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  • 化簡(jiǎn)[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x

    化簡(jiǎn)[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x
    數(shù)學(xué)人氣:322 ℃時(shí)間:2020-03-21 04:51:54
    優(yōu)質(zhì)解答
    sin^4x-sin^2xcos^2x+cos^4x
    =sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x
    =(sin^2x+cos^2x)^2-3sin^2xcos^2x
    =1-3sin^2xcos^2x
    所以分子=3sin^2xcos^2x
    所以原式=3sin^2xcos^2x/[sin^2x]+3sin^2x
    =3cos^2x+3sin^2x
    =3(sin^2x+cos^2x)
    =3
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