已知f(x)＝sin(π-x)cos(2π-x)tan(-x＋π)/cos(－π/2＋x),求f(-31π/3)的值_数学解答

已知f(x)＝sin(π-x)cos(2π-x)tan(-x＋π)/cos(－π/2＋x),求f(-31π/3)的值

人气：472 ℃　时间：2020-05-15 22:06:37

解答

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