若9π/4＜θ＜5π/2.,则cosθ,sinθ,tanθ大小关系正确的是A.cosθ＜sinθ＜tanθ B.sinθ＜cosθ＜tanθC.tanθ＜cosθ＜sinθ D.tanθ＜sinθ＜cosθ知道很简单,但求详细说明.

若9π/4＜θ＜5π/2.,则cosθ,sinθ,tanθ大小关系正确的是A.cosθ＜sinθ＜tanθ B.sinθ＜cosθ＜tanθC.tanθ＜cosθ＜sinθ D.tanθ＜sinθ＜cosθ知道很简单,但求详细说明. 已知π＜θ＜3π/2,cosθ=-4/5,求cosθ/2. 已知sinθsinβ=-4/7,则cosθcosβ∈已知tan(π-θ)=3,求下式的值 (5xin^3θ+cosθ)/(2cos^3θ+sin^2θcosθ) 若2kπ+π＜θ＜2kπ+(k∈Z),则sinθ,cosθ,tanθ的大小关系是( ) 若2kπ+π＜θ＜2kπ+5π/4(k∈Z),则sinθ,cosθ,tanθ的大小关系是( ) A.sinθ＜cosθ＜tanθ B.cosθ＜tanθ＜sinθC.cosθ＜sinθ＜tanθ D.sinθ＜tanθ＜cosθ 已知：θ∈(0,π/2) 证明：sin（cosθ)＜cosθ＜cos(sinθ) √(1+cosθ/1-cosθ)-√(1-cosθ/1+cosθ）（π＜θ＜3/2π） 若f(cosθ)=sin^2-3cosθ,则f(2cos π/3 )=_____ 化简cosθ+cos(θ+2π/3)+cos(θ+4π/3) 若cosθ=-5/13,且θ∈(π/2.π),求sin(π-θ)+cos(-θ)的值 若cos(π/4+θ)cos(π/4-θ)=1/4,则sin4θ+cos4θ等于 若cos(π/4-θ)cos(π/4+θ)=√2/6,且0 设0＜θ＜π/2,向量a=(sin2θ,cos),b=(cosθ,1),若a平行b,则tanθ= 化简根号下1+cos(3π-θ)/2 (3π/2＜θ＜2π) sin θ/2 怎算?化简：sinα*cos^5α -cosαsin^5化简根号下1+cos(3π-θ)/2 (3π/2＜θ＜2π) sin θ/2 怎算?化简：sinα*cos^5α -cosαsin^5α ..1/4 sin4α 、 计算：cos（π/5)+cos（2π/5)+cos(3π/5)+cos(4π/5) 求值:cos(π/11)cos(2π/11)cos(3π/11)cos(4π/11)cos(5π/11) 求值：cos(π/11)-cos(2π/11)+cos(3π/11)-cos(4π/11)+cos(5π/11) cos(π/11)cos(2π/11)cos(3π/11)cos(4π/11)cos(5π/11) 已知sinα+cosα=4/5且3π/22,若sinα+sinβ+sinγ=0,cosα+cosβ+cosγ=0,则cos(α-β)的值