cos(2π/3+2α)= - cos（π-(2π/3+2α)）
=- cos(π/3-2α)
=- cos(2α-π/3) (公式cos2α=1-2(sinα)^2）
=- (1-2(sin(α-π/6))^2)
因为 sin(π/6-α)=1/3 所以（sin(α-π/6))^2=1/9
故cos(2π/3+2α)= -(1-2*1/9)
=-7/9
是否可以解决您的问题？
cos(2π/3+2α)
=cos[2(π/3+α)]
=2cos^2(π/3+α)-1
=2cos^2[π/2-(π/6-α)]-1
=2sin^2(π/6-α)-1
=2*(1/3)^2-1
=2/9-1
=-7/9
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