已知cos(π/6-a)=2/3 则sin(a-2π/3)=?

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已知cos(π/6-a)=2/3 则sin(a-2π/3)=?
已知cos(π/6-a)=2/3 则sin(a-2π/3)=?

已知cos(π/6-a)=2/3 则sin(a-2π/3)=?
sin(a-2π/3)
=-sin[π+(a-2π/3)]
=-sin(a+π/3)
=-cos[π/2-(a+π/3)]
=-cos(π/6-a)
=-2/3

cos(π/6-a)=2/3
sin(a-2π/3)=-sin(2π/3-a)=-sin(π/6-a+π/2)=-cos(π/6-a)=-2/3

cos(π/6-a)=2/3,则
sin(π/2+π/6-a)=sin(2π/3-a)=2/3
sin(a-2π/3)=-2/3