若(sinθ+cosθ)/(sinθ-cosθ)=2,则sin(θ-5π)*sin(3π/2-θ)等于?
若(sinθ+cosθ)/(sinθ-cosθ)=2,则sin(θ-5π)*sin(3π/2-θ)等于?
sin(π-θ)+cos(2π-θ)/cos(5π/2-θ)+sin(3π/2+θ)=2,则sinθcosθ=_____
求证:(1+cosθ+cosθ/2) /(sinθ+sinθ/2)=sinθ/1-cosθ
参数方程化为普通方程 x=(sinθ+cosθ)/(2sinθ+3cosθ) y=sinθ/(2sinθ+3cosθ)
求证(1+sinθ+cosθ)/(1+sinθ-cosθ)+(1-cosθ+sinθ)/(1+cosθ+sinθ)=2/
1.已知2sin(3π+θ)=cos(π+θ),求2sin
若sinα+cosαsinα−cosα=2,则sin(α-5π)•sin(3π2-α)等于( )
若2sin(π4+α)=sin θ+cos θ,2sin2β=sin 2θ,求证:sin&
已知tanθ=根号2,求(1)(cosθ+sinθ)/(cosθ-sinθ);(2)sin²θ-sinθcos
证明下列恒等式: (1)2sin(2/π+x)cos(2/π-x)*cosθ+(2cos^2x-1)*sinθ=sin(
(sinθ+cosθ)/(sinθ-cosθ)=2,则sin(θ-5π)*sin(3π/2-θ)=
(sinθcosπ/3)-(cosθsinπ/3)怎么化简得sin(θ-π/3)