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Very simple

  1. Jun 6, 2008 #1
    suppose p and q are positive rational numbers with the condition : 0<x<Pi/2
    find the minimum y=Tan(x)^p+Cot(x)^q
     
  2. jcsd
  3. Jun 6, 2008 #2

    arildno

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    Well set u=Tan(x),
    Hence, y=u(x)^p+(1/u)^q, along with [tex]\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}[/tex]
    should settle it nicely.
     
  4. Jun 6, 2008 #3
    you did not use the condition
     
  5. Jun 6, 2008 #4

    tiny-tim

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    Hi hadi! :smile:

    arildno left that to you!

    If 0 < x < π/2, and u = tanx, then the condition on u is … ? :smile:
     
  6. Jun 6, 2008 #5
    i found this problems in a book which was just talking about trigonometry and that book was empty of calculus
     
  7. Jun 6, 2008 #6

    tiny-tim

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    ah … you put this in the Calculus & Analysis sub-forum, so we assumed you wanted a calculus answer! :smile:

    I really have no idea how to do this with trignonometry. :redface:
     
  8. Jun 6, 2008 #7
    you are right
     
  9. Jun 6, 2008 #8
    please tell me where is the appropriate sub-forum
     
  10. Jun 6, 2008 #9

    tiny-tim

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  11. Jun 6, 2008 #10

    HallsofIvy

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    [tex] tan(x)= \frac{sin(x)}{cos(x)}[/tex]
    and
    [tex] cot(x)= \frac{cos(x)}{sin(x)}[/tex]
    so
    [tex]tan^p(x)+ cot^q(x)= \frac{sin^p(x)}{cos^p(x)}+ \frac{cos^q(x)}{sin^q(x)}= \frac{sin^{p+q}(x)+ cos^{p+q}(x)}{sin^q(x)cos^p(x)}[/tex]
     
  12. Jun 7, 2008 #11
    Are you sure that you have found the minimum of this problem?
    or changing tan to sin/cos and .....
     
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