sin(θ-5π)cos[(-π/2)-θ]cos(8π-θ)}/sin[θ-(3π/2)]sin(-θ-4π)我晕，正的还是负的

θ∈(0,π/2),比较cosθ、sin(cosθ)、cos(sinθ)的大小 sin(π-θ）+cos(2π-θ）/cos(5π/2-θ)+sin(3π/2+θ)=2,则sinθcosθ=______ 求值:已知sinθ+cosθ=1/5,已知θ∈(0,π).求(1)sinθ*cosθ(2)sinθ-cosθ(3)tanθ (sinθ+cosθ)/(sinθ-cosθ)=2,则sin(θ-5π)*sin(3π/2-θ)= 若（sinθ+cosθ）/（sinθ-cosθ）=2,则sin（θ-5π）*sin（3π/2-θ）等于? 若sin（θ-π）=2cos（2π-θ）求（1）sinθ+5cosθ/sinθ-3cosθ (2)sinθcosθ的值 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) 已知tanθ=3,求(3cosθ-5sin^2θcosθ)/sin(π-θ)的值 已知 sin(θ+kπ)=-2cos (θ+kπ) 求 ⑴4sinθ-2cosθ/5cosθ+3sinθ; ⑵(1/4)sin平方θ+(2/5)cos平方θ已知sin(θ+kπ)=-2cos (θ+kπ)求⑴ 4sinθ-2cosθ/5cosθ+3sinθ⑵(1/4)sin平方θ+(2/5)cos平方θ 若[sin(π-θ)+cos(2π-θ)]/[cos(5π/2-θ)+sin(3π/2+θ)]=2,求sinθ*cosθ的值若[sin（π-θ）+cos（2π-θ）]/[cos(5π/2-θ)+sin(3π/2+θ）]=2,求sinθ*cosθ的值 已知sin(θ+kπ)=-2cos(θ+kπ),k∈Z,求4sinθ-2cosθ/5cosθ+3sinθ①4sinθ-2cosθ/5cosθ+3sinθ ②sin²θ+2/5cos²θ 求证：-sinθ=cos(π/2+θ)-sinθ=cos(π/2+θ) cot(π+θ)=2,则(3sinθ-cosθ)/(sinθ+cosθ)=? 利用和差角公式化简 (2)sin(π/3+α)+sin(π/3-α)(2)sin(π/3+α)+sin(π/3-α)(3)cos(π/4+α)-cos(π/4-α)(4)cos(60°+α)+cos(60°-α)(5)sin(α-β)cosβ+cos(α-β)sinβ(6)cos(α+β)cosβ+sin(α+β)sinβ sin(θ-5π)cos[(-π/2)-θ]cos(8π-θ)}/sin[θ-(3π/2)]sin(-θ-4π)我晕，正的还是负的 已知：θ∈(0,π/2) 证明：sin（cosθ)＜cosθ＜cos(sinθ) sin(a-π)+5cos(2π-a)+4cos(π+a)/cos(π-a)-sin(-a) 求 [sin(5π-α)cos(-π-α)]/[cos(α-π)cos(π/2+α)]化简