cosθ+cos(120°-θ)+cos(120°+θ)怎么化简,急用,
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/08 06:34:56
cosθ+cos(120°-θ)+cos(120°+θ)怎么化简,急用,
cosθ+cos(120°-θ)+cos(120°+θ)怎么化简,急用,
cosθ+cos(120°-θ)+cos(120°+θ)怎么化简,急用,
cosθ+cos(120°-θ)+cos(120°+θ)
=cosθ+cos120°cosθ+sin120°sinθ+cos120°cosθ-sin120°sinθ
=cosθ+2cos120°cosθ
=cosθ+2x(-1/2)cosθ
=0
cosθ+cos(120°-θ)+cos(120°+θ)怎么化简,急用,
cos
cos²θ-2cos-3/cos²θ+cosθ为什么等于cosθ-3/cosθ
|cosθ|=-cosθ,tanθ
cosθ>0,cos(θ/2)
cos²α+cos²(120°-α)+cosαcos(120°-α)化简cos²α+cos²(120°-α)+cosαcos(120°-α)
cos²α+cos²(120°-α)+cosαcos(120°-α) cos²α+cos²(120°-α)+cosαcos(120°-α)
根号(1-cosθ/1+cosθ)+根号(1+cosθ/1-cosθ)θ属于(π/2,π)
化简cosα+cos(120°-α)+cos(120+α)
cos^2x+cos^2(x+120°)+cos^2(x+240°)
cos θ/(1-tan θ)怎样化成cos θ/[(cosθ-sinθ)/cosθ]
已知cosθ
1+cos(2θ)+cos(4θ)+cos(6θ)=4cosθcos(2θ)cos(3θ)证明相等.
求证:(1+cosθ+cosθ/2) /(sinθ+sinθ/2)=sinθ/1-cosθ
求证:(1+sinθ-cosθ)/(1+sinθ+cosθ)=sinθ/(cosθ+1)
已知F(θ)=cos^2θ+cos^2(θ+α)+cos^2(θ+β)问是否满足0
已知sin(3π+θ)=lg1/(10开3次方0{cos(π+θ)/cosθ[cos(π-θ)-1]}+{ cos(θ-π)/cos(π-θ)+cos(θ已知sin(3π+θ)=lg1/(10开3次方)求{cos(π+θ)/cosθ[cos(π-θ)-1]}+{ cos(θ-2π)/cos(π-θ)+cos(θ-2π)}(最后结果
已知sin(3π+θ)=lg1/(10开3次方),求值cos(π+θ)/cosθ{cos(π-θ)-1} + cos(θ-π)/cos(π-θ)+cos(更正:已知sin(3π+θ)=lg1/(10开3次方),求值cos(π+θ)/cosθ{cos(π-θ)-1} + cos(θ-π)/cos(π-θ)+cos(θ-2π) 高一数学