How to rewrite $cos2x$ in terms of $frac{1-sin x}{ 1+sin x}$












0












$begingroup$


Would you please tell me how to rewrite $cos2x$ in terms of



$$frac{1-sin(x)}{ 1+sin(x)}$$
I had rewritten $cos(2x)$ in terms of $tan(x)$. But no results!
I did this job! Here:
$$cos2x=frac{cos(x)-sin(x)}{ cos(x)+sin(x) }$$






functions trigonometry







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  • 1




    $begingroup$
    If you got $cos(2x)=1$ (which is the fraction you wrote), you made a mistake.
    $endgroup$
    – Christoph
    Dec 9 '18 at 18:29
















0












$begingroup$


Would you please tell me how to rewrite $cos2x$ in terms of



$$frac{1-sin(x)}{ 1+sin(x)}$$
I had rewritten $cos(2x)$ in terms of $tan(x)$. But no results!
I did this job! Here:
$$cos2x=frac{cos(x)-sin(x)}{ cos(x)+sin(x) }$$






functions trigonometry







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If you got $cos(2x)=1$ (which is the fraction you wrote), you made a mistake.
    $endgroup$
    – Christoph
    Dec 9 '18 at 18:29














0












0








0





$begingroup$


Would you please tell me how to rewrite $cos2x$ in terms of



$$frac{1-sin(x)}{ 1+sin(x)}$$
I had rewritten $cos(2x)$ in terms of $tan(x)$. But no results!
I did this job! Here:
$$cos2x=frac{cos(x)-sin(x)}{ cos(x)+sin(x) }$$






functions trigonometry







share|cite|improve this question











$endgroup$




Would you please tell me how to rewrite $cos2x$ in terms of



$$frac{1-sin(x)}{ 1+sin(x)}$$
I had rewritten $cos(2x)$ in terms of $tan(x)$. But no results!
I did this job! Here:
$$cos2x=frac{cos(x)-sin(x)}{ cos(x)+sin(x) }$$






functions trigonometry




functions trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 18:35







user602338

















asked Dec 9 '18 at 18:23









user602338user602338

1607




1607








  • 1




    $begingroup$
    If you got $cos(2x)=1$ (which is the fraction you wrote), you made a mistake.
    $endgroup$
    – Christoph
    Dec 9 '18 at 18:29














  • 1




    $begingroup$
    If you got $cos(2x)=1$ (which is the fraction you wrote), you made a mistake.
    $endgroup$
    – Christoph
    Dec 9 '18 at 18:29








1




1




$begingroup$
If you got $cos(2x)=1$ (which is the fraction you wrote), you made a mistake.
$endgroup$
– Christoph
Dec 9 '18 at 18:29




$begingroup$
If you got $cos(2x)=1$ (which is the fraction you wrote), you made a mistake.
$endgroup$
– Christoph
Dec 9 '18 at 18:29










2 Answers
2






active

oldest

votes


















3












$begingroup$

Hint:



If $dfrac{1-sin x}{1+sin x}=y,sin x=?$



Now use $cos2x=1-2sin^2x$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Hint :



    $$cos(2x) = cos^2(x) - sin^2(x)$$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Would you explain your hint more bro?
      $endgroup$
      – user602338
      Dec 9 '18 at 18:32











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Hint:



    If $dfrac{1-sin x}{1+sin x}=y,sin x=?$



    Now use $cos2x=1-2sin^2x$






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      Hint:



      If $dfrac{1-sin x}{1+sin x}=y,sin x=?$



      Now use $cos2x=1-2sin^2x$






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        Hint:



        If $dfrac{1-sin x}{1+sin x}=y,sin x=?$



        Now use $cos2x=1-2sin^2x$






        share|cite|improve this answer









        $endgroup$



        Hint:



        If $dfrac{1-sin x}{1+sin x}=y,sin x=?$



        Now use $cos2x=1-2sin^2x$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 9 '18 at 18:33









        lab bhattacharjeelab bhattacharjee

        225k15157275




        225k15157275























            1












            $begingroup$

            Hint :



            $$cos(2x) = cos^2(x) - sin^2(x)$$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Would you explain your hint more bro?
              $endgroup$
              – user602338
              Dec 9 '18 at 18:32
















            1












            $begingroup$

            Hint :



            $$cos(2x) = cos^2(x) - sin^2(x)$$






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              Would you explain your hint more bro?
              $endgroup$
              – user602338
              Dec 9 '18 at 18:32














            1












            1








            1





            $begingroup$

            Hint :



            $$cos(2x) = cos^2(x) - sin^2(x)$$






            share|cite|improve this answer









            $endgroup$



            Hint :



            $$cos(2x) = cos^2(x) - sin^2(x)$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 9 '18 at 18:28









            RebellosRebellos

            14.6k31247




            14.6k31247








            • 1




              $begingroup$
              Would you explain your hint more bro?
              $endgroup$
              – user602338
              Dec 9 '18 at 18:32














            • 1




              $begingroup$
              Would you explain your hint more bro?
              $endgroup$
              – user602338
              Dec 9 '18 at 18:32








            1




            1




            $begingroup$
            Would you explain your hint more bro?
            $endgroup$
            – user602338
            Dec 9 '18 at 18:32




            $begingroup$
            Would you explain your hint more bro?
            $endgroup$
            – user602338
            Dec 9 '18 at 18:32


















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