Simplifying $(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x$












0












$begingroup$


$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x =$



Right Answer: $0$, but I could not solve this question. Help me please.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Welcome to MathSE. Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Feb 3 at 12:55










  • $begingroup$
    $cos^2x=1-sin^2x$
    $endgroup$
    – Lord Shark the Unknown
    Feb 3 at 12:56










  • $begingroup$
    $$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x = sin^4x (sin^2x - 1) + cos^4x (cos^2x - 1) + sin^2x cos^2x. $$ Can you take it from here?
    $endgroup$
    – HerrWarum
    Feb 3 at 12:57


















0












$begingroup$


$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x =$



Right Answer: $0$, but I could not solve this question. Help me please.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Welcome to MathSE. Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Feb 3 at 12:55










  • $begingroup$
    $cos^2x=1-sin^2x$
    $endgroup$
    – Lord Shark the Unknown
    Feb 3 at 12:56










  • $begingroup$
    $$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x = sin^4x (sin^2x - 1) + cos^4x (cos^2x - 1) + sin^2x cos^2x. $$ Can you take it from here?
    $endgroup$
    – HerrWarum
    Feb 3 at 12:57
















0












0








0





$begingroup$


$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x =$



Right Answer: $0$, but I could not solve this question. Help me please.










share|cite|improve this question











$endgroup$




$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x =$



Right Answer: $0$, but I could not solve this question. Help me please.







trigonometry weierstrass-factorization






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 3 at 12:52









N. F. Taussig

45.4k103358




45.4k103358










asked Feb 3 at 12:47









GeoerkeamGeoerkeam

33




33








  • 1




    $begingroup$
    Welcome to MathSE. Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Feb 3 at 12:55










  • $begingroup$
    $cos^2x=1-sin^2x$
    $endgroup$
    – Lord Shark the Unknown
    Feb 3 at 12:56










  • $begingroup$
    $$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x = sin^4x (sin^2x - 1) + cos^4x (cos^2x - 1) + sin^2x cos^2x. $$ Can you take it from here?
    $endgroup$
    – HerrWarum
    Feb 3 at 12:57
















  • 1




    $begingroup$
    Welcome to MathSE. Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Feb 3 at 12:55










  • $begingroup$
    $cos^2x=1-sin^2x$
    $endgroup$
    – Lord Shark the Unknown
    Feb 3 at 12:56










  • $begingroup$
    $$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x = sin^4x (sin^2x - 1) + cos^4x (cos^2x - 1) + sin^2x cos^2x. $$ Can you take it from here?
    $endgroup$
    – HerrWarum
    Feb 3 at 12:57










1




1




$begingroup$
Welcome to MathSE. Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Feb 3 at 12:55




$begingroup$
Welcome to MathSE. Please edit your question to show what you have attempted and explain where you are stuck so that you receive responses that address the specific difficulties you are encountering. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Feb 3 at 12:55












$begingroup$
$cos^2x=1-sin^2x$
$endgroup$
– Lord Shark the Unknown
Feb 3 at 12:56




$begingroup$
$cos^2x=1-sin^2x$
$endgroup$
– Lord Shark the Unknown
Feb 3 at 12:56












$begingroup$
$$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x = sin^4x (sin^2x - 1) + cos^4x (cos^2x - 1) + sin^2x cos^2x. $$ Can you take it from here?
$endgroup$
– HerrWarum
Feb 3 at 12:57






$begingroup$
$$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x = sin^4x (sin^2x - 1) + cos^4x (cos^2x - 1) + sin^2x cos^2x. $$ Can you take it from here?
$endgroup$
– HerrWarum
Feb 3 at 12:57












3 Answers
3






active

oldest

votes


















0












$begingroup$

$$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x $$
$$= sin^6x + cos^6x - sin^4x -cos^4x+sin^2xcos^2x$$
$$= cos^6x(tan^6x + 1 - tan^4xsec^2{x} -sec^2x+tan^2xsec^2x)$$
$$= cos^6x(tan^6x + 1 - tan^4x(1+tan^2x) -(1+tan^2x)+tan^2x(1+tan^2x))$$
$$= cos^6x(tan^6x + 1 - tan^4x-tan^6x -1-tan^2x+tan^2x+tan^4x)$$
$$= cos^6x(0)$$
$$= 0$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    With $c:=cos^2 x,,s:=sin^2 x$, it suffices to note $$c^3+s^3-c^2-s^2+cs=(c+s-1)(c^2-cs+s^2).$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Could you explain how you would come up with that expression assuming you had never seen it before? Is there an algebra trick I'm missing out?
      $endgroup$
      – Elie Louis
      Feb 4 at 12:51












    • $begingroup$
      @ElieLouis It's natural to divide $c+s-1$ into the original expression to see whether the remainder is $0$. In this case, the factorisation of $c^3+s^3$ as $(c+s)(c^2-cs+s^2)$ proves helpful.
      $endgroup$
      – J.G.
      Feb 4 at 13:00



















    0












    $begingroup$

    hint: $sin^2 x + cos^2 x = 1$; if you square that, you get
    $$
    sin^4 x + 2sin^2 x cos^2 x + cos^4 x = 1,
    $$

    which you can turn into
    $$
    sin^4 x + cos^4 x = 1 - 2sin^2 x cos^2 x
    $$

    and that lets you get rid of the 4th powers in the left hand side.



    Can you run with that idea?






    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%2f3098526%2fsimplifying-sin6x-cos6x-sin4x-cos4x-sin2x-cos2x%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









      0












      $begingroup$

      $$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x $$
      $$= sin^6x + cos^6x - sin^4x -cos^4x+sin^2xcos^2x$$
      $$= cos^6x(tan^6x + 1 - tan^4xsec^2{x} -sec^2x+tan^2xsec^2x)$$
      $$= cos^6x(tan^6x + 1 - tan^4x(1+tan^2x) -(1+tan^2x)+tan^2x(1+tan^2x))$$
      $$= cos^6x(tan^6x + 1 - tan^4x-tan^6x -1-tan^2x+tan^2x+tan^4x)$$
      $$= cos^6x(0)$$
      $$= 0$$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        $$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x $$
        $$= sin^6x + cos^6x - sin^4x -cos^4x+sin^2xcos^2x$$
        $$= cos^6x(tan^6x + 1 - tan^4xsec^2{x} -sec^2x+tan^2xsec^2x)$$
        $$= cos^6x(tan^6x + 1 - tan^4x(1+tan^2x) -(1+tan^2x)+tan^2x(1+tan^2x))$$
        $$= cos^6x(tan^6x + 1 - tan^4x-tan^6x -1-tan^2x+tan^2x+tan^4x)$$
        $$= cos^6x(0)$$
        $$= 0$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          $$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x $$
          $$= sin^6x + cos^6x - sin^4x -cos^4x+sin^2xcos^2x$$
          $$= cos^6x(tan^6x + 1 - tan^4xsec^2{x} -sec^2x+tan^2xsec^2x)$$
          $$= cos^6x(tan^6x + 1 - tan^4x(1+tan^2x) -(1+tan^2x)+tan^2x(1+tan^2x))$$
          $$= cos^6x(tan^6x + 1 - tan^4x-tan^6x -1-tan^2x+tan^2x+tan^4x)$$
          $$= cos^6x(0)$$
          $$= 0$$






          share|cite|improve this answer









          $endgroup$



          $$(sin^6x + cos^6x) - (sin^4x + cos^4x) + sin^2xcos^2x $$
          $$= sin^6x + cos^6x - sin^4x -cos^4x+sin^2xcos^2x$$
          $$= cos^6x(tan^6x + 1 - tan^4xsec^2{x} -sec^2x+tan^2xsec^2x)$$
          $$= cos^6x(tan^6x + 1 - tan^4x(1+tan^2x) -(1+tan^2x)+tan^2x(1+tan^2x))$$
          $$= cos^6x(tan^6x + 1 - tan^4x-tan^6x -1-tan^2x+tan^2x+tan^4x)$$
          $$= cos^6x(0)$$
          $$= 0$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 3 at 12:58









          Peter ForemanPeter Foreman

          7,5331320




          7,5331320























              1












              $begingroup$

              With $c:=cos^2 x,,s:=sin^2 x$, it suffices to note $$c^3+s^3-c^2-s^2+cs=(c+s-1)(c^2-cs+s^2).$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Could you explain how you would come up with that expression assuming you had never seen it before? Is there an algebra trick I'm missing out?
                $endgroup$
                – Elie Louis
                Feb 4 at 12:51












              • $begingroup$
                @ElieLouis It's natural to divide $c+s-1$ into the original expression to see whether the remainder is $0$. In this case, the factorisation of $c^3+s^3$ as $(c+s)(c^2-cs+s^2)$ proves helpful.
                $endgroup$
                – J.G.
                Feb 4 at 13:00
















              1












              $begingroup$

              With $c:=cos^2 x,,s:=sin^2 x$, it suffices to note $$c^3+s^3-c^2-s^2+cs=(c+s-1)(c^2-cs+s^2).$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Could you explain how you would come up with that expression assuming you had never seen it before? Is there an algebra trick I'm missing out?
                $endgroup$
                – Elie Louis
                Feb 4 at 12:51












              • $begingroup$
                @ElieLouis It's natural to divide $c+s-1$ into the original expression to see whether the remainder is $0$. In this case, the factorisation of $c^3+s^3$ as $(c+s)(c^2-cs+s^2)$ proves helpful.
                $endgroup$
                – J.G.
                Feb 4 at 13:00














              1












              1








              1





              $begingroup$

              With $c:=cos^2 x,,s:=sin^2 x$, it suffices to note $$c^3+s^3-c^2-s^2+cs=(c+s-1)(c^2-cs+s^2).$$






              share|cite|improve this answer









              $endgroup$



              With $c:=cos^2 x,,s:=sin^2 x$, it suffices to note $$c^3+s^3-c^2-s^2+cs=(c+s-1)(c^2-cs+s^2).$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Feb 3 at 13:11









              J.G.J.G.

              33.5k23252




              33.5k23252












              • $begingroup$
                Could you explain how you would come up with that expression assuming you had never seen it before? Is there an algebra trick I'm missing out?
                $endgroup$
                – Elie Louis
                Feb 4 at 12:51












              • $begingroup$
                @ElieLouis It's natural to divide $c+s-1$ into the original expression to see whether the remainder is $0$. In this case, the factorisation of $c^3+s^3$ as $(c+s)(c^2-cs+s^2)$ proves helpful.
                $endgroup$
                – J.G.
                Feb 4 at 13:00


















              • $begingroup$
                Could you explain how you would come up with that expression assuming you had never seen it before? Is there an algebra trick I'm missing out?
                $endgroup$
                – Elie Louis
                Feb 4 at 12:51












              • $begingroup$
                @ElieLouis It's natural to divide $c+s-1$ into the original expression to see whether the remainder is $0$. In this case, the factorisation of $c^3+s^3$ as $(c+s)(c^2-cs+s^2)$ proves helpful.
                $endgroup$
                – J.G.
                Feb 4 at 13:00
















              $begingroup$
              Could you explain how you would come up with that expression assuming you had never seen it before? Is there an algebra trick I'm missing out?
              $endgroup$
              – Elie Louis
              Feb 4 at 12:51






              $begingroup$
              Could you explain how you would come up with that expression assuming you had never seen it before? Is there an algebra trick I'm missing out?
              $endgroup$
              – Elie Louis
              Feb 4 at 12:51














              $begingroup$
              @ElieLouis It's natural to divide $c+s-1$ into the original expression to see whether the remainder is $0$. In this case, the factorisation of $c^3+s^3$ as $(c+s)(c^2-cs+s^2)$ proves helpful.
              $endgroup$
              – J.G.
              Feb 4 at 13:00




              $begingroup$
              @ElieLouis It's natural to divide $c+s-1$ into the original expression to see whether the remainder is $0$. In this case, the factorisation of $c^3+s^3$ as $(c+s)(c^2-cs+s^2)$ proves helpful.
              $endgroup$
              – J.G.
              Feb 4 at 13:00











              0












              $begingroup$

              hint: $sin^2 x + cos^2 x = 1$; if you square that, you get
              $$
              sin^4 x + 2sin^2 x cos^2 x + cos^4 x = 1,
              $$

              which you can turn into
              $$
              sin^4 x + cos^4 x = 1 - 2sin^2 x cos^2 x
              $$

              and that lets you get rid of the 4th powers in the left hand side.



              Can you run with that idea?






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                hint: $sin^2 x + cos^2 x = 1$; if you square that, you get
                $$
                sin^4 x + 2sin^2 x cos^2 x + cos^4 x = 1,
                $$

                which you can turn into
                $$
                sin^4 x + cos^4 x = 1 - 2sin^2 x cos^2 x
                $$

                and that lets you get rid of the 4th powers in the left hand side.



                Can you run with that idea?






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  hint: $sin^2 x + cos^2 x = 1$; if you square that, you get
                  $$
                  sin^4 x + 2sin^2 x cos^2 x + cos^4 x = 1,
                  $$

                  which you can turn into
                  $$
                  sin^4 x + cos^4 x = 1 - 2sin^2 x cos^2 x
                  $$

                  and that lets you get rid of the 4th powers in the left hand side.



                  Can you run with that idea?






                  share|cite|improve this answer









                  $endgroup$



                  hint: $sin^2 x + cos^2 x = 1$; if you square that, you get
                  $$
                  sin^4 x + 2sin^2 x cos^2 x + cos^4 x = 1,
                  $$

                  which you can turn into
                  $$
                  sin^4 x + cos^4 x = 1 - 2sin^2 x cos^2 x
                  $$

                  and that lets you get rid of the 4th powers in the left hand side.



                  Can you run with that idea?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Feb 3 at 12:57









                  John HughesJohn Hughes

                  65.5k24293




                  65.5k24293






























                      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%2f3098526%2fsimplifying-sin6x-cos6x-sin4x-cos4x-sin2x-cos2x%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

                      'app-layout' is not a known element: how to share Component with different Modules

                      android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

                      WPF add header to Image with URL pettitions [duplicate]