已知tanθ=(sin α-cos α)/(sin α+cos α) a,θ(0,π/2) 求证cos(3/2兀+ α) -sin(5π/2-α)=根号2sin(θ-4π)
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![已知tanθ=(sin α-cos α)/(sin α+cos α) a,θ(0,π/2) 求证cos(3/2兀+ α) -sin(5π/2-α)=根号2sin(θ-4π)](/uploads/image/z/1063622-38-2.jpg?t=%E5%B7%B2%E7%9F%A5tan%CE%B8%3D%28sin+%CE%B1-cos+%CE%B1%29%2F%28sin+%CE%B1%2Bcos+%CE%B1%29+a%2C%CE%B8%280%2C%CF%80%2F2%29+%E6%B1%82%E8%AF%81cos%283%2F2%E5%85%80%2B+%CE%B1%29+-sin%285%CF%80%2F2-%CE%B1%EF%BC%89%3D%E6%A0%B9%E5%8F%B72sin%28%CE%B8-4%CF%80%EF%BC%89)
已知tanθ=(sin α-cos α)/(sin α+cos α) a,θ(0,π/2) 求证cos(3/2兀+ α) -sin(5π/2-α)=根号2sin(θ-4π)
已知tanθ=(sin α-cos α)/(sin α+cos α) a,θ(0,π/2) 求证cos(3/2兀+ α) -sin(5π/2-α)=根号2sin(θ-4π)
已知tanθ=(sin α-cos α)/(sin α+cos α) a,θ(0,π/2) 求证cos(3/2兀+ α) -sin(5π/2-α)=根号2sin(θ-4π)
tanθ=(sinα-cosα)/(sinα+cosα)
sinθ/cosθ=(sinα-cosα)/(sinα+cosα)
sinθ(sinα+cosα)=cosθ(sinα-cosα)
sinθsinα+sinθcosα=cosθsinα-cosθcosα
sinθcosα-cosθsinα=-(cosθcosα+sinθsinα)
即sin(θ-α)=-cos(θ-α)
那么tan(θ-α)=-1
又α,θ∈(0,π/2)
θ-α=-π/4
θ=α-π/4
cos(3/2π+α)-sin(5/2π-α)
=sinα-cosα
=√2(√2/2sinα-√2/2cosα)
=√sin(α-π/4)
=√2sinθ
=√2sin(θ-4π)
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