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Math Help - finding the exact trigonometric ratios of an equation

  1. #1
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    finding the exact trigonometric ratios of an equation

    Find the exact trigonometric roots of  16x^5 - 20x^3 + 5x + 1 = 0
    given that  cos5\theta = 16cos^5\theta - 20cos^3\theta + 5cos\theta

    heres how I did it

    let x= cos@
    cos5@ = 16x^5 - 20x^3 + 5x + 1
    cos5@ = 0, 16x^5 -20x^3 + 5x + 1 = 0

    5@ = 2(pi)k +/- pi/2 where k is any integer
    = (2(pi)k +/- pi)/10

    x = cos@
    = cos {2(pi)k +/- pi]/10}
    where k is any integer

    but how do I proceed to find the five roots of the equation. and is the above correct
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  2. #2
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    Hello differentiate
    Quote Originally Posted by differentiate View Post
    Find the exact trigonometric roots of  16x^5 - 20x^3 + 5x + 1 = 0
    given that  cos5\theta = 16cos^5\theta - 20cos^3\theta + 5cos\theta

    heres how I did it

    let x= cos@
    cos5@ = 16x^5 - 20x^3 + 5x + 1
    cos5@ = 0, 16x^5 -20x^3 + 5x + 1 = 0

    5@ = 2(pi)k +/- pi/2 where k is any integer
    = (2(pi)k +/- pi)/10

    x = cos@
    = cos {2(pi)k +/- pi]/10}
    where k is any integer

    but how do I proceed to find the five roots of the equation. and is the above correct
    You have more-or-less the right idea, but are wrong in one or two details.

    Let x = \cos\theta

    Then \cos5\theta = 16x^5 -20x^3+5x (not, as you said, 16x^5 -20x^3+5x+1)

    So 16x^5 -20x^3+5x+1=0

    \Rightarrow 16x^5 -20x^3+5x=-1

    \Rightarrow \cos5\theta = -1

    \Rightarrow 5\theta = (2n+1)\pi, n \in \mathbb{Z}

    \Rightarrow \theta = \frac{(2n+1)\pi}{5}

    So the 5 roots are \cos\left(\frac{(2n+1)\pi}{5}\right), n = 0,1, 2, 3, 4

    (Note that these are not all distinct. The equation has two pairs of equal roots, because \cos\left( \frac{\pi}{5}\right)=\cos\left( \frac{9\pi}{5}\right) and \cos\left( \frac{3\pi}{5}\right)=\cos\left( \frac{7\pi}{5}\right).)

    Grandad
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