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  • 證明(cosα)的八次方-(sinα)的八次方-cos2α=-1/4sin2αsin4α

    證明(cosα)的八次方-(sinα)的八次方-cos2α=-1/4sin2αsin4α
    數(shù)學(xué)人氣:812 ℃時間:2019-08-20 16:43:14
    優(yōu)質(zhì)解答
    證明:
    左=(cosα)^8-(sinα)^8-cos2α
    =[(cosα)^4+(sinα)^4][(cosα)^4-(sinα)^4]-cos2α
    =[(cosα)^4+(sinα)^4][(cosα)^2-(sinα)^2][(cosα)^2+(sinα)^2]-cos2α
    =cos2α[(cosα)^4+(sinα)^4]-cos2α
    =cos2α{[cosα)^2+(sinα)^2]^2-2(sinα)^2(cosα)^2}-cos2α
    =cos2α[1-(sin2α)^2/2]-cos2α
    =cos2α-cos2α(sin2α)^2/2-cos2α
    =-cos2α(sin2α)^2/2
    =-cos2α[(1-cos4α)/2]/2
    =-1/4*cos2α(1-cos4α)
    =-1/4(cos2α-cos2αcos4α)
    =-1/4[cos(4α-2α)-cos2αcos4α]
    =-1/4[cos4αcos2α+sin4αsin2α-cos2αcos4α]
    =-1/4sin2αsin4α
    =右
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