- #1

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find the minimum y=Tan(x)^p+Cot(x)^q

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- Thread starter hadi amiri 4
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- #1

- 98

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find the minimum y=Tan(x)^p+Cot(x)^q

- #2

arildno

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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

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you did not use the condition

- #4

tiny-tim

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you did not use the condition

Hi hadi!

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

- #5

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- #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!

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

- #7

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you are right

- #8

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please tell me where is the appropriate sub-forum

- #9

tiny-tim

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That would be

- #10

HallsofIvy

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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

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Are you sure that you have found the minimum of this problem?

or changing tan to sin/cos and .....

or changing tan to sin/cos and .....

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