Question

已知函数f（x）=sin（x+θ）+sin（x-θ）-2sinθ x=R时f（x）大于等于0 θ属于（0,3/2π）且t

Answer 1

f（x）=sin（x+θ）+sin（x-θ）-2sinθ

=2[sin(x+θ+x-θ)/2*cos(x+θ-x+θ)/2]-2sinθ

=2sinx*cosθ-2sinθ=√(4sin^2x+4)cos(α+θ)≥0

θ∈(0,π/2)

tan2θ=-3/4

2θ∈(π/2,π)

sin2θ=-3/4cos2θ

(sin2θ)^2+(cos2θ)^2=25(cos2θ)^2/16=1

cos2θ=-4/5

cos2θ=2(cosθ)^2-1=-4/5

cosθ=√10/10