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Math Help - Maclaurin expansion

  1. #1
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    Maclaurin expansion

    Find the Maclaurin expansion for  sin^4x using the double angle identity for cos2x and cos4x.

    So...
    This is how i worked out so far, yet i'm not sure whether it is correct or not

    Identity of cos4x = cos(2x + 2x) = cos2x.cos2x + sin2x.sin2x
    using the identity of  cos2x = (1 - 2sin^2x)(1-2sin^2x) + sin2x.sin2x

    which equals to

    <br />
cos4x = 1 - 4sin^2x + 4sin^4x + sin2x.sin2x <br />

    hence
    <br />
sin^4x = \frac{cos4x + 4sin^2x - sin2x.sin2x - 1}{4} = sin^4x<br />

    If it is correct i'm not sure whether to continue with the problem or not, this question has 8 marks...and i'm puzzled.. if anyone could help it would be greatly appreciated!
    thanks in advance
    kleyzam
    Last edited by kleyzam; November 6th 2008 at 03:35 PM.
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  2. #2
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    Hello, kleyzam!

    Find the Maclaurin expansion for \sin^4\!x
    using the double-angle identity for \cos2x and \cos4x.
    I'd approach it like this . . .

    (\sin^2\!x)^2 \;=\;\left(\frac{1-\cos2x}{2}\right)^2 \;=\;\tfrac{1}{4}\left(1 - 2\cos2x + \cos^2\!2x\right)

    . . = \;\tfrac{1}{4}\left(1 - 2\cos2x + \frac{1+\cos4x}{2}\right) \;=\;\tfrac{1}{8}\left(\cos4x - 4\cos2x + 3\right)


    Then: . \sin^4\!x \;=\;\frac{1}{8}\begin{Bmatrix}\cos4x \\ -4\cos2x \\ + 3 \end{Bmatrix} \;= . \begin{Bmatrix}1 - \frac{(4x)^2}{2!} + \frac{(4x)^4}{4!} - \frac{(4x)^6}{6!} + \hdots \\ \\[-4mm]<br />
\text{-}4\left[1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \frac{(2x)^6}{6!} + \hdots \right] \\ \\[-4mm] + 3 \end{Bmatrix} . . . . etc.


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