(1+sinθ+cosθ)/(1+sinθ-cosθ)=1/2
求cos2θ

解：∵tan(θ/2)=[sin(θ/2)]/[cos(θ/2)]
=[2cos(θ/2)]sin(θ/2)]/[2cos(θ/2)cos(θ/2)]
=(sinθ)/(1+cosθ),
tan(θ/2)=[sin(θ/2)]/[cos(θ/2)]
=[2sin(θ/2)sin(θ/2)]/[2sin(θ/2)cos(θ/2)]
=(1-cosθ)/(sinθ),
∴tan(θ/2)=(sinθ)/(1+cosθ)=(1-cosθ)/(sinθ)
=(1+sinθ-cosθ)/(1+cosθ+sinθ),(比例的等比性质)
=2.
∴tanθ=[2tan(θ/2)]/{1-[tan(θ/2)]^2}=(2×2)/(1-2^2)=-4/3.
∴cos2θ=[1-(tanθ)^2]/[1+(tanθ)^2]=[1-(-4/3)^2]/[1+(-4/3)^2]
=-7/25.