Question

若sinθ+cosθ=1/2,且0<θ<π,求sin2θ,cos2θ,tan2θ的值

Answer 1

sinθ+cosθ=1/2
两边平方：
(Sinθ)^2+(cosθ)^2+2sinθcosθ=1/4
1+2sinθcosθ=1/4
2sinθcosθ=－3/4
∵0<θ<π
∴sinθ>0,
∵2sinθcosθ=－3/4<0
∴cosθ<0
∴π/2<θ<π
又∵sinθ+cosθ=1/2>0
∴sinθ绝对值>cosθ绝对值
∴π/2<θ<3π/4
∴π<2θ<3π/2

sin2θ＝2sinθcosθ=－3/4
cos2θ=－√[1－(sin2θ)^2]＝－√[1－(-3/4)^2]=－√5/4
tan2θ=sin2θ/cos2θ=(-3/4)/(－√5/4)=3√5/5

Answer 2

(sinθ+cosθ)^2 = 1+sin2θ = 1/4
sin2θ = -3/4
cos2θ = -sqrt(7)/4
tan2θ = 3/sqrt(7)