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    cassmelnags's Avatar
    cassmelnags Posts: 1, Reputation: 1
    New Member
     
    #1

    Dec 15, 2011, 05:21 PM
    hardest identity problem EVER!
    I have to get this answer by tomorrow or I get an F, HELP!
    here's the problem:

    1-(2sinxcosx cos2x)^2
    ---------------------- <- over
    2sin2xcos2x

    (yes, it is a fraction)

    the answer is supposed to be one and I have to prove it

    I've gotten this far:

    2-4sin^2xcox^2x 8sinxcos^6x-4sinxcosx 4cos^4x-4cosx
    ----------------------------------------------------
    2sin2xcos2x

    the rest is lost on me
    Aurora2000's Avatar
    Aurora2000 Posts: 111, Reputation: 16
    Junior Member
     
    #2

    Dec 16, 2011, 12:14 AM
    The display is quite strange, I will interpret as

    (if you meant another expression, repost using "math" environment)

    The problem is lot easier than what seems, just try to simplify as much as you can before going with pure computations. Otherwise you get an unmanageable expression.

    Use first



    Numerator:


    and using ,


    Denominator is , thus the original expression


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