cosθ^2+cos(θ+2π/3)^2+cos(θ-2π/3)^2 计算

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cosθ^2+cos(θ+2π/3)^2+cos(θ-2π/3)^2 计算
cosθ^2+cos(θ+2π/3)^2+cos(θ-2π/3)^2 计算

cosθ^2+cos(θ+2π/3)^2+cos(θ-2π/3)^2 计算
cos^2θ+cos^2(θ+2π/3)+cos^2(θ-2π/3)
=(1+cos2θ)/2+[1+cos(2θ+4π/3)]/2+[1+cos(2θ-4π/3)]/2
=3/2 +[cos2θ+cos(2θ+4π/3)+cos(2θ-4π/3)]/2
cos2θ+cos(2θ+4π/3)+cos(2θ-4π/3)
=cos2θ+cos2θcos(4π/3)-sin2θsin(4π/3)+cos2θcos(4π/3)+sin2θsin(4π/3)
=cos2θ+2cos2θcos(4π/3)
=0
所以原式=3/2

(cosθ)^2+[cos(θ+2π/3)]^2+[cos(θ-2π/3)]^2
=(1+cos2θ)/2+[1+cos(2θ+4π/3)]/2+[1+cos(2θ-4π/3)]/2
=3/2+1/2*[cos2θ+2cos2θcos(4π/3)]
=3/2+1/2*[cos2θ-cos2θ]
=3/2 。

cos²θ-2cos-3/cos²θ+cosθ为什么等于cosθ-3/cosθ cosθ^2+cos(θ+π/3)^2-cosθcos(θ+π/3)化简如何化简 cosθ^2+cos(θ+2π/3)^2+cos(θ-2π/3)^2 计算化简cosθ+cos(θ+2π/3)+cos(θ+4π/3) cosθ>0,cos(θ/2) 1+cos(2θ)+cos(4θ)+cos(6θ)=4cosθcos(2θ)cos(3θ)证明相等. cos(θ-π/2)等于cos(π/2-θ) 为什么cos(θ-2π)=cosθ, 根号(1-cosθ/1+cosθ)+根号（1+cosθ/1-cosθ)θ属于（π/2,π） θ∈(0,π/2),比较cosθ、sin(cosθ)、cos(sinθ)的大小计算：cos（π/5)+cos（2π/5)+cos(3π/5)+cos(4π/5) cos^2α+cos^2(α+π/3)-cosαcos(α+π/3)化简已知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）｝（最后结果计算cos²θ+cos²（θ+2π/3）+cos²（θ-2π/3）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）高一数学若f(cosθ)=sin^2-3cosθ,则f(2cos π/3 )=_____ √(1+cosθ/1-cosθ)-√(1-cosθ/1+cosθ）（π＜θ＜3/2π）求值:cos(π/11)cos(2π/11)cos(3π/11)cos(4π/11)cos(5π/11)