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Math Help - [SOLVED] prove equality using double angle identity

  1. #1
    Junior Member
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    [SOLVED] prove equality using double angle identity

    The equation is
    8sin^2xcos^2X = 1-cos4x

    I wasn't sure what to do with the right side, so I did all my work on the left. I factored out the 8
    8(sin^2xcos^2x)
    which is
    2(2sinxcosx)(2sinxcosx)
    since 2sinxcosx is the double angle identity for sine I subbed that in
    2(sin2x)(sin2x)
    which gets turned into
    2sin^2x

    I may have just gone down the wrong path as I'm not sure how that's supposed to turn into 1-cos4x. Any pointers?
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  2. #2
    MHF Contributor
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    West Malaysia
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    Quote Originally Posted by satis View Post
    The equation is
    8sin^2xcos^2X = 1-cos4x

    I wasn't sure what to do with the right side, so I did all my work on the left. I factored out the 8
    8(sin^2xcos^2x)
    which is
    2(2sinxcosx)(2sinxcosx)
    since 2sinxcosx is the double angle identity for sine I subbed that in
    2(sin2x)(sin2x)
    which gets turned into
    2sin^2x

    I may have just gone down the wrong path as I'm not sure how that's supposed to turn into 1-cos4x. Any pointers?
    HI

    lets start from RHS .

    1-\cos 4x=1-\cos 2(2x)

    =1-[\cos^2 2x-\sin^2 2x]

    =\cos^2 2x+ \sin^2 2x-\cos^2 2x+\sin^2 2x

    =2\sin^2 2x

    =2(2\sin x\cos x)^2

    try continuing from here .
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  3. #3
    Junior Member
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    Got it, thank you very much.
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