Simplified form of $cos^{-1}big[frac{3}{5}cdotcos x+frac{4}{5}cdotsin xbig]$, where...












1












$begingroup$



Find the simplified form of $cos^{-1}bigg[dfrac{3}{5}cdotcos x+dfrac{4}{5}cdotsin xbigg]$, where $xinBig[dfrac{-3pi}{4},dfrac{3pi}{4}Big]$




My reference gives the solution $tan^{-1}frac43-x$, but is it a complete solution ?



My Attempt



Let $alpha=cos^{-1}dfrac{3}{5}implies dfrac{3}{5}=cosalpha,;dfrac{4}{5}=sinalpha$
$$
cos^{-1}bigg[dfrac{3}{5}cdotcos x+dfrac{4}{5}cdotsin xbigg]=cos^{-1}bigg[cosalphacdotcos x+sinalphacdotsin xbigg]\
=cos^{-1}bigg[cosBig(alpha-xBig)bigg]=2npipm(alpha-x)=2npipmBig(tan^{-1}frac{4}{3}-xBig)\
=tan^{-1}frac{4}{3}-xquadtext{iff }tan^{-1}frac{4}{3}-xin[0,pi]
$$
$$
-xinBig[dfrac{-3pi}{4},dfrac{3pi}{4}Big]quad&quadalpha=tan^{-1}frac{4}{3}inBig(0,frac{pi}{2}Big)\
impliesalpha-xinbig[frac{-3pi}{4},frac{5pi}{4}big]notsubset[0,pi]
$$






trigonometry inverse-function







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    At the very end, you mean $notsubset$ ("not a subset") instead of $notin$ ("not an element").
    $endgroup$
    – Barry Cipra
    Dec 10 '18 at 12:34
















1












$begingroup$



Find the simplified form of $cos^{-1}bigg[dfrac{3}{5}cdotcos x+dfrac{4}{5}cdotsin xbigg]$, where $xinBig[dfrac{-3pi}{4},dfrac{3pi}{4}Big]$




My reference gives the solution $tan^{-1}frac43-x$, but is it a complete solution ?



My Attempt



Let $alpha=cos^{-1}dfrac{3}{5}implies dfrac{3}{5}=cosalpha,;dfrac{4}{5}=sinalpha$
$$
cos^{-1}bigg[dfrac{3}{5}cdotcos x+dfrac{4}{5}cdotsin xbigg]=cos^{-1}bigg[cosalphacdotcos x+sinalphacdotsin xbigg]\
=cos^{-1}bigg[cosBig(alpha-xBig)bigg]=2npipm(alpha-x)=2npipmBig(tan^{-1}frac{4}{3}-xBig)\
=tan^{-1}frac{4}{3}-xquadtext{iff }tan^{-1}frac{4}{3}-xin[0,pi]
$$
$$
-xinBig[dfrac{-3pi}{4},dfrac{3pi}{4}Big]quad&quadalpha=tan^{-1}frac{4}{3}inBig(0,frac{pi}{2}Big)\
impliesalpha-xinbig[frac{-3pi}{4},frac{5pi}{4}big]notsubset[0,pi]
$$






trigonometry inverse-function







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    At the very end, you mean $notsubset$ ("not a subset") instead of $notin$ ("not an element").
    $endgroup$
    – Barry Cipra
    Dec 10 '18 at 12:34














1












1








1


0



$begingroup$



Find the simplified form of $cos^{-1}bigg[dfrac{3}{5}cdotcos x+dfrac{4}{5}cdotsin xbigg]$, where $xinBig[dfrac{-3pi}{4},dfrac{3pi}{4}Big]$




My reference gives the solution $tan^{-1}frac43-x$, but is it a complete solution ?



My Attempt



Let $alpha=cos^{-1}dfrac{3}{5}implies dfrac{3}{5}=cosalpha,;dfrac{4}{5}=sinalpha$
$$
cos^{-1}bigg[dfrac{3}{5}cdotcos x+dfrac{4}{5}cdotsin xbigg]=cos^{-1}bigg[cosalphacdotcos x+sinalphacdotsin xbigg]\
=cos^{-1}bigg[cosBig(alpha-xBig)bigg]=2npipm(alpha-x)=2npipmBig(tan^{-1}frac{4}{3}-xBig)\
=tan^{-1}frac{4}{3}-xquadtext{iff }tan^{-1}frac{4}{3}-xin[0,pi]
$$
$$
-xinBig[dfrac{-3pi}{4},dfrac{3pi}{4}Big]quad&quadalpha=tan^{-1}frac{4}{3}inBig(0,frac{pi}{2}Big)\
impliesalpha-xinbig[frac{-3pi}{4},frac{5pi}{4}big]notsubset[0,pi]
$$






trigonometry inverse-function







share|cite|improve this question











$endgroup$





Find the simplified form of $cos^{-1}bigg[dfrac{3}{5}cdotcos x+dfrac{4}{5}cdotsin xbigg]$, where $xinBig[dfrac{-3pi}{4},dfrac{3pi}{4}Big]$




My reference gives the solution $tan^{-1}frac43-x$, but is it a complete solution ?



My Attempt



Let $alpha=cos^{-1}dfrac{3}{5}implies dfrac{3}{5}=cosalpha,;dfrac{4}{5}=sinalpha$
$$
cos^{-1}bigg[dfrac{3}{5}cdotcos x+dfrac{4}{5}cdotsin xbigg]=cos^{-1}bigg[cosalphacdotcos x+sinalphacdotsin xbigg]\
=cos^{-1}bigg[cosBig(alpha-xBig)bigg]=2npipm(alpha-x)=2npipmBig(tan^{-1}frac{4}{3}-xBig)\
=tan^{-1}frac{4}{3}-xquadtext{iff }tan^{-1}frac{4}{3}-xin[0,pi]
$$
$$
-xinBig[dfrac{-3pi}{4},dfrac{3pi}{4}Big]quad&quadalpha=tan^{-1}frac{4}{3}inBig(0,frac{pi}{2}Big)\
impliesalpha-xinbig[frac{-3pi}{4},frac{5pi}{4}big]notsubset[0,pi]
$$






trigonometry inverse-function




trigonometry inverse-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 12:54







ss1729

















asked Dec 10 '18 at 11:21









ss1729ss1729

1,9211723




1,9211723








  • 2




    $begingroup$
    At the very end, you mean $notsubset$ ("not a subset") instead of $notin$ ("not an element").
    $endgroup$
    – Barry Cipra
    Dec 10 '18 at 12:34














  • 2




    $begingroup$
    At the very end, you mean $notsubset$ ("not a subset") instead of $notin$ ("not an element").
    $endgroup$
    – Barry Cipra
    Dec 10 '18 at 12:34








2




2




$begingroup$
At the very end, you mean $notsubset$ ("not a subset") instead of $notin$ ("not an element").
$endgroup$
– Barry Cipra
Dec 10 '18 at 12:34




$begingroup$
At the very end, you mean $notsubset$ ("not a subset") instead of $notin$ ("not an element").
$endgroup$
– Barry Cipra
Dec 10 '18 at 12:34










2 Answers
2






active

oldest

votes


















1












$begingroup$

Your book is wrong. $xinBig[-3pi/4, 3pi/4Big]$, which is an interval of length $3pi/2$. Whatever be the value of $alpha, alpha-x$ will belong to an interval of length $3pi/2$, which means $alpha-x$ is not confined to $[0, pi].$



So the answer is $begin{cases}2pi-alpha+x,&xin[-3pi/4,alpha-pi)\alpha-x,&xin[alpha-pi,alpha]\x-alpha,&xin(alpha, 3pi/4]end{cases}$






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    $$-dfrac{3pi}4le xledfrac{3pi}4$$



    $$iff-dfrac{3pi}4-cos^{-1}dfrac35le x-cos^{-1}dfrac35ledfrac{3pi}4-cos^{-1}dfrac35$$



    Now $dfrac{3pi}4-cos^{-1}dfrac35lepi$ as $cos^{-1}dfrac35>0>dfrac{3pi}4-pi$



    So, $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=x-cos^{-1}dfrac35$ if $x-cos^{-1}dfrac35ge0iff xgecos^{-1}dfrac35$



    Again we can prove $-2pi<-dfrac{3pi}4-cos^{-1}dfrac35<-pi$



    For $-pi<x-cos^{-1}dfrac35<0,$ $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=-left(x-cos^{-1}dfrac35right)$



    For $-2pi<x-cos^{-1}dfrac35<-pi,$ $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=2pi+x-cos^{-1}dfrac35$






    share|cite|improve this answer









    $endgroup$













      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%2f3033793%2fsimplified-form-of-cos-1-big-frac35-cdot-cos-x-frac45-cdot-sin-x%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Your book is wrong. $xinBig[-3pi/4, 3pi/4Big]$, which is an interval of length $3pi/2$. Whatever be the value of $alpha, alpha-x$ will belong to an interval of length $3pi/2$, which means $alpha-x$ is not confined to $[0, pi].$



      So the answer is $begin{cases}2pi-alpha+x,&xin[-3pi/4,alpha-pi)\alpha-x,&xin[alpha-pi,alpha]\x-alpha,&xin(alpha, 3pi/4]end{cases}$






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        Your book is wrong. $xinBig[-3pi/4, 3pi/4Big]$, which is an interval of length $3pi/2$. Whatever be the value of $alpha, alpha-x$ will belong to an interval of length $3pi/2$, which means $alpha-x$ is not confined to $[0, pi].$



        So the answer is $begin{cases}2pi-alpha+x,&xin[-3pi/4,alpha-pi)\alpha-x,&xin[alpha-pi,alpha]\x-alpha,&xin(alpha, 3pi/4]end{cases}$






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          Your book is wrong. $xinBig[-3pi/4, 3pi/4Big]$, which is an interval of length $3pi/2$. Whatever be the value of $alpha, alpha-x$ will belong to an interval of length $3pi/2$, which means $alpha-x$ is not confined to $[0, pi].$



          So the answer is $begin{cases}2pi-alpha+x,&xin[-3pi/4,alpha-pi)\alpha-x,&xin[alpha-pi,alpha]\x-alpha,&xin(alpha, 3pi/4]end{cases}$






          share|cite|improve this answer











          $endgroup$



          Your book is wrong. $xinBig[-3pi/4, 3pi/4Big]$, which is an interval of length $3pi/2$. Whatever be the value of $alpha, alpha-x$ will belong to an interval of length $3pi/2$, which means $alpha-x$ is not confined to $[0, pi].$



          So the answer is $begin{cases}2pi-alpha+x,&xin[-3pi/4,alpha-pi)\alpha-x,&xin[alpha-pi,alpha]\x-alpha,&xin(alpha, 3pi/4]end{cases}$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 '18 at 12:00

























          answered Dec 10 '18 at 11:32









          Shubham JohriShubham Johri

          5,122717




          5,122717























              0












              $begingroup$

              $$-dfrac{3pi}4le xledfrac{3pi}4$$



              $$iff-dfrac{3pi}4-cos^{-1}dfrac35le x-cos^{-1}dfrac35ledfrac{3pi}4-cos^{-1}dfrac35$$



              Now $dfrac{3pi}4-cos^{-1}dfrac35lepi$ as $cos^{-1}dfrac35>0>dfrac{3pi}4-pi$



              So, $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=x-cos^{-1}dfrac35$ if $x-cos^{-1}dfrac35ge0iff xgecos^{-1}dfrac35$



              Again we can prove $-2pi<-dfrac{3pi}4-cos^{-1}dfrac35<-pi$



              For $-pi<x-cos^{-1}dfrac35<0,$ $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=-left(x-cos^{-1}dfrac35right)$



              For $-2pi<x-cos^{-1}dfrac35<-pi,$ $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=2pi+x-cos^{-1}dfrac35$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $$-dfrac{3pi}4le xledfrac{3pi}4$$



                $$iff-dfrac{3pi}4-cos^{-1}dfrac35le x-cos^{-1}dfrac35ledfrac{3pi}4-cos^{-1}dfrac35$$



                Now $dfrac{3pi}4-cos^{-1}dfrac35lepi$ as $cos^{-1}dfrac35>0>dfrac{3pi}4-pi$



                So, $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=x-cos^{-1}dfrac35$ if $x-cos^{-1}dfrac35ge0iff xgecos^{-1}dfrac35$



                Again we can prove $-2pi<-dfrac{3pi}4-cos^{-1}dfrac35<-pi$



                For $-pi<x-cos^{-1}dfrac35<0,$ $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=-left(x-cos^{-1}dfrac35right)$



                For $-2pi<x-cos^{-1}dfrac35<-pi,$ $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=2pi+x-cos^{-1}dfrac35$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $$-dfrac{3pi}4le xledfrac{3pi}4$$



                  $$iff-dfrac{3pi}4-cos^{-1}dfrac35le x-cos^{-1}dfrac35ledfrac{3pi}4-cos^{-1}dfrac35$$



                  Now $dfrac{3pi}4-cos^{-1}dfrac35lepi$ as $cos^{-1}dfrac35>0>dfrac{3pi}4-pi$



                  So, $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=x-cos^{-1}dfrac35$ if $x-cos^{-1}dfrac35ge0iff xgecos^{-1}dfrac35$



                  Again we can prove $-2pi<-dfrac{3pi}4-cos^{-1}dfrac35<-pi$



                  For $-pi<x-cos^{-1}dfrac35<0,$ $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=-left(x-cos^{-1}dfrac35right)$



                  For $-2pi<x-cos^{-1}dfrac35<-pi,$ $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=2pi+x-cos^{-1}dfrac35$






                  share|cite|improve this answer









                  $endgroup$



                  $$-dfrac{3pi}4le xledfrac{3pi}4$$



                  $$iff-dfrac{3pi}4-cos^{-1}dfrac35le x-cos^{-1}dfrac35ledfrac{3pi}4-cos^{-1}dfrac35$$



                  Now $dfrac{3pi}4-cos^{-1}dfrac35lepi$ as $cos^{-1}dfrac35>0>dfrac{3pi}4-pi$



                  So, $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=x-cos^{-1}dfrac35$ if $x-cos^{-1}dfrac35ge0iff xgecos^{-1}dfrac35$



                  Again we can prove $-2pi<-dfrac{3pi}4-cos^{-1}dfrac35<-pi$



                  For $-pi<x-cos^{-1}dfrac35<0,$ $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=-left(x-cos^{-1}dfrac35right)$



                  For $-2pi<x-cos^{-1}dfrac35<-pi,$ $cos^{-1}bigg[cosBig(x-cos^{-1}dfrac35Big)bigg]=2pi+x-cos^{-1}dfrac35$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 10 '18 at 11:51









                  lab bhattacharjeelab bhattacharjee

                  225k15157275




                  225k15157275






























                      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%2f3033793%2fsimplified-form-of-cos-1-big-frac35-cdot-cos-x-frac45-cdot-sin-x%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

                      Ellipse (mathématiques)

                      Image
                      Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                      Read more

                      Quarter-circle Tiles

                      Image
                      up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
                      Read more

                      Mont Emei

                      Image
                      Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
                      Read more