look at the other quadrants clearly.
    $endgroup$
    – KM101
    Dec 5 '18 at 15:05












  • $begingroup$
    Sorry for unclear statement. The range of $theta$ is limited to the first quadrant, and what I want to prove is there is no other case rather than the two cases I mentioned.
    $endgroup$
    – Jeongu Kim
    Dec 5 '18 at 15:10












  • $begingroup$
    Hint: $2sin^2(theta)=1-cos(2theta)$
    $endgroup$
    – Ingix
    Dec 5 '18 at 15:45












  • $begingroup$
    Wow, great idea!
    $endgroup$
    – Jeongu Kim
    Dec 5 '18 at 15:54
















0












$begingroup$


For $0 le theta le pi/2$, when are both $theta/pi$ and $sqrt2sintheta$ rational?



I think $theta=0, pi/4$ is the only cases. This problem seems to be related to Niven's theorem, but I cannot prove this.






trigonometry irrational-numbers rational-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: sine is periodic
    $endgroup$
    – Sorfosh
    Dec 5 '18 at 14:57










  • $begingroup$
    In addition to the hint provided, you can also look at the other quadrants clearly.
    $endgroup$
    – KM101
    Dec 5 '18 at 15:05












  • $begingroup$
    Sorry for unclear statement. The range of $theta$ is limited to the first quadrant, and what I want to prove is there is no other case rather than the two cases I mentioned.
    $endgroup$
    – Jeongu Kim
    Dec 5 '18 at 15:10












  • $begingroup$
    Hint: $2sin^2(theta)=1-cos(2theta)$
    $endgroup$
    – Ingix
    Dec 5 '18 at 15:45












  • $begingroup$
    Wow, great idea!
    $endgroup$
    – Jeongu Kim
    Dec 5 '18 at 15:54














0












0








0





$begingroup$


For $0 le theta le pi/2$, when are both $theta/pi$ and $sqrt2sintheta$ rational?



I think $theta=0, pi/4$ is the only cases. This problem seems to be related to Niven's theorem, but I cannot prove this.






trigonometry irrational-numbers rational-numbers







share|cite|improve this question











$endgroup$




For $0 le theta le pi/2$, when are both $theta/pi$ and $sqrt2sintheta$ rational?



I think $theta=0, pi/4$ is the only cases. This problem seems to be related to Niven's theorem, but I cannot prove this.






trigonometry irrational-numbers rational-numbers




trigonometry irrational-numbers rational-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 15:08







Jeongu Kim

















asked Dec 5 '18 at 14:52









Jeongu KimJeongu Kim

734




734








  • 1




    $begingroup$
    Hint: sine is periodic
    $endgroup$
    – Sorfosh
    Dec 5 '18 at 14:57










  • $begingroup$
    In addition to the hint provided, you can also look at the other quadrants clearly.
    $endgroup$
    – KM101
    Dec 5 '18 at 15:05












  • $begingroup$
    Sorry for unclear statement. The range of $theta$ is limited to the first quadrant, and what I want to prove is there is no other case rather than the two cases I mentioned.
    $endgroup$
    – Jeongu Kim
    Dec 5 '18 at 15:10












  • $begingroup$
    Hint: $2sin^2(theta)=1-cos(2theta)$
    $endgroup$
    – Ingix
    Dec 5 '18 at 15:45












  • $begingroup$
    Wow, great idea!
    $endgroup$
    – Jeongu Kim
    Dec 5 '18 at 15:54














  • 1




    $begingroup$
    Hint: sine is periodic
    $endgroup$
    – Sorfosh
    Dec 5 '18 at 14:57










  • $begingroup$
    In addition to the hint provided, you can also look at the other quadrants clearly.
    $endgroup$
    – KM101
    Dec 5 '18 at 15:05












  • $begingroup$
    Sorry for unclear statement. The range of $theta$ is limited to the first quadrant, and what I want to prove is there is no other case rather than the two cases I mentioned.
    $endgroup$
    – Jeongu Kim
    Dec 5 '18 at 15:10












  • $begingroup$
    Hint: $2sin^2(theta)=1-cos(2theta)$
    $endgroup$
    – Ingix
    Dec 5 '18 at 15:45












  • $begingroup$
    Wow, great idea!
    $endgroup$
    – Jeongu Kim
    Dec 5 '18 at 15:54








1




1




$begingroup$
Hint: sine is periodic
$endgroup$
– Sorfosh
Dec 5 '18 at 14:57




$begingroup$
Hint: sine is periodic
$endgroup$
– Sorfosh
Dec 5 '18 at 14:57












$begingroup$
In addition to the hint provided, you can also look at the other quadrants clearly.
$endgroup$
– KM101
Dec 5 '18 at 15:05






$begingroup$
In addition to the hint provided, you can also look at the other quadrants clearly.
$endgroup$
– KM101
Dec 5 '18 at 15:05














$begingroup$
Sorry for unclear statement. The range of $theta$ is limited to the first quadrant, and what I want to prove is there is no other case rather than the two cases I mentioned.
$endgroup$
– Jeongu Kim
Dec 5 '18 at 15:10






$begingroup$
Sorry for unclear statement. The range of $theta$ is limited to the first quadrant, and what I want to prove is there is no other case rather than the two cases I mentioned.
$endgroup$
– Jeongu Kim
Dec 5 '18 at 15:10














$begingroup$
Hint: $2sin^2(theta)=1-cos(2theta)$
$endgroup$
– Ingix
Dec 5 '18 at 15:45






$begingroup$
Hint: $2sin^2(theta)=1-cos(2theta)$
$endgroup$
– Ingix
Dec 5 '18 at 15:45














$begingroup$
Wow, great idea!
$endgroup$
– Jeongu Kim
Dec 5 '18 at 15:54




$begingroup$
Wow, great idea!
$endgroup$
– Jeongu Kim
Dec 5 '18 at 15:54










1 Answer
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Let us assume that $thetainpimathbb{Q}$ and $sqrt{2}sinthetainmathbb{Q}$. If $sinfrac{pi p}{q}=cosleft(frac{pi q}{2q}-frac{2pi p}{2q}right)=cosleft(frac{2pi|q-2p|}{4q}right)$ is an algebraic number of degree $2$ over $mathbb{Q}$, then we must have $frac{1}{2}varphi(4q)=2$ or $varphi(4q)=8$, so $qin{2,3,4,5,6}$. Now a manual inspection completes the job, or $cosfrac{pi}{5}inmathbb{Q}(sqrt{5})setminusmathbb{Q}$ and $cosfrac{pi}{6}inmathbb{Q}(sqrt{3})setminusmathbb{Q}$.






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    $begingroup$

    Let us assume that $thetainpimathbb{Q}$ and $sqrt{2}sinthetainmathbb{Q}$. If $sinfrac{pi p}{q}=cosleft(frac{pi q}{2q}-frac{2pi p}{2q}right)=cosleft(frac{2pi|q-2p|}{4q}right)$ is an algebraic number of degree $2$ over $mathbb{Q}$, then we must have $frac{1}{2}varphi(4q)=2$ or $varphi(4q)=8$, so $qin{2,3,4,5,6}$. Now a manual inspection completes the job, or $cosfrac{pi}{5}inmathbb{Q}(sqrt{5})setminusmathbb{Q}$ and $cosfrac{pi}{6}inmathbb{Q}(sqrt{3})setminusmathbb{Q}$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Let us assume that $thetainpimathbb{Q}$ and $sqrt{2}sinthetainmathbb{Q}$. If $sinfrac{pi p}{q}=cosleft(frac{pi q}{2q}-frac{2pi p}{2q}right)=cosleft(frac{2pi|q-2p|}{4q}right)$ is an algebraic number of degree $2$ over $mathbb{Q}$, then we must have $frac{1}{2}varphi(4q)=2$ or $varphi(4q)=8$, so $qin{2,3,4,5,6}$. Now a manual inspection completes the job, or $cosfrac{pi}{5}inmathbb{Q}(sqrt{5})setminusmathbb{Q}$ and $cosfrac{pi}{6}inmathbb{Q}(sqrt{3})setminusmathbb{Q}$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Let us assume that $thetainpimathbb{Q}$ and $sqrt{2}sinthetainmathbb{Q}$. If $sinfrac{pi p}{q}=cosleft(frac{pi q}{2q}-frac{2pi p}{2q}right)=cosleft(frac{2pi|q-2p|}{4q}right)$ is an algebraic number of degree $2$ over $mathbb{Q}$, then we must have $frac{1}{2}varphi(4q)=2$ or $varphi(4q)=8$, so $qin{2,3,4,5,6}$. Now a manual inspection completes the job, or $cosfrac{pi}{5}inmathbb{Q}(sqrt{5})setminusmathbb{Q}$ and $cosfrac{pi}{6}inmathbb{Q}(sqrt{3})setminusmathbb{Q}$.






        share|cite|improve this answer









        $endgroup$



        Let us assume that $thetainpimathbb{Q}$ and $sqrt{2}sinthetainmathbb{Q}$. If $sinfrac{pi p}{q}=cosleft(frac{pi q}{2q}-frac{2pi p}{2q}right)=cosleft(frac{2pi|q-2p|}{4q}right)$ is an algebraic number of degree $2$ over $mathbb{Q}$, then we must have $frac{1}{2}varphi(4q)=2$ or $varphi(4q)=8$, so $qin{2,3,4,5,6}$. Now a manual inspection completes the job, or $cosfrac{pi}{5}inmathbb{Q}(sqrt{5})setminusmathbb{Q}$ and $cosfrac{pi}{6}inmathbb{Q}(sqrt{3})setminusmathbb{Q}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 '18 at 23:26









        Jack D'AurizioJack D'Aurizio

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