Very simple

  • #1

Main Question or Discussion Point

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
 

Answers and Replies

  • #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.
 
  • #3
you did not use the condition
 
  • #4
tiny-tim
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you did not use the condition
Hi hadi! :smile:

arildno left that to you!

If 0 < x < π/2, and u = tanx, then the condition on u is … ? :smile:
 
  • #5
i found this problems in a book which was just talking about trigonometry and that book was empty of calculus
 
  • #6
tiny-tim
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i found this problems in a book which was just talking about trigonometry and that book was empty of calculus
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:
 
  • #7
you are right
 
  • #8
please tell me where is the appropriate sub-forum
 
  • #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]
 
  • #11
Are you sure that you have found the minimum of this problem?
or changing tan to sin/cos and .....
 

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