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












1














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







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    1














    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



























      1












      1








      1







      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















      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 at 18:23









      John Hughes

      62.3k24090




      62.3k24090










      asked Dec 8 at 18:12









      user602338

      1587




      1587






















          3 Answers
          3






          active

          oldest

          votes


















          5














          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





























            2














            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





























              0














              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





















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










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












              • 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 :)
                – Shaun
                Dec 8 at 18:28










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










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











              Your Answer





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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              5














              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


























                5














                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
























                  5












                  5








                  5






                  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












                  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 at 18:22









                  Nanayajitzuki

                  1885




                  1885























                      2














                      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


























                        2














                        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
























                          2












                          2








                          2






                          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












                          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 at 18:21









                          Ivo Terek

                          45.2k951139




                          45.2k951139























                              0














                              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





















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










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












                              • 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 :)
                                – Shaun
                                Dec 8 at 18:28










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










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
















                              0














                              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





















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










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












                              • 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 :)
                                – Shaun
                                Dec 8 at 18:28










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










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














                              0












                              0








                              0






                              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












                              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 at 18:18









                              Andrei

                              11k21025




                              11k21025












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










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












                              • 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 :)
                                – Shaun
                                Dec 8 at 18:28










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










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


















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










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












                              • 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 :)
                                – Shaun
                                Dec 8 at 18:28










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










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
















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




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












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






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














                              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 :)
                              – Shaun
                              Dec 8 at 18:28




                              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 :)
                              – Shaun
                              Dec 8 at 18:28












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




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












                              Ah, okay; thanks again :)
                              – Shaun
                              Dec 8 at 18:31




                              Ah, okay; thanks again :)
                              – Shaun
                              Dec 8 at 18:31


















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