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)
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/02 16:29:31
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)
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)
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
=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]
(1)将cos^2换成1-sin^2,全换掉,即可化简
(2)(1-cosα^4-sinα^4)/(1-cosα^6-sinα^6)=[(1-cosα^2)(1+cosα^2)-sinα^4]/……=[sinα^2(1+cosα^2-sinα^2)]/……=sinα^2(2cosα^2)/……=0.5sin2α^2/……
省略号的部分用立方差公式,同理展开,方法基本相同。
1.=sin^2a-(1-sin^2a)^2+sin^2acos^2a=sin^2a+(1-sina^2)^2+sin^2a(1-sin^2a)=sin^2a+(1-sin^2a)=1
2.=[1-(cos^2a+sin^2a)^2+2sin^2acos^2a]/{[1-(cos^2a)^3]-sin^4a}
=[2sin^2acos^2a]/[(1-cos^2a)(1+cos^4...
全部展开
1.=sin^2a-(1-sin^2a)^2+sin^2acos^2a=sin^2a+(1-sina^2)^2+sin^2a(1-sin^2a)=sin^2a+(1-sin^2a)=1
2.=[1-(cos^2a+sin^2a)^2+2sin^2acos^2a]/{[1-(cos^2a)^3]-sin^4a}
=[2sin^2acos^2a]/[(1-cos^2a)(1+cos^4a+cos^2a)-(sin^2)^2]
=[2sin^2acos^2a]/[(sin^2a)(1+cos^4a+cos^2a)-(sin^2)^2]
=[2sin^2acos^2a]/[(sin^2a)(1+cos^4a+cos^2a-sin^2a)]
=[2sin^2acos^2a]/[(1-cos^2a)(1+cos^4a+cos^2a)-(sin^2)^2]
=2/(2+cos^2)
收起