(1-sin^6θ-cos^6θ)/(1-sin^4θ-cos^4θ)

1个回答

  • 1-sin^6θ-cos^6θ

    =1-[(sin^2θ)^3+(cos^2θ)^3]

    =1-(sin^2θ+cos^2θ)(sin^4θ-sin^2θcos^2θ+cos^4θ)

    =1-(sin^4θ-sin^2θcos^2θ+cos^4θ)

    =1-(sin^4θ+2sin^2θcos^2θ+cos^4θ-3sin^2θcos^2θ)

    =1-[(sin^2θ+cos^2θ)^2-3sin^2θcos^2θ]

    =3sin^2θcos^2θ

    1-sin^4θ-cos^4θ

    =1-(sin^4θ+cos^4θ)

    =1-[(sin^4θ+2sin^2θcos^2θ+cos^4θ-2sin^2θcos^2θ)]

    =1-[(sin^2θ+cos^2θ)^2-2sin^2θcos^2θ]

    =2sin^2θcos^2θ

    所以:(1-sin^6θ-cos^6θ)/(1-sin^4θ-cos^4θ)

    =(3sin^2θcos^2θ)/(2sin^2θcos^2θ)

    =3/2.