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Math Help - Double-Angle Formula For Sin Question Problem

  1. #1
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    Double-Angle Formula For Sin Question Problem

    I've been asked the prove that
    sin^{2}xcos^{2}x=\frac{1}{8}(1-cos(4x))
    by using the double-angle for sin(2\theta) and cos(2\theta) in turn.
    I'm stuck at the first section of the solution given for the question, I think if I can understand this first part I will be OK after that. The first part of the solution is:

    From the double-angle formula for sin(2\theta), with \theta=x, we have
    sin^{2}xcos^{2}x=(sinxcosx)^{2}=\frac{1}{4}sin^{2}  (2x).

    Where has all this come from?

    The only formula I've been given for the double-angle formula for sin(2\theta) is
    sin(2\theta)=2sin(\theta)cos(\theta)
    Any help clearing this up would be greatly appreciated.
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  2. #2
    Super Member Quacky's Avatar
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    Try going in reverse, starting with Sin(2x):

    Sin(2x)=2Sin(x)Cos(x)

    Therefore Sin^2(2x)=(2Sin(x)Cos(x))^2

    =4Sin^2(x)Cos^2(x)

    You have Sin^2(x)Cos^2(x)

    =\frac{1}{4}Sin^2(2x)
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  3. #3
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    \displaystyle \begin{align*}\frac{1}{8}\left[1 - \cos{(4x)}\right] &= \frac{1}{8}\left\{1- \left[\cos^2{(2x)} - \sin^2{(2x)}\right]\right\} \\ &= \frac{1}{8}\left[1 - \cos^2{(2x)} + \sin^2{(2x)} \right] \\ &= \frac{1}{8}\left[\sin^2{(2x)} + \sin^2{(2x)}\right] \\ &= \frac{1}{8}\left[2\sin^2{(2x)}\right] \\ &= \frac{1}{4}\sin^2{(2x)} \\ &= \frac{1}{4}\left[\sin{(2x)}\right]^2 \\ &= \frac{1}{4}\left[2\sin{(x)}\cos{(x)}\right]^2 \\ &= \frac{1}{4}\left[4\sin^2{(x)}\cos^2{(x)}\right] \\ &= \sin^2{(x)}\cos^2{(x)}\end{align*}
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  4. #4
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    Many thanks for that.

    I understand both solutions now.

    I wish text books were more thorugh with their solutions, at least for the first year of studying or so anyway.

    Thanks again, I really appreciate it.
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