若sin(π/6-α)=1/3,则cos(2π/3+2α)的值?请针对每一步做说明,清楚一点.

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)= - cos（π-(2π/3+2α)）=- cos(π/3-2α)这一步是如何变化过来的=- cos(2α-π/3)这一步是如何变化过来的cos(2π/3+2α)= - cos（π-(2π/3+2α)）//注：直接将括号内的展开就可以得到下面的=- cos(π/3-2α)//注：根据公式cos(-β)=cosβ,这里有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