Solve for common points in a circle and curve

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











up vote
2
down vote

favorite
1












[problem]



I got following equation of a circle..



$x^2+y^2=9 ..(1)$



and a equation of curve as follows..



$tan^4x + cot^4 x + 2 =4sin^2y....(2)$



I have to find points which satisfy both the equation..i.e points which are on the curve and inscribed inside or on the circle in common



[ My approach]



For solving following equation I first equate for trignometric function,So that i can obtain equation(2) in terms of x and y...Mathmetically I targeted for this.



$y=frac arcsin sqrt(tan^4x + cot^4 x + 2 ) 2$
(from equation (2))



I am trying to equate $sqrt(tan^4x + cot^4 x + 2 )$ in terms of $sin a$ where "a" is any polynomial expression ..



[HELP]



How should i do so..I cant break that in terms or sin function..It's going lengthy and complex . Or if any other less costly approach could be made










share|cite|improve this question



























    up vote
    2
    down vote

    favorite
    1












    [problem]



    I got following equation of a circle..



    $x^2+y^2=9 ..(1)$



    and a equation of curve as follows..



    $tan^4x + cot^4 x + 2 =4sin^2y....(2)$



    I have to find points which satisfy both the equation..i.e points which are on the curve and inscribed inside or on the circle in common



    [ My approach]



    For solving following equation I first equate for trignometric function,So that i can obtain equation(2) in terms of x and y...Mathmetically I targeted for this.



    $y=frac arcsin sqrt(tan^4x + cot^4 x + 2 ) 2$
    (from equation (2))



    I am trying to equate $sqrt(tan^4x + cot^4 x + 2 )$ in terms of $sin a$ where "a" is any polynomial expression ..



    [HELP]



    How should i do so..I cant break that in terms or sin function..It's going lengthy and complex . Or if any other less costly approach could be made










    share|cite|improve this question

























      up vote
      2
      down vote

      favorite
      1









      up vote
      2
      down vote

      favorite
      1






      1





      [problem]



      I got following equation of a circle..



      $x^2+y^2=9 ..(1)$



      and a equation of curve as follows..



      $tan^4x + cot^4 x + 2 =4sin^2y....(2)$



      I have to find points which satisfy both the equation..i.e points which are on the curve and inscribed inside or on the circle in common



      [ My approach]



      For solving following equation I first equate for trignometric function,So that i can obtain equation(2) in terms of x and y...Mathmetically I targeted for this.



      $y=frac arcsin sqrt(tan^4x + cot^4 x + 2 ) 2$
      (from equation (2))



      I am trying to equate $sqrt(tan^4x + cot^4 x + 2 )$ in terms of $sin a$ where "a" is any polynomial expression ..



      [HELP]



      How should i do so..I cant break that in terms or sin function..It's going lengthy and complex . Or if any other less costly approach could be made










      share|cite|improve this question















      [problem]



      I got following equation of a circle..



      $x^2+y^2=9 ..(1)$



      and a equation of curve as follows..



      $tan^4x + cot^4 x + 2 =4sin^2y....(2)$



      I have to find points which satisfy both the equation..i.e points which are on the curve and inscribed inside or on the circle in common



      [ My approach]



      For solving following equation I first equate for trignometric function,So that i can obtain equation(2) in terms of x and y...Mathmetically I targeted for this.



      $y=frac arcsin sqrt(tan^4x + cot^4 x + 2 ) 2$
      (from equation (2))



      I am trying to equate $sqrt(tan^4x + cot^4 x + 2 )$ in terms of $sin a$ where "a" is any polynomial expression ..



      [HELP]



      How should i do so..I cant break that in terms or sin function..It's going lengthy and complex . Or if any other less costly approach could be made







      trigonometry quadratics






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Sep 5 at 10:54









      Kevin

      5,138722




      5,138722










      asked Sep 5 at 10:52









      Shivanshu Raj

      184




      184




















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          1
          down vote













          Hint:



          For real $x,$



          $$dfractan^4x+cot^4x2gesqrttan^4xcot^4x=?$$



          $$impliessin^2yge1impliessin^2y=1$$ for real$y$






          share|cite|improve this answer




















          • did you left + sign in equation of inequilaty or multiplying tan cot??
            – Shivanshu Raj
            Sep 5 at 15:19










          • @Raj, I have used Inequality_of_arithmetic_and_geometric_means
            – lab bhattacharjee
            Sep 5 at 15:47










          • I still can't get till here....I am still confusing lot
            – Shivanshu Raj
            Sep 6 at 10:38










          • @ShivanshuRaj, $$(tan^2x-cot^2x)^2ge0$$ for real positive $tan^2x,cot^2x$. Following this logic, we need $$sin^2y=1impliescos y=0$$ right? So, $y=?$
            – lab bhattacharjee
            Sep 6 at 10:45










          • y became a linear equation
            – Shivanshu Raj
            Sep 7 at 2:42










          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%2f2906129%2fsolve-for-common-points-in-a-circle-and-curve%23new-answer', 'question_page');

          );

          Post as a guest






























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote













          Hint:



          For real $x,$



          $$dfractan^4x+cot^4x2gesqrttan^4xcot^4x=?$$



          $$impliessin^2yge1impliessin^2y=1$$ for real$y$






          share|cite|improve this answer




















          • did you left + sign in equation of inequilaty or multiplying tan cot??
            – Shivanshu Raj
            Sep 5 at 15:19










          • @Raj, I have used Inequality_of_arithmetic_and_geometric_means
            – lab bhattacharjee
            Sep 5 at 15:47










          • I still can't get till here....I am still confusing lot
            – Shivanshu Raj
            Sep 6 at 10:38










          • @ShivanshuRaj, $$(tan^2x-cot^2x)^2ge0$$ for real positive $tan^2x,cot^2x$. Following this logic, we need $$sin^2y=1impliescos y=0$$ right? So, $y=?$
            – lab bhattacharjee
            Sep 6 at 10:45










          • y became a linear equation
            – Shivanshu Raj
            Sep 7 at 2:42














          up vote
          1
          down vote













          Hint:



          For real $x,$



          $$dfractan^4x+cot^4x2gesqrttan^4xcot^4x=?$$



          $$impliessin^2yge1impliessin^2y=1$$ for real$y$






          share|cite|improve this answer




















          • did you left + sign in equation of inequilaty or multiplying tan cot??
            – Shivanshu Raj
            Sep 5 at 15:19










          • @Raj, I have used Inequality_of_arithmetic_and_geometric_means
            – lab bhattacharjee
            Sep 5 at 15:47










          • I still can't get till here....I am still confusing lot
            – Shivanshu Raj
            Sep 6 at 10:38










          • @ShivanshuRaj, $$(tan^2x-cot^2x)^2ge0$$ for real positive $tan^2x,cot^2x$. Following this logic, we need $$sin^2y=1impliescos y=0$$ right? So, $y=?$
            – lab bhattacharjee
            Sep 6 at 10:45










          • y became a linear equation
            – Shivanshu Raj
            Sep 7 at 2:42












          up vote
          1
          down vote










          up vote
          1
          down vote









          Hint:



          For real $x,$



          $$dfractan^4x+cot^4x2gesqrttan^4xcot^4x=?$$



          $$impliessin^2yge1impliessin^2y=1$$ for real$y$






          share|cite|improve this answer












          Hint:



          For real $x,$



          $$dfractan^4x+cot^4x2gesqrttan^4xcot^4x=?$$



          $$impliessin^2yge1impliessin^2y=1$$ for real$y$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 5 at 14:10









          lab bhattacharjee

          216k14153266




          216k14153266











          • did you left + sign in equation of inequilaty or multiplying tan cot??
            – Shivanshu Raj
            Sep 5 at 15:19










          • @Raj, I have used Inequality_of_arithmetic_and_geometric_means
            – lab bhattacharjee
            Sep 5 at 15:47










          • I still can't get till here....I am still confusing lot
            – Shivanshu Raj
            Sep 6 at 10:38










          • @ShivanshuRaj, $$(tan^2x-cot^2x)^2ge0$$ for real positive $tan^2x,cot^2x$. Following this logic, we need $$sin^2y=1impliescos y=0$$ right? So, $y=?$
            – lab bhattacharjee
            Sep 6 at 10:45










          • y became a linear equation
            – Shivanshu Raj
            Sep 7 at 2:42
















          • did you left + sign in equation of inequilaty or multiplying tan cot??
            – Shivanshu Raj
            Sep 5 at 15:19










          • @Raj, I have used Inequality_of_arithmetic_and_geometric_means
            – lab bhattacharjee
            Sep 5 at 15:47










          • I still can't get till here....I am still confusing lot
            – Shivanshu Raj
            Sep 6 at 10:38










          • @ShivanshuRaj, $$(tan^2x-cot^2x)^2ge0$$ for real positive $tan^2x,cot^2x$. Following this logic, we need $$sin^2y=1impliescos y=0$$ right? So, $y=?$
            – lab bhattacharjee
            Sep 6 at 10:45










          • y became a linear equation
            – Shivanshu Raj
            Sep 7 at 2:42















          did you left + sign in equation of inequilaty or multiplying tan cot??
          – Shivanshu Raj
          Sep 5 at 15:19




          did you left + sign in equation of inequilaty or multiplying tan cot??
          – Shivanshu Raj
          Sep 5 at 15:19












          @Raj, I have used Inequality_of_arithmetic_and_geometric_means
          – lab bhattacharjee
          Sep 5 at 15:47




          @Raj, I have used Inequality_of_arithmetic_and_geometric_means
          – lab bhattacharjee
          Sep 5 at 15:47












          I still can't get till here....I am still confusing lot
          – Shivanshu Raj
          Sep 6 at 10:38




          I still can't get till here....I am still confusing lot
          – Shivanshu Raj
          Sep 6 at 10:38












          @ShivanshuRaj, $$(tan^2x-cot^2x)^2ge0$$ for real positive $tan^2x,cot^2x$. Following this logic, we need $$sin^2y=1impliescos y=0$$ right? So, $y=?$
          – lab bhattacharjee
          Sep 6 at 10:45




          @ShivanshuRaj, $$(tan^2x-cot^2x)^2ge0$$ for real positive $tan^2x,cot^2x$. Following this logic, we need $$sin^2y=1impliescos y=0$$ right? So, $y=?$
          – lab bhattacharjee
          Sep 6 at 10:45












          y became a linear equation
          – Shivanshu Raj
          Sep 7 at 2:42




          y became a linear equation
          – Shivanshu Raj
          Sep 7 at 2:42

















           

          draft saved


          draft discarded















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2906129%2fsolve-for-common-points-in-a-circle-and-curve%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?

          Mutual Information Always Non-negative

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