作业帮一、已知TAN2θ =-2根号2 θ∈(π/2,π) (1)求TANθ的值 (2)求[2COS^2(θ/2-SINθ-1)]/[根号2SIN(θ+π/4)]的值.
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![一、已知TAN2θ =-2根号2 θ∈(π/2,π) (1)求TANθ的值 (2)求[2COS^2(θ/2-SINθ-1)]/[根号2SIN(θ+π/4)]的值.](/uploads/image/z/5183380-28-0.jpg?t=%E4%B8%80%E3%80%81%E5%B7%B2%E7%9F%A5TAN2%CE%B8+%3D-2%E6%A0%B9%E5%8F%B72+%CE%B8%E2%88%88%28%CF%80%2F2%2C%CF%80%29+%281%29%E6%B1%82TAN%CE%B8%E7%9A%84%E5%80%BC+%282%29%E6%B1%82%5B2COS%5E2%28%CE%B8%2F2-SIN%CE%B8-1%29%5D%2F%5B%E6%A0%B9%E5%8F%B72SIN%28%CE%B8%2B%CF%80%2F4%29%5D%E7%9A%84%E5%80%BC.)
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一、已知TAN2θ =-2根号2 θ∈(π/2,π) (1)求TANθ的值 (2)求[2COS^2(θ/2-SINθ-1)]/[根号2SIN(θ+π/4)]的值.
一、已知TAN2θ =-2根号2 θ∈(π/2,π) (1)求TANθ的值 (2)求[2COS^2(θ/2-SINθ-1)]/[根号2SIN(θ+π/4)]的值.
一、已知TAN2θ =-2根号2 θ∈(π/2,π) (1)求TANθ的值 (2)求[2COS^2(θ/2-SINθ-1)]/[根号2SIN(θ+π/4)]的值.
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
一、已知TAN2θ =-2根号2 θ∈(π/2,π) (1)求TANθ的值 (2)求[2COS^2(θ/2-SINθ-1)]/[根号2SIN(θ+π/4)]的值.
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