show this identity with trigometric












4












$begingroup$


I sent a post earlier. Follow is an original problem. I got an error identity from a previous calculation error. Now there should be no problem.



Problem::



let $x,yin (0,dfrac{pi}{2})$. show that
$$dfrac{sin{(x+y)}tan{x}-cos{(x+y)}}{sin{(x+y)}tan{y}-cos{(x+y)}}=dfrac{cos{(2x+y)}cos{y}}{cos{(x+2y)}cos{x}}$$



This identity comes from the fact that I deal with a geometric problem and use trigonometric functions to calculate an identity that needs to be proved.Thanks






trigonometry







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    This version seems to be true. :) (Note: You should not delete a question that has received an answer. Doing so is inconsiderate to the answerer who has taken valuable time to respond to your question.)
    $endgroup$
    – Blue
    4 hours ago


















4












$begingroup$


I sent a post earlier. Follow is an original problem. I got an error identity from a previous calculation error. Now there should be no problem.



Problem::



let $x,yin (0,dfrac{pi}{2})$. show that
$$dfrac{sin{(x+y)}tan{x}-cos{(x+y)}}{sin{(x+y)}tan{y}-cos{(x+y)}}=dfrac{cos{(2x+y)}cos{y}}{cos{(x+2y)}cos{x}}$$



This identity comes from the fact that I deal with a geometric problem and use trigonometric functions to calculate an identity that needs to be proved.Thanks






trigonometry







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    This version seems to be true. :) (Note: You should not delete a question that has received an answer. Doing so is inconsiderate to the answerer who has taken valuable time to respond to your question.)
    $endgroup$
    – Blue
    4 hours ago
















4












4








4


2



$begingroup$


I sent a post earlier. Follow is an original problem. I got an error identity from a previous calculation error. Now there should be no problem.



Problem::



let $x,yin (0,dfrac{pi}{2})$. show that
$$dfrac{sin{(x+y)}tan{x}-cos{(x+y)}}{sin{(x+y)}tan{y}-cos{(x+y)}}=dfrac{cos{(2x+y)}cos{y}}{cos{(x+2y)}cos{x}}$$



This identity comes from the fact that I deal with a geometric problem and use trigonometric functions to calculate an identity that needs to be proved.Thanks






trigonometry







share|cite|improve this question









$endgroup$




I sent a post earlier. Follow is an original problem. I got an error identity from a previous calculation error. Now there should be no problem.



Problem::



let $x,yin (0,dfrac{pi}{2})$. show that
$$dfrac{sin{(x+y)}tan{x}-cos{(x+y)}}{sin{(x+y)}tan{y}-cos{(x+y)}}=dfrac{cos{(2x+y)}cos{y}}{cos{(x+2y)}cos{x}}$$



This identity comes from the fact that I deal with a geometric problem and use trigonometric functions to calculate an identity that needs to be proved.Thanks






trigonometry




trigonometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 4 hours ago









function sugfunction sug

3381438




3381438








  • 1




    $begingroup$
    This version seems to be true. :) (Note: You should not delete a question that has received an answer. Doing so is inconsiderate to the answerer who has taken valuable time to respond to your question.)
    $endgroup$
    – Blue
    4 hours ago
















  • 1




    $begingroup$
    This version seems to be true. :) (Note: You should not delete a question that has received an answer. Doing so is inconsiderate to the answerer who has taken valuable time to respond to your question.)
    $endgroup$
    – Blue
    4 hours ago










1




1




$begingroup$
This version seems to be true. :) (Note: You should not delete a question that has received an answer. Doing so is inconsiderate to the answerer who has taken valuable time to respond to your question.)
$endgroup$
– Blue
4 hours ago






$begingroup$
This version seems to be true. :) (Note: You should not delete a question that has received an answer. Doing so is inconsiderate to the answerer who has taken valuable time to respond to your question.)
$endgroup$
– Blue
4 hours ago












1 Answer
1






active

oldest

votes


















5












$begingroup$

Hint:



$$begin{align}
sin(x+y)tan x - cos(x+y) &= phantom{-}frac{1}{cos x}left(;sin(x+y) sin x - cos(x+y)cos x;right) \[4pt]
&= -frac1{cos x}cosleft((x+y)+xright) \[4pt]
&= -frac1{cos x}cosleft(2x+yright)
end{align}$$






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    5












    $begingroup$

    Hint:



    $$begin{align}
    sin(x+y)tan x - cos(x+y) &= phantom{-}frac{1}{cos x}left(;sin(x+y) sin x - cos(x+y)cos x;right) \[4pt]
    &= -frac1{cos x}cosleft((x+y)+xright) \[4pt]
    &= -frac1{cos x}cosleft(2x+yright)
    end{align}$$






    share|cite|improve this answer









    $endgroup$


















      5












      $begingroup$

      Hint:



      $$begin{align}
      sin(x+y)tan x - cos(x+y) &= phantom{-}frac{1}{cos x}left(;sin(x+y) sin x - cos(x+y)cos x;right) \[4pt]
      &= -frac1{cos x}cosleft((x+y)+xright) \[4pt]
      &= -frac1{cos x}cosleft(2x+yright)
      end{align}$$






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        Hint:



        $$begin{align}
        sin(x+y)tan x - cos(x+y) &= phantom{-}frac{1}{cos x}left(;sin(x+y) sin x - cos(x+y)cos x;right) \[4pt]
        &= -frac1{cos x}cosleft((x+y)+xright) \[4pt]
        &= -frac1{cos x}cosleft(2x+yright)
        end{align}$$






        share|cite|improve this answer









        $endgroup$



        Hint:



        $$begin{align}
        sin(x+y)tan x - cos(x+y) &= phantom{-}frac{1}{cos x}left(;sin(x+y) sin x - cos(x+y)cos x;right) \[4pt]
        &= -frac1{cos x}cosleft((x+y)+xright) \[4pt]
        &= -frac1{cos x}cosleft(2x+yright)
        end{align}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 4 hours ago









        BlueBlue

        49k870156




        49k870156






























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