作业帮化简：（1）sin^2·α+cos^4·α+sin^2·αcos^2·α（2）（1-cos^4·α-sin^4·α）/(
来源：学生作业帮 编辑：作业帮 分类：数学作业 时间：2024/05/29 09:58:35
化简：
（1）sin^2·α+cos^4·α+sin^2·αcos^2·α
（2）（1-cos^4·α-sin^4·α）/(1-cos^6·α-sin^4·α)
（1）sin^2·α+cos^4·α+sin^2·αcos^2·α
（2）（1-cos^4·α-sin^4·α）/(1-cos^6·α-sin^4·α)
（1）sin^2·α+cos^4·α+sin^2·αcos^2·α
=sin^2·α+cos^2·α(cos^2·α+sin^2·α)
= sin^2·α+cos^2·α
=1
（2）（1-cos^4·α-sin^4·α）/(1-cos^6·α-sin^4·α)
=[(1+sin^2·α)(1-sin^2·α)-cos^4·α]/[(1+sin^2·α)(1-sin^2·α)-cos^6·α]
=[cos^2·α+cos^2·αsin^2·α-cos^4·α]/[cos^2·α+cos^2·αsin^2·α-cos^6·α]
=[cos^2·α(1-cos^2·α)+cos^2·αsin^2·α]/[cos^2·α(1-cos^4·α)+cos^2·αsin^2·α]
=[2cos^2·αsin^2·α]/[cos^2·α(1-cos^2·α)(1+cos^2·α)+cos^2·αsin^2·α]
=[2cos^2·αsin^2·α]/[cos^2·αsin^2·α(1+cos^2·α)+cos^2·αsin^2·α]
=[2cos^2·αsin^2·α]/[2cos^2·αsin^2·α+sin^2·αcos^4·α]
=2/[2+cos^2·α]
=sin^2·α+cos^2·α(cos^2·α+sin^2·α)
= sin^2·α+cos^2·α
=1
（2）（1-cos^4·α-sin^4·α）/(1-cos^6·α-sin^4·α)
=[(1+sin^2·α)(1-sin^2·α)-cos^4·α]/[(1+sin^2·α)(1-sin^2·α)-cos^6·α]
=[cos^2·α+cos^2·αsin^2·α-cos^4·α]/[cos^2·α+cos^2·αsin^2·α-cos^6·α]
=[cos^2·α(1-cos^2·α)+cos^2·αsin^2·α]/[cos^2·α(1-cos^4·α)+cos^2·αsin^2·α]
=[2cos^2·αsin^2·α]/[cos^2·α(1-cos^2·α)(1+cos^2·α)+cos^2·αsin^2·α]
=[2cos^2·αsin^2·α]/[cos^2·αsin^2·α(1+cos^2·α)+cos^2·αsin^2·α]
=[2cos^2·αsin^2·α]/[2cos^2·αsin^2·α+sin^2·αcos^4·α]
=2/[2+cos^2·α]
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