已知θ∈(π/2,π),cosθ/2=a,则根号下1+cosθ/2+根号下1-sinθ 答案是根号下（1-a²

(1+cosθ)/2=[1+2cos²(θ/2)-1]/2=cos²(θ/2)
1-sinθ=sin²(θ/2)+cos²(θ/2)-2sin(θ/2)cos(θ/2)=[sin(θ/2)-cos(θ/2)]²
∵ θ∈(π/2,π)
∴ θ/2∈(π/4,π/2)
∴ cos(θ/2)>0,sin(θ/2)>cos(θ/2)
∴ 根号下1+cosθ/2+根号下1-sinθ
=|cos(θ/2)|+|sin(θ/2)-cos(θ/2)|
=cos(θ/2)+[sin(θ/2)-cos(θ/2)]
=sin(θ/2)
∵ sin²(θ/2)=1-cos²(θ/2)=1-a²
∵ sin(θ/2)>0
∴ sin(θ/2)=√(1-a²)
∴ 根号下1+cosθ/2+根号下1-sinθ =sin(θ/2)=√(1-a²)