作业帮化简cos2α/tan(π/4+α)
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/29 03:10:09
x){3bsK47hۨiTGVΆn\ř`#m&y#8[?YF Y*(gn^6T-X}#.@~qAb [6`%jBD4B ] a
s!j `Y2$ ?
化简cos2α/tan(π/4+α)
化简cos2α/tan(π/4+α)
化简cos2α/tan(π/4+α)
cos2α/tan(π/4+α)
=sin(π/2+2α)/tan(π/4+α)
=2sin(π/4+α)cos(π/4+α)/[sin(π/4+α)/cos(π/4+α)]
=2cos^2(π/4+α)
=2[1+cos(π/2+2α)]/2
=1-sin2α
cos2a/tan(π/4+a)=cos2a/(tanπ/4-tana)/(1+tanπ*tana)=cos2a*(1+tana)/(1-tana)
化简cos2α/tan(π/4+α)
cos2α/tan(α+π/4)=
化简cos2(-α)-tan(2π+α)/sin(-α)
tan（π/4+α）=k 求cos2α
化学cos2α/tan(π/4-a)如题
化简[tan(π/4+α)cos2α]/[2cos^2(π/4-α)]
化简cos2(α-π/4)+cos2(α+π/4)
已知tan(π/4+α)=1/2 求（1）tanα的值 （2）sin2α-cos2(平方）α/1+cos2α
化简:sin2α/(1-cos2α)-1/tanα=
化简 2cos2α+sin2αtanα
已知tanα=-3/4,则cos2α=
已知tan(π/4+α)=2,求tanα和sin2α+sinα+cos2α的值
tan(α+π/4)=2,则cos2α+3sin^2α+tan2α=?
2tan(π/4-α)sin2（π/4+α）/2cos2α-1
已知tan(π /4)=1/ 2 求cos2α/sin2α+cos^2α
已知sin2α=a,cos2α=b,求tan(α+π/4)的值.
已知sin2α=a,cos2α=b,求tan(α+π/4)的值
tan(π/4+α)=2010,那么1/cos2α+tan2α=