FGvanessa
幼苗
共回答了17个问题采纳率:82.4% 举报
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.
1年前
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