2(sinθ+cosθ)= 2√2sin(θ+π/4)为什么?; (-π/2
2(sinθ+cosθ)= 2√2sin(θ+π/4)为什么?; (-π/2
为什么sin2θ+sinθ=2sinθcosθ+sinθ=sinθ(2cosθ+1)
已知 sin(θ+kπ)=-2cos (θ+kπ) 求 ⑴4sinθ-2cosθ/5cosθ+3sinθ; ⑵(1/4)
θ∈(0,π/2),比较cosθ、sin(cosθ)、cos(sinθ)的大小
已知(4sinθ-2cosθ)/(3sinθ+5cosθ)=6/11,求5cos^2θ/(sin^2θ+2sinθcos
求证sinθ/(1+cosθ)+(1+cosθ)/sinθ=2/sinθ
(sinθ+cosθ)/(sinθ-cosθ)=2,则sin(θ-5π)*sin(3π/2-θ)=
若(sinθ+cosθ)/(sinθ-cosθ)=2,则sin(θ-5π)*sin(3π/2-θ)等于?
若2sin(π4+α)=sin θ+cos θ,2sin2β=sin 2θ,求证:sin&
化简:1+sinθ+cosθ+2sinθcosθ /1+sinθ+cosθ
sin(π-θ)+cos(2π-θ)/cos(5π/2-θ)+sin(3π/2+θ)=2,则sinθcosθ=_____
2sinθ+cosθ/sinθ-3cosθ=-5,求cos2θ+4sinθ