Evaluate and Simplify $cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$












1














I am trying to evaluate and simplify $cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$.



I am getting $frac{11}{10}$ but the answer is $frac{3-4sqrt{3}}{10}$



My Process:



$cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$



$cos[cos^{-1}(frac{3}{5})] + cos(frac{pi}{3})$



$(frac{2}{2}) cdot frac{3}{5} + frac{1}{2} cdot (frac{5}{5})$



$frac{6}{10} + frac{5}{10}$



$frac{11}{10}$






algebra-precalculus trigonometry







share|cite|improve this question


















  • 1




    Your $frac{11}{10}$ can't possibly be the cosine of anything!
    – TonyK
    Nov 22 '18 at 18:05






  • 1




    @TonyK Not for real ;^)
    – Michael Hoppe
    Nov 22 '18 at 19:28
















1














I am trying to evaluate and simplify $cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$.



I am getting $frac{11}{10}$ but the answer is $frac{3-4sqrt{3}}{10}$



My Process:



$cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$



$cos[cos^{-1}(frac{3}{5})] + cos(frac{pi}{3})$



$(frac{2}{2}) cdot frac{3}{5} + frac{1}{2} cdot (frac{5}{5})$



$frac{6}{10} + frac{5}{10}$



$frac{11}{10}$






algebra-precalculus trigonometry







share|cite|improve this question


















  • 1




    Your $frac{11}{10}$ can't possibly be the cosine of anything!
    – TonyK
    Nov 22 '18 at 18:05






  • 1




    @TonyK Not for real ;^)
    – Michael Hoppe
    Nov 22 '18 at 19:28














1












1








1







I am trying to evaluate and simplify $cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$.



I am getting $frac{11}{10}$ but the answer is $frac{3-4sqrt{3}}{10}$



My Process:



$cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$



$cos[cos^{-1}(frac{3}{5})] + cos(frac{pi}{3})$



$(frac{2}{2}) cdot frac{3}{5} + frac{1}{2} cdot (frac{5}{5})$



$frac{6}{10} + frac{5}{10}$



$frac{11}{10}$






algebra-precalculus trigonometry







share|cite|improve this question













I am trying to evaluate and simplify $cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$.



I am getting $frac{11}{10}$ but the answer is $frac{3-4sqrt{3}}{10}$



My Process:



$cos[cos^{-1}(frac{3}{5}) + frac{pi}{3}]$



$cos[cos^{-1}(frac{3}{5})] + cos(frac{pi}{3})$



$(frac{2}{2}) cdot frac{3}{5} + frac{1}{2} cdot (frac{5}{5})$



$frac{6}{10} + frac{5}{10}$



$frac{11}{10}$






algebra-precalculus trigonometry




algebra-precalculus trigonometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 22 '18 at 17:56









LuminousNutriaLuminousNutria

1709




1709








  • 1




    Your $frac{11}{10}$ can't possibly be the cosine of anything!
    – TonyK
    Nov 22 '18 at 18:05






  • 1




    @TonyK Not for real ;^)
    – Michael Hoppe
    Nov 22 '18 at 19:28














  • 1




    Your $frac{11}{10}$ can't possibly be the cosine of anything!
    – TonyK
    Nov 22 '18 at 18:05






  • 1




    @TonyK Not for real ;^)
    – Michael Hoppe
    Nov 22 '18 at 19:28








1




1




Your $frac{11}{10}$ can't possibly be the cosine of anything!
– TonyK
Nov 22 '18 at 18:05




Your $frac{11}{10}$ can't possibly be the cosine of anything!
– TonyK
Nov 22 '18 at 18:05




1




1




@TonyK Not for real ;^)
– Michael Hoppe
Nov 22 '18 at 19:28




@TonyK Not for real ;^)
– Michael Hoppe
Nov 22 '18 at 19:28










5 Answers
5






active

oldest

votes


















3














You cannot separate out the $cos$ function as you have done in step two.



You can remember this identity.



$$cos(A+B)=cos(A)cos(B)-sin(A)sin(B)$$



Here $arccos(frac{3}{5})$ is an angle ( i.e., approximately $53$ degrees)



Using thus result you should get the desired answer.






share|cite|improve this answer



















  • 1




    (+1) ... for discussing the error in the OP
    – Mark Viola
    Nov 22 '18 at 18:05










  • @MarkViola thanks for your upvote.
    – Akash Roy
    Nov 22 '18 at 18:06










  • Thanks once again for your valuable edits @Mark, will bear these in mind.
    – Akash Roy
    Nov 22 '18 at 18:08










  • Thank you! I get it now. I just have to simplify $cos[cos^{-1}(frac{3}{5})] cdot cos(frac{3}{5}) - sin[cos^{-1}(frac{3}{5})] cdot sin(frac{pi}{3})$
    – LuminousNutria
    Nov 22 '18 at 18:09












  • @Luminous you can accept my answer if you understood.
    – Akash Roy
    Nov 22 '18 at 18:10



















1














The $cos $ function doesn't behaves in linear way. In the 2nd step in your calculation you have treated it like a linear function. Just use the formula: $$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$$.






share|cite|improve this answer

















  • 1




    (+1) for mentioning the error in the OP
    – Mark Viola
    Nov 22 '18 at 18:10



















0














$$
cos(a+b)=cos acos b-sin asin b
$$
and hence
$$
cosleft[cos^{-1}(frac{3}{5}) + frac{pi}{3}right]=cosleft(cos^{-1}left(frac{3}{5}right)right)cosleft(frac{pi}{3}right)-sinleft(cos^{-1}left(frac{3}{5}right)right)sinleft(frac{pi}{3}right)\=frac{3}{5}cdotfrac{1}{2}-sqrt{1-frac{3^2}{5^2}}cdotfrac{sqrt{3}}{2}=cdots
$$






share|cite|improve this answer





























    0














    By compound-angle formula,
    begin{eqnarray*}
    cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & cosleft(cos^{-1}(frac{3}{5})right)cosfrac{pi}{3}-sinleft(cos^{-1}(frac{3}{5})right)sin(frac{pi}{3})\
    & = & frac{3}{5}cdotfrac{1}{2}-sinleft(cos^{-1}(frac{3}{5})right)cdotfrac{sqrt{3}}{2}.
    end{eqnarray*}



    To evaluate $sinleft(cos^{-1}(frac{3}{5})right)$, we let $theta=cos^{-1}(frac{3}{5})$.
    Then $costheta=frac{3}{5}$. Recall that $sin^{2}theta+cos^{2}theta=1$
    and observe that $0<theta<frac{pi}{2}$. Therefore $sintheta>0$ and
    $sintheta$ is given by $sintheta=sqrt{1-cos^{2}theta}=frac{4}{5}$.
    It follows that
    begin{eqnarray*}
    cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & frac{3}{5}cdotfrac{1}{2}-frac{4}{5}cdotfrac{sqrt{3}}{2}\
    & = & frac{3}{10}-frac{2sqrt{3}}{5}
    end{eqnarray*}






    share|cite|improve this answer





























      -1














      Cos^-1(3/5) is 37°. So we get cos(37°+60°) which is equal to -0.39 same as the term you have mentioned as the answer.






      share|cite|improve this answer





















      • The first sentence in this answer is false: $arccosfrac 35$ is not $37^circ$.
        – Saucy O'Path
        Nov 24 '18 at 10:09












      • ... and $cos 97^circ$ is not $-0.39$.
        – Saucy O'Path
        Nov 24 '18 at 11:17











      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%2f3009421%2fevaluate-and-simplify-cos-cos-1-frac35-frac-pi3%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3














      You cannot separate out the $cos$ function as you have done in step two.



      You can remember this identity.



      $$cos(A+B)=cos(A)cos(B)-sin(A)sin(B)$$



      Here $arccos(frac{3}{5})$ is an angle ( i.e., approximately $53$ degrees)



      Using thus result you should get the desired answer.






      share|cite|improve this answer



















      • 1




        (+1) ... for discussing the error in the OP
        – Mark Viola
        Nov 22 '18 at 18:05










      • @MarkViola thanks for your upvote.
        – Akash Roy
        Nov 22 '18 at 18:06










      • Thanks once again for your valuable edits @Mark, will bear these in mind.
        – Akash Roy
        Nov 22 '18 at 18:08










      • Thank you! I get it now. I just have to simplify $cos[cos^{-1}(frac{3}{5})] cdot cos(frac{3}{5}) - sin[cos^{-1}(frac{3}{5})] cdot sin(frac{pi}{3})$
        – LuminousNutria
        Nov 22 '18 at 18:09












      • @Luminous you can accept my answer if you understood.
        – Akash Roy
        Nov 22 '18 at 18:10
















      3














      You cannot separate out the $cos$ function as you have done in step two.



      You can remember this identity.



      $$cos(A+B)=cos(A)cos(B)-sin(A)sin(B)$$



      Here $arccos(frac{3}{5})$ is an angle ( i.e., approximately $53$ degrees)



      Using thus result you should get the desired answer.






      share|cite|improve this answer



















      • 1




        (+1) ... for discussing the error in the OP
        – Mark Viola
        Nov 22 '18 at 18:05










      • @MarkViola thanks for your upvote.
        – Akash Roy
        Nov 22 '18 at 18:06










      • Thanks once again for your valuable edits @Mark, will bear these in mind.
        – Akash Roy
        Nov 22 '18 at 18:08










      • Thank you! I get it now. I just have to simplify $cos[cos^{-1}(frac{3}{5})] cdot cos(frac{3}{5}) - sin[cos^{-1}(frac{3}{5})] cdot sin(frac{pi}{3})$
        – LuminousNutria
        Nov 22 '18 at 18:09












      • @Luminous you can accept my answer if you understood.
        – Akash Roy
        Nov 22 '18 at 18:10














      3












      3








      3






      You cannot separate out the $cos$ function as you have done in step two.



      You can remember this identity.



      $$cos(A+B)=cos(A)cos(B)-sin(A)sin(B)$$



      Here $arccos(frac{3}{5})$ is an angle ( i.e., approximately $53$ degrees)



      Using thus result you should get the desired answer.






      share|cite|improve this answer














      You cannot separate out the $cos$ function as you have done in step two.



      You can remember this identity.



      $$cos(A+B)=cos(A)cos(B)-sin(A)sin(B)$$



      Here $arccos(frac{3}{5})$ is an angle ( i.e., approximately $53$ degrees)



      Using thus result you should get the desired answer.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Nov 22 '18 at 18:07









      Mark Viola

      130k1274170




      130k1274170










      answered Nov 22 '18 at 18:00









      Akash RoyAkash Roy

      1




      1








      • 1




        (+1) ... for discussing the error in the OP
        – Mark Viola
        Nov 22 '18 at 18:05










      • @MarkViola thanks for your upvote.
        – Akash Roy
        Nov 22 '18 at 18:06










      • Thanks once again for your valuable edits @Mark, will bear these in mind.
        – Akash Roy
        Nov 22 '18 at 18:08










      • Thank you! I get it now. I just have to simplify $cos[cos^{-1}(frac{3}{5})] cdot cos(frac{3}{5}) - sin[cos^{-1}(frac{3}{5})] cdot sin(frac{pi}{3})$
        – LuminousNutria
        Nov 22 '18 at 18:09












      • @Luminous you can accept my answer if you understood.
        – Akash Roy
        Nov 22 '18 at 18:10














      • 1




        (+1) ... for discussing the error in the OP
        – Mark Viola
        Nov 22 '18 at 18:05










      • @MarkViola thanks for your upvote.
        – Akash Roy
        Nov 22 '18 at 18:06










      • Thanks once again for your valuable edits @Mark, will bear these in mind.
        – Akash Roy
        Nov 22 '18 at 18:08










      • Thank you! I get it now. I just have to simplify $cos[cos^{-1}(frac{3}{5})] cdot cos(frac{3}{5}) - sin[cos^{-1}(frac{3}{5})] cdot sin(frac{pi}{3})$
        – LuminousNutria
        Nov 22 '18 at 18:09












      • @Luminous you can accept my answer if you understood.
        – Akash Roy
        Nov 22 '18 at 18:10








      1




      1




      (+1) ... for discussing the error in the OP
      – Mark Viola
      Nov 22 '18 at 18:05




      (+1) ... for discussing the error in the OP
      – Mark Viola
      Nov 22 '18 at 18:05












      @MarkViola thanks for your upvote.
      – Akash Roy
      Nov 22 '18 at 18:06




      @MarkViola thanks for your upvote.
      – Akash Roy
      Nov 22 '18 at 18:06












      Thanks once again for your valuable edits @Mark, will bear these in mind.
      – Akash Roy
      Nov 22 '18 at 18:08




      Thanks once again for your valuable edits @Mark, will bear these in mind.
      – Akash Roy
      Nov 22 '18 at 18:08












      Thank you! I get it now. I just have to simplify $cos[cos^{-1}(frac{3}{5})] cdot cos(frac{3}{5}) - sin[cos^{-1}(frac{3}{5})] cdot sin(frac{pi}{3})$
      – LuminousNutria
      Nov 22 '18 at 18:09






      Thank you! I get it now. I just have to simplify $cos[cos^{-1}(frac{3}{5})] cdot cos(frac{3}{5}) - sin[cos^{-1}(frac{3}{5})] cdot sin(frac{pi}{3})$
      – LuminousNutria
      Nov 22 '18 at 18:09














      @Luminous you can accept my answer if you understood.
      – Akash Roy
      Nov 22 '18 at 18:10




      @Luminous you can accept my answer if you understood.
      – Akash Roy
      Nov 22 '18 at 18:10











      1














      The $cos $ function doesn't behaves in linear way. In the 2nd step in your calculation you have treated it like a linear function. Just use the formula: $$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$$.






      share|cite|improve this answer

















      • 1




        (+1) for mentioning the error in the OP
        – Mark Viola
        Nov 22 '18 at 18:10
















      1














      The $cos $ function doesn't behaves in linear way. In the 2nd step in your calculation you have treated it like a linear function. Just use the formula: $$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$$.






      share|cite|improve this answer

















      • 1




        (+1) for mentioning the error in the OP
        – Mark Viola
        Nov 22 '18 at 18:10














      1












      1








      1






      The $cos $ function doesn't behaves in linear way. In the 2nd step in your calculation you have treated it like a linear function. Just use the formula: $$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$$.






      share|cite|improve this answer












      The $cos $ function doesn't behaves in linear way. In the 2nd step in your calculation you have treated it like a linear function. Just use the formula: $$cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Nov 22 '18 at 17:59









      tarit goswamitarit goswami

      1,7381421




      1,7381421








      • 1




        (+1) for mentioning the error in the OP
        – Mark Viola
        Nov 22 '18 at 18:10














      • 1




        (+1) for mentioning the error in the OP
        – Mark Viola
        Nov 22 '18 at 18:10








      1




      1




      (+1) for mentioning the error in the OP
      – Mark Viola
      Nov 22 '18 at 18:10




      (+1) for mentioning the error in the OP
      – Mark Viola
      Nov 22 '18 at 18:10











      0














      $$
      cos(a+b)=cos acos b-sin asin b
      $$
      and hence
      $$
      cosleft[cos^{-1}(frac{3}{5}) + frac{pi}{3}right]=cosleft(cos^{-1}left(frac{3}{5}right)right)cosleft(frac{pi}{3}right)-sinleft(cos^{-1}left(frac{3}{5}right)right)sinleft(frac{pi}{3}right)\=frac{3}{5}cdotfrac{1}{2}-sqrt{1-frac{3^2}{5^2}}cdotfrac{sqrt{3}}{2}=cdots
      $$






      share|cite|improve this answer


























        0














        $$
        cos(a+b)=cos acos b-sin asin b
        $$
        and hence
        $$
        cosleft[cos^{-1}(frac{3}{5}) + frac{pi}{3}right]=cosleft(cos^{-1}left(frac{3}{5}right)right)cosleft(frac{pi}{3}right)-sinleft(cos^{-1}left(frac{3}{5}right)right)sinleft(frac{pi}{3}right)\=frac{3}{5}cdotfrac{1}{2}-sqrt{1-frac{3^2}{5^2}}cdotfrac{sqrt{3}}{2}=cdots
        $$






        share|cite|improve this answer
























          0












          0








          0






          $$
          cos(a+b)=cos acos b-sin asin b
          $$
          and hence
          $$
          cosleft[cos^{-1}(frac{3}{5}) + frac{pi}{3}right]=cosleft(cos^{-1}left(frac{3}{5}right)right)cosleft(frac{pi}{3}right)-sinleft(cos^{-1}left(frac{3}{5}right)right)sinleft(frac{pi}{3}right)\=frac{3}{5}cdotfrac{1}{2}-sqrt{1-frac{3^2}{5^2}}cdotfrac{sqrt{3}}{2}=cdots
          $$






          share|cite|improve this answer












          $$
          cos(a+b)=cos acos b-sin asin b
          $$
          and hence
          $$
          cosleft[cos^{-1}(frac{3}{5}) + frac{pi}{3}right]=cosleft(cos^{-1}left(frac{3}{5}right)right)cosleft(frac{pi}{3}right)-sinleft(cos^{-1}left(frac{3}{5}right)right)sinleft(frac{pi}{3}right)\=frac{3}{5}cdotfrac{1}{2}-sqrt{1-frac{3^2}{5^2}}cdotfrac{sqrt{3}}{2}=cdots
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 '18 at 18:03









          Yiorgos S. SmyrlisYiorgos S. Smyrlis

          62.8k1383163




          62.8k1383163























              0














              By compound-angle formula,
              begin{eqnarray*}
              cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & cosleft(cos^{-1}(frac{3}{5})right)cosfrac{pi}{3}-sinleft(cos^{-1}(frac{3}{5})right)sin(frac{pi}{3})\
              & = & frac{3}{5}cdotfrac{1}{2}-sinleft(cos^{-1}(frac{3}{5})right)cdotfrac{sqrt{3}}{2}.
              end{eqnarray*}



              To evaluate $sinleft(cos^{-1}(frac{3}{5})right)$, we let $theta=cos^{-1}(frac{3}{5})$.
              Then $costheta=frac{3}{5}$. Recall that $sin^{2}theta+cos^{2}theta=1$
              and observe that $0<theta<frac{pi}{2}$. Therefore $sintheta>0$ and
              $sintheta$ is given by $sintheta=sqrt{1-cos^{2}theta}=frac{4}{5}$.
              It follows that
              begin{eqnarray*}
              cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & frac{3}{5}cdotfrac{1}{2}-frac{4}{5}cdotfrac{sqrt{3}}{2}\
              & = & frac{3}{10}-frac{2sqrt{3}}{5}
              end{eqnarray*}






              share|cite|improve this answer


























                0














                By compound-angle formula,
                begin{eqnarray*}
                cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & cosleft(cos^{-1}(frac{3}{5})right)cosfrac{pi}{3}-sinleft(cos^{-1}(frac{3}{5})right)sin(frac{pi}{3})\
                & = & frac{3}{5}cdotfrac{1}{2}-sinleft(cos^{-1}(frac{3}{5})right)cdotfrac{sqrt{3}}{2}.
                end{eqnarray*}



                To evaluate $sinleft(cos^{-1}(frac{3}{5})right)$, we let $theta=cos^{-1}(frac{3}{5})$.
                Then $costheta=frac{3}{5}$. Recall that $sin^{2}theta+cos^{2}theta=1$
                and observe that $0<theta<frac{pi}{2}$. Therefore $sintheta>0$ and
                $sintheta$ is given by $sintheta=sqrt{1-cos^{2}theta}=frac{4}{5}$.
                It follows that
                begin{eqnarray*}
                cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & frac{3}{5}cdotfrac{1}{2}-frac{4}{5}cdotfrac{sqrt{3}}{2}\
                & = & frac{3}{10}-frac{2sqrt{3}}{5}
                end{eqnarray*}






                share|cite|improve this answer
























                  0












                  0








                  0






                  By compound-angle formula,
                  begin{eqnarray*}
                  cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & cosleft(cos^{-1}(frac{3}{5})right)cosfrac{pi}{3}-sinleft(cos^{-1}(frac{3}{5})right)sin(frac{pi}{3})\
                  & = & frac{3}{5}cdotfrac{1}{2}-sinleft(cos^{-1}(frac{3}{5})right)cdotfrac{sqrt{3}}{2}.
                  end{eqnarray*}



                  To evaluate $sinleft(cos^{-1}(frac{3}{5})right)$, we let $theta=cos^{-1}(frac{3}{5})$.
                  Then $costheta=frac{3}{5}$. Recall that $sin^{2}theta+cos^{2}theta=1$
                  and observe that $0<theta<frac{pi}{2}$. Therefore $sintheta>0$ and
                  $sintheta$ is given by $sintheta=sqrt{1-cos^{2}theta}=frac{4}{5}$.
                  It follows that
                  begin{eqnarray*}
                  cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & frac{3}{5}cdotfrac{1}{2}-frac{4}{5}cdotfrac{sqrt{3}}{2}\
                  & = & frac{3}{10}-frac{2sqrt{3}}{5}
                  end{eqnarray*}






                  share|cite|improve this answer












                  By compound-angle formula,
                  begin{eqnarray*}
                  cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & cosleft(cos^{-1}(frac{3}{5})right)cosfrac{pi}{3}-sinleft(cos^{-1}(frac{3}{5})right)sin(frac{pi}{3})\
                  & = & frac{3}{5}cdotfrac{1}{2}-sinleft(cos^{-1}(frac{3}{5})right)cdotfrac{sqrt{3}}{2}.
                  end{eqnarray*}



                  To evaluate $sinleft(cos^{-1}(frac{3}{5})right)$, we let $theta=cos^{-1}(frac{3}{5})$.
                  Then $costheta=frac{3}{5}$. Recall that $sin^{2}theta+cos^{2}theta=1$
                  and observe that $0<theta<frac{pi}{2}$. Therefore $sintheta>0$ and
                  $sintheta$ is given by $sintheta=sqrt{1-cos^{2}theta}=frac{4}{5}$.
                  It follows that
                  begin{eqnarray*}
                  cos(cos^{-1}(frac{3}{5})+frac{pi}{3}) & = & frac{3}{5}cdotfrac{1}{2}-frac{4}{5}cdotfrac{sqrt{3}}{2}\
                  & = & frac{3}{10}-frac{2sqrt{3}}{5}
                  end{eqnarray*}







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 22 '18 at 18:05









                  Danny Pak-Keung ChanDanny Pak-Keung Chan

                  2,25038




                  2,25038























                      -1














                      Cos^-1(3/5) is 37°. So we get cos(37°+60°) which is equal to -0.39 same as the term you have mentioned as the answer.






                      share|cite|improve this answer





















                      • The first sentence in this answer is false: $arccosfrac 35$ is not $37^circ$.
                        – Saucy O'Path
                        Nov 24 '18 at 10:09












                      • ... and $cos 97^circ$ is not $-0.39$.
                        – Saucy O'Path
                        Nov 24 '18 at 11:17
















                      -1














                      Cos^-1(3/5) is 37°. So we get cos(37°+60°) which is equal to -0.39 same as the term you have mentioned as the answer.






                      share|cite|improve this answer





















                      • The first sentence in this answer is false: $arccosfrac 35$ is not $37^circ$.
                        – Saucy O'Path
                        Nov 24 '18 at 10:09












                      • ... and $cos 97^circ$ is not $-0.39$.
                        – Saucy O'Path
                        Nov 24 '18 at 11:17














                      -1












                      -1








                      -1






                      Cos^-1(3/5) is 37°. So we get cos(37°+60°) which is equal to -0.39 same as the term you have mentioned as the answer.






                      share|cite|improve this answer












                      Cos^-1(3/5) is 37°. So we get cos(37°+60°) which is equal to -0.39 same as the term you have mentioned as the answer.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 24 '18 at 9:59









                      Siddharth DhimanSiddharth Dhiman

                      1




                      1












                      • The first sentence in this answer is false: $arccosfrac 35$ is not $37^circ$.
                        – Saucy O'Path
                        Nov 24 '18 at 10:09












                      • ... and $cos 97^circ$ is not $-0.39$.
                        – Saucy O'Path
                        Nov 24 '18 at 11:17


















                      • The first sentence in this answer is false: $arccosfrac 35$ is not $37^circ$.
                        – Saucy O'Path
                        Nov 24 '18 at 10:09












                      • ... and $cos 97^circ$ is not $-0.39$.
                        – Saucy O'Path
                        Nov 24 '18 at 11:17
















                      The first sentence in this answer is false: $arccosfrac 35$ is not $37^circ$.
                      – Saucy O'Path
                      Nov 24 '18 at 10:09






                      The first sentence in this answer is false: $arccosfrac 35$ is not $37^circ$.
                      – Saucy O'Path
                      Nov 24 '18 at 10:09














                      ... and $cos 97^circ$ is not $-0.39$.
                      – Saucy O'Path
                      Nov 24 '18 at 11:17




                      ... and $cos 97^circ$ is not $-0.39$.
                      – Saucy O'Path
                      Nov 24 '18 at 11:17


















                      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.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • 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.


                      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%2f3009421%2fevaluate-and-simplify-cos-cos-1-frac35-frac-pi3%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

                      mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

                      Image
                      0 I just uploaded a new plugin to my wordpress site and I got this warning: mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924 Usually if I have a problem, I can find answers on google but this one seems pretty particular and I'm not finding any info on it except for this forum. I'm not terribly techie but if someone can tell me how to read this and how to change it I can follow directions pretty well. Thanks All! PS If I don't have the proper tags here could someone add the appropriate tags? php mysql wordpress share | improve this question asked Nov 16 '18 at 20:41 ...
                      Read more

                      How to change which sound is reproduced for terminal bell?

                      Image
                      3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
                      Read more

                      Can I use Tabulator js library in my java Spring + Thymeleaf project?

                      Image
                      0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
                      Read more