sin(π-θ)+cos(2π-θ)/cos(5π/2-θ)+sin(3π/2+θ)=2,则sinθcosθ=_____
sin(π-θ)+cos(2π-θ)/cos(5π/2-θ)+sin(3π/2+θ)=2,则sinθcosθ=_____
若(sinθ+cosθ)/(sinθ-cosθ)=2,则sin(θ-5π)*sin(3π/2-θ)等于?
(sinθ+cosθ)/(sinθ-cosθ)=2,则sin(θ-5π)*sin(3π/2-θ)=
θ∈(0,π/2),比较cosθ、sin(cosθ)、cos(sinθ)的大小
已知 sin(θ+kπ)=-2cos (θ+kπ) 求 ⑴4sinθ-2cosθ/5cosθ+3sinθ; ⑵(1/4)
证明下列恒等式: (1)2sin(2/π+x)cos(2/π-x)*cosθ+(2cos^2x-1)*sinθ=sin(
求证:(1+cosθ+cosθ/2) /(sinθ+sinθ/2)=sinθ/1-cosθ
1.已知2sin(3π+θ)=cos(π+θ),求2sin
求证(1+sinθ+cosθ)/(1+sinθ-cosθ)+(1-cosθ+sinθ)/(1+cosθ+sinθ)=2/
已知tanθ=3,求(3cosθ-5sin^2θcosθ)/sin(π-θ)的值
已知(4sinθ-2cosθ)/(3sinθ+5cosθ)=6/11,求5cos^2θ/(sin^2θ+2sinθcos
参数方程化为普通方程 x=(sinθ+cosθ)/(2sinθ+3cosθ) y=sinθ/(2sinθ+3cosθ)