1.
已知tan2θ=-2√2
则tan2θ=2tanθ/(1-tan²θ)
θ∈(π/2,π)
,则tan
θ>0
2tanθ=(1-tan²θ)(-2√2)
√2tan²θ-tanθ-√2=0
(√2tanθ+1)(tanθ-√2)=0
解得tanθ=√2
2.
[2COS^2(θ/2)-SINθ-1]/[根号2SIN(θ+π/4)]
=(cosθ-sinθ)/(sinθ+cosθ)
=(1-tanθ)/(1+tanθ)
=(1-√2)/(1+√2)
=(1-√2)²/(1-2)
=2√2-3
希望可以帮到你,望采纳,谢谢。