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Math Help - lowering powers help please

  1. #1
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    Post lowering powers help please

    Use the formulas for lowering powers to rewrite: cos^4x sin^4x in terms of the first power of cosine?
    help me please I dont know how to do it.
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  2. #2
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    Re: lowering powers help please

    Hello, aser03!

    We are expected to know these identities:

    . . \sin^2\!\theta + \cos^2\!\theta \:=\:1 \quad\Rightarrow\quad \sin^2\!\theta \:=\:1-\cos^2\!\theta

    . . \sin2\theta \,=\,2\sin\theta\cos\theta

    . . \cos^2\!\theta \:=\:\frac{1+\cos2\theta}{2}



    \text{Use the formulas for lowering powers to rewrite }\cos^4\!x\sin^4\!x
    . . \text{ in terms of the first power of cosine.}

    \cos^4\!x\sin^4\!x \;=\;(\sin x\cos x)^4

    . . . . . . . . =\;\tfrac{1}{16}(2\sin x\cos x)^4

    . . . . . . . . =\;\tfrac{1}{16}(\sin2x)^4

    . . . . . . . . =\;\tfrac{1}{16}(\sin^2\!2x)^2

    . . . . . . . . =\;\tfrac{1}{16}(1-\cos^2\!2x)^2

    . . . . . . . . =\;\tfrac{1}{16}\left(1 - \tfrac{1\:+\:\cos4x}{2}\right)^2

    . . . . . . . . =\;\tfrac{1}{16}\left(\tfrac{2\,-\,(1\,+\,\cos4x)}{2}\right)^2

    . . . . . . . . =\;\tfrac{1}{16}\left(\tfrac{1\,-\,\cos4x}{2}\right)^2

    . . . . . . . . =\;\tfrac{1}{16}\cdot\tfrac{1}{4}(1-\cos4x)^2

    . . . . . . . . =\;\tfrac{1}{64}(1 - 2\cos4x + \cos^2\!4x)

    . . . . . . . . =\;\tfrac{1}{64}\left(1 - 2\cos4x + \tfrac{1 + \cos8x}{2}\right)

    . . . . . . . . =\;\tfrac{1}{64}\left(1 - 2\cos4x + \tfrac{1}{2} + \tfrac{1}{2}\cos8x\right)

    . . . . . . . . =\;\tfrac{1}{64}\left(\tfrac{3}{2} - 2\cos4x + \tfrac{1}{2}\cos8x\right)

    . . . . . . . . =\;\tfrac{1}{128}(3 - 4\cos4x + \cos8x)
    Thanks from Gusbob
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