有两个函数f(x)=asin(kx+π/3),g(x)=bcos(2kx-π/3)（k＞0）,它们

T1+T2=2π/|k|+2π/|2k|=3π/k=3π/2k=2f(x)=asin(2x-π/3)g(x)=bcos(4x-π/6)f(π/2)=asin(2π/3)=asin(π-π/3)=asin(π/3)=acos(π/2-π/6)=acos(π/6)=g(π/2)=bcos(-π/6)=bcos(π/6)所以a=bf(π/4)=asin(π/6)=-√3g(π/4)-1=-√3bcos(5π/6)-1所以a/2=-√3*b(-√3/2)-1=3b/2-1a=3b-2a=b所以a=b=1f(x)=sin(2x-π/3)g(x)=cos(4x-π/6)