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

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

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

等于-7/9

cos(2π/3+2x)=cos2(π/3+x)=2[cos(π/3+x)]^2-1
=2[sin(π/2-π/3-x)]^2-1
=2[sin(π/6-x)]^2-1
=2/9-1
=-7/9

cos(2π/3+2x)=2(cos(π/3+x))^2-1=2(sin(π/2-π/3-x))^2-1=2*(1/3)^2-1= -7/9