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
已知 sin a-cos a=1/2,则 sin^3 a-cos^3 a
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