若sin(π/6-α)=1/3,则cos(2π/3十2α)
问题描述：
若sin(π/6-α)=1/3,则cos(2π/3十2α)
答
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