sin((π/6)-α)=1/3,
而sin((π/6)-α)=cos[π/2-((π/6)-α))]
= cos((π/3)+α),
∴cos((π/3)+α)=1/3.
所以cos((2π/3)+2α)= cos[2((π/3)+α)]
= 2cos²((π/3)+α)-1
=-7/9.
sin((π/6)-α)=1/3,
而sin((π/6)-α)=cos[π/2-((π/6)-α))]
= cos((π/3)+α),
∴cos((π/3)+α)=1/3.
所以cos((2π/3)+2α)= cos[2((π/3)+α)]
= 2cos²((π/3)+α)-1
=-7/9.