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Math Help - Double angle formulas

  1. #1
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    Double angle formulas

    Hi

    I'm having some problems seeing how they've come to an answer in the course book, here we go.

    sin^2x \ cos^2x = \frac{1}{4}\ sin^2(2x)

    and then through several stages, some skipped, we get

    =\frac{1}{8}\ (1-cos(4x))

    Any help would be appreciated.

    Thanks
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  2. #2
    MHF Contributor red_dog's Avatar
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    We have \sin 2x=2\sin x\cos x and \cos 4x=1-2\sin^22x

    Now \sin^2x\cos^2x=(\sin x\cos x)^2=\left(\frac{\sin 2x}{2}\right)^2=\frac{1}{4}\sin^22x=

    =\frac{1}{8}\cdot 2\sin^22x=\frac{1}{8}(1-1+2\sin^22x)=\frac{1}{8}[1-(1-2\sin^22x)]=\frac{1}{8}(1-\cos 4x)
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  3. #3
    Junior Member darence's Avatar
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    Wink

    look here
    Some formulae may be helpfull
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  4. #4
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    use cos2x formula
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  5. #5
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    <br /> <br />
cos(4x) = cos(2x + 2x) = .... = 8cos^4 x - 8cos^2 x + 1<br />

    <br />
cos^4 x = (1-sin^2 x)^2 = 1 - 2sin^2 x + sin^4 x<br />
    ------------------------------------------
    <br />
sin^2 xcos^2 x = \frac{1}{8} (1 - cos(4x))<br />

    <br />
sin^2 xcos^2 x = \frac {1}{8} (1 - 8cos^4 x + 8cos^2 x - 1)<br />

    <br />
 sin^2 xcos^2 x = -cos^4 x + cos^2 x <br />

    <br />
 sin^2 xcos^2 x = -1 + 2sin^2x - sin^4 x + (1- sin^2 x)<br />

    <br />
 sin^2 xcos^2 x = sin^2 x - sin^4 x<br />

    <br />
  sin^2 xcos^2 x = sin^2 x (1 - sin^2 x)<br />

    <br />
cos^2 x =  1 - sin^2 x<br />

    <br />
cos^2 x =  cos^2 x<br />

    sorry for the long explanation :P

    EDIT: just realized this wasn't really about proving the equalities.. meh I'll leave it unedited anyways )
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