化简（1-sin6θ-cos6θ）/（1-sin4θ-cos4θ）如题 谢谢了
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.
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