(1+sinθ+cosθ)/(1+sinθ-cosθ)=1/2求cos2θ

熱心網友

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),(比例的等比性質)=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。

熱心網友

由 高1數學題3 可知(1+sinθ+cosθ)/(1+sinθ-cosθ)=cotθ/2=1/2即tanθ/2=2tanθ=2(tanθ/2)/[1-(tanθ/2)^2]=-4/3cos2θ=[1+(tanθ/2)^2]/[1--(tanθ/2)^2]=-25/7.