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    tali31's Avatar
    tali31 Posts: 1, Reputation: 1
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    #1

    Oct 22, 2007, 06:41 PM
    Proving Trig Identities
    how would you prove sin4x - cos4x = sin2x-cos2x as well as sin4x + 2sin2xcos2x +cos4x = 1
    :confused:
    s_cianci's Avatar
    s_cianci Posts: 5,472, Reputation: 760
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    #2

    Oct 22, 2007, 06:46 PM
    In both cases, you'll want to factor the left side of the equation. Treat sin^4 x - cos^4 x as a difference of two squares and treat sin^4 x + 2sin^2x cos^2x + cos^4x as a perfect-square trinomial. Also the 1st. Pythagorean Identity will come into play ; sin^2x + cos^2x = 1.
    terryg752's Avatar
    terryg752 Posts: 197, Reputation: 4
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    #3

    Oct 22, 2007, 11:03 PM
    I think you have written the question incorrectly.

    These identities are not correct.

    I think you should have written:

    1. (Sinx)^4 - (Cos x)^4

    = [ (sinx)^2 - (cos)^2 ] [ (sinx)^2 + (Cos x)^2) ]

    = (sinx)^2 - (cos)^2

    2. sin^4 x + 2 sin^2 co^2 x + cos^4 x

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

    = 1^2

    = 1
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