作业帮设tan2θ =-2根号2,2θ∈（π/2,π）求(2cos^2θ-2-sinθ-1)/(sinθ+cosθ)
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设tan2θ =-2根号2,2θ∈（π/2,π）求(2cos^2θ-2-sinθ-1)/(sinθ+cosθ)
设tan2θ =-2根号2,2θ∈（π/2,π）求(2cos^2θ-2-sinθ-1)/(sinθ+cosθ)
设tan2θ =-2根号2,2θ∈（π/2,π）求(2cos^2θ-2-sinθ-1)/(sinθ+cosθ)
tan2θ=2√2,θ∈（π/2,π）求 [2cos²(θ/2)-sinθ-1]/(sinθ+cosθ )
tan2θ=2tanθ/(1-tan²θ)=2√2
tanθ=(√2)(1-tan²θ)
(√2)tan²θ+tanθ-√2=0
tanθ=(-1±3)/(2√2)
θ∈（π/2,π）
∴tanθ＜0
∴tanθ=(-1-3)/(2√2) =-根号2
[2cos²(θ/2)-sinθ-1]/(sinθ+cosθ )
= { [2cos²(θ/2)-1] -sinθ } / (sinθ+cosθ )
= (cosθ-sinθ)/(sinθ+cosθ)
=(1-tanθ)/(1+tanθ)
=(1+√2)/(1-√2)
=-(1+√2)^2
=-3 -2√2
设tan2θ =-2根号2,2θ∈（π/2,π）求(2cos^2θ-2-sinθ-1)/(sinθ+cosθ)
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