f(x)=√3sin(π-wx)coswx-cos^2wx+1/2
=√3sinwxcoswx-(cos2wx+1)/2+1/2
=√3/2sin2wx-1/2cos2wx
=sin(2wx-π/6)
T=π/2 , T=2π/2w=π/2 , w=2第二问呢？f(a/4+π/6)=12/13,求tan(a+π/4)的值f(x)=sin(4x-π/6)12/13=sin[4(a/4+π/6)-π/6]=sin(a+π/2)=cosa , sina=-5/13,tana=-5/12 tan(a+π/4)=(tana+1)/(1-tana)=(7/12)/(17/12)=7/17