已知tanθ=1/2,则sin2θ+sinθ^2=
已知2sin^2θ+sin2θ/1+tanθ=k,0
证明 (2sinα-sin2α)/(2sinα+sin2α)=tan²(θ/2)
sin2θ+sinθ/2cos2θ+2sin^θ+cosθ=tanθ 数学题
求证 (sinθ+cosθ-1)(sinθ-cosθ+1)) /sin2θ=tanθ/2
求证(sinθ+cosθ-1)(sinθ-cosθ+1)/sin2θ=tanθ/2
已知cosθ=-3/5,θ∈(π/2,π) 求2/sin2θ-cosθ/sinθ+tan^2(θ/2)+sin^4(θ/
已知(sinθ+cosθ)/(sinθ-cosθ)=2,则sin2θ
已知(2sin^2α+sin2α)/(1+tanα)=k,(π/4
已知2sin^α+sin2α/1+tanα=k(π/4
已知tanα=2 求sin2α-3sinαcosα+1
已知tanθ=2,求sin2θ-cos2θ/1+cot²θ的值
求证:sin2θ+sinθ/2cos2θ+2sin^2θ+cosθ=tanθ