sin(a-π/3)=1/3,则cos(π/6+a)=

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sin(a-π/3)=1/3,则cos(π/6+a)=sin(a-π/3)=1/3,则cos(π/6+a)=sin(a-π/3)=1/3,则cos(π/6+a)=cos(π/6+a)=sin[π/2-

sin(a-π/3)=1/3,则cos(π/6+a)=
sin(a-π/3)=1/3,则cos(π/6+a)=

sin(a-π/3)=1/3,则cos(π/6+a)=
cos(π/6+a)
=sin[π/2-(a+π/6)]
=sin(π/3-a)
=-sin(a-π/3)
=-1/3

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