How Does the Inequality (cos(x))^p ≤ cos(px) Work?

[medinap]

Homework Statement

(cos(x))^p \leq cos(px)

0\leqx\leqpi half

and p, 0\leq(not equal) p \leq(not equal) 1

i need help, if some one can tell me how to started, what should i used i will really apreciate it! (sorry for my english )

Homework Equations

The Attempt at a Solution

 

[Nanie]

http://img218.imageshack.us/img218/6451/ine111es8.jpg

 

[tiny-tim]

without LaTeX

http://img218.imageshack.us/img218/6451/ine111es8.jpg

[/URL]

Hi Nanie!

Or, without LaTeX:

(cosθ)^p ≤ cos(pθ)

0 ≤ θ ≤ π/2

0 < p < 1.

 

[Nanie]

jeje thanks

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I'm friend of medinap...we really need help with this exercise!

First the professor...gave us (cos\theta)^p\leq cos(p\theta) but he don't gave us the values of p (we asume that n is for all natural p=2(2n+1)...but yesterday! he made a correction that 0< p< 1

I don't know what to do

and sorry for my english too.

 

[HallsofIvy]

Your English is excellent!

Perhaps the simplest way to prove this is to use a Taylor's polynomial. cos(\theta)= 1- (1/2)\theta^2+ (1/4!)\theta^4-\cdot\cdot\cdot
so cos(\theta)\le 1- (1/2)\theta^2. Now use the extended binomial theorem to take that to the p power. You only need the first two terms.

 

[Nanie]

i don't know how to used it, explain me please!

 

[Nanie]

i know how to used the binomial theorem, but i don't know how to used it with taylor's polynomial i don't even know what that is..

 

[tiny-tim]

Hi medinap! Hi Nanie!

Hint: differentiate (cosθ)^p and cos(pθ), and remember that cos ≤ 1.

(btw, you both need to be more careful about using the past tense …

it's "if some one can tell me how to start, what should i use …"

and "he didn't gave us the values of p (we assumed that …" )​

 

[Nanie]

ok thanks... I will try! jeje...I have one hour to think !

 

[Nanie]

F(\theta) = cos (\theta) ^p- cos (p\theta)

How I use this?

iah...I don't Know...

F(theta) = cos (theta)^p- cos (p(theta))

 

[tiny-tim]

F(\theta) = cos (\theta) ^p- cos (p\theta)

How I use this?

Hi Nanie!

(what happened to that θ i gave you? )

What is F'(θ)? Is it positive or negative?

 

[Nanie]

Thanks!

F(\theta) = p (sen\theta/cos^{1-p}) - sen(p\theta) > 0 ...however ... cos (p\theta) > cos (\theta)^{p}



F'(\theta) = p (sen\theta/cos^{1-p} - sen p (\theta) > 0 (positive)


We did it...?