已知tan2θ=3/4(π/2<θ<π),求(2cos² ×θ/2+sinθ-1)/√2cos(θ+π/4)的值
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![已知tan2θ=3/4(π/2<θ<π),求(2cos² ×θ/2+sinθ-1)/√2cos(θ+π/4)的值](/uploads/image/z/1780211-11-1.jpg?t=%E5%B7%B2%E7%9F%A5tan2%CE%B8%3D3%2F4%28%CF%80%2F2%EF%BC%9C%CE%B8%EF%BC%9C%CF%80%29%2C%E6%B1%82%282cos%26%23178%3B+%C3%97%CE%B8%2F2%2Bsin%CE%B8-1%29%2F%E2%88%9A2cos%28%CE%B8%2B%CF%80%2F4%29%E7%9A%84%E5%80%BC)
已知tan2θ=3/4(π/2<θ<π),求(2cos² ×θ/2+sinθ-1)/√2cos(θ+π/4)的值
已知tan2θ=3/4(π/2<θ<π),求(2cos² ×θ/2+sinθ-1)/√2cos(θ+π/4)的值
已知tan2θ=3/4(π/2<θ<π),求(2cos² ×θ/2+sinθ-1)/√2cos(θ+π/4)的值
解
tan2θ=(2tanθ)/(1-tan^2θ)=3/4
即3(1-tan^2θ)=8tanθ
即3tan^2θ+8tanθ-3=0
(3tanθ-1)(tanθ+3)=0
∵π/2<θ<π
∴tanθ<0
∴tanθ=-3
∴(2cos^2θ/2+sinθ-1)/(√2cos(θ+π/4))
=(cosθ+sinθ)/(cosθ-sinθ)
=(1+tanθ)/(1-tanθ)
=(1-3)/(1+3)
=-1/2
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