sin(π/6)-a=1/3,则cos(2π/3)+2a=?

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sin(π/6)-a=1/3,则cos(2π/3)+2a=?sin(π/6)-a=1/3,则cos(2π/3)+2a=?sin(π/6)-a=1/3,则cos(2π/3)+2a=?令x=π/6-aa=

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

sin(π/6)-a=1/3,则cos(2π/3)+2a=?
令x=π/6-a
a=π/6-x
2π/3+2a=π-2x
原式sinx=1/3
原式=cos(π-2x)
=-cos2x
=-(1-2sin²x)
=-7/9