解
f(x)=[sinxcos(-x)tan(-x)]/cos[-(π/2-x)]
=[-sinxcosxtanx]/cos(π/2-x)
=(-sin²x)/sinx
=-sinx
f(-31π/3)=-sin(-31π/3)=sin(31π/3)=sin(10π+π/3)=sinπ/3=√3/2