解：f(x)=cos(π/3+x)cos(π/3-x)=1/2[cos(π/3+x+π/3-x)+cos(π/3+x-π/3+x)=1/2[cos(2π/3)+cos(2x)] =1/2[cos(π-π/3)+cos(2x)]=1/2[-cos(π/3)+cos(2x)]=1/2[cos(2x)-1/2]=cos(2x)/2-1/4
