Simplify trigonometric expression using trigonometric identities












3












$begingroup$


I have the trigonometric expression: $$2sin x +2sin left(frac{pi} {3} -xright) $$
and it should simplified in: $$sin x + sqrt 3 cos x$$ but I do not know what formulas to apply. Could you tell me how to simplify it?










share|cite|improve this question











$endgroup$












  • $begingroup$
    mathworld.wolfram.com/ProsthaphaeresisFormulas.html
    $endgroup$
    – lab bhattacharjee
    Apr 1 at 15:27










  • $begingroup$
    Just between us: IMO the two expressions are of the same complexity.
    $endgroup$
    – Yves Daoust
    Apr 1 at 15:30










  • $begingroup$
    = 2 cos(x-π/6).
    $endgroup$
    – Anton Sherwood
    Apr 1 at 22:26
















3












$begingroup$


I have the trigonometric expression: $$2sin x +2sin left(frac{pi} {3} -xright) $$
and it should simplified in: $$sin x + sqrt 3 cos x$$ but I do not know what formulas to apply. Could you tell me how to simplify it?










share|cite|improve this question











$endgroup$












  • $begingroup$
    mathworld.wolfram.com/ProsthaphaeresisFormulas.html
    $endgroup$
    – lab bhattacharjee
    Apr 1 at 15:27










  • $begingroup$
    Just between us: IMO the two expressions are of the same complexity.
    $endgroup$
    – Yves Daoust
    Apr 1 at 15:30










  • $begingroup$
    = 2 cos(x-π/6).
    $endgroup$
    – Anton Sherwood
    Apr 1 at 22:26














3












3








3





$begingroup$


I have the trigonometric expression: $$2sin x +2sin left(frac{pi} {3} -xright) $$
and it should simplified in: $$sin x + sqrt 3 cos x$$ but I do not know what formulas to apply. Could you tell me how to simplify it?










share|cite|improve this question











$endgroup$




I have the trigonometric expression: $$2sin x +2sin left(frac{pi} {3} -xright) $$
and it should simplified in: $$sin x + sqrt 3 cos x$$ but I do not know what formulas to apply. Could you tell me how to simplify it?







trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 1 at 15:55









MarianD

2,2611618




2,2611618










asked Apr 1 at 15:19









zaz9999zaz9999

162




162












  • $begingroup$
    mathworld.wolfram.com/ProsthaphaeresisFormulas.html
    $endgroup$
    – lab bhattacharjee
    Apr 1 at 15:27










  • $begingroup$
    Just between us: IMO the two expressions are of the same complexity.
    $endgroup$
    – Yves Daoust
    Apr 1 at 15:30










  • $begingroup$
    = 2 cos(x-π/6).
    $endgroup$
    – Anton Sherwood
    Apr 1 at 22:26


















  • $begingroup$
    mathworld.wolfram.com/ProsthaphaeresisFormulas.html
    $endgroup$
    – lab bhattacharjee
    Apr 1 at 15:27










  • $begingroup$
    Just between us: IMO the two expressions are of the same complexity.
    $endgroup$
    – Yves Daoust
    Apr 1 at 15:30










  • $begingroup$
    = 2 cos(x-π/6).
    $endgroup$
    – Anton Sherwood
    Apr 1 at 22:26
















$begingroup$
mathworld.wolfram.com/ProsthaphaeresisFormulas.html
$endgroup$
– lab bhattacharjee
Apr 1 at 15:27




$begingroup$
mathworld.wolfram.com/ProsthaphaeresisFormulas.html
$endgroup$
– lab bhattacharjee
Apr 1 at 15:27












$begingroup$
Just between us: IMO the two expressions are of the same complexity.
$endgroup$
– Yves Daoust
Apr 1 at 15:30




$begingroup$
Just between us: IMO the two expressions are of the same complexity.
$endgroup$
– Yves Daoust
Apr 1 at 15:30












$begingroup$
= 2 cos(x-π/6).
$endgroup$
– Anton Sherwood
Apr 1 at 22:26




$begingroup$
= 2 cos(x-π/6).
$endgroup$
– Anton Sherwood
Apr 1 at 22:26










3 Answers
3






active

oldest

votes


















7












$begingroup$

Using the formula for $sin (alpha - beta)$ you obtain



begin{align}
2&sin x +2sin left(frac{pi} {3} -xright)\
= 2&sin x +2left[sin left(frac{pi} {3}right) cos x - cosleft(frac{pi} {3}right) sin xright]\
= 2&sin x + 2left[{sqrt 3 over 2} cos x - frac 1 2 sin xright]\[1ex]
= 2&sin x + sqrt 3 cos x - sin x\[1em] = , &color{red}{sin x + sqrt 3 cos x}
end{align}






share|cite|improve this answer











$endgroup$





















    4












    $begingroup$

    Use that $$sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$$






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Hint: $sin(frac{pi}{3}-x)=sin frac{pi}{3}cos x-cos frac{pi}{3}sin x$






      share|cite|improve this answer









      $endgroup$














        Your Answer








        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%2f3170733%2fsimplify-trigonometric-expression-using-trigonometric-identities%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









        7












        $begingroup$

        Using the formula for $sin (alpha - beta)$ you obtain



        begin{align}
        2&sin x +2sin left(frac{pi} {3} -xright)\
        = 2&sin x +2left[sin left(frac{pi} {3}right) cos x - cosleft(frac{pi} {3}right) sin xright]\
        = 2&sin x + 2left[{sqrt 3 over 2} cos x - frac 1 2 sin xright]\[1ex]
        = 2&sin x + sqrt 3 cos x - sin x\[1em] = , &color{red}{sin x + sqrt 3 cos x}
        end{align}






        share|cite|improve this answer











        $endgroup$


















          7












          $begingroup$

          Using the formula for $sin (alpha - beta)$ you obtain



          begin{align}
          2&sin x +2sin left(frac{pi} {3} -xright)\
          = 2&sin x +2left[sin left(frac{pi} {3}right) cos x - cosleft(frac{pi} {3}right) sin xright]\
          = 2&sin x + 2left[{sqrt 3 over 2} cos x - frac 1 2 sin xright]\[1ex]
          = 2&sin x + sqrt 3 cos x - sin x\[1em] = , &color{red}{sin x + sqrt 3 cos x}
          end{align}






          share|cite|improve this answer











          $endgroup$
















            7












            7








            7





            $begingroup$

            Using the formula for $sin (alpha - beta)$ you obtain



            begin{align}
            2&sin x +2sin left(frac{pi} {3} -xright)\
            = 2&sin x +2left[sin left(frac{pi} {3}right) cos x - cosleft(frac{pi} {3}right) sin xright]\
            = 2&sin x + 2left[{sqrt 3 over 2} cos x - frac 1 2 sin xright]\[1ex]
            = 2&sin x + sqrt 3 cos x - sin x\[1em] = , &color{red}{sin x + sqrt 3 cos x}
            end{align}






            share|cite|improve this answer











            $endgroup$



            Using the formula for $sin (alpha - beta)$ you obtain



            begin{align}
            2&sin x +2sin left(frac{pi} {3} -xright)\
            = 2&sin x +2left[sin left(frac{pi} {3}right) cos x - cosleft(frac{pi} {3}right) sin xright]\
            = 2&sin x + 2left[{sqrt 3 over 2} cos x - frac 1 2 sin xright]\[1ex]
            = 2&sin x + sqrt 3 cos x - sin x\[1em] = , &color{red}{sin x + sqrt 3 cos x}
            end{align}







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Apr 1 at 15:59

























            answered Apr 1 at 15:44









            MarianDMarianD

            2,2611618




            2,2611618























                4












                $begingroup$

                Use that $$sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$$






                share|cite|improve this answer









                $endgroup$


















                  4












                  $begingroup$

                  Use that $$sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$$






                  share|cite|improve this answer









                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    Use that $$sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$$






                    share|cite|improve this answer









                    $endgroup$



                    Use that $$sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Apr 1 at 15:26









                    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                    79k42867




                    79k42867























                        1












                        $begingroup$

                        Hint: $sin(frac{pi}{3}-x)=sin frac{pi}{3}cos x-cos frac{pi}{3}sin x$






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Hint: $sin(frac{pi}{3}-x)=sin frac{pi}{3}cos x-cos frac{pi}{3}sin x$






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Hint: $sin(frac{pi}{3}-x)=sin frac{pi}{3}cos x-cos frac{pi}{3}sin x$






                            share|cite|improve this answer









                            $endgroup$



                            Hint: $sin(frac{pi}{3}-x)=sin frac{pi}{3}cos x-cos frac{pi}{3}sin x$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Apr 1 at 15:25









                            VasyaVasya

                            4,4671618




                            4,4671618






























                                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%2f3170733%2fsimplify-trigonometric-expression-using-trigonometric-identities%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

                                Paul Cézanne

                                UIScrollView CustomStickyHeader Resize height generates problems when scroll is too fast

                                Angular material date-picker (MatDatepicker) auto completes the date on focus out