2sinθcosθ+2sinθ-cosθ-1 = 0 0
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2sinθcosθ+2sinθ-cosθ-1 = 0 0
2sinθcosθ+2sinθ-cosθ-1 = 0 0
2sinθcosθ+2sinθ-cosθ-1 = 0 0
2sinθcosθ+2sinθ-cosθ-1
=2sinθ(cosθ+1)-(cosθ+1)
=(cosθ+1)(2sinθ-1)=0
cosθ=-1 or sinθ=1/2
θ=π or π /6 or 2π /3
化简:1+sinθ+cosθ+2sinθcosθ /1+sinθ+cosθ
θ∈(0,π/2),比较cosθ、sin(cosθ)、cos(sinθ)的大小
求证sinθ/(1+cosθ)+(1+cosθ)/sinθ=2/sinθ
2sinθ-cosθ=1 (sinθ+cosθ+1)/(sinθ-cosθ+1)如题已知2sinθ-cosθ=1 求(sinθ+cosθ+1)/(sinθ-cosθ+1)
求证:(1+cosθ+cosθ/2) /(sinθ+sinθ/2)=sinθ/1-cosθ
若sin θ-cos θ 分之sin θ+cos θ=2 则sin θcos θ 是
若(sinθ+cosθ)/(sinθ-cosθ)=2,则sinθcosθ的值是
sinθ /2cosθ /2 化简
化简:sin* θ/2cos* θ/2
1-cosθ/(2sinθ/2)
sinθ/2 乘cosθ/2
sinθ/2cosθ/2=
证明(1-2sinθcosθ)/(cos^2θ-sin^2θ)=(cos^2θ-sin^2θ)/(1-2sinθcosθ)
求证 (1-2sinθcosθ)/(cos^2θ-sin^2θ)=(cos^θ-sin^2θ)/(1+2sinθcosθ)
求证(1-sinθcosθ)除以(cos^2θ-sin^2θ)=(cos^2θ-sin^2θ)除以(1+2sinθcosθ)
化简2sin θcosθ结果是
2/sinθ+3/cosθ(0
原题是(sinθ+cosθ)^2