来自何发智的问题
已知tan2θ=2根号2,π<2θ<2π,求(2(cosθ/2)^2-sinθ-1)/(根号2倍sin(π/4+θ))
已知tan2θ=2根号2,π<2θ<2π,求(2(cosθ/2)^2-sinθ-1)/(根号2倍sin(π/4+θ))
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2020-12-08 22:15
已知tan2θ=2根号2,π<2θ<2π,求(2(cosθ/2)^2-sinθ-1)/(根号2倍sin(π/4+θ))
已知tan2θ=2根号2,π<2θ<2π,求(2(cosθ/2)^2-sinθ-1)/(根号2倍sin(π/4+θ))
(2(cosθ/2)^2-sinθ-1)/(根号2倍sin(π/4+θ))=(cosθ-sinθ)/√2*(√2/2cosθ+√2/2sinθ)=(cosθ-sinθ)/(cosθ+sinθ)=(1-tanθ)/(1+tanθ)tan2θ==2tanθ/(1-tan^2θ)=2√2tanθ=√2/2或tanθ=-√2代入tan...