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Math Help - complex exponential

  1. #1
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    complex exponential

    Hi,
    I need help using the complex exponential to express
    cos6x as a polynomial in terms of cosx

    I don't know how to get it in terms of cosx without sinx.

    Thankyou,
    ArTiCk
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ArTiCK View Post
    Hi,
    I need help using the complex exponential to express
    cos6x as a polynomial in terms of cosx

    I don't know how to get it in terms of cosx without sinx.

    Thankyou,
    ArTiCk
    cos(6x) = Re[e^{6ix}] = Re[(cos(x) + i~sin(x))^6]
    where "Re" is the real part of the expression.

    -Dan
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  3. #3
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    Quote Originally Posted by ArTiCK View Post
    Hi,
    I need help using the complex exponential to express
    cos6x as a polynomial in terms of cosx

    I don't know how to get it in terms of cosx without sinx.

    Thankyou,
    ArTiCk
    \cos (6x) = Re\left[e^{i(6x)}\right] = Re \left[ (e^{ix})^6 \right]

     = Re [ (\cos x + i \sin x)^6]

     = Re[ \cos^6 x + 6i \cos^5 x \sin x - 15 \cos^4 x \sin^2 x - 20 i \cos^3 x \sin^3 x + 15 \cos^2 x \sin^4 x + 6i \cos x \sin^5 x - \sin^6 x]

    where the expansion is done using the binomial theorem

    = Re[(\cos^6 x - 15 \cos^4 x \sin^2 x + 15 \cos^2 x \sin^4 x - \sin^6 x) + i ( .......)]

     = \cos^6 x - 15 \cos^4 x \sin^2 x + 15 \cos^2 x \sin^4 x - \sin^6 x

    = \cos^6 x - 15 \cos^4 x (1 - \cos^2 x) + 15 \cos^2 x (1 - \cos^2 x)^2 - (1 - \cos^2 x)^3

    and I'll leave the expansion and subsequent simplification for you.

    I get 18 \cos^6 x - 20 \cos^4 x + 4 \cos^2 x - 1 ......

    You should check that your answer gives 1 when you substitute x = 0. (Or that you get -1 when you substitute x = \frac{\pi}{6} )


    ps: I reserve the right for at least one error in this reply.
    Last edited by mr fantastic; April 11th 2008 at 04:46 AM. Reason: Added two missing ]'s (not that anyone would notice!) and moved some stuff to a new line.
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