已知cos(π/6-α),則cos(5π/6+α)×sin(2π/3-α)=

已知cos(π/6-α),則cos(5π/6+α)×sin(2π/3-α)=

原式=cos(5π/6+α)*sin(2π/3-α)
=cos[(α+π/3)+π/2]*sin[π-(α+π/3)]
=-sin(α+π/3)*sin(α+π/3)
=-sin²;(α+π/3)
=-cos²;(α-π/6)