若sin(π/6-α)=1/3,则cos(2/3π+2α)=?A-7/9 B-1/3 C7/9 D1/3
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若sin(π/6-α)=1/3,则cos(2/3π+2α)=?A-7/9B-1/3C7/9D1/3若sin(π/6-α)=1/3,则cos(2/3π+2α)=?A-7/9B-1/3C7/9D1/3若s
若sin(π/6-α)=1/3,则cos(2/3π+2α)=?A-7/9 B-1/3 C7/9 D1/3
若sin(π/6-α)=1/3,则cos(2/3π+2α)=?A-7/9 B-1/3 C7/9 D1/3
若sin(π/6-α)=1/3,则cos(2/3π+2α)=?A-7/9 B-1/3 C7/9 D1/3
A-7/9
cos(2π/3+2α)
=-cos[π-(2π/3+2α)]
=-cos(π/3-2α)
=-cos[2(π/6-α)]
=-{1-2[sin(π/6-α)]^2}
=-(1-2/9)
=-7/9
sin(π/6-A)=1/3
yinwei cos2A=1-2sin^2 A
suoyi cos(pi/3-2A)=7/9
cos(2/3π+2α)=cos(-2/3π-2α)= -cos(pi+(-2/3π-2α))=-cos(pi/3-2A)=-7/9
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