Very simple

hadi amiri 4 · Jun 6, 2008

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

arildno · Jun 6, 2008

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.

hadi amiri 4 · Jun 6, 2008

you did not use the condition

tiny-tim · Jun 6, 2008

hadi amiri 4 said: ↑

you did not use the condition

Hi hadi!

arildno left that to you!

If 0 < x < π/2, and u = tanx, then the condition on u is … ?

hadi amiri 4 · Jun 6, 2008

i found this problems in a book which was just talking about trigonometry and that book was empty of calculus

tiny-tim · Jun 6, 2008

hadi amiri 4 said: ↑

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!

I really have no idea how to do this with trignonometry.

hadi amiri 4 · Jun 6, 2008

you are right

hadi amiri 4 · Jun 6, 2008

please tell me where is the appropriate sub-forum

tiny-tim · Jun 6, 2008

Hi hadi!

That would be Precalculus Mathematics, at https://www.physicsforums.com/forumdisplay.php?f=155

HallsofIvy · Jun 6, 2008

[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]

hadi amiri 4 · Jun 7, 2008

Are you sure that you have found the minimum of this problem?
or changing tan to sin/cos and .....