Rational values for sine, cosine, and tangent












2












$begingroup$


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



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?










share|cite|improve this question









New contributor




Shahul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 3




    $begingroup$
    Any Pythagorean triple will give such values.
    $endgroup$
    – Blue
    57 mins ago
















2












$begingroup$


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



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?










share|cite|improve this question









New contributor




Shahul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 3




    $begingroup$
    Any Pythagorean triple will give such values.
    $endgroup$
    – Blue
    57 mins ago














2












2








2





$begingroup$


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



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?










share|cite|improve this question









New contributor




Shahul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




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



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?







algebra-precalculus trigonometry rational-numbers rationality-testing






share|cite|improve this question









New contributor




Shahul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Shahul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 43 mins ago









Eevee Trainer

6,1731936




6,1731936






New contributor




Shahul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 1 hour ago









ShahulShahul

161




161




New contributor




Shahul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Shahul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Shahul is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 3




    $begingroup$
    Any Pythagorean triple will give such values.
    $endgroup$
    – Blue
    57 mins ago














  • 3




    $begingroup$
    Any Pythagorean triple will give such values.
    $endgroup$
    – Blue
    57 mins ago








3




3




$begingroup$
Any Pythagorean triple will give such values.
$endgroup$
– Blue
57 mins ago




$begingroup$
Any Pythagorean triple will give such values.
$endgroup$
– Blue
57 mins ago










3 Answers
3






active

oldest

votes


















4












$begingroup$

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
$$
frac{u^2-v^2}{u^2+v^2}text{ and }frac{2uv}{u^2+v^2}
$$






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 $sin x=frac{2ab}{a^2+b^2},,cos x=frac{a^2-b^2}{a^2+b^2},,tan x=frac{2ab}{a^2-b^2}$ with $a,,binmathbb{Z},,anepm b$.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Note that



      $$frac{sin(x)}{cos(x)} = tan(x)$$



      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



      $$sin(x) = frac{b}{c} = frac{m^2 - n^2}{m^2 + n^2} ;;; cos(x) = frac{2mn}{m^2 + n^2} ;;; tan(x) = frac{m^2 - n^2}{2mn}$$



      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









      $endgroup$













        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "69"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });






        Shahul is a new contributor. Be nice, and check out our Code of Conduct.










        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3119802%2frational-values-for-sine-cosine-and-tangent%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4












        $begingroup$

        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
        $$
        frac{u^2-v^2}{u^2+v^2}text{ and }frac{2uv}{u^2+v^2}
        $$






        share|cite|improve this answer











        $endgroup$


















          4












          $begingroup$

          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
          $$
          frac{u^2-v^2}{u^2+v^2}text{ and }frac{2uv}{u^2+v^2}
          $$






          share|cite|improve this answer











          $endgroup$
















            4












            4








            4





            $begingroup$

            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
            $$
            frac{u^2-v^2}{u^2+v^2}text{ and }frac{2uv}{u^2+v^2}
            $$






            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
            $$
            frac{u^2-v^2}{u^2+v^2}text{ and }frac{2uv}{u^2+v^2}
            $$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 29 mins ago

























            answered 53 mins ago









            ArthurArthur

            115k7116198




            115k7116198























                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 $sin x=frac{2ab}{a^2+b^2},,cos x=frac{a^2-b^2}{a^2+b^2},,tan x=frac{2ab}{a^2-b^2}$ with $a,,binmathbb{Z},,anepm b$.






                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 $sin x=frac{2ab}{a^2+b^2},,cos x=frac{a^2-b^2}{a^2+b^2},,tan x=frac{2ab}{a^2-b^2}$ with $a,,binmathbb{Z},,anepm b$.






                  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 $sin x=frac{2ab}{a^2+b^2},,cos x=frac{a^2-b^2}{a^2+b^2},,tan x=frac{2ab}{a^2-b^2}$ with $a,,binmathbb{Z},,anepm b$.






                    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 $sin x=frac{2ab}{a^2+b^2},,cos x=frac{a^2-b^2}{a^2+b^2},,tan x=frac{2ab}{a^2-b^2}$ with $a,,binmathbb{Z},,anepm b$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 52 mins ago









                    J.G.J.G.

                    27.3k22843




                    27.3k22843























                        1












                        $begingroup$

                        Note that



                        $$frac{sin(x)}{cos(x)} = tan(x)$$



                        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



                        $$sin(x) = frac{b}{c} = frac{m^2 - n^2}{m^2 + n^2} ;;; cos(x) = frac{2mn}{m^2 + n^2} ;;; tan(x) = frac{m^2 - n^2}{2mn}$$



                        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



                          $$frac{sin(x)}{cos(x)} = tan(x)$$



                          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



                          $$sin(x) = frac{b}{c} = frac{m^2 - n^2}{m^2 + n^2} ;;; cos(x) = frac{2mn}{m^2 + n^2} ;;; tan(x) = frac{m^2 - n^2}{2mn}$$



                          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



                            $$frac{sin(x)}{cos(x)} = tan(x)$$



                            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



                            $$sin(x) = frac{b}{c} = frac{m^2 - n^2}{m^2 + n^2} ;;; cos(x) = frac{2mn}{m^2 + n^2} ;;; tan(x) = frac{m^2 - n^2}{2mn}$$



                            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



                            $$frac{sin(x)}{cos(x)} = tan(x)$$



                            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



                            $$sin(x) = frac{b}{c} = frac{m^2 - n^2}{m^2 + n^2} ;;; cos(x) = frac{2mn}{m^2 + n^2} ;;; tan(x) = frac{m^2 - n^2}{2mn}$$



                            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 48 mins ago









                            Eevee TrainerEevee Trainer

                            6,1731936




                            6,1731936






















                                Shahul is a new contributor. Be nice, and check out our Code of Conduct.










                                draft saved

                                draft discarded


















                                Shahul is a new contributor. Be nice, and check out our Code of Conduct.













                                Shahul is a new contributor. Be nice, and check out our Code of Conduct.












                                Shahul is a new contributor. Be nice, and check out our Code of Conduct.
















                                Thanks for contributing an answer to Mathematics Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3119802%2frational-values-for-sine-cosine-and-tangent%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                Popular posts from this blog

                                Liste der Baudenkmale in Friedland (Mecklenburg)

                                Single-Malt-Whisky

                                Czorneboh