Evaluating $dfrac1sin(2x) + dfrac1sin(4x) + dfrac1sin(8x) + dfrac1sin(16x)$

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
2
down vote

favorite













Evaluate



$$dfrac1sin(2x) + dfrac1sin(4x) + dfrac1sin(8x) + dfrac1sin(16x)$$




It would be tough for us to solve it using trigonometric identities. There should be strictly an easy trick to proceed.



Rewriting and using trigonometric identities



$$dfrac1sin(2x) + dfrac1sin(2x) cos (2x) + dfrac1 2big [2sin (2x)cos (2x)cos (4x)big ] + dfrac1sin(16x)$$



What am I missing?



Regards










share|cite|improve this question





















  • See math.stackexchange.com/questions/1591220/…
    – lab bhattacharjee
    Sep 6 at 11:27














up vote
2
down vote

favorite













Evaluate



$$dfrac1sin(2x) + dfrac1sin(4x) + dfrac1sin(8x) + dfrac1sin(16x)$$




It would be tough for us to solve it using trigonometric identities. There should be strictly an easy trick to proceed.



Rewriting and using trigonometric identities



$$dfrac1sin(2x) + dfrac1sin(2x) cos (2x) + dfrac1 2big [2sin (2x)cos (2x)cos (4x)big ] + dfrac1sin(16x)$$



What am I missing?



Regards










share|cite|improve this question





















  • See math.stackexchange.com/questions/1591220/…
    – lab bhattacharjee
    Sep 6 at 11:27












up vote
2
down vote

favorite









up vote
2
down vote

favorite












Evaluate



$$dfrac1sin(2x) + dfrac1sin(4x) + dfrac1sin(8x) + dfrac1sin(16x)$$




It would be tough for us to solve it using trigonometric identities. There should be strictly an easy trick to proceed.



Rewriting and using trigonometric identities



$$dfrac1sin(2x) + dfrac1sin(2x) cos (2x) + dfrac1 2big [2sin (2x)cos (2x)cos (4x)big ] + dfrac1sin(16x)$$



What am I missing?



Regards










share|cite|improve this question














Evaluate



$$dfrac1sin(2x) + dfrac1sin(4x) + dfrac1sin(8x) + dfrac1sin(16x)$$




It would be tough for us to solve it using trigonometric identities. There should be strictly an easy trick to proceed.



Rewriting and using trigonometric identities



$$dfrac1sin(2x) + dfrac1sin(2x) cos (2x) + dfrac1 2big [2sin (2x)cos (2x)cos (4x)big ] + dfrac1sin(16x)$$



What am I missing?



Regards







trigonometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 6 at 11:11









Busi

325110




325110











  • See math.stackexchange.com/questions/1591220/…
    – lab bhattacharjee
    Sep 6 at 11:27
















  • See math.stackexchange.com/questions/1591220/…
    – lab bhattacharjee
    Sep 6 at 11:27















See math.stackexchange.com/questions/1591220/…
– lab bhattacharjee
Sep 6 at 11:27




See math.stackexchange.com/questions/1591220/…
– lab bhattacharjee
Sep 6 at 11:27










3 Answers
3






active

oldest

votes

















up vote
3
down vote













$$sin(A-B)=sin A cos B-cos A sin B$$
$$frac1sin 2x=dfracsin(2x-x)sin2xsin x=cot x-cot2x$$
$$frac1sin 4x=dfracsin(4x-2x)sin4xsin 2x=cot 2x-cot4x$$
$$frac1sin 8x=dfracsin(8x-4x)sin8xsin 4x=cot 4x-cot8x$$



$$frac1sin 16x=dfracsin(16x-8x)sin16xsin 8x=cot 8x-cot16x$$






share|cite|improve this answer






















  • Where did that $cot$ come from?
    – Busi
    Sep 6 at 12:23










  • apply $sin(a-b)$ formula
    – Deepesh Meena
    Sep 6 at 13:42










  • @DeepeshMeena, Have you noticed the link?
    – lab bhattacharjee
    Sep 7 at 7:44










  • which link can you provide me?
    – Deepesh Meena
    Sep 7 at 9:42

















up vote
2
down vote













$$frac1sin2x=fracsin xsin 2x sin x=fracsin (2x-x)sin 2x sin x=fracsin 2x cos x - cos 2x sin xsin 2x sin x=cot x - cot2x$$



$$frac1sin4x=cot 2x - cot4x$$



$$frac1sin8x=cot 4x - cot8x$$



$$frac1sin16x=cot 8x - cot16x$$



$$frac1sin2x+frac1sin4x+frac1sin8x+frac1sin16x=cot x-cot 16x$$






share|cite|improve this answer



























    up vote
    1
    down vote













    Hint : Notice that $frac1sin(2x))$ is a common factor of all terms of the sum.






    share|cite|improve this answer




















    • Then notice that $frac1cos2x$ is a common factor of everything that's left except the 1 in front. Then you'll end up with easy $1/a + 1/b$ problems.
      – PackSciences
      Sep 6 at 11:26










    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',
    convertImagesToLinks: true,
    noModals: false,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    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%2f2907344%2fevaluating-dfrac1-sin2x-dfrac1-sin4x-dfrac1-sin8x-d%23new-answer', 'question_page');

    );

    Post as a guest






























    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote













    $$sin(A-B)=sin A cos B-cos A sin B$$
    $$frac1sin 2x=dfracsin(2x-x)sin2xsin x=cot x-cot2x$$
    $$frac1sin 4x=dfracsin(4x-2x)sin4xsin 2x=cot 2x-cot4x$$
    $$frac1sin 8x=dfracsin(8x-4x)sin8xsin 4x=cot 4x-cot8x$$



    $$frac1sin 16x=dfracsin(16x-8x)sin16xsin 8x=cot 8x-cot16x$$






    share|cite|improve this answer






















    • Where did that $cot$ come from?
      – Busi
      Sep 6 at 12:23










    • apply $sin(a-b)$ formula
      – Deepesh Meena
      Sep 6 at 13:42










    • @DeepeshMeena, Have you noticed the link?
      – lab bhattacharjee
      Sep 7 at 7:44










    • which link can you provide me?
      – Deepesh Meena
      Sep 7 at 9:42














    up vote
    3
    down vote













    $$sin(A-B)=sin A cos B-cos A sin B$$
    $$frac1sin 2x=dfracsin(2x-x)sin2xsin x=cot x-cot2x$$
    $$frac1sin 4x=dfracsin(4x-2x)sin4xsin 2x=cot 2x-cot4x$$
    $$frac1sin 8x=dfracsin(8x-4x)sin8xsin 4x=cot 4x-cot8x$$



    $$frac1sin 16x=dfracsin(16x-8x)sin16xsin 8x=cot 8x-cot16x$$






    share|cite|improve this answer






















    • Where did that $cot$ come from?
      – Busi
      Sep 6 at 12:23










    • apply $sin(a-b)$ formula
      – Deepesh Meena
      Sep 6 at 13:42










    • @DeepeshMeena, Have you noticed the link?
      – lab bhattacharjee
      Sep 7 at 7:44










    • which link can you provide me?
      – Deepesh Meena
      Sep 7 at 9:42












    up vote
    3
    down vote










    up vote
    3
    down vote









    $$sin(A-B)=sin A cos B-cos A sin B$$
    $$frac1sin 2x=dfracsin(2x-x)sin2xsin x=cot x-cot2x$$
    $$frac1sin 4x=dfracsin(4x-2x)sin4xsin 2x=cot 2x-cot4x$$
    $$frac1sin 8x=dfracsin(8x-4x)sin8xsin 4x=cot 4x-cot8x$$



    $$frac1sin 16x=dfracsin(16x-8x)sin16xsin 8x=cot 8x-cot16x$$






    share|cite|improve this answer














    $$sin(A-B)=sin A cos B-cos A sin B$$
    $$frac1sin 2x=dfracsin(2x-x)sin2xsin x=cot x-cot2x$$
    $$frac1sin 4x=dfracsin(4x-2x)sin4xsin 2x=cot 2x-cot4x$$
    $$frac1sin 8x=dfracsin(8x-4x)sin8xsin 4x=cot 4x-cot8x$$



    $$frac1sin 16x=dfracsin(16x-8x)sin16xsin 8x=cot 8x-cot16x$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Sep 15 at 0:13

























    answered Sep 6 at 12:04









    Deepesh Meena

    3,8912825




    3,8912825











    • Where did that $cot$ come from?
      – Busi
      Sep 6 at 12:23










    • apply $sin(a-b)$ formula
      – Deepesh Meena
      Sep 6 at 13:42










    • @DeepeshMeena, Have you noticed the link?
      – lab bhattacharjee
      Sep 7 at 7:44










    • which link can you provide me?
      – Deepesh Meena
      Sep 7 at 9:42
















    • Where did that $cot$ come from?
      – Busi
      Sep 6 at 12:23










    • apply $sin(a-b)$ formula
      – Deepesh Meena
      Sep 6 at 13:42










    • @DeepeshMeena, Have you noticed the link?
      – lab bhattacharjee
      Sep 7 at 7:44










    • which link can you provide me?
      – Deepesh Meena
      Sep 7 at 9:42















    Where did that $cot$ come from?
    – Busi
    Sep 6 at 12:23




    Where did that $cot$ come from?
    – Busi
    Sep 6 at 12:23












    apply $sin(a-b)$ formula
    – Deepesh Meena
    Sep 6 at 13:42




    apply $sin(a-b)$ formula
    – Deepesh Meena
    Sep 6 at 13:42












    @DeepeshMeena, Have you noticed the link?
    – lab bhattacharjee
    Sep 7 at 7:44




    @DeepeshMeena, Have you noticed the link?
    – lab bhattacharjee
    Sep 7 at 7:44












    which link can you provide me?
    – Deepesh Meena
    Sep 7 at 9:42




    which link can you provide me?
    – Deepesh Meena
    Sep 7 at 9:42










    up vote
    2
    down vote













    $$frac1sin2x=fracsin xsin 2x sin x=fracsin (2x-x)sin 2x sin x=fracsin 2x cos x - cos 2x sin xsin 2x sin x=cot x - cot2x$$



    $$frac1sin4x=cot 2x - cot4x$$



    $$frac1sin8x=cot 4x - cot8x$$



    $$frac1sin16x=cot 8x - cot16x$$



    $$frac1sin2x+frac1sin4x+frac1sin8x+frac1sin16x=cot x-cot 16x$$






    share|cite|improve this answer
























      up vote
      2
      down vote













      $$frac1sin2x=fracsin xsin 2x sin x=fracsin (2x-x)sin 2x sin x=fracsin 2x cos x - cos 2x sin xsin 2x sin x=cot x - cot2x$$



      $$frac1sin4x=cot 2x - cot4x$$



      $$frac1sin8x=cot 4x - cot8x$$



      $$frac1sin16x=cot 8x - cot16x$$



      $$frac1sin2x+frac1sin4x+frac1sin8x+frac1sin16x=cot x-cot 16x$$






      share|cite|improve this answer






















        up vote
        2
        down vote










        up vote
        2
        down vote









        $$frac1sin2x=fracsin xsin 2x sin x=fracsin (2x-x)sin 2x sin x=fracsin 2x cos x - cos 2x sin xsin 2x sin x=cot x - cot2x$$



        $$frac1sin4x=cot 2x - cot4x$$



        $$frac1sin8x=cot 4x - cot8x$$



        $$frac1sin16x=cot 8x - cot16x$$



        $$frac1sin2x+frac1sin4x+frac1sin8x+frac1sin16x=cot x-cot 16x$$






        share|cite|improve this answer












        $$frac1sin2x=fracsin xsin 2x sin x=fracsin (2x-x)sin 2x sin x=fracsin 2x cos x - cos 2x sin xsin 2x sin x=cot x - cot2x$$



        $$frac1sin4x=cot 2x - cot4x$$



        $$frac1sin8x=cot 4x - cot8x$$



        $$frac1sin16x=cot 8x - cot16x$$



        $$frac1sin2x+frac1sin4x+frac1sin8x+frac1sin16x=cot x-cot 16x$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 6 at 12:08









        Oldboy

        3,3201323




        3,3201323




















            up vote
            1
            down vote













            Hint : Notice that $frac1sin(2x))$ is a common factor of all terms of the sum.






            share|cite|improve this answer




















            • Then notice that $frac1cos2x$ is a common factor of everything that's left except the 1 in front. Then you'll end up with easy $1/a + 1/b$ problems.
              – PackSciences
              Sep 6 at 11:26














            up vote
            1
            down vote













            Hint : Notice that $frac1sin(2x))$ is a common factor of all terms of the sum.






            share|cite|improve this answer




















            • Then notice that $frac1cos2x$ is a common factor of everything that's left except the 1 in front. Then you'll end up with easy $1/a + 1/b$ problems.
              – PackSciences
              Sep 6 at 11:26












            up vote
            1
            down vote










            up vote
            1
            down vote









            Hint : Notice that $frac1sin(2x))$ is a common factor of all terms of the sum.






            share|cite|improve this answer












            Hint : Notice that $frac1sin(2x))$ is a common factor of all terms of the sum.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Sep 6 at 11:19









            PackSciences

            41616




            41616











            • Then notice that $frac1cos2x$ is a common factor of everything that's left except the 1 in front. Then you'll end up with easy $1/a + 1/b$ problems.
              – PackSciences
              Sep 6 at 11:26
















            • Then notice that $frac1cos2x$ is a common factor of everything that's left except the 1 in front. Then you'll end up with easy $1/a + 1/b$ problems.
              – PackSciences
              Sep 6 at 11:26















            Then notice that $frac1cos2x$ is a common factor of everything that's left except the 1 in front. Then you'll end up with easy $1/a + 1/b$ problems.
            – PackSciences
            Sep 6 at 11:26




            Then notice that $frac1cos2x$ is a common factor of everything that's left except the 1 in front. Then you'll end up with easy $1/a + 1/b$ problems.
            – PackSciences
            Sep 6 at 11:26

















             

            draft saved


            draft discarded















































             


            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2907344%2fevaluating-dfrac1-sin2x-dfrac1-sin4x-dfrac1-sin8x-d%23new-answer', 'question_page');

            );

            Post as a guest













































































            這個網誌中的熱門文章

            How to combine Bézier curves to a surface?

            Carbon dioxide

            Why am i infinitely getting the same tweet with the Twitter Search API?