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已知tan2θ=2√2,且0
人气:230 ℃ 时间:2020-03-27 19:05:38
解答
tan2θ=2tanθ/[1-(tanθ)^2]=2√2
√2(tanθ)^2+tanθ-√2=0,tanθ=-√2或tanθ=√2/2
[sinθ+(2sin²θ/2-1)]/[√2cos(π/4-θ)]
=(sinθ-cosθ)/(sinθ+cosθ)
=(tanθ-1)/(tanθ+1)
=2√2+3或2√2-3
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