若sin(π/6-α)=1/3,则cos(2π/3+2α)=
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cos(2π/3+2α)

=cos[2(π/3+α)]

=2cos^2(π/3+α)-1

=2cos^2[π/2-(π/6-α)]-1

=2sin^2(π/6-α)-1

=2*(1/3)^2-1

=2/9-1

=-7/9

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