If $cos^4 alpha+4sin^4 beta-4sqrt{2}cos alpha sin beta +2=0$, then find $alpha$, $beta$ in $(0,fracpi2)$












0












$begingroup$



If $cos^4 alpha+4sin^4 beta-4sqrt{2}cos alpha sin beta +2=0$,



where $displaystyle alpha, beta in bigg(0,frac{pi}{2}bigg)$. Then value of $alpha,beta$ are




Try: I am trying to convert it into sum of square of quantity
like $$(cos^2 alpha)^2+(2sin^2 beta)^2-2cdot cos^2 alpha cdot 2sin^2 beta-4sqrt{2}cos alpha sin beta +2+4cos^2 alpha cdot sin^2 beta$$



$$(cos^2 alpha--2sin^2 beta)^2-4sqrt{2}cos alpha sin beta+4cos^2 alpha cdot sin^2 beta$$



Now i did not know how to solve it, could some help me






trigonometry







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  • $begingroup$
    $cos^4 alpha+4sin^4alpha-4sqrt{2}cos alpha cdot sin alpha +2=0 implies (cos^2 alpha-2sin^2alpha)^2+4cos^2alpha sin^2 alpha-4sqrt{2}cos alpha sin alpha+2=0implies (cos^2alpha-2sin^2alpha)^2+2(sqrt{2}acos alpha sin alpha-1)^2=0$
    $endgroup$
    – DXT
    Jan 24 at 12:49


















0












$begingroup$



If $cos^4 alpha+4sin^4 beta-4sqrt{2}cos alpha sin beta +2=0$,



where $displaystyle alpha, beta in bigg(0,frac{pi}{2}bigg)$. Then value of $alpha,beta$ are




Try: I am trying to convert it into sum of square of quantity
like $$(cos^2 alpha)^2+(2sin^2 beta)^2-2cdot cos^2 alpha cdot 2sin^2 beta-4sqrt{2}cos alpha sin beta +2+4cos^2 alpha cdot sin^2 beta$$



$$(cos^2 alpha--2sin^2 beta)^2-4sqrt{2}cos alpha sin beta+4cos^2 alpha cdot sin^2 beta$$



Now i did not know how to solve it, could some help me






trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    $cos^4 alpha+4sin^4alpha-4sqrt{2}cos alpha cdot sin alpha +2=0 implies (cos^2 alpha-2sin^2alpha)^2+4cos^2alpha sin^2 alpha-4sqrt{2}cos alpha sin alpha+2=0implies (cos^2alpha-2sin^2alpha)^2+2(sqrt{2}acos alpha sin alpha-1)^2=0$
    $endgroup$
    – DXT
    Jan 24 at 12:49
















0












0








0





$begingroup$



If $cos^4 alpha+4sin^4 beta-4sqrt{2}cos alpha sin beta +2=0$,



where $displaystyle alpha, beta in bigg(0,frac{pi}{2}bigg)$. Then value of $alpha,beta$ are




Try: I am trying to convert it into sum of square of quantity
like $$(cos^2 alpha)^2+(2sin^2 beta)^2-2cdot cos^2 alpha cdot 2sin^2 beta-4sqrt{2}cos alpha sin beta +2+4cos^2 alpha cdot sin^2 beta$$



$$(cos^2 alpha--2sin^2 beta)^2-4sqrt{2}cos alpha sin beta+4cos^2 alpha cdot sin^2 beta$$



Now i did not know how to solve it, could some help me






trigonometry







share|cite|improve this question











$endgroup$





If $cos^4 alpha+4sin^4 beta-4sqrt{2}cos alpha sin beta +2=0$,



where $displaystyle alpha, beta in bigg(0,frac{pi}{2}bigg)$. Then value of $alpha,beta$ are




Try: I am trying to convert it into sum of square of quantity
like $$(cos^2 alpha)^2+(2sin^2 beta)^2-2cdot cos^2 alpha cdot 2sin^2 beta-4sqrt{2}cos alpha sin beta +2+4cos^2 alpha cdot sin^2 beta$$



$$(cos^2 alpha--2sin^2 beta)^2-4sqrt{2}cos alpha sin beta+4cos^2 alpha cdot sin^2 beta$$



Now i did not know how to solve it, could some help me






trigonometry




trigonometry






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share|cite|improve this question













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edited Jan 17 at 10:47









Blue

48.4k870154




48.4k870154










asked Jan 17 at 10:01









DXTDXT

5,6892630




5,6892630












  • $begingroup$
    $cos^4 alpha+4sin^4alpha-4sqrt{2}cos alpha cdot sin alpha +2=0 implies (cos^2 alpha-2sin^2alpha)^2+4cos^2alpha sin^2 alpha-4sqrt{2}cos alpha sin alpha+2=0implies (cos^2alpha-2sin^2alpha)^2+2(sqrt{2}acos alpha sin alpha-1)^2=0$
    $endgroup$
    – DXT
    Jan 24 at 12:49




















  • $begingroup$
    $cos^4 alpha+4sin^4alpha-4sqrt{2}cos alpha cdot sin alpha +2=0 implies (cos^2 alpha-2sin^2alpha)^2+4cos^2alpha sin^2 alpha-4sqrt{2}cos alpha sin alpha+2=0implies (cos^2alpha-2sin^2alpha)^2+2(sqrt{2}acos alpha sin alpha-1)^2=0$
    $endgroup$
    – DXT
    Jan 24 at 12:49


















$begingroup$
$cos^4 alpha+4sin^4alpha-4sqrt{2}cos alpha cdot sin alpha +2=0 implies (cos^2 alpha-2sin^2alpha)^2+4cos^2alpha sin^2 alpha-4sqrt{2}cos alpha sin alpha+2=0implies (cos^2alpha-2sin^2alpha)^2+2(sqrt{2}acos alpha sin alpha-1)^2=0$
$endgroup$
– DXT
Jan 24 at 12:49






$begingroup$
$cos^4 alpha+4sin^4alpha-4sqrt{2}cos alpha cdot sin alpha +2=0 implies (cos^2 alpha-2sin^2alpha)^2+4cos^2alpha sin^2 alpha-4sqrt{2}cos alpha sin alpha+2=0implies (cos^2alpha-2sin^2alpha)^2+2(sqrt{2}acos alpha sin alpha-1)^2=0$
$endgroup$
– DXT
Jan 24 at 12:49












2 Answers
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$begingroup$

Hint:



For real $cos^2alpha,sin^2beta$



$$dfrac{cos^4alpha+4sin^4beta+1+1}4gesqrt[4]{cos^4alphacdot4sin^4beta}$$



The equality will occur if $cos^4alpha=4sin^4beta=1$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    See artofproblemsolving.com/wiki/index.php/…
    $endgroup$
    – lab bhattacharjee
    Jan 17 at 11:01



















1












$begingroup$

If you just call $x=cos(alpha)$ and $y=sin(beta)$ you get the 4th degree polynomial equation $x^4-4sqrt{2}xy + y^4 +2 =0$. I don't see a clever way to solve this directly but this can be solved algebraically. So I would put this equation into mathematica or some similar software and then look a the 4 solutions.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






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    oldest

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    active

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    active

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    2












    $begingroup$

    Hint:



    For real $cos^2alpha,sin^2beta$



    $$dfrac{cos^4alpha+4sin^4beta+1+1}4gesqrt[4]{cos^4alphacdot4sin^4beta}$$



    The equality will occur if $cos^4alpha=4sin^4beta=1$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      See artofproblemsolving.com/wiki/index.php/…
      $endgroup$
      – lab bhattacharjee
      Jan 17 at 11:01
















    2












    $begingroup$

    Hint:



    For real $cos^2alpha,sin^2beta$



    $$dfrac{cos^4alpha+4sin^4beta+1+1}4gesqrt[4]{cos^4alphacdot4sin^4beta}$$



    The equality will occur if $cos^4alpha=4sin^4beta=1$






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      See artofproblemsolving.com/wiki/index.php/…
      $endgroup$
      – lab bhattacharjee
      Jan 17 at 11:01














    2












    2








    2





    $begingroup$

    Hint:



    For real $cos^2alpha,sin^2beta$



    $$dfrac{cos^4alpha+4sin^4beta+1+1}4gesqrt[4]{cos^4alphacdot4sin^4beta}$$



    The equality will occur if $cos^4alpha=4sin^4beta=1$






    share|cite|improve this answer









    $endgroup$



    Hint:



    For real $cos^2alpha,sin^2beta$



    $$dfrac{cos^4alpha+4sin^4beta+1+1}4gesqrt[4]{cos^4alphacdot4sin^4beta}$$



    The equality will occur if $cos^4alpha=4sin^4beta=1$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 17 at 10:33









    lab bhattacharjeelab bhattacharjee

    225k15157275




    225k15157275








    • 1




      $begingroup$
      See artofproblemsolving.com/wiki/index.php/…
      $endgroup$
      – lab bhattacharjee
      Jan 17 at 11:01














    • 1




      $begingroup$
      See artofproblemsolving.com/wiki/index.php/…
      $endgroup$
      – lab bhattacharjee
      Jan 17 at 11:01








    1




    1




    $begingroup$
    See artofproblemsolving.com/wiki/index.php/…
    $endgroup$
    – lab bhattacharjee
    Jan 17 at 11:01




    $begingroup$
    See artofproblemsolving.com/wiki/index.php/…
    $endgroup$
    – lab bhattacharjee
    Jan 17 at 11:01











    1












    $begingroup$

    If you just call $x=cos(alpha)$ and $y=sin(beta)$ you get the 4th degree polynomial equation $x^4-4sqrt{2}xy + y^4 +2 =0$. I don't see a clever way to solve this directly but this can be solved algebraically. So I would put this equation into mathematica or some similar software and then look a the 4 solutions.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      If you just call $x=cos(alpha)$ and $y=sin(beta)$ you get the 4th degree polynomial equation $x^4-4sqrt{2}xy + y^4 +2 =0$. I don't see a clever way to solve this directly but this can be solved algebraically. So I would put this equation into mathematica or some similar software and then look a the 4 solutions.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        If you just call $x=cos(alpha)$ and $y=sin(beta)$ you get the 4th degree polynomial equation $x^4-4sqrt{2}xy + y^4 +2 =0$. I don't see a clever way to solve this directly but this can be solved algebraically. So I would put this equation into mathematica or some similar software and then look a the 4 solutions.






        share|cite|improve this answer









        $endgroup$



        If you just call $x=cos(alpha)$ and $y=sin(beta)$ you get the 4th degree polynomial equation $x^4-4sqrt{2}xy + y^4 +2 =0$. I don't see a clever way to solve this directly but this can be solved algebraically. So I would put this equation into mathematica or some similar software and then look a the 4 solutions.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 17 at 10:29









        quaraguequarague

        353110




        353110






























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