Question

求助！！高一数学！！已知tan2θ=2根号2,0<θ<π,求(sinθ+2sin^2θ/2-1)/[根号2cos(π/4-θ)]

已知tan2θ=2根号2,0<θ<π,求(sinθ+2sin^2θ/2-1)/[根号2cos(π/4-θ)]
怎么做阿？？尽快！！
e~就是[sinθ+2sin^2(θ/2)-1]/[(根号2)cos(π/4-θ)]
这样可以了吗

Answer 1

tan2θ=2根号2,0<θ<π
tan2θ=2tanθ/(1-tan^2θ)=2根号2
tanθ=根号2(1-tan^2θ)
根号2 tan^2θ+tanθ-根号2=0
tanθ=(-1±3)/(2根号2) =-根号2，或根号2/2

cosθ=1-2sin^2θ/2
2sin^2θ/2-1=-cosθ

(sinθ+2sin^2θ/2-1)/[根号2cos(π/4-θ)]
=(sinθ-cosθ) / {根号2[cosπ/4cosθ)+sinπ/4sinθ]}
=(sinθ-cosθ) /(cosθ+sinθ)
=(1-tanθ)/(1+tanθ)

tanθ=-根号2时：
=(1+根号2)/(1-根号2)
=-(1+根号2)^2
=-3-2根号2

tanθ=根号2/2 时：
=(1-根号2/2)/(1+根号2/2)
=(2-根号2)/(2+根号2)
=(2-根号2)^2/2
=3-2根号2

Answer 2

建议贴个图上来

大概思路是先化简，求出tanθ,再化简所求式子，利用弦化切
嗯，降幂
2sin^2(θ/2)=1-coθ,所以sinθ+2sin^2(θ/2)-1=sinθ-cosθ=（根号2)sin(π/4-θ)]

根号2)sin(θ-π/4)]/[(根号2)cos(π/4-θ)]=tan(π/4-θ),
tan2θ=2可以求出tanθ,这样明了吧？