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Math Help - De Moivre, and trigonometric identities

  1. #1
    Junior Member Evales's Avatar
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    De Moivre, and trigonometric identities

    Question:
    Use De Moivre's Theorem to express cos6x and use the result to solve 32x^6-48x^4+18x^2=1.


    My problems:
    I was under the impression that De Moivre's theorem was something like (cisx)^n=cisnx. I don't see how it applies.

    My working:
    cos6x = cos^23x - sin^23x
    cos6x = (16cos^6x + 9cos^2 - 24cos^4x) - (16sin^6x + 9sin^2 - 24sin^4x



    What is this? lol
    How do I use DeMoivre's Theorem?
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  2. #2
    Flow Master
    mr fantastic's Avatar
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    Quote Originally Posted by Evales View Post
    Question:
    Use De Moivre's Theorem to express cos6x and use the result to solve 32x^6-48x^4+18x^2=1.

    My problems:
    I was under the impression that De Moivre's theorem was something like (cisx)^n=cisnx. I don't see how it applies.

    My working:
    cos6x = cos^23x - sin^23x
    cos6x = (16cos^6x + 9cos^2 - 24cos^4x) - (16sin^6x + 9sin^2 - 24sin^4x


    What is this? lol
    How do I use DeMoivre's Theorem?
    (\text{cis} x)^6 = \text{cis} (6x) = \cos (6x) + i \sin (6x) .... (1)

    (\text{cis} x)^6 = (\cos x + i \sin x)^6 = (\cos^6 x - 15 \cos^4 x \sin^2 x + 15 \cos^2 x \sin^4 x - \sin^6 x) + i (6 \cos^5 x \sin x - 20 \cos^3 x \sin^3 x + 6 \cos x \sin^5 x) .... (2)

    Equate the real components of (1) and (2) and you get an expression for \cos (6x).

    Write this expression in terms of powers of \cos x. Strangely enough, you get

    \cos (6 x) = 32 \cos^6 x - 48 \cos^4 x + 18 \cos^2 x - 1.

    So the equation 32 x^6 - 48 x^4 + 18 x^2 - 1 = 0 is equivalent to the equation \cos (6 \theta) = 0 where x = \cos \theta \, ....
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