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)
因為 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)= - cos（π-(2π/3+2α)）=- cos(π/3-2α)這一步是如何變化過來的=- cos(2α-π/3)這一步是如何變化過來的cos(2π/3+2α)= - cos（π-(2π/3+2α)）//注：直接將括號內(nèi)的展開就可以得到下面的=- cos(π/3-2α)//注：根據(jù)公式cos(-β)=cosβ,這里有cos(π/3-2α)=cos(-(2α-π/3))=cos(2α-π/3）=- cos(2α-π/3)(公式cos2α=1-2(sinα)^2）=- (1-2(sin(α-π/6))^2)因為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