若sin(π/6-α)=1/3,则cos(2π/3+α)=

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若sin(π/6-α)=1/3,则cos(2π/3+α)=
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若sin(π/6-α)=1/3,则cos(2π/3+α)=
若sin(π/6-α)=1/3,则cos(2π/3+α)=

若sin(π/6-α)=1/3,则cos(2π/3+α)=
∵sin(π/6+α)=1/3
∴cos(2π/3+α)=-cos[π-(2π/3+α)]=-cos(π/3-α)=-1/3.
解题思路：
主要运用正、余弦的公式变换：
cosα=sin(π/2-α),cosα=-cos(π-α).
不懂的欢迎追问,

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