Mathematical solution for intuitive trigonometric equation












1












$begingroup$


I derived the following trigonometrical equation from a real triangle, knowing that the angle $alpha$ is an acute one:



$sin 3alpha = 2sinalpha$



Just by eye-balling the equation, and remembering the trigonometric unit circle, I know that:



$sin 90 = 1 = 2sin 30 = 2 cdot 0.5 = 1$



Therefore, $alpha = 30^circ$ is a possible solution.

I am unaware of any trigonometric identity to help me simplify this equation in order to get all possible solutions for $alpha$, and this intuitive solution is the best I can come up with. I plugged this equation into symbolab.com, but their solution seems very long-winded, and I am hoping for the possibility that a simpler one exists.



How can I solve this type of problem when the intuitive approach fails?






trigonometry







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I derived the following trigonometrical equation from a real triangle, knowing that the angle $alpha$ is an acute one:



    $sin 3alpha = 2sinalpha$



    Just by eye-balling the equation, and remembering the trigonometric unit circle, I know that:



    $sin 90 = 1 = 2sin 30 = 2 cdot 0.5 = 1$



    Therefore, $alpha = 30^circ$ is a possible solution.

    I am unaware of any trigonometric identity to help me simplify this equation in order to get all possible solutions for $alpha$, and this intuitive solution is the best I can come up with. I plugged this equation into symbolab.com, but their solution seems very long-winded, and I am hoping for the possibility that a simpler one exists.



    How can I solve this type of problem when the intuitive approach fails?






    trigonometry







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I derived the following trigonometrical equation from a real triangle, knowing that the angle $alpha$ is an acute one:



      $sin 3alpha = 2sinalpha$



      Just by eye-balling the equation, and remembering the trigonometric unit circle, I know that:



      $sin 90 = 1 = 2sin 30 = 2 cdot 0.5 = 1$



      Therefore, $alpha = 30^circ$ is a possible solution.

      I am unaware of any trigonometric identity to help me simplify this equation in order to get all possible solutions for $alpha$, and this intuitive solution is the best I can come up with. I plugged this equation into symbolab.com, but their solution seems very long-winded, and I am hoping for the possibility that a simpler one exists.



      How can I solve this type of problem when the intuitive approach fails?






      trigonometry







      share|cite|improve this question











      $endgroup$




      I derived the following trigonometrical equation from a real triangle, knowing that the angle $alpha$ is an acute one:



      $sin 3alpha = 2sinalpha$



      Just by eye-balling the equation, and remembering the trigonometric unit circle, I know that:



      $sin 90 = 1 = 2sin 30 = 2 cdot 0.5 = 1$



      Therefore, $alpha = 30^circ$ is a possible solution.

      I am unaware of any trigonometric identity to help me simplify this equation in order to get all possible solutions for $alpha$, and this intuitive solution is the best I can come up with. I plugged this equation into symbolab.com, but their solution seems very long-winded, and I am hoping for the possibility that a simpler one exists.



      How can I solve this type of problem when the intuitive approach fails?






      trigonometry




      trigonometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 8 '18 at 15:32







      daedsidog

















      asked Dec 8 '18 at 15:26









      daedsidogdaedsidog

      29017




      29017






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          If triple angle relation for sine is known, letting $ sin alpha =s$,



          $$3 s -4 s^3= 2 srightarrow s =(0, pm frac12), quad alpha=(0, pm 30^{circ}, 150^{circ}pm30^{circ} ) ..$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Forgive my ignorance, but I don't understand what this syntax is meant to represent.
            $endgroup$
            – daedsidog
            Dec 8 '18 at 15:48










          • $begingroup$
            Sorry I should have given it fully, shall edit it in the answer.
            $endgroup$
            – Narasimham
            Dec 8 '18 at 16:14





















          1












          $begingroup$

          Here is a lead.



          $$sin(3x) = sin(2x + x)
          = sin(2x)cos(x) + cos(2x)sin(x).$$
          Now invoke some double angle magig to get
          $$sin(3x) = 2sin(x)cos^2(x) + (1 - 2sin^2(x))sin(x).$$
          Next, use the pythagorean identities to gete
          $$sin(3x) = 2sin(x)(1 - sin^2(x)) + (1 - 2sin^2(x))sin(x). $$
          Can you use this?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This never occurred to me, but it seems to me that there's a lot more work to be done in order to derive the actual angles, isn't there?
            $endgroup$
            – daedsidog
            Dec 8 '18 at 15:47











          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
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3031233%2fmathematical-solution-for-intuitive-trigonometric-equation%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          If triple angle relation for sine is known, letting $ sin alpha =s$,



          $$3 s -4 s^3= 2 srightarrow s =(0, pm frac12), quad alpha=(0, pm 30^{circ}, 150^{circ}pm30^{circ} ) ..$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Forgive my ignorance, but I don't understand what this syntax is meant to represent.
            $endgroup$
            – daedsidog
            Dec 8 '18 at 15:48










          • $begingroup$
            Sorry I should have given it fully, shall edit it in the answer.
            $endgroup$
            – Narasimham
            Dec 8 '18 at 16:14


















          2












          $begingroup$

          If triple angle relation for sine is known, letting $ sin alpha =s$,



          $$3 s -4 s^3= 2 srightarrow s =(0, pm frac12), quad alpha=(0, pm 30^{circ}, 150^{circ}pm30^{circ} ) ..$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Forgive my ignorance, but I don't understand what this syntax is meant to represent.
            $endgroup$
            – daedsidog
            Dec 8 '18 at 15:48










          • $begingroup$
            Sorry I should have given it fully, shall edit it in the answer.
            $endgroup$
            – Narasimham
            Dec 8 '18 at 16:14
















          2












          2








          2





          $begingroup$

          If triple angle relation for sine is known, letting $ sin alpha =s$,



          $$3 s -4 s^3= 2 srightarrow s =(0, pm frac12), quad alpha=(0, pm 30^{circ}, 150^{circ}pm30^{circ} ) ..$$






          share|cite|improve this answer











          $endgroup$



          If triple angle relation for sine is known, letting $ sin alpha =s$,



          $$3 s -4 s^3= 2 srightarrow s =(0, pm frac12), quad alpha=(0, pm 30^{circ}, 150^{circ}pm30^{circ} ) ..$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 8 '18 at 16:14

























          answered Dec 8 '18 at 15:37









          NarasimhamNarasimham

          20.7k52158




          20.7k52158












          • $begingroup$
            Forgive my ignorance, but I don't understand what this syntax is meant to represent.
            $endgroup$
            – daedsidog
            Dec 8 '18 at 15:48










          • $begingroup$
            Sorry I should have given it fully, shall edit it in the answer.
            $endgroup$
            – Narasimham
            Dec 8 '18 at 16:14




















          • $begingroup$
            Forgive my ignorance, but I don't understand what this syntax is meant to represent.
            $endgroup$
            – daedsidog
            Dec 8 '18 at 15:48










          • $begingroup$
            Sorry I should have given it fully, shall edit it in the answer.
            $endgroup$
            – Narasimham
            Dec 8 '18 at 16:14


















          $begingroup$
          Forgive my ignorance, but I don't understand what this syntax is meant to represent.
          $endgroup$
          – daedsidog
          Dec 8 '18 at 15:48




          $begingroup$
          Forgive my ignorance, but I don't understand what this syntax is meant to represent.
          $endgroup$
          – daedsidog
          Dec 8 '18 at 15:48












          $begingroup$
          Sorry I should have given it fully, shall edit it in the answer.
          $endgroup$
          – Narasimham
          Dec 8 '18 at 16:14






          $begingroup$
          Sorry I should have given it fully, shall edit it in the answer.
          $endgroup$
          – Narasimham
          Dec 8 '18 at 16:14













          1












          $begingroup$

          Here is a lead.



          $$sin(3x) = sin(2x + x)
          = sin(2x)cos(x) + cos(2x)sin(x).$$
          Now invoke some double angle magig to get
          $$sin(3x) = 2sin(x)cos^2(x) + (1 - 2sin^2(x))sin(x).$$
          Next, use the pythagorean identities to gete
          $$sin(3x) = 2sin(x)(1 - sin^2(x)) + (1 - 2sin^2(x))sin(x). $$
          Can you use this?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This never occurred to me, but it seems to me that there's a lot more work to be done in order to derive the actual angles, isn't there?
            $endgroup$
            – daedsidog
            Dec 8 '18 at 15:47
















          1












          $begingroup$

          Here is a lead.



          $$sin(3x) = sin(2x + x)
          = sin(2x)cos(x) + cos(2x)sin(x).$$
          Now invoke some double angle magig to get
          $$sin(3x) = 2sin(x)cos^2(x) + (1 - 2sin^2(x))sin(x).$$
          Next, use the pythagorean identities to gete
          $$sin(3x) = 2sin(x)(1 - sin^2(x)) + (1 - 2sin^2(x))sin(x). $$
          Can you use this?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This never occurred to me, but it seems to me that there's a lot more work to be done in order to derive the actual angles, isn't there?
            $endgroup$
            – daedsidog
            Dec 8 '18 at 15:47














          1












          1








          1





          $begingroup$

          Here is a lead.



          $$sin(3x) = sin(2x + x)
          = sin(2x)cos(x) + cos(2x)sin(x).$$
          Now invoke some double angle magig to get
          $$sin(3x) = 2sin(x)cos^2(x) + (1 - 2sin^2(x))sin(x).$$
          Next, use the pythagorean identities to gete
          $$sin(3x) = 2sin(x)(1 - sin^2(x)) + (1 - 2sin^2(x))sin(x). $$
          Can you use this?






          share|cite|improve this answer









          $endgroup$



          Here is a lead.



          $$sin(3x) = sin(2x + x)
          = sin(2x)cos(x) + cos(2x)sin(x).$$
          Now invoke some double angle magig to get
          $$sin(3x) = 2sin(x)cos^2(x) + (1 - 2sin^2(x))sin(x).$$
          Next, use the pythagorean identities to gete
          $$sin(3x) = 2sin(x)(1 - sin^2(x)) + (1 - 2sin^2(x))sin(x). $$
          Can you use this?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 '18 at 15:39









          ncmathsadistncmathsadist

          42.7k260103




          42.7k260103












          • $begingroup$
            This never occurred to me, but it seems to me that there's a lot more work to be done in order to derive the actual angles, isn't there?
            $endgroup$
            – daedsidog
            Dec 8 '18 at 15:47


















          • $begingroup$
            This never occurred to me, but it seems to me that there's a lot more work to be done in order to derive the actual angles, isn't there?
            $endgroup$
            – daedsidog
            Dec 8 '18 at 15:47
















          $begingroup$
          This never occurred to me, but it seems to me that there's a lot more work to be done in order to derive the actual angles, isn't there?
          $endgroup$
          – daedsidog
          Dec 8 '18 at 15:47




          $begingroup$
          This never occurred to me, but it seems to me that there's a lot more work to be done in order to derive the actual angles, isn't there?
          $endgroup$
          – daedsidog
          Dec 8 '18 at 15:47


















          draft saved

          draft discarded




















































          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%2f3031233%2fmathematical-solution-for-intuitive-trigonometric-equation%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

          Ellipse (mathématiques)

          Image
          Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
          Read more

          Quarter-circle Tiles

          Image
          up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
          Read more

          Mont Emei

          Image
          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
          Read more