sin(7π／6-α)
=cos[π/2-(7π/6-α)]
=cos(-2π/3+α)
=cos(2π/3-α)
=-cos[π-(2π/3-α)]
=-cos(π/3+α)
=1/3
cos(π/3+α)=-1/3
cos(2π/3+2α)
=cos[2(π/3+α)]
=2[cos(π/3+α)]^2-1
=-7/9
sin(7π／6-α)=1/3,则cos（2π/3+2α）等于
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