sin^2θ-3sinθcosθ+1
化简:1+sinθ+cosθ+2sinθcosθ /1+sinθ+cosθ
求证sinθ/(1+cosθ)+(1+cosθ)/sinθ=2/sinθ
sinθ-cosθ=1/2,则sin^3θ-cos^3θ=?.
为什么sin2θ+sinθ=2sinθcosθ+sinθ=sinθ(2cosθ+1)
求证:(1+cosθ+cosθ/2) /(sinθ+sinθ/2)=sinθ/1-cosθ
sin^2θ/sinθ-cosθ + cosθ/1-tanθ = sin^2θ/sinθ-cosθ + cosθ/1-(
求证(1+sinθ+cosθ)/(1+sinθ-cosθ)+(1-cosθ+sinθ)/(1+cosθ+sinθ)=2/
求证(1-sinθcosθ)除以(cos^2θ-sin^2θ)=(cos^2θ-sin^2θ)除以(1+2sinθcos
已知(4sinθ-2cosθ)/(3sinθ+5cosθ)=6/11,求5cos^2θ/(sin^2θ+2sinθcos
已知向量a=(sinθ,cosθ-2sinθ),向量b=(1,2) 求tanθ 求sinθ*cosθ-3cos^2θ
已知θ为第三象限角,1-sinθcosθ-3cos^2=0则5sin^2θ+3sinθcosθ=?
已知 sin(θ+kπ)=-2cos (θ+kπ) 求 ⑴4sinθ-2cosθ/5cosθ+3sinθ; ⑵(1/4)