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(cos^2x-sin^2x)^2=(1-sin2x)(1+sin2x)?

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  • MyRank
    Lv 6
    2 months ago

    (cos²x-sin²x)² = (1-sin2x) (1+sin2x)

    L.H.S (cos²x-sin²x)²

    = cos^4x + sin^4x - 2sinx cosx

    = (cos²x+sin²x)² - 2sin²x cos²x - 2sinx cosx

    = (1)² - 2sin²x cos²x - 2sinx cosx

    = 1 - 2sinx cosx (1+ sinx cosx )

    =1 - sin2x (1+ sinx cosx )

    =1 - sin2x (1+ sinx cosx )

     

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  • 2 months ago

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

    ((1 - sin^2x) - sin^2x = 

    (1 - sin x)(1 + sin x) - sin^2x

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  • (1 - sin(2x)) * (1 + sin(2x)) =>

    1 - sin(2x)^2 =>

    1 - (1/2) * (1 - cos(4x)) =>

    (1/2) * (2 - 1 + cos(4x)) =>

    (1/2) * (1 + cos(4x))

    (cos(x)^2 - sin(x)^2)^2 =>

    (cos(2x))^2 =>

    (1/2) * (1 + cos(4x))

    So

    (cos(x)^2 - sin(x)^2)^2 =>

    (cos(2x))^2 =>

    (1/2) * (1 + cos(4x)) =>

    (1/2) * (2 - 1 + cos(4x)) =>

    1 - (1/2) * (1 - cos(4x)) =>

    1 - sin(2x)^2 =>

    (1 - sin(2x)) * (1 + sin(2x))

    See how that works?  Nothing more than half-angle and double-angle identities.

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  • Joseph
    Lv 7
    2 months ago

    You need to do your own work, not ask random strangers for (possibly wrong) answers.

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