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）

题目

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）

答案

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)