搜搜考试网(1-sin^6θ-cos^6θ)/(1-sin^4θ-cos^4θ)
来源：学生作业帮 编辑：搜搜考试网作业帮 分类：数学作业 时间：2024/05/05 02:54:37
(1-sin^6θ-cos^6θ)/(1-sin^4θ-cos^4θ)
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-[(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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