作业帮急.设f(θ)=[2cos2θ+sin2(2π-θ)+sin(π/2+θ)-3]/[2+2cos2(π+θ)+cos(-
来源：学生作业帮 编辑：拍题作业网作业帮 分类：数学作业 时间：2024/05/11 13:43:08
急.设f(θ)=[2cos2θ+sin2(2π-θ)+sin(π/2+θ)-3]/[2+2cos2(π+θ)+cos(-θ)]求f(π/3)
设f(θ)=[2cos2θ+sin2(2π-θ)+sin(π/2+θ)-3]/[2+2cos2(π+θ)+cos(-θ)]
求f(π/3)
设f(θ)=[2cos2θ+sin2(2π-θ)+sin(π/2+θ)-3]/[2+2cos2(π+θ)+cos(-θ)]
求f(π/3)
f(θ)=[2cos2θ+sin2(2π-θ)+sin(π/2+θ)-3]/[2+2cos2(π+θ)+cos(-θ)]
=[2cos2θ-sin2θ+cosθ-3]/[2+2cos2θ+cosθ]
故f(π/3)=[2cos2π/3-sin2π/3+cosπ/3-3]/[2+2cos2π/3+cosπ/3]
=[-1-根号3/2+1/2-3]/[2-1+1/2]
=[-根号3-7]/3
=[2cos2θ-sin2θ+cosθ-3]/[2+2cos2θ+cosθ]
故f(π/3)=[2cos2π/3-sin2π/3+cosπ/3-3]/[2+2cos2π/3+cosπ/3]
=[-1-根号3/2+1/2-3]/[2-1+1/2]
=[-根号3-7]/3
急.设f(θ)=[2cos2θ+sin2(2π-θ)+sin(π/2+θ)-3]/[2+2cos2(π+θ)+cos(-
若sin(π/4+α)=sinθ+cosθ,2sin^2β=sin2θ,求证：sin2θ+2cos2β=3
：求证:(1-2sinθcosθ)/(cos2θ-sin2θ)=(cos2θ-sin2θ)/(1+2sinθcosθ).
sin2θ+sinθ/2cos2θ+2sin^θ+cosθ=tanθ 数学题
sinθ+2cosθ=0,则2cos2θ-sin2θ=
若2sin(π/4+a）=sinθ+cosθ,2sin^2β=sin2θ,求证sin2a+(1/2)cos2β=0.
若2sin(π/4+a）=sinθ+cosθ,2sin^2β=sin2θ,求证sin2a+(1/2)cos2β=0
已知sinθ-2cosθ=0,求sin2θ-cos2θ/1=sin2θ
已知sinθ-2cosθ=0,求sin2θ-cos2θ/sinθ*cosθ 的值
已知sinθ+cosθ=√2/2,0＜θ＜3π/4,求sin2θ,cos2θ的值.
若1/3sinθ-cosθ=0,则2cos2θ+1/2sin2θ=?
已知2cosθ-sinθ=0 求3cos2θ+4sin2θ的值