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  • 化簡:

    化簡:
    (1)sin^2·α+cos^4·α+sin^2·αcos^2·α
    (2)(1-cos^4·α-sin^4·α)/(1-cos^6·α-sin^4·α)
    數(shù)學(xué)人氣:995 ℃時間:2020-06-25 03:02:28
    優(yōu)質(zhì)解答
    (1)sin^2·α+cos^4·α+sin^2·αcos^2·α
    =sin^2·α+cos^2·α(cos^2·α+sin^2·α)
    = sin^2·α+cos^2·α
    =1
    (2)(1-cos^4·α-sin^4·α)/(1-cos^6·α-sin^4·α)
    =[(1+sin^2·α)(1-sin^2·α)-cos^4·α]/[(1+sin^2·α)(1-sin^2·α)-cos^6·α]
    =[cos^2·α+cos^2·αsin^2·α-cos^4·α]/[cos^2·α+cos^2·αsin^2·α-cos^6·α]
    =[cos^2·α(1-cos^2·α)+cos^2·αsin^2·α]/[cos^2·α(1-cos^4·α)+cos^2·αsin^2·α]
    =[2cos^2·αsin^2·α]/[cos^2·α(1-cos^2·α)(1+cos^2·α)+cos^2·αsin^2·α]
    =[2cos^2·αsin^2·α]/[cos^2·αsin^2·α(1+cos^2·α)+cos^2·αsin^2·α]
    =[2cos^2·αsin^2·α]/[2cos^2·αsin^2·α+sin^2·αcos^4·α]
    =2/[2+cos^2·α]
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