已知cos(θ+π/2)=-1/2,求cos(θ+π)/sin(π/2-θ)[cos(3π-θ)-1]+cos(θ-2π)/cos(-θ)*cos(π-θ)+sin(θ+5π/2)的值
网友回答
cos(θ+π)/sin(π/2-θ)[cos(3π-θ）-1]+cos(θ-2π）/cos（-θ）*cos（π-θ）+sin（θ+5π/2）
=(-cosθ)/(cosθ) [-cosθ-1]+(cosθ/cosθ）*(-cosθ）+cosθ
=cosθ+1+(-cosθ)+cosθ
=cosθ+1
=√3/2+1 或 =-√3/2+1
cos(θ+π/2)=-1/2
sinθ=cos(π/2-θ)=1/2, cosθ=√3/2 或 cosθ=-√3/2