若sin(π/6-啊)=1/3,则cos(2π/3+2a)=?

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

若sin(π/6-啊)=1/3,则cos(2π/3+2a)=?
sin(π/6-a)=1/3
sin(π/6-a)=cos[π/2-(π/6-a)]=cos(π/3+a)=1/3
cos(2π/3+2a)=cos[2(π/3+a)]=2[cos(π/3+a)]^2-1=-7/9

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