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:大师作文网作业帮 分类:数学作业 时间:2024/10/07 12:20:17
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]
化简:(1)sin^2·α+cos^4·α+sin^2·αcos^2·α(2)(1-cos^4·α-sin^4·α)/(
sin(π/2+α)·cos(π/2-α)/cos(π+α)+sin(π-α)·cos(π/2+α)/sin(π+α)=
化简:(1)sin(α+β)−2sinαcosβ2sinαsinβ+cos(α+β)
【sin^2-2sinα·cosα-cos^2α】/4cos^2-3sin^2α 已知tanα=3 求值 P.S那个分子
若(sinα+cosα)/(sinα-cosα)=2 则sin(α-5π)·sin(3π/2-α)=?
若tanα=根号下2 求1)(sinα+cosα)/(cosα-sinα) 2)2sin^α-sinαcosα+cos^
若tanα=根号2,求值(1)cosα+sinα/cosα-sinα;(2)2sin平方α-sinαcosα+cos平方
已知3sinα-2cosα=0,求(cosα-sinα)/(cosα+sinα)+(cosα+sinα)/(cosα-s
(sinα+cosα)^2化简
求证:(3-sin^4α-cos^4α)/2cos^2α+1+tan^2α+sin^2α
求证:(1-sinα+cosα)/(1+sinα+cosα)=tan(π/4-α/2)
求证:1-2sinαcosα/cos²α-sin²α=tan(π/4-α)