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Math Help - lowering powers

  1. #1
    Newbie
    Joined
    May 2010
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    lowering powers

    rewrite the expression in terms of the first power of cosine

    sin^4x

    so far i have
    (sin^2x)^2

    = (1-cos2x/2)^2

    = (1-cos2x/2)(1-cos2x/2)

    =1-cos^2.2x/4

    so my question is, is it right? or is there more
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, reino17!

    Your algebra is off . . .


    Rewrite the expression in terms of the first power of cosine: . \sin^4x

    \sin^4\!x \;\;=\;\;\left(\sin^2\!x\right)^2 \;\;=\;\;\left(\frac{1-\cos2x}{2}\right)^2 \;\;= . \frac{1-2\cos2x + \cos^2\!2x}{4}\;\;=\;\;\frac{1 - 2\cos2x + \frac{1 + \cos4x}{2}}{4}


    Multiply by \tfrac{2}{2}:\quad \frac{2-4\cos2x + 1 + \cos4x}{8} \;\;=\;\;\tfrac{1}{8}(3 - 4\cos 2x + \cos4x)

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  3. #3
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    May 2010
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    ahh ok i see where i screwed up. thanks dude
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