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