已知α是第三象限角,且f(α)=[sin(α-π/2)cos(3π/2+α)tan(π-α)]/[tan( -α-π)s
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/22 01:36:49
已知α是第三象限角,且f(α)=[sin(α-π/2)cos(3π/2+α)tan(π-α)]/[tan( -α-π)sin(-π-α)
(1)化简f(α)
(2)若cos(α-3π/2)=1/5,求f(α)的值
(1)化简f(α)
(2)若cos(α-3π/2)=1/5,求f(α)的值
(1)化简f(α)
f(α)=sin(α-π/2)cos(3π/2+α)tan(π-α)除以tan(-α-π)sin(-α-π)
=-cosαsinα(-tanα)÷[(-tanα)sinα]
=-cosα
(2)若cos(α-3π/2)=1/5,求f(α)的值
cos(α-3π/2)=cos(π/2+α)=-sinα=1/5
cos²α=1-1/25=24/25
cosα=-2√6/5
f(α)=-cosα=2√6/5
f(α)=sin(α-π/2)cos(3π/2+α)tan(π-α)除以tan(-α-π)sin(-α-π)
=-cosαsinα(-tanα)÷[(-tanα)sinα]
=-cosα
(2)若cos(α-3π/2)=1/5,求f(α)的值
cos(α-3π/2)=cos(π/2+α)=-sinα=1/5
cos²α=1-1/25=24/25
cosα=-2√6/5
f(α)=-cosα=2√6/5
②若cos(已知α是第三象限角f(α)=sin(π-α)cos(2π-α)tan(-α-π)/tan(-α)sin(-π
已知α是第三象限角,且f(α)=sin(π-α)cos(2π-α)tan(-α+3π/2)/cos(-α -π) (1)
已知α是第三象限角,且f(α)=[sin(α-π/2)cos(3π/2+α)tan(π-α)]/[tan( -α-π)s
已知α是第三象限角,且f(α)=[sin(π+α)cos(2π-α)tan(-α+3π/2)tan(-α-π)]/sin
α是第三象限角,f(α)={sin(α-π/2)cos(3π/2+α)tan(π-α)}/{tan(-α-π)sin(-
已知α为第三象限角,且f(α)=sin(π-α)cos(2π-α)tan(-α+π)/sin(π+α)tan(2π-α)
已知α是第三象限的角,且f(α)=sin(π−α)cos(2π−α)tan(3π2−α)cos(−π−α).
已知tanα=2,且α是第三象限角,求sin(kπ -α)+cos(kπ+α)的值
已知α是第三象限角,f(α)=[sin(π-a)cos(2π-a)tan(1.5π-a)tan(-a-π)]/sin(-
已知f(α)=sin(π-α)cos(2π-α)tan(π-α)/-tan(-π-α)sin(-π-α),若α是第三象限
已知sinα=2/3,cosβ=-3/4,α∈(π/2,π),β是第三象限角,求tan(α+β)
已知角a是第三象限角,且f(a)=[sin(a-π/2)cos(3π/2+a)tan(π-a)]除以tan(-π-a)s