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Thread: 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 $\displaystyle cos6x$ and use the result to solve $\displaystyle 32x^6-48x^4+18x^2=1$.


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

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



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

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

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


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

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

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

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

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