Minimum of Tan^p + Cot^q for 0 < x < Pi/2

[hadi amiri 4]

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]

Well set u=Tan(x),
Hence, y=u(x)^p+(1/u)^q, along with $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$$
should settle it nicely.

 

[hadi amiri 4]

you did not use the condition

 

[tiny-tim]

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]

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

 

[tiny-tim]

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]

you are right

 

[hadi amiri 4]

please tell me where is the appropriate sub-forum

 

[tiny-tim]

Hi hadi!

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

 

[HallsofIvy]

$$tan(x)= \frac{sin(x)}{cos(x)}$$
and
$$cot(x)= \frac{cos(x)}{sin(x)}$$
so
$$tan^p(x)+ cot^q(x)= \frac{sin^p(x)}{cos^p(x)}+ \frac{cos^q(x)}{sin^q(x)}= \frac{sin^{p+q}(x)+ cos^{p+q}(x)}{sin^q(x)cos^p(x)}$$

 

[hadi amiri 4]

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