Rational with finite decimals values for sine, cosine, and tangent

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What are the possible combinations of sine, cosine, and tangent values such that all three are simultaneously rational with finite decimals?



I am aware of the below two cases.



  • $sin(x) = 0, cos(x) = 1, tan(x) = 0$ is a trivial case for angle measure zero


  • $sin(x) = 0.6, cos(x) = 0.8, tan(x) = 0.75$ is another case, for an angle of $37$ degrees (approximately)


Are there any other combinations that satisfy the condition? If not, is it possible to prove that these are the only possible combinations?



Edit: Sorry, I missed one detail when posting the question. I am looking for sin, cos and tan values being rational numbers with finite decimals. For example, (7, 24, 25) is a Pythagorean triple that does not satisfy the condition because tan value would be 0.291666....










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  • 4




    $begingroup$
    Any Pythagorean triple will give such values.
    $endgroup$
    – Blue
    Feb 20 at 6:55










  • $begingroup$
    With your edit, any positive example will show a power of $25$ minus a power of $4$ being a square number, and you have essentially already found $25^0-4^0=0^2$ and $25^1-4^2=3^2$. Can you find any others?
    $endgroup$
    – Henry
    Feb 20 at 8:36






  • 4




    $begingroup$
    Restricting it to terminating decimals makes the question much less interesting. There's no great mathematical significance to the fact that modern civilization tends to use decimal (and of course our computers use binary / hexadecimal).
    $endgroup$
    – PM 2Ring
    Feb 20 at 8:58










  • $begingroup$
    Apart from the examples you have, I doubt there are any essentially different examples with fewer than ten decimal places, and there may be none
    $endgroup$
    – Henry
    Feb 20 at 10:13






  • 1




    $begingroup$
    @PM2Ring What is interesting is a matter of opinion. If you wish, you can work on figuring out an algorithm for generating solutions in an arbitrary base, and then set the base to 10.
    $endgroup$
    – Acccumulation
    Feb 20 at 16:25















6












$begingroup$


What are the possible combinations of sine, cosine, and tangent values such that all three are simultaneously rational with finite decimals?



I am aware of the below two cases.



  • $sin(x) = 0, cos(x) = 1, tan(x) = 0$ is a trivial case for angle measure zero


  • $sin(x) = 0.6, cos(x) = 0.8, tan(x) = 0.75$ is another case, for an angle of $37$ degrees (approximately)


Are there any other combinations that satisfy the condition? If not, is it possible to prove that these are the only possible combinations?



Edit: Sorry, I missed one detail when posting the question. I am looking for sin, cos and tan values being rational numbers with finite decimals. For example, (7, 24, 25) is a Pythagorean triple that does not satisfy the condition because tan value would be 0.291666....










share|cite|improve this question











$endgroup$







  • 4




    $begingroup$
    Any Pythagorean triple will give such values.
    $endgroup$
    – Blue
    Feb 20 at 6:55










  • $begingroup$
    With your edit, any positive example will show a power of $25$ minus a power of $4$ being a square number, and you have essentially already found $25^0-4^0=0^2$ and $25^1-4^2=3^2$. Can you find any others?
    $endgroup$
    – Henry
    Feb 20 at 8:36






  • 4




    $begingroup$
    Restricting it to terminating decimals makes the question much less interesting. There's no great mathematical significance to the fact that modern civilization tends to use decimal (and of course our computers use binary / hexadecimal).
    $endgroup$
    – PM 2Ring
    Feb 20 at 8:58










  • $begingroup$
    Apart from the examples you have, I doubt there are any essentially different examples with fewer than ten decimal places, and there may be none
    $endgroup$
    – Henry
    Feb 20 at 10:13






  • 1




    $begingroup$
    @PM2Ring What is interesting is a matter of opinion. If you wish, you can work on figuring out an algorithm for generating solutions in an arbitrary base, and then set the base to 10.
    $endgroup$
    – Acccumulation
    Feb 20 at 16:25













6












6








6


2



$begingroup$


What are the possible combinations of sine, cosine, and tangent values such that all three are simultaneously rational with finite decimals?



I am aware of the below two cases.



  • $sin(x) = 0, cos(x) = 1, tan(x) = 0$ is a trivial case for angle measure zero


  • $sin(x) = 0.6, cos(x) = 0.8, tan(x) = 0.75$ is another case, for an angle of $37$ degrees (approximately)


Are there any other combinations that satisfy the condition? If not, is it possible to prove that these are the only possible combinations?



Edit: Sorry, I missed one detail when posting the question. I am looking for sin, cos and tan values being rational numbers with finite decimals. For example, (7, 24, 25) is a Pythagorean triple that does not satisfy the condition because tan value would be 0.291666....










share|cite|improve this question











$endgroup$




What are the possible combinations of sine, cosine, and tangent values such that all three are simultaneously rational with finite decimals?



I am aware of the below two cases.



  • $sin(x) = 0, cos(x) = 1, tan(x) = 0$ is a trivial case for angle measure zero


  • $sin(x) = 0.6, cos(x) = 0.8, tan(x) = 0.75$ is another case, for an angle of $37$ degrees (approximately)


Are there any other combinations that satisfy the condition? If not, is it possible to prove that these are the only possible combinations?



Edit: Sorry, I missed one detail when posting the question. I am looking for sin, cos and tan values being rational numbers with finite decimals. For example, (7, 24, 25) is a Pythagorean triple that does not satisfy the condition because tan value would be 0.291666....







algebra-precalculus trigonometry rational-numbers rationality-testing






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share|cite|improve this question













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edited Feb 20 at 7:58







Shahul

















asked Feb 20 at 6:50









ShahulShahul

363




363







  • 4




    $begingroup$
    Any Pythagorean triple will give such values.
    $endgroup$
    – Blue
    Feb 20 at 6:55










  • $begingroup$
    With your edit, any positive example will show a power of $25$ minus a power of $4$ being a square number, and you have essentially already found $25^0-4^0=0^2$ and $25^1-4^2=3^2$. Can you find any others?
    $endgroup$
    – Henry
    Feb 20 at 8:36






  • 4




    $begingroup$
    Restricting it to terminating decimals makes the question much less interesting. There's no great mathematical significance to the fact that modern civilization tends to use decimal (and of course our computers use binary / hexadecimal).
    $endgroup$
    – PM 2Ring
    Feb 20 at 8:58










  • $begingroup$
    Apart from the examples you have, I doubt there are any essentially different examples with fewer than ten decimal places, and there may be none
    $endgroup$
    – Henry
    Feb 20 at 10:13






  • 1




    $begingroup$
    @PM2Ring What is interesting is a matter of opinion. If you wish, you can work on figuring out an algorithm for generating solutions in an arbitrary base, and then set the base to 10.
    $endgroup$
    – Acccumulation
    Feb 20 at 16:25












  • 4




    $begingroup$
    Any Pythagorean triple will give such values.
    $endgroup$
    – Blue
    Feb 20 at 6:55










  • $begingroup$
    With your edit, any positive example will show a power of $25$ minus a power of $4$ being a square number, and you have essentially already found $25^0-4^0=0^2$ and $25^1-4^2=3^2$. Can you find any others?
    $endgroup$
    – Henry
    Feb 20 at 8:36






  • 4




    $begingroup$
    Restricting it to terminating decimals makes the question much less interesting. There's no great mathematical significance to the fact that modern civilization tends to use decimal (and of course our computers use binary / hexadecimal).
    $endgroup$
    – PM 2Ring
    Feb 20 at 8:58










  • $begingroup$
    Apart from the examples you have, I doubt there are any essentially different examples with fewer than ten decimal places, and there may be none
    $endgroup$
    – Henry
    Feb 20 at 10:13






  • 1




    $begingroup$
    @PM2Ring What is interesting is a matter of opinion. If you wish, you can work on figuring out an algorithm for generating solutions in an arbitrary base, and then set the base to 10.
    $endgroup$
    – Acccumulation
    Feb 20 at 16:25







4




4




$begingroup$
Any Pythagorean triple will give such values.
$endgroup$
– Blue
Feb 20 at 6:55




$begingroup$
Any Pythagorean triple will give such values.
$endgroup$
– Blue
Feb 20 at 6:55












$begingroup$
With your edit, any positive example will show a power of $25$ minus a power of $4$ being a square number, and you have essentially already found $25^0-4^0=0^2$ and $25^1-4^2=3^2$. Can you find any others?
$endgroup$
– Henry
Feb 20 at 8:36




$begingroup$
With your edit, any positive example will show a power of $25$ minus a power of $4$ being a square number, and you have essentially already found $25^0-4^0=0^2$ and $25^1-4^2=3^2$. Can you find any others?
$endgroup$
– Henry
Feb 20 at 8:36




4




4




$begingroup$
Restricting it to terminating decimals makes the question much less interesting. There's no great mathematical significance to the fact that modern civilization tends to use decimal (and of course our computers use binary / hexadecimal).
$endgroup$
– PM 2Ring
Feb 20 at 8:58




$begingroup$
Restricting it to terminating decimals makes the question much less interesting. There's no great mathematical significance to the fact that modern civilization tends to use decimal (and of course our computers use binary / hexadecimal).
$endgroup$
– PM 2Ring
Feb 20 at 8:58












$begingroup$
Apart from the examples you have, I doubt there are any essentially different examples with fewer than ten decimal places, and there may be none
$endgroup$
– Henry
Feb 20 at 10:13




$begingroup$
Apart from the examples you have, I doubt there are any essentially different examples with fewer than ten decimal places, and there may be none
$endgroup$
– Henry
Feb 20 at 10:13




1




1




$begingroup$
@PM2Ring What is interesting is a matter of opinion. If you wish, you can work on figuring out an algorithm for generating solutions in an arbitrary base, and then set the base to 10.
$endgroup$
– Acccumulation
Feb 20 at 16:25




$begingroup$
@PM2Ring What is interesting is a matter of opinion. If you wish, you can work on figuring out an algorithm for generating solutions in an arbitrary base, and then set the base to 10.
$endgroup$
– Acccumulation
Feb 20 at 16:25










4 Answers
4






active

oldest

votes


















7












$begingroup$

Any Pythagorean triple will give you rational $sin, cos, tan$ values, and any triple of rational trig values will give you a Pythagorean triple.



The Pythagorean triples have been completely classified. Take two natural numbers $u>v$. Then
$$
(u^2-v^2)^2 + (2uv)^2=(u^2+v^2)^2
$$

and any Pythagorean triple is either of this form or a multiple of such a triple. So the possible rational $sin$ and $cos$ pairs are all possible values of
$$
fracu^2-v^2u^2+v^2text and frac2uvu^2+v^2
$$






share|cite|improve this answer











$endgroup$












  • $begingroup$
    You are right, But I missed a detail in my question - it is now edited.
    $endgroup$
    – Shahul
    Feb 20 at 8:00










  • $begingroup$
    @Shahul That's an easy fix. Make sure $u^2 + v^2$ is a divisor of $10^n$ for some $n$. In other words that the only prime numbers in the prime decomposition of $u^2 + v^2$ are $2$ and $5$.
    $endgroup$
    – Arthur
    Feb 20 at 8:02







  • 2




    $begingroup$
    In order to guarantee $tan$ has finite decimal expansion too, you need either $2uv$ or $u^2-v^2$ to be a divisor of a power of $10$ too. Taking $u,v$ relatively prime and exactly one even (we can always assume that for Pythagorean triples) you can say much more - for instance, $2uv$ and $u^2+v^2$ are relatively prime, so if they both divide $10^n$, they are a power of $2$ and $5$ respectively.
    $endgroup$
    – Wojowu
    Feb 20 at 9:20






  • 1




    $begingroup$
    @Wojowu And hence wlog $u=1$ and $v=2^k$ for some $kgeq0$, and $1+4^k=5^l$ so $5^l-4^k=1$, which has the unique integer solution $l=k=1$.
    $endgroup$
    – Servaes
    Feb 20 at 9:37











  • $begingroup$
    I searched all u and v from 1 through 100,000. The only solutions were u=3,v=1 and multiples of those, making sin=0.6, cos=0.8 and tan=0.75. Could this be the only solution?
    $endgroup$
    – user1008646
    Feb 21 at 6:20


















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A hypotenuse-$1$ right-angled triangle containing an angle $x$ has side length $sin x$ opposite it and $cos x$ adjacent to it, and if these are rational so is their ratio $tan x$. But these lengths are rational iff the triangle can be scaled to have integer side lengths, so the rational choices for $sin x,,cos x,,tan x$ with $x$ acute are precisely those obtained from Pythagorean triples. Extending to arbitrary angles so some functions may be negative, the general solution is $sin x=frac2aba^2+b^2,,cos x=fraca^2-b^2a^2+b^2,,tan x=frac2aba^2-b^2$ with $a,,binmathbbZ,,anepm b$.






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    1












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    Note that



    $$fracsin(x)cos(x) = tan(x)$$



    Therefore, whenever sine and cosine are rational, and cosine nonzero, tangent is also rational. Since sine and cosine are also continuous for all real numbers $x$ and are what we call "onto" or "surjective" on the interval $[-1,1]$, for each number in that interval we can find $x$ such that $sin(x)$ and $cos(x)$ are that number.



    We also know that the rationals are "dense" in the reals - that is, in between any two real numbers, you can always find a rational number. It ergo goes that we can find infinitely many $x$ combos so that $sin(x)$ and $cos(x)$ (independent of each other) are rational.




    This doesn't really address the case of their ratio or them being rational at the same time, though.



    For this, as noted by Blue in the comments, Pythagorean triples will always work. Pythagorean triples define three sides of a right triangle, specifically positive integer solutions to $a^2 + b^2 = c^2$. Because the triangle is a right triangle, we can also note that any $x$ we use as the angle here (opposite a leg) will be between $0$ and $pi/2$ in measure, noninclusive.



    There is a method to generate infinitely many Pythagorean triples too! Take positive, unequal integers $m,n$ - any two you want. Then $a = 2mn, b = m^2 - n^2, c = m^2 + n^2$. Since $m,n$ are integers, then $a,b,c$ are integers. Then in turn $sin(x) = b/c$ (or $a/c$, depending on the choice of angle), and $cos(x) = a/c$ (or $b/c$), giving those two are rational. From that, we see that $tan(x) = b/a$ (or $a/b$) and thus is also ratio.



    Thus, for each pair of unequal, positive integers $m,n$ we can have



    $$sin(x) = fracbc = fracm^2 - n^2m^2 + n^2 ;;; cos(x) = frac2mnm^2 + n^2 ;;; tan(x) = fracm^2 - n^22mn$$



    giving infinitely many solutions such that $sin(x),cos(x),tan(x)$ are all rational. From there, finding the corresponding $x$ is trivial.



    As for whether these are the only solutions, I do not know.






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      -1












      $begingroup$

      Have you tried using base $6$? I have heard that doing so can result in more accurate results when using $pi$. It may also work better in this case, as the desired result are finite decimals.






      share|cite|improve this answer











      $endgroup$








      • 1




        $begingroup$
        Welcome to Math.SE. How is this relevant to the question. Please elaborate
        $endgroup$
        – Shailesh
        Feb 21 at 0:37










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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      7












      $begingroup$

      Any Pythagorean triple will give you rational $sin, cos, tan$ values, and any triple of rational trig values will give you a Pythagorean triple.



      The Pythagorean triples have been completely classified. Take two natural numbers $u>v$. Then
      $$
      (u^2-v^2)^2 + (2uv)^2=(u^2+v^2)^2
      $$

      and any Pythagorean triple is either of this form or a multiple of such a triple. So the possible rational $sin$ and $cos$ pairs are all possible values of
      $$
      fracu^2-v^2u^2+v^2text and frac2uvu^2+v^2
      $$






      share|cite|improve this answer











      $endgroup$












      • $begingroup$
        You are right, But I missed a detail in my question - it is now edited.
        $endgroup$
        – Shahul
        Feb 20 at 8:00










      • $begingroup$
        @Shahul That's an easy fix. Make sure $u^2 + v^2$ is a divisor of $10^n$ for some $n$. In other words that the only prime numbers in the prime decomposition of $u^2 + v^2$ are $2$ and $5$.
        $endgroup$
        – Arthur
        Feb 20 at 8:02







      • 2




        $begingroup$
        In order to guarantee $tan$ has finite decimal expansion too, you need either $2uv$ or $u^2-v^2$ to be a divisor of a power of $10$ too. Taking $u,v$ relatively prime and exactly one even (we can always assume that for Pythagorean triples) you can say much more - for instance, $2uv$ and $u^2+v^2$ are relatively prime, so if they both divide $10^n$, they are a power of $2$ and $5$ respectively.
        $endgroup$
        – Wojowu
        Feb 20 at 9:20






      • 1




        $begingroup$
        @Wojowu And hence wlog $u=1$ and $v=2^k$ for some $kgeq0$, and $1+4^k=5^l$ so $5^l-4^k=1$, which has the unique integer solution $l=k=1$.
        $endgroup$
        – Servaes
        Feb 20 at 9:37











      • $begingroup$
        I searched all u and v from 1 through 100,000. The only solutions were u=3,v=1 and multiples of those, making sin=0.6, cos=0.8 and tan=0.75. Could this be the only solution?
        $endgroup$
        – user1008646
        Feb 21 at 6:20















      7












      $begingroup$

      Any Pythagorean triple will give you rational $sin, cos, tan$ values, and any triple of rational trig values will give you a Pythagorean triple.



      The Pythagorean triples have been completely classified. Take two natural numbers $u>v$. Then
      $$
      (u^2-v^2)^2 + (2uv)^2=(u^2+v^2)^2
      $$

      and any Pythagorean triple is either of this form or a multiple of such a triple. So the possible rational $sin$ and $cos$ pairs are all possible values of
      $$
      fracu^2-v^2u^2+v^2text and frac2uvu^2+v^2
      $$






      share|cite|improve this answer











      $endgroup$












      • $begingroup$
        You are right, But I missed a detail in my question - it is now edited.
        $endgroup$
        – Shahul
        Feb 20 at 8:00










      • $begingroup$
        @Shahul That's an easy fix. Make sure $u^2 + v^2$ is a divisor of $10^n$ for some $n$. In other words that the only prime numbers in the prime decomposition of $u^2 + v^2$ are $2$ and $5$.
        $endgroup$
        – Arthur
        Feb 20 at 8:02







      • 2




        $begingroup$
        In order to guarantee $tan$ has finite decimal expansion too, you need either $2uv$ or $u^2-v^2$ to be a divisor of a power of $10$ too. Taking $u,v$ relatively prime and exactly one even (we can always assume that for Pythagorean triples) you can say much more - for instance, $2uv$ and $u^2+v^2$ are relatively prime, so if they both divide $10^n$, they are a power of $2$ and $5$ respectively.
        $endgroup$
        – Wojowu
        Feb 20 at 9:20






      • 1




        $begingroup$
        @Wojowu And hence wlog $u=1$ and $v=2^k$ for some $kgeq0$, and $1+4^k=5^l$ so $5^l-4^k=1$, which has the unique integer solution $l=k=1$.
        $endgroup$
        – Servaes
        Feb 20 at 9:37











      • $begingroup$
        I searched all u and v from 1 through 100,000. The only solutions were u=3,v=1 and multiples of those, making sin=0.6, cos=0.8 and tan=0.75. Could this be the only solution?
        $endgroup$
        – user1008646
        Feb 21 at 6:20













      7












      7








      7





      $begingroup$

      Any Pythagorean triple will give you rational $sin, cos, tan$ values, and any triple of rational trig values will give you a Pythagorean triple.



      The Pythagorean triples have been completely classified. Take two natural numbers $u>v$. Then
      $$
      (u^2-v^2)^2 + (2uv)^2=(u^2+v^2)^2
      $$

      and any Pythagorean triple is either of this form or a multiple of such a triple. So the possible rational $sin$ and $cos$ pairs are all possible values of
      $$
      fracu^2-v^2u^2+v^2text and frac2uvu^2+v^2
      $$






      share|cite|improve this answer











      $endgroup$



      Any Pythagorean triple will give you rational $sin, cos, tan$ values, and any triple of rational trig values will give you a Pythagorean triple.



      The Pythagorean triples have been completely classified. Take two natural numbers $u>v$. Then
      $$
      (u^2-v^2)^2 + (2uv)^2=(u^2+v^2)^2
      $$

      and any Pythagorean triple is either of this form or a multiple of such a triple. So the possible rational $sin$ and $cos$ pairs are all possible values of
      $$
      fracu^2-v^2u^2+v^2text and frac2uvu^2+v^2
      $$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Feb 20 at 7:23

























      answered Feb 20 at 7:00









      ArthurArthur

      118k7118201




      118k7118201











      • $begingroup$
        You are right, But I missed a detail in my question - it is now edited.
        $endgroup$
        – Shahul
        Feb 20 at 8:00










      • $begingroup$
        @Shahul That's an easy fix. Make sure $u^2 + v^2$ is a divisor of $10^n$ for some $n$. In other words that the only prime numbers in the prime decomposition of $u^2 + v^2$ are $2$ and $5$.
        $endgroup$
        – Arthur
        Feb 20 at 8:02







      • 2




        $begingroup$
        In order to guarantee $tan$ has finite decimal expansion too, you need either $2uv$ or $u^2-v^2$ to be a divisor of a power of $10$ too. Taking $u,v$ relatively prime and exactly one even (we can always assume that for Pythagorean triples) you can say much more - for instance, $2uv$ and $u^2+v^2$ are relatively prime, so if they both divide $10^n$, they are a power of $2$ and $5$ respectively.
        $endgroup$
        – Wojowu
        Feb 20 at 9:20






      • 1




        $begingroup$
        @Wojowu And hence wlog $u=1$ and $v=2^k$ for some $kgeq0$, and $1+4^k=5^l$ so $5^l-4^k=1$, which has the unique integer solution $l=k=1$.
        $endgroup$
        – Servaes
        Feb 20 at 9:37











      • $begingroup$
        I searched all u and v from 1 through 100,000. The only solutions were u=3,v=1 and multiples of those, making sin=0.6, cos=0.8 and tan=0.75. Could this be the only solution?
        $endgroup$
        – user1008646
        Feb 21 at 6:20
















      • $begingroup$
        You are right, But I missed a detail in my question - it is now edited.
        $endgroup$
        – Shahul
        Feb 20 at 8:00










      • $begingroup$
        @Shahul That's an easy fix. Make sure $u^2 + v^2$ is a divisor of $10^n$ for some $n$. In other words that the only prime numbers in the prime decomposition of $u^2 + v^2$ are $2$ and $5$.
        $endgroup$
        – Arthur
        Feb 20 at 8:02







      • 2




        $begingroup$
        In order to guarantee $tan$ has finite decimal expansion too, you need either $2uv$ or $u^2-v^2$ to be a divisor of a power of $10$ too. Taking $u,v$ relatively prime and exactly one even (we can always assume that for Pythagorean triples) you can say much more - for instance, $2uv$ and $u^2+v^2$ are relatively prime, so if they both divide $10^n$, they are a power of $2$ and $5$ respectively.
        $endgroup$
        – Wojowu
        Feb 20 at 9:20






      • 1




        $begingroup$
        @Wojowu And hence wlog $u=1$ and $v=2^k$ for some $kgeq0$, and $1+4^k=5^l$ so $5^l-4^k=1$, which has the unique integer solution $l=k=1$.
        $endgroup$
        – Servaes
        Feb 20 at 9:37











      • $begingroup$
        I searched all u and v from 1 through 100,000. The only solutions were u=3,v=1 and multiples of those, making sin=0.6, cos=0.8 and tan=0.75. Could this be the only solution?
        $endgroup$
        – user1008646
        Feb 21 at 6:20















      $begingroup$
      You are right, But I missed a detail in my question - it is now edited.
      $endgroup$
      – Shahul
      Feb 20 at 8:00




      $begingroup$
      You are right, But I missed a detail in my question - it is now edited.
      $endgroup$
      – Shahul
      Feb 20 at 8:00












      $begingroup$
      @Shahul That's an easy fix. Make sure $u^2 + v^2$ is a divisor of $10^n$ for some $n$. In other words that the only prime numbers in the prime decomposition of $u^2 + v^2$ are $2$ and $5$.
      $endgroup$
      – Arthur
      Feb 20 at 8:02





      $begingroup$
      @Shahul That's an easy fix. Make sure $u^2 + v^2$ is a divisor of $10^n$ for some $n$. In other words that the only prime numbers in the prime decomposition of $u^2 + v^2$ are $2$ and $5$.
      $endgroup$
      – Arthur
      Feb 20 at 8:02





      2




      2




      $begingroup$
      In order to guarantee $tan$ has finite decimal expansion too, you need either $2uv$ or $u^2-v^2$ to be a divisor of a power of $10$ too. Taking $u,v$ relatively prime and exactly one even (we can always assume that for Pythagorean triples) you can say much more - for instance, $2uv$ and $u^2+v^2$ are relatively prime, so if they both divide $10^n$, they are a power of $2$ and $5$ respectively.
      $endgroup$
      – Wojowu
      Feb 20 at 9:20




      $begingroup$
      In order to guarantee $tan$ has finite decimal expansion too, you need either $2uv$ or $u^2-v^2$ to be a divisor of a power of $10$ too. Taking $u,v$ relatively prime and exactly one even (we can always assume that for Pythagorean triples) you can say much more - for instance, $2uv$ and $u^2+v^2$ are relatively prime, so if they both divide $10^n$, they are a power of $2$ and $5$ respectively.
      $endgroup$
      – Wojowu
      Feb 20 at 9:20




      1




      1




      $begingroup$
      @Wojowu And hence wlog $u=1$ and $v=2^k$ for some $kgeq0$, and $1+4^k=5^l$ so $5^l-4^k=1$, which has the unique integer solution $l=k=1$.
      $endgroup$
      – Servaes
      Feb 20 at 9:37





      $begingroup$
      @Wojowu And hence wlog $u=1$ and $v=2^k$ for some $kgeq0$, and $1+4^k=5^l$ so $5^l-4^k=1$, which has the unique integer solution $l=k=1$.
      $endgroup$
      – Servaes
      Feb 20 at 9:37













      $begingroup$
      I searched all u and v from 1 through 100,000. The only solutions were u=3,v=1 and multiples of those, making sin=0.6, cos=0.8 and tan=0.75. Could this be the only solution?
      $endgroup$
      – user1008646
      Feb 21 at 6:20




      $begingroup$
      I searched all u and v from 1 through 100,000. The only solutions were u=3,v=1 and multiples of those, making sin=0.6, cos=0.8 and tan=0.75. Could this be the only solution?
      $endgroup$
      – user1008646
      Feb 21 at 6:20











      3












      $begingroup$

      A hypotenuse-$1$ right-angled triangle containing an angle $x$ has side length $sin x$ opposite it and $cos x$ adjacent to it, and if these are rational so is their ratio $tan x$. But these lengths are rational iff the triangle can be scaled to have integer side lengths, so the rational choices for $sin x,,cos x,,tan x$ with $x$ acute are precisely those obtained from Pythagorean triples. Extending to arbitrary angles so some functions may be negative, the general solution is $sin x=frac2aba^2+b^2,,cos x=fraca^2-b^2a^2+b^2,,tan x=frac2aba^2-b^2$ with $a,,binmathbbZ,,anepm b$.






      share|cite|improve this answer









      $endgroup$

















        3












        $begingroup$

        A hypotenuse-$1$ right-angled triangle containing an angle $x$ has side length $sin x$ opposite it and $cos x$ adjacent to it, and if these are rational so is their ratio $tan x$. But these lengths are rational iff the triangle can be scaled to have integer side lengths, so the rational choices for $sin x,,cos x,,tan x$ with $x$ acute are precisely those obtained from Pythagorean triples. Extending to arbitrary angles so some functions may be negative, the general solution is $sin x=frac2aba^2+b^2,,cos x=fraca^2-b^2a^2+b^2,,tan x=frac2aba^2-b^2$ with $a,,binmathbbZ,,anepm b$.






        share|cite|improve this answer









        $endgroup$















          3












          3








          3





          $begingroup$

          A hypotenuse-$1$ right-angled triangle containing an angle $x$ has side length $sin x$ opposite it and $cos x$ adjacent to it, and if these are rational so is their ratio $tan x$. But these lengths are rational iff the triangle can be scaled to have integer side lengths, so the rational choices for $sin x,,cos x,,tan x$ with $x$ acute are precisely those obtained from Pythagorean triples. Extending to arbitrary angles so some functions may be negative, the general solution is $sin x=frac2aba^2+b^2,,cos x=fraca^2-b^2a^2+b^2,,tan x=frac2aba^2-b^2$ with $a,,binmathbbZ,,anepm b$.






          share|cite|improve this answer









          $endgroup$



          A hypotenuse-$1$ right-angled triangle containing an angle $x$ has side length $sin x$ opposite it and $cos x$ adjacent to it, and if these are rational so is their ratio $tan x$. But these lengths are rational iff the triangle can be scaled to have integer side lengths, so the rational choices for $sin x,,cos x,,tan x$ with $x$ acute are precisely those obtained from Pythagorean triples. Extending to arbitrary angles so some functions may be negative, the general solution is $sin x=frac2aba^2+b^2,,cos x=fraca^2-b^2a^2+b^2,,tan x=frac2aba^2-b^2$ with $a,,binmathbbZ,,anepm b$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 20 at 7:00









          J.G.J.G.

          30.3k23148




          30.3k23148





















              1












              $begingroup$

              Note that



              $$fracsin(x)cos(x) = tan(x)$$



              Therefore, whenever sine and cosine are rational, and cosine nonzero, tangent is also rational. Since sine and cosine are also continuous for all real numbers $x$ and are what we call "onto" or "surjective" on the interval $[-1,1]$, for each number in that interval we can find $x$ such that $sin(x)$ and $cos(x)$ are that number.



              We also know that the rationals are "dense" in the reals - that is, in between any two real numbers, you can always find a rational number. It ergo goes that we can find infinitely many $x$ combos so that $sin(x)$ and $cos(x)$ (independent of each other) are rational.




              This doesn't really address the case of their ratio or them being rational at the same time, though.



              For this, as noted by Blue in the comments, Pythagorean triples will always work. Pythagorean triples define three sides of a right triangle, specifically positive integer solutions to $a^2 + b^2 = c^2$. Because the triangle is a right triangle, we can also note that any $x$ we use as the angle here (opposite a leg) will be between $0$ and $pi/2$ in measure, noninclusive.



              There is a method to generate infinitely many Pythagorean triples too! Take positive, unequal integers $m,n$ - any two you want. Then $a = 2mn, b = m^2 - n^2, c = m^2 + n^2$. Since $m,n$ are integers, then $a,b,c$ are integers. Then in turn $sin(x) = b/c$ (or $a/c$, depending on the choice of angle), and $cos(x) = a/c$ (or $b/c$), giving those two are rational. From that, we see that $tan(x) = b/a$ (or $a/b$) and thus is also ratio.



              Thus, for each pair of unequal, positive integers $m,n$ we can have



              $$sin(x) = fracbc = fracm^2 - n^2m^2 + n^2 ;;; cos(x) = frac2mnm^2 + n^2 ;;; tan(x) = fracm^2 - n^22mn$$



              giving infinitely many solutions such that $sin(x),cos(x),tan(x)$ are all rational. From there, finding the corresponding $x$ is trivial.



              As for whether these are the only solutions, I do not know.






              share|cite|improve this answer









              $endgroup$

















                1












                $begingroup$

                Note that



                $$fracsin(x)cos(x) = tan(x)$$



                Therefore, whenever sine and cosine are rational, and cosine nonzero, tangent is also rational. Since sine and cosine are also continuous for all real numbers $x$ and are what we call "onto" or "surjective" on the interval $[-1,1]$, for each number in that interval we can find $x$ such that $sin(x)$ and $cos(x)$ are that number.



                We also know that the rationals are "dense" in the reals - that is, in between any two real numbers, you can always find a rational number. It ergo goes that we can find infinitely many $x$ combos so that $sin(x)$ and $cos(x)$ (independent of each other) are rational.




                This doesn't really address the case of their ratio or them being rational at the same time, though.



                For this, as noted by Blue in the comments, Pythagorean triples will always work. Pythagorean triples define three sides of a right triangle, specifically positive integer solutions to $a^2 + b^2 = c^2$. Because the triangle is a right triangle, we can also note that any $x$ we use as the angle here (opposite a leg) will be between $0$ and $pi/2$ in measure, noninclusive.



                There is a method to generate infinitely many Pythagorean triples too! Take positive, unequal integers $m,n$ - any two you want. Then $a = 2mn, b = m^2 - n^2, c = m^2 + n^2$. Since $m,n$ are integers, then $a,b,c$ are integers. Then in turn $sin(x) = b/c$ (or $a/c$, depending on the choice of angle), and $cos(x) = a/c$ (or $b/c$), giving those two are rational. From that, we see that $tan(x) = b/a$ (or $a/b$) and thus is also ratio.



                Thus, for each pair of unequal, positive integers $m,n$ we can have



                $$sin(x) = fracbc = fracm^2 - n^2m^2 + n^2 ;;; cos(x) = frac2mnm^2 + n^2 ;;; tan(x) = fracm^2 - n^22mn$$



                giving infinitely many solutions such that $sin(x),cos(x),tan(x)$ are all rational. From there, finding the corresponding $x$ is trivial.



                As for whether these are the only solutions, I do not know.






                share|cite|improve this answer









                $endgroup$















                  1












                  1








                  1





                  $begingroup$

                  Note that



                  $$fracsin(x)cos(x) = tan(x)$$



                  Therefore, whenever sine and cosine are rational, and cosine nonzero, tangent is also rational. Since sine and cosine are also continuous for all real numbers $x$ and are what we call "onto" or "surjective" on the interval $[-1,1]$, for each number in that interval we can find $x$ such that $sin(x)$ and $cos(x)$ are that number.



                  We also know that the rationals are "dense" in the reals - that is, in between any two real numbers, you can always find a rational number. It ergo goes that we can find infinitely many $x$ combos so that $sin(x)$ and $cos(x)$ (independent of each other) are rational.




                  This doesn't really address the case of their ratio or them being rational at the same time, though.



                  For this, as noted by Blue in the comments, Pythagorean triples will always work. Pythagorean triples define three sides of a right triangle, specifically positive integer solutions to $a^2 + b^2 = c^2$. Because the triangle is a right triangle, we can also note that any $x$ we use as the angle here (opposite a leg) will be between $0$ and $pi/2$ in measure, noninclusive.



                  There is a method to generate infinitely many Pythagorean triples too! Take positive, unequal integers $m,n$ - any two you want. Then $a = 2mn, b = m^2 - n^2, c = m^2 + n^2$. Since $m,n$ are integers, then $a,b,c$ are integers. Then in turn $sin(x) = b/c$ (or $a/c$, depending on the choice of angle), and $cos(x) = a/c$ (or $b/c$), giving those two are rational. From that, we see that $tan(x) = b/a$ (or $a/b$) and thus is also ratio.



                  Thus, for each pair of unequal, positive integers $m,n$ we can have



                  $$sin(x) = fracbc = fracm^2 - n^2m^2 + n^2 ;;; cos(x) = frac2mnm^2 + n^2 ;;; tan(x) = fracm^2 - n^22mn$$



                  giving infinitely many solutions such that $sin(x),cos(x),tan(x)$ are all rational. From there, finding the corresponding $x$ is trivial.



                  As for whether these are the only solutions, I do not know.






                  share|cite|improve this answer









                  $endgroup$



                  Note that



                  $$fracsin(x)cos(x) = tan(x)$$



                  Therefore, whenever sine and cosine are rational, and cosine nonzero, tangent is also rational. Since sine and cosine are also continuous for all real numbers $x$ and are what we call "onto" or "surjective" on the interval $[-1,1]$, for each number in that interval we can find $x$ such that $sin(x)$ and $cos(x)$ are that number.



                  We also know that the rationals are "dense" in the reals - that is, in between any two real numbers, you can always find a rational number. It ergo goes that we can find infinitely many $x$ combos so that $sin(x)$ and $cos(x)$ (independent of each other) are rational.




                  This doesn't really address the case of their ratio or them being rational at the same time, though.



                  For this, as noted by Blue in the comments, Pythagorean triples will always work. Pythagorean triples define three sides of a right triangle, specifically positive integer solutions to $a^2 + b^2 = c^2$. Because the triangle is a right triangle, we can also note that any $x$ we use as the angle here (opposite a leg) will be between $0$ and $pi/2$ in measure, noninclusive.



                  There is a method to generate infinitely many Pythagorean triples too! Take positive, unequal integers $m,n$ - any two you want. Then $a = 2mn, b = m^2 - n^2, c = m^2 + n^2$. Since $m,n$ are integers, then $a,b,c$ are integers. Then in turn $sin(x) = b/c$ (or $a/c$, depending on the choice of angle), and $cos(x) = a/c$ (or $b/c$), giving those two are rational. From that, we see that $tan(x) = b/a$ (or $a/b$) and thus is also ratio.



                  Thus, for each pair of unequal, positive integers $m,n$ we can have



                  $$sin(x) = fracbc = fracm^2 - n^2m^2 + n^2 ;;; cos(x) = frac2mnm^2 + n^2 ;;; tan(x) = fracm^2 - n^22mn$$



                  giving infinitely many solutions such that $sin(x),cos(x),tan(x)$ are all rational. From there, finding the corresponding $x$ is trivial.



                  As for whether these are the only solutions, I do not know.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 20 at 7:04









                  Eevee TrainerEevee Trainer

                  7,91521439




                  7,91521439





















                      -1












                      $begingroup$

                      Have you tried using base $6$? I have heard that doing so can result in more accurate results when using $pi$. It may also work better in this case, as the desired result are finite decimals.






                      share|cite|improve this answer











                      $endgroup$








                      • 1




                        $begingroup$
                        Welcome to Math.SE. How is this relevant to the question. Please elaborate
                        $endgroup$
                        – Shailesh
                        Feb 21 at 0:37















                      -1












                      $begingroup$

                      Have you tried using base $6$? I have heard that doing so can result in more accurate results when using $pi$. It may also work better in this case, as the desired result are finite decimals.






                      share|cite|improve this answer











                      $endgroup$








                      • 1




                        $begingroup$
                        Welcome to Math.SE. How is this relevant to the question. Please elaborate
                        $endgroup$
                        – Shailesh
                        Feb 21 at 0:37













                      -1












                      -1








                      -1





                      $begingroup$

                      Have you tried using base $6$? I have heard that doing so can result in more accurate results when using $pi$. It may also work better in this case, as the desired result are finite decimals.






                      share|cite|improve this answer











                      $endgroup$



                      Have you tried using base $6$? I have heard that doing so can result in more accurate results when using $pi$. It may also work better in this case, as the desired result are finite decimals.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Feb 21 at 1:10









                      dantopa

                      6,64942245




                      6,64942245










                      answered Feb 21 at 0:16









                      cboscbos

                      11




                      11







                      • 1




                        $begingroup$
                        Welcome to Math.SE. How is this relevant to the question. Please elaborate
                        $endgroup$
                        – Shailesh
                        Feb 21 at 0:37












                      • 1




                        $begingroup$
                        Welcome to Math.SE. How is this relevant to the question. Please elaborate
                        $endgroup$
                        – Shailesh
                        Feb 21 at 0:37







                      1




                      1




                      $begingroup$
                      Welcome to Math.SE. How is this relevant to the question. Please elaborate
                      $endgroup$
                      – Shailesh
                      Feb 21 at 0:37




                      $begingroup$
                      Welcome to Math.SE. How is this relevant to the question. Please elaborate
                      $endgroup$
                      – Shailesh
                      Feb 21 at 0:37

















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