Results 1 to 3 of 3

Math Help - Sec-calcul

  1. #1
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    534

    Sec-calcul

    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 />
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by dhiab View Post
    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.
    Last edited by Opalg; September 13th 2009 at 12:36 PM. Reason: Corrected arithmetic mistakes
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member pacman's Avatar
    Joined
    Jul 2009
    Posts
    448
    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?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Complex -calcul
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: December 24th 2009, 06:57 AM
  2. [SOLVED] simple &amp; complicated calcul
    Posted in the Algebra Forum
    Replies: 4
    Last Post: December 6th 2009, 05:49 PM
  3. [SOLVED] calcul riddle
    Posted in the Algebra Forum
    Replies: 9
    Last Post: November 6th 2009, 01:59 PM
  4. [SOLVED] calcul riddle
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 6th 2009, 01:14 PM
  5. Roots and calcul
    Posted in the Algebra Forum
    Replies: 6
    Last Post: September 8th 2009, 05:12 AM

Search Tags


/mathhelpforum @mathhelpforum