Sec-calcul

**dhiab** · September 12th 2009, 08:31 AM

Calculate : $\sec ^8 \left( {\frac{\pi }{9}} \right) + \sec ^8 \left( {\frac{{5\pi }}{9}} \right) + \sec ^8 \left( {\frac{{7\pi }}{9}} \right)<br />$

**Opalg** · September 12th 2009, 12:13 PM

Originally Posted by dhiab

Calculate : $\sec ^8 \left( {\frac{\pi }{9}} \right) + \sec ^8 \left( {\frac{{5\pi }}{9}} \right) + \sec ^8 \left( {\frac{{7\pi }}{9}} \right)<br />$

$\frac\pi9,\ \frac{5\pi}9,\ \frac{7\pi}9$ are the roots of the equation $\cos(3\theta) = \tfrac12$ . But $\cos(3\theta) = 4\cos^3\theta-3\cos\theta$ . Therefore $\cos\Bigl(\frac\pi9\Bigr),\ \cos\Bigl(\frac{5\pi}9\Bigr),\ \cos\Bigl(\frac{7\pi}9\Bigr)$ are the roots of the equation $4x^3-3x = \tfrac12$ . Writing y = 1/x, you see that $\sec\Bigl(\frac\pi9\Bigr),\ \sec\Bigl(\frac{5\pi}9\Bigr),\ \sec\Bigl(\frac{7\pi}9\Bigr)$ are the roots of the equation $y^3 + 6y^2 - 8 = 0$ .

If we call the roots of that equation $\alpha,\ \beta,\ \gamma$ then we know that $\textstyle\sum\alpha = -6$ , $\textstyle\sum\beta\gamma = 0$ and $\alpha\beta\gamma = 8$ . Now successively compute that
$\textstyle\sum\alpha^2 = \bigl(\sum\alpha\bigr)^2 - 2\sum\beta\gamma = 36$ ,
$\textstyle\sum\beta^2\gamma^2 = \bigl(\sum\beta\gamma\bigr)^2 - 2\alpha\beta\gamma\sum\alpha = 96$ ,
$\textstyle\sum\alpha^4 = \bigl(\sum\alpha^2\bigr)^2 - 2\sum\beta^2\gamma^2 = 36^2 - 192 = 1104$ ,
$\textstyle\sum\beta^4\gamma^4 = \bigl(\sum\beta^2\gamma^2\bigr)^2 - 2(\alpha\beta\gamma)^2\sum\alpha^2 = 96^2 - 128\times36 = 4608$ ,
$\textstyle\sum\alpha^8 = \bigl(\sum\alpha^4\bigr)^2 - 2\sum\beta^4\gamma^4 = 1104^2 - 2\times4608 = \boxed{1,209,600}$ .

That looks like a large number, but if you compute the 8th powers of $\sec\Bigl(\frac\pi9\Bigr)\approx1.064$ , $\sec\Bigl(\frac{5\pi}9\Bigr)\approx-5.759$ and $\sec\Bigl(\frac{7\pi}9\Bigr)\approx-1.305$ , you'll find that the sum comes out right.

**pacman** · September 13th 2009, 10:48 PM

My calculator gives this value: sec^8 (pi/9) + sec^8 (5pi/9) + sec^8 (7pi/9) > 1.2096

How to get a closed form answer for this?

dhiab, any hint?

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