若sin(π/6-α)=1/3,则cos(2π/3十2α)
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若sin(π/6-α)=1/3,则cos(2π/3十2α)若sin(π/6-α)=1/3,则cos(2π/3十2α)若sin(π/6-α)=1/3,则cos(2π/3十2α)答:sin(π/6-a)=
若sin(π/6-α)=1/3,则cos(2π/3十2α)
若sin(π/6-α)=1/3,则cos(2π/3十2α)
若sin(π/6-α)=1/3,则cos(2π/3十2α)
答:
sin(π/6-a)=1/3
sin(a-π/6)=-1/3
cos(2π/3+2a)
=cos(π-π/3+2a)
=-cos(π/3-2a)
=-cos[2(a-π/6)]
=-1+2*{sin(a-π/6) ]^2
=-1+2*(-1/3)^2
=-1+2/9
=-7/9
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