已知sin(π/3-a)=1/3,则cos(2π/3-2a)=( )

问题描述:

已知sin(π/3-a)=1/3,则cos(2π/3-2a)=( )

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