若1/cosθ-1/sinθ=1则sin2θ的值为

问题描述:

若1/cosθ-1/sinθ=1则sin2θ的值为

1/cosθ-1/sinθ=1
(sinθ-cosθ)/sinθcosθ=1
sinθ-cosθ=sinθcosθ
(sinθ)^2+(cosθ)^2-2sinθcosθ=(sinθcosθ)^2
1-2sinθcosθ=(sinθcosθ)^2
(sinθcosθ)^2+2sinθcosθ-1=0
sinθcosθ=t
t^2+2t-1=0
b^2-4ac=4-4(-1)=8
t=(-2±2√2)/2=-1±√2
sin2θ=2sinθcosθ=2(-1±√2)
-1≤sin2θ≤1
sin2θ=2(-1+√2)