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sin(π/3 -α)=cos(π/2 - π/6 - α)
=sin[π/2 - (π/6 + α)]
=cos(π/6 + α)=1/3
则cos(π/3 + 2α)=cos[2(π/6 + α)]
=2cos²(π/6 + α) - 1
=2×(1/3)² - 1
=-7/9
=sin[π/2 - (π/6 + α)]
=cos(π/6 + α)=1/3
则cos(π/3 + 2α)=cos[2(π/6 + α)]
=2cos²(π/6 + α) - 1
=2×(1/3)² - 1
=-7/9
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