Cos(pi/7) and cos (pi/9)

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Cos(pi/7) and cos (pi/9)

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#1 h May 20, 2014, 12:10 PM • 2 Y

Y by Adventure10, Mango247

Prove that:
$\frac{1}{\sqrt[3]{\cos \frac{\pi}{7}}} - \frac{1}{\sqrt[3]{\cos \frac{2\pi}{7}}} +\frac{1}{\sqrt[3]{\cos \frac{3\pi}{7}}} =\sqrt[3]{6\sqrt[3]{7}-8}$
$\frac{1}{\sqrt[3]{\cos \frac{\pi}{9}}} - \frac{1}{\sqrt[3]{\cos \frac{2\pi}{9}}} +\frac{1}{\sqrt[3]{\cos \frac{5\pi}{9}}} =\sqrt[3]{6-6\sqrt[3]{9}}$

This post has been edited 1 time. Last edited by _Mark_01, May 20, 2014, 7:46 PM

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nsato

#2 h May 20, 2014, 2:44 PM • 3 Y

Y by _Mark_01, Ha_ha_ha, Adventure10

Your second equation
$\frac{1}{\sqrt[3]{\cos\frac{\pi}{9}}}-\frac{1}{\sqrt[3]{\cos\frac{2\pi}{9}}}+\frac{1}{\sqrt[3]{\cos\frac{3\pi}{9}}}=\sqrt[3]{6-6\sqrt[3]{9}}$
is not true. The left-hand side is positive, but the right-hand side is negative.

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arqady

#3 h Aug 28, 2014, 1:01 AM • 5 Y

Y by kprepaf, lalaman125, Ha_ha_ha, Adventure10, Mango247

_Mark_01 wrote:

Prove that:
$\frac{1}{\sqrt[3]{\cos \frac{\pi}{9}}} - \frac{1}{\sqrt[3]{\cos \frac{2\pi}{9}}} +\frac{1}{\sqrt[3]{\cos \frac{5\pi}{9}}} =\sqrt[3]{6-6\sqrt[3]{9}}$

Id est, we need to prove that $\frac{1}{\sqrt[3]{\cos \frac{2\pi}{9}}} + \frac{1}{\sqrt[3]{\cos \frac{4\pi}{9}}} +\frac{1}{\sqrt[3]{\cos \frac{8\pi}{9}}} =\sqrt[3]{6\sqrt[3]{9}-6}$ .
We obtain:
$\cos\frac{2\pi}{9}+\cos\frac{4\pi}{9}+\cos\frac{8\pi}{9}=2\cos\frac{2\pi}{3}\cos\frac{\pi}{9}-\cos\frac{\pi}{9}=0$ .
$\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}+\cos\frac{2\pi}{9}\cos\frac{8\pi}{9}+\cos\frac{4\pi}{9}\cos\frac{8\pi}{9}=$
$=\frac{1}{2}\left(\cos\frac{2\pi}{3}+\cos\frac{2\pi}{9}+\cos\frac{8\pi}{9}+\cos\frac{2\pi}{3}+\cos\frac{2\pi}{3}+\cos\frac{4\pi}{9}\right)=-\frac{3}{4}$ .
$\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}\cos\frac{8\pi}{9}=\frac{8\sin\frac{2\pi}{9}\cos\frac{2\pi}{9}\cos\frac{4\pi}{9}\cos\frac{8\pi}{9}}{8\sin\frac{2\pi}{9}}=\frac{\sin\frac{16\pi}{9}}{8\sin\frac{2\pi}{9}}=-\frac{1}{8}$ .
Let $\cos\frac{2\pi}{9}=a$ , $\cos\frac{4\pi}{9}=b$ , $\cos\frac{8\pi}{9}=c$ , $\sqrt[3]a+\sqrt[3]b+\sqrt[3]c=x$ and $\sqrt[3]{ab}+\sqrt[3]{ac}+\sqrt[3]{bc}=y$ .
Hence, $a+b+c=0$ , $ab+ac+bc=-\frac{3}{4}$ , $abc=-\frac{1}{8}$ and we need to prove that $y=\frac{1}{2}\sqrt[3]{6-6\sqrt[3]{9}}$ .
Indeed, $x^3=a+b+c+3xy-3\sqrt[3]{abc}=3xy+\frac{3}{2}$ and
$y^3=ab+ac+bc+3xy\sqrt[3]{abc}-3\sqrt[3]{a^2b^2c^2}=-\frac{3}{2}xy-\frac{3}{2}$ .
Thus, $x^3y^3=\left(3xy+\frac{3}{2}\right)\left(-\frac{3}{2}xy-\frac{3}{2}\right)=-\frac{9}{2}x^2y^2-\frac{27}{2}xy-\frac{9}{4}$
or $8x^3y^3+18x^2y^2+54xy+18=0$
or $(2xy+3)^3=9$
or $xy=\frac{1}{2}\left(\sqrt[3]9-3\right)$ .
Id est, $y=\frac{1}{2}\sqrt[3]{6-6\sqrt[3]{9}}$ and we are done.
The first equality we can prove by the same way.

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arqady

#4 h Aug 28, 2014, 1:07 AM • 4 Y

Y by kprepaf, Ha_ha_ha, Adventure10, Mango247

For the collection.
Let $a$ , $b$ and $c$ be roots of the equation $x^3-16x^2-57x+1=0$ . Prove that:
$\sqrt[5]a+\sqrt[5]b+\sqrt[5]c=1$

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#5 h Aug 28, 2014, 1:11 PM • 3 Y

Y by Ha_ha_ha, Adventure10, Mango247

How can this be proved arqady? Have you a solytion?

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arqady

#6 h Aug 28, 2014, 4:31 PM • 3 Y

Y by Ha_ha_ha, Adventure10, Mango247

Iliopoulos wrote:

How can this be proved arqady?

Try to guess a roots.

Iliopoulos wrote:

Have you a solytion?

Of course! It's my problem.

This post has been edited 1 time. Last edited by arqady, Aug 28, 2014, 4:34 PM

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#8 h Aug 29, 2014, 6:11 PM • 2 Y

Y by Ha_ha_ha, Adventure10

The polynomial $x^3-16x^2-57x+1$ admits $3$ real roots: one negative ( $a$ ) and two positive ( $b,c$ ). Then $a'=a^{1/5}<0,b'=b^{1/5}>0,c'=c^{1/5}>0$ . $a',b',c'$ are some roots of the polynomial $u^{15}-16u^{10}-57u^5+1=(u^3-u^2-2u+1)P_{12}(u)$ where $P_{12}$ is a polynomial of degree $12$ without any real roots. Then $a',b',c'$ are the roots of $u^3-u^2-2u+1$ . Finally $a'+b'+c'=1$ and we are done

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arqady

#9 h Aug 30, 2014, 3:58 AM • 3 Y

Y by Ha_ha_ha, Adventure10, Mango247

loup blanc wrote:

$u^{15}-16u^{10}-57u^5+1=(u^3-u^2-2u+1)P_{12}(u)$ where $P_{12}$ is a polynomial of degree $12$ without any real roots.

How we can get it during a competition without computer?

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#10 h Aug 30, 2014, 9:32 AM • 3 Y

Y by Ha_ha_ha, Adventure10, Mango247

Yes Michael, you are right.
Let $f=u^{15}-16u^{10}-57u^5+1$ . $f''$ presents a sole signum change in $\alpha>1$ . Then $f'$ has $2$ real roots $\beta>1,\gamma<0$ . Finally $f$ has at most $3$ real roots.
EDIT: I'm going too fast. It remains to find, with hand, the factor $u^3-u^2-2u+1$ .

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#11 h Aug 30, 2014, 11:18 AM • 2 Y

Y by Ha_ha_ha, Adventure10

I did not understand why $a',b',c'$ are conjugate over $\mathbb{Z}$ . The reason is because $a,b,c$ are fifth powers:
$(-4u^2+2)^5,(-2u)^5,(4u^2+2u-1)^5$ where $u=\cos(2\pi/7)$ .

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