If $(fcirc g)(x)=tan^2x$ and $g(x)=sqrt{cos2x}$ then find $f(x)=?$












1












$begingroup$


We've given :
$$(fcirc g)(x)=tan^2x$$
and
$$g(x)=sqrt{cos 2x}$$
Then how to find the function $f(x)$?
I know that
$$(fcirc g)(x)=f(g(x))= f( sqrt{cos2x})$$
But I do not know how to find $f(x)$!
Please help me!






algebra-precalculus functions function-and-relation-composition







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    We've given :
    $$(fcirc g)(x)=tan^2x$$
    and
    $$g(x)=sqrt{cos 2x}$$
    Then how to find the function $f(x)$?
    I know that
    $$(fcirc g)(x)=f(g(x))= f( sqrt{cos2x})$$
    But I do not know how to find $f(x)$!
    Please help me!






    algebra-precalculus functions function-and-relation-composition







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      We've given :
      $$(fcirc g)(x)=tan^2x$$
      and
      $$g(x)=sqrt{cos 2x}$$
      Then how to find the function $f(x)$?
      I know that
      $$(fcirc g)(x)=f(g(x))= f( sqrt{cos2x})$$
      But I do not know how to find $f(x)$!
      Please help me!






      algebra-precalculus functions function-and-relation-composition







      share|cite|improve this question











      $endgroup$




      We've given :
      $$(fcirc g)(x)=tan^2x$$
      and
      $$g(x)=sqrt{cos 2x}$$
      Then how to find the function $f(x)$?
      I know that
      $$(fcirc g)(x)=f(g(x))= f( sqrt{cos2x})$$
      But I do not know how to find $f(x)$!
      Please help me!






      algebra-precalculus functions function-and-relation-composition




      algebra-precalculus functions function-and-relation-composition






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 8 '18 at 18:23









      John Hughes

      62.9k24090




      62.9k24090










      asked Dec 8 '18 at 18:12









      user602338user602338

      1707




      1707






















          3 Answers
          3






          active

          oldest

          votes


















          5












          $begingroup$

          So you can find



          $$cos x=frac{cos2x+1}{2}=frac{g^2(x)+1}{2}$$



          and



          $$f(g(x))=tan^2 x=frac{1}{cos^2x}-1=frac{1-g^2(x)}{1+g^2(x)}$$



          obviously



          $$f(x)=frac{1-x^2}{1+x^2}$$






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            Think "what do I have to do to $sqrt{cos(2x)}$ to get $tan^2x$?".



            Note that $$cos(2x) = 2cos^2x-1qquadmbox{and}qquad tan^2x=sec^2x-1,$$so that $$tan^2x=frac{2}{cos(2x)+1}-1$$



            Meaning that if you start with $sqrt{cos(2x)}$, you have to first square it and then apply the above. $$f(x)=frac{2}{x^2+1}-1. $$






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              You say $y=sqrt{cos 2x}$. Then you calculate $tan^2 x$ in terms of $y$. You square $y$, then you calculate $sin^2x$. Then $$tan^2 x=frac{sin^2 x}{1-sin^2 x}$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                I don't see how this gives $f(x)$.
                $endgroup$
                – Shaun
                Dec 8 '18 at 18:23










              • $begingroup$
                $f(y)=$ some expression in terms of $y$ only. Just replace $y$ with $x$ (at the end)
                $endgroup$
                – Andrei
                Dec 8 '18 at 18:25












              • $begingroup$
                Nah, I still don't see it. (I'm a little rusty on this stuff. Sorry.) I'll take your word for it for now, then perhaps think about it later when I have more time. Thank you anyway :)
                $endgroup$
                – Shaun
                Dec 8 '18 at 18:28










              • $begingroup$
                Just look at the other answers. They explicitly follow the steps that I've described
                $endgroup$
                – Andrei
                Dec 8 '18 at 18:30










              • $begingroup$
                Ah, okay; thanks again :)
                $endgroup$
                – Shaun
                Dec 8 '18 at 18:31











              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%2f3031429%2fif-f-circ-gx-tan2x-and-gx-sqrt-cos2x-then-find-fx%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









              5












              $begingroup$

              So you can find



              $$cos x=frac{cos2x+1}{2}=frac{g^2(x)+1}{2}$$



              and



              $$f(g(x))=tan^2 x=frac{1}{cos^2x}-1=frac{1-g^2(x)}{1+g^2(x)}$$



              obviously



              $$f(x)=frac{1-x^2}{1+x^2}$$






              share|cite|improve this answer









              $endgroup$


















                5












                $begingroup$

                So you can find



                $$cos x=frac{cos2x+1}{2}=frac{g^2(x)+1}{2}$$



                and



                $$f(g(x))=tan^2 x=frac{1}{cos^2x}-1=frac{1-g^2(x)}{1+g^2(x)}$$



                obviously



                $$f(x)=frac{1-x^2}{1+x^2}$$






                share|cite|improve this answer









                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  So you can find



                  $$cos x=frac{cos2x+1}{2}=frac{g^2(x)+1}{2}$$



                  and



                  $$f(g(x))=tan^2 x=frac{1}{cos^2x}-1=frac{1-g^2(x)}{1+g^2(x)}$$



                  obviously



                  $$f(x)=frac{1-x^2}{1+x^2}$$






                  share|cite|improve this answer









                  $endgroup$



                  So you can find



                  $$cos x=frac{cos2x+1}{2}=frac{g^2(x)+1}{2}$$



                  and



                  $$f(g(x))=tan^2 x=frac{1}{cos^2x}-1=frac{1-g^2(x)}{1+g^2(x)}$$



                  obviously



                  $$f(x)=frac{1-x^2}{1+x^2}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 8 '18 at 18:22









                  NanayajitzukiNanayajitzuki

                  3185




                  3185























                      2












                      $begingroup$

                      Think "what do I have to do to $sqrt{cos(2x)}$ to get $tan^2x$?".



                      Note that $$cos(2x) = 2cos^2x-1qquadmbox{and}qquad tan^2x=sec^2x-1,$$so that $$tan^2x=frac{2}{cos(2x)+1}-1$$



                      Meaning that if you start with $sqrt{cos(2x)}$, you have to first square it and then apply the above. $$f(x)=frac{2}{x^2+1}-1. $$






                      share|cite|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        Think "what do I have to do to $sqrt{cos(2x)}$ to get $tan^2x$?".



                        Note that $$cos(2x) = 2cos^2x-1qquadmbox{and}qquad tan^2x=sec^2x-1,$$so that $$tan^2x=frac{2}{cos(2x)+1}-1$$



                        Meaning that if you start with $sqrt{cos(2x)}$, you have to first square it and then apply the above. $$f(x)=frac{2}{x^2+1}-1. $$






                        share|cite|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          Think "what do I have to do to $sqrt{cos(2x)}$ to get $tan^2x$?".



                          Note that $$cos(2x) = 2cos^2x-1qquadmbox{and}qquad tan^2x=sec^2x-1,$$so that $$tan^2x=frac{2}{cos(2x)+1}-1$$



                          Meaning that if you start with $sqrt{cos(2x)}$, you have to first square it and then apply the above. $$f(x)=frac{2}{x^2+1}-1. $$






                          share|cite|improve this answer









                          $endgroup$



                          Think "what do I have to do to $sqrt{cos(2x)}$ to get $tan^2x$?".



                          Note that $$cos(2x) = 2cos^2x-1qquadmbox{and}qquad tan^2x=sec^2x-1,$$so that $$tan^2x=frac{2}{cos(2x)+1}-1$$



                          Meaning that if you start with $sqrt{cos(2x)}$, you have to first square it and then apply the above. $$f(x)=frac{2}{x^2+1}-1. $$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 8 '18 at 18:21









                          Ivo TerekIvo Terek

                          45.4k952141




                          45.4k952141























                              0












                              $begingroup$

                              You say $y=sqrt{cos 2x}$. Then you calculate $tan^2 x$ in terms of $y$. You square $y$, then you calculate $sin^2x$. Then $$tan^2 x=frac{sin^2 x}{1-sin^2 x}$$






                              share|cite|improve this answer









                              $endgroup$













                              • $begingroup$
                                I don't see how this gives $f(x)$.
                                $endgroup$
                                – Shaun
                                Dec 8 '18 at 18:23










                              • $begingroup$
                                $f(y)=$ some expression in terms of $y$ only. Just replace $y$ with $x$ (at the end)
                                $endgroup$
                                – Andrei
                                Dec 8 '18 at 18:25












                              • $begingroup$
                                Nah, I still don't see it. (I'm a little rusty on this stuff. Sorry.) I'll take your word for it for now, then perhaps think about it later when I have more time. Thank you anyway :)
                                $endgroup$
                                – Shaun
                                Dec 8 '18 at 18:28










                              • $begingroup$
                                Just look at the other answers. They explicitly follow the steps that I've described
                                $endgroup$
                                – Andrei
                                Dec 8 '18 at 18:30










                              • $begingroup$
                                Ah, okay; thanks again :)
                                $endgroup$
                                – Shaun
                                Dec 8 '18 at 18:31
















                              0












                              $begingroup$

                              You say $y=sqrt{cos 2x}$. Then you calculate $tan^2 x$ in terms of $y$. You square $y$, then you calculate $sin^2x$. Then $$tan^2 x=frac{sin^2 x}{1-sin^2 x}$$






                              share|cite|improve this answer









                              $endgroup$













                              • $begingroup$
                                I don't see how this gives $f(x)$.
                                $endgroup$
                                – Shaun
                                Dec 8 '18 at 18:23










                              • $begingroup$
                                $f(y)=$ some expression in terms of $y$ only. Just replace $y$ with $x$ (at the end)
                                $endgroup$
                                – Andrei
                                Dec 8 '18 at 18:25












                              • $begingroup$
                                Nah, I still don't see it. (I'm a little rusty on this stuff. Sorry.) I'll take your word for it for now, then perhaps think about it later when I have more time. Thank you anyway :)
                                $endgroup$
                                – Shaun
                                Dec 8 '18 at 18:28










                              • $begingroup$
                                Just look at the other answers. They explicitly follow the steps that I've described
                                $endgroup$
                                – Andrei
                                Dec 8 '18 at 18:30










                              • $begingroup$
                                Ah, okay; thanks again :)
                                $endgroup$
                                – Shaun
                                Dec 8 '18 at 18:31














                              0












                              0








                              0





                              $begingroup$

                              You say $y=sqrt{cos 2x}$. Then you calculate $tan^2 x$ in terms of $y$. You square $y$, then you calculate $sin^2x$. Then $$tan^2 x=frac{sin^2 x}{1-sin^2 x}$$






                              share|cite|improve this answer









                              $endgroup$



                              You say $y=sqrt{cos 2x}$. Then you calculate $tan^2 x$ in terms of $y$. You square $y$, then you calculate $sin^2x$. Then $$tan^2 x=frac{sin^2 x}{1-sin^2 x}$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 8 '18 at 18:18









                              AndreiAndrei

                              11.6k21026




                              11.6k21026












                              • $begingroup$
                                I don't see how this gives $f(x)$.
                                $endgroup$
                                – Shaun
                                Dec 8 '18 at 18:23










                              • $begingroup$
                                $f(y)=$ some expression in terms of $y$ only. Just replace $y$ with $x$ (at the end)
                                $endgroup$
                                – Andrei
                                Dec 8 '18 at 18:25












                              • $begingroup$
                                Nah, I still don't see it. (I'm a little rusty on this stuff. Sorry.) I'll take your word for it for now, then perhaps think about it later when I have more time. Thank you anyway :)
                                $endgroup$
                                – Shaun
                                Dec 8 '18 at 18:28










                              • $begingroup$
                                Just look at the other answers. They explicitly follow the steps that I've described
                                $endgroup$
                                – Andrei
                                Dec 8 '18 at 18:30










                              • $begingroup$
                                Ah, okay; thanks again :)
                                $endgroup$
                                – Shaun
                                Dec 8 '18 at 18:31


















                              • $begingroup$
                                I don't see how this gives $f(x)$.
                                $endgroup$
                                – Shaun
                                Dec 8 '18 at 18:23










                              • $begingroup$
                                $f(y)=$ some expression in terms of $y$ only. Just replace $y$ with $x$ (at the end)
                                $endgroup$
                                – Andrei
                                Dec 8 '18 at 18:25












                              • $begingroup$
                                Nah, I still don't see it. (I'm a little rusty on this stuff. Sorry.) I'll take your word for it for now, then perhaps think about it later when I have more time. Thank you anyway :)
                                $endgroup$
                                – Shaun
                                Dec 8 '18 at 18:28










                              • $begingroup$
                                Just look at the other answers. They explicitly follow the steps that I've described
                                $endgroup$
                                – Andrei
                                Dec 8 '18 at 18:30










                              • $begingroup$
                                Ah, okay; thanks again :)
                                $endgroup$
                                – Shaun
                                Dec 8 '18 at 18:31
















                              $begingroup$
                              I don't see how this gives $f(x)$.
                              $endgroup$
                              – Shaun
                              Dec 8 '18 at 18:23




                              $begingroup$
                              I don't see how this gives $f(x)$.
                              $endgroup$
                              – Shaun
                              Dec 8 '18 at 18:23












                              $begingroup$
                              $f(y)=$ some expression in terms of $y$ only. Just replace $y$ with $x$ (at the end)
                              $endgroup$
                              – Andrei
                              Dec 8 '18 at 18:25






                              $begingroup$
                              $f(y)=$ some expression in terms of $y$ only. Just replace $y$ with $x$ (at the end)
                              $endgroup$
                              – Andrei
                              Dec 8 '18 at 18:25














                              $begingroup$
                              Nah, I still don't see it. (I'm a little rusty on this stuff. Sorry.) I'll take your word for it for now, then perhaps think about it later when I have more time. Thank you anyway :)
                              $endgroup$
                              – Shaun
                              Dec 8 '18 at 18:28




                              $begingroup$
                              Nah, I still don't see it. (I'm a little rusty on this stuff. Sorry.) I'll take your word for it for now, then perhaps think about it later when I have more time. Thank you anyway :)
                              $endgroup$
                              – Shaun
                              Dec 8 '18 at 18:28












                              $begingroup$
                              Just look at the other answers. They explicitly follow the steps that I've described
                              $endgroup$
                              – Andrei
                              Dec 8 '18 at 18:30




                              $begingroup$
                              Just look at the other answers. They explicitly follow the steps that I've described
                              $endgroup$
                              – Andrei
                              Dec 8 '18 at 18:30












                              $begingroup$
                              Ah, okay; thanks again :)
                              $endgroup$
                              – Shaun
                              Dec 8 '18 at 18:31




                              $begingroup$
                              Ah, okay; thanks again :)
                              $endgroup$
                              – Shaun
                              Dec 8 '18 at 18:31


















                              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%2f3031429%2fif-f-circ-gx-tan2x-and-gx-sqrt-cos2x-then-find-fx%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

                              Berounka

                              Image
                              Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
                              Read more

                              Sphinx de Gizeh

                              Image
                              Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
                              Read more

                              Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

                              Image
                              3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
                              Read more