设sinθ+cosθ=√2/3,π/2
设sinθ+cosθ=√2/3,π/2
设θ是第三象限的角,sin(θ/2+3π/2)>0,则[√(1-sinθ)]/[cos(θ/2)-sin(θ/2)]的值
设f(α)=2sinαcosα+cosα/1+sin²α+cos(3π/2+α)-sin²(π/2+
设向量a=(3/2,sin θ),b=(cosθ,1/3),其中0
设tan2θ =-2根号2,2θ∈(π/2,π)求(2cos^2θ-2-sinθ-1)/(sinθ+cosθ)
设0≤θ≤2π,已知两个向量OP1=(cosθ,sinθ),OP2 = (2+sinθ,2-cosθ ),则向量P1P2
设0≤θ≤2π,已知两个向量OP1 = (cosθ,sinθ),OP2 = (2+sinθ,2-cosθ ),则向量P1
★ 设函数f(x) = [ (sinθ / 3) * x^3 ] + [ ((√3)cosθ / 2) * x^2 ]
已知(4sinθ-2cosθ)/(3sinθ+5cosθ)=6/11,求5cos^2θ/(sin^2θ+2sinθcos
sin(π-θ)+cos(2π-θ)/cos(5π/2-θ)+sin(3π/2+θ)=2,则sinθcosθ=_____
已知 sin(θ+kπ)=-2cos (θ+kπ) 求 ⑴4sinθ-2cosθ/5cosθ+3sinθ; ⑵(1/4)
(sinθ+cosθ)/(sinθ-cosθ)=2,则sin(θ-5π)*sin(3π/2-θ)=