若sin(π/6-α)=1/3,则cos(2/3π+2α)=?A-7/9 B-1/3 C7/9 D1/3
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A-7/9

cos(2π/3+2α)

=-cos[π-(2π/3+2α)]

=-cos(π/3-2α)

=-cos[2(π/6-α)]

=-{1-2[sin(π/6-α)]^2}

=-(1-2/9)

=-7/9

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