Cosine and Inequality

kunny

Prove the following inequality.

$\cos^{4006}\left(\frac{1}{2}\right)\cos^{4005}\left(\frac{1}{3}\right)\cos^{4004}\left(\frac{1}{4}\right)\cdot\cdot\cdot\cos^...$

Kent Merryfield

One quick reaction and a place not to go: I was thinking about variations on Jensen or AM-GM, but the inequality sign is pointing in the wrong direction for that.

**Posted:** Thu Oct 21, 2004 12:10 pm

kunny

Thank you again, Kent Merryfield! Razz

I look forward to your solution.

kunny

**Posted:** Thu Oct 21, 2004 12:18 pm

kunny

Can anyone solve this problem?

**Posted:** Sun Dec 12, 2004 6:36 pm

Moubinool

I wonder if you get this inequality by comparing

$\cos^{2n}\left(\frac{1}{2}\right)\cos^{2n-1}\left(\frac{1}{3}\right)\cos^{2n-2}\left(\frac{1}{4}\right)\cdot\cdot\cdot\cos^{2...$

and

$\frac{\sqrt{n+1}}{2^{n}}$

**Posted:** Fri Dec 17, 2004 2:37 pm

kunny

Yes.Tha'ts right!

**Posted:** Fri Dec 17, 2004 4:39 pm

Moubinool

I tried different ideas, but no proof.
Can you give an hint for this one kunny ?

**Posted:** Mon Dec 27, 2004 2:41 pm

kunny

O.K., Moubinool. Smile

The hint is as follows.

You can use the inequality $x\geqq |\sin x|\ (x\geqq 0)$ .

kunny

**Posted:** Mon Dec 27, 2004 2:58 pm

Moubinool

(1) $|\cos x|\geq \sqrt{1-x^2}$ for $x=1/2;...;1/(2n+1)$

Call the product of cosinus, $P_n$ , you get with (1)

$P_n\geq Q_n=\prod_{k=2}^{2n+1} ( 1-\frac{1}{k^2})^{\frac{2n-k+2}{2}}$

From this point simplify $Q_n$ and get the minoration $\frac{\sqrt{n+1}}{2^n}$

**Posted:** Tue Dec 28, 2004 2:24 am