化简:[(1+sinθ+cosθ)(sinθ/2-cosθ/2)]/√(2+2cosθ)(0<θ<π)
化简:1+sinθ+cosθ+2sinθcosθ /1+sinθ+cosθ
化简:[(1+sinθ+cosθ)(sinθ/2-cosθ/2)]/√(2+2cosθ)(0<θ<π)
θ∈(0,π/2),比较cosθ、sin(cosθ)、cos(sinθ)的大小
求证:(1+cosθ+cosθ/2) /(sinθ+sinθ/2)=sinθ/1-cosθ
化简2cosθ/√1-sin^2θ+√1-cos^2θ/sinθ
化简【1+sinθ-cosθ/1+sinθ+cosθ】+cot(θ/2)
求证(1+sinθ+cosθ)/(1+sinθ-cosθ)+(1-cosθ+sinθ)/(1+cosθ+sinθ)=2/
三角恒等变换,化简:(1+sinθ+cosθ)(sinθ/2-cosθ/2)/√2+2cosθ
求证sinθ/(1+cosθ)+(1+cosθ)/sinθ=2/sinθ
sin^2θ/sinθ-cosθ + cosθ/1-tanθ = sin^2θ/sinθ-cosθ + cosθ/1-(
求证(1-sinθcosθ)除以(cos^2θ-sin^2θ)=(cos^2θ-sin^2θ)除以(1+2sinθcos
已知tanθ=根号2,求(1)(cosθ+sinθ)/(cosθ-sinθ);(2)sin²θ-sinθcos