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Thread: trigo-sum

  1. #1
    Super Member dhiab's Avatar
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    Thumbs up trigo-sum

    Calculate this sum :
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    Last edited by dhiab; Jun 16th 2009 at 09:36 PM.
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  2. #2
    Master Of Puppets
    pickslides's Avatar
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    you need to evaluate $\displaystyle cos(\frac{\pi}{9})$

    To do this you need to consider $\displaystyle cos(a\pm b) = cos(a)\times cos(b)\mp sin(a)\times sin(b)$

    where $\displaystyle cos(\frac{\pi}{9}) = cos(a\pm b)$

    only after finding this will you be able to get an exact value
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  3. #3
    Flow Master
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    Quote Originally Posted by pickslides View Post
    you need to evaluate $\displaystyle cos(\frac{\pi}{9})$

    To do this you need to consider $\displaystyle cos(a\pm b) = cos(a)\times cos(b)\mp sin(a)\times sin(b)$

    where $\displaystyle cos(\frac{\pi}{9}) = cos(a\pm b)$

    only after finding this will you be able to get an exact value
    This technique will not work in this instance.


    Edit: The off-topic discussion regarding the value of questions like this one can be found here: http://www.mathhelpforum.com/math-he...t-century.html. Any further discussion of this nature should be continued at that thread.
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  4. #4
    Super Member dhiab's Avatar
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    help

    help: use relation
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  5. #5
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by dhiab View Post
    help: use relation
    Or you can try

    $\displaystyle
    (a_1^2+a_2^2+a_3^2+a_4^2)(b_1^2+b_2^2+b_3^2+b_4^2) =
    $

    $\displaystyle
    (a_1 b_1 - a_2 b_2 - a_3 b_3 - a_4 b_4)^2 +
    (a_1 b_2 + a_2 b_1 + a_3 b_4 - a_4 b_3)^2 +
    $

    $\displaystyle (a_1 b_3 - a_2 b_4 + a_3 b_1 + a_4 b_2)^2 +
    (a_1 b_4 + a_2 b_3 - a_3 b_2 + a_4 b_1)^2$

    although I don't promise anything.








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  6. #6
    Flow Master
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    Quote Originally Posted by dhiab View Post
    help: use relation
    That thought had crossed my mind but it will then be necessary to solve the cubic equation $\displaystyle 4 t^3 - 3t - \frac{1}{2} = 0$ and I think that will be no mean feat (the method of Cardano could be used I suppose).

    There will be a clever trick but I don't have time right.
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  7. #7
    Super Member malaygoel's Avatar
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    Let us consider the equation
    $\displaystyle cos3\theta=cos\frac{\pi}{3}$

    there are infinite solutions of this equation
    in the domain $\displaystyle [0,2\pi]$, the solutions are

    $\displaystyle \theta=\frac{\pi}{9},\frac{5\pi}{9},\frac{7\pi}{9} ,\frac{11\pi}{9},\frac{13\pi}{9},\frac{17\pi}{9}.. ...............(a)$

    Equation1:
    $\displaystyle cos3\theta=cos\frac{\pi}{3}$
    Equation 2:
    $\displaystyle 4cos^{3}x-3cosx=1/2$
    The above two equations are identical. Every value of $\displaystyle \theta$ that satisfies equation 1 satisfies equation 2. If $\displaystyle \theta$ is a solution of equation 1, the $\displaystyle cos\theta$ is a root of the second equation.

    For above solutions of $\displaystyle \theta$

    $\displaystyle cos\theta=cos\frac{\pi}{9},cos\frac{5\pi}{9},cos\f rac{7\pi}{9},cos\frac{11\pi}{9},cos\frac{13\pi}{9} ,cos\frac{17\pi}{9}$
    the last three of the above six are just the repetition of the first three. So the roots of Equation 2 can be

    $\displaystyle cos\frac{\pi}{9}(=cos\frac{17\pi}{9}),cos\frac{5\p i}{9}(=cos\frac{13\pi}{9}),cos\frac{7\pi}{9}(=cos\ frac{11\pi}{9})$

    Hence, we can say that $\displaystyle cos\frac{\pi}{9},cos\frac{5\pi}{9},cos\frac{7\pi}{ 9}$ are the three roots of equation 2.

    Multipying equation 2 by $\displaystyle sec^{3}\theta$
    we get

    $\displaystyle sec^{3}\theta+6sec^{2}\theta-8=0$

    roots are$\displaystyle sec\frac{\pi}{9},sec\frac{5\pi}{9},sec\frac{7\pi}{ 9}$

    $\displaystyle sec\frac{\pi}{9}+sec\frac{5\pi}{9}+sec\frac{7\pi}{ 9}=-6$

    $\displaystyle sec\frac{\pi}{9}.sec\frac{5\pi}{9}+sec\frac{7\pi}{ 9}.sec\frac{\pi}{9}+sec\frac{5\pi}{9}.sec\frac{7\p i}{9}=0$

    the question of this thread is
    $\displaystyle sec^{2}\frac{\pi}{9}+sec^{2}\frac{5\pi}{9}+sec^{2} \frac{7\pi}{9}+4$

    $\displaystyle =(sec\frac{\pi}{9}+sec\frac{5\pi}{9}+sec\frac{7\pi }{9})^{2}-2(sec\frac{\pi}{9}.sec\frac{5\pi}{9}+sec\frac{7\pi }{9}.sec\frac{\pi}{9}+sec\frac{5\pi}{9}.sec\frac{7 \pi}{9})+4$

    $\displaystyle =(-6)^{2}+2(0)+4$

    $\displaystyle =40$
    Last edited by malaygoel; Jun 19th 2009 at 01:30 AM.
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  8. #8
    Super Member dhiab's Avatar
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    Quote Originally Posted by mr fantastic View Post
    This technique will not work in this instance.


    Edit: The off-topic discussion regarding the value of questions like this one can be found here: http://www.mathhelpforum.com/math-he...t-century.html. Any further discussion of this nature should be continued at that thread.
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