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

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

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

已知cos(π/6-α),则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)

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