若sin(π /6-a)=1/3,则cos(2π /3+2a)=__________
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若sin(π/6-a)=1/3,则cos(2π/3+2a)=__________若sin(π/6-a)=1/3,则cos(2π/3+2a)=__________若sin(π/6-a)=1/3,则cos
若sin(π /6-a)=1/3,则cos(2π /3+2a)=__________
若sin(π /6-a)=1/3,则cos(2π /3+2a)=__________
若sin(π /6-a)=1/3,则cos(2π /3+2a)=__________
cos(2π/3+2a)
=cos[2(π/3+a)]
=2cos²(π/3+a)-1
=2sin²[π/2-(π/3+a)]-1
=2sin²(π/6-a)-1
=-7/9
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