Question

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

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

Answer 1

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

Answer 2

解：
sin（π-x）=sinx，
cos（2π-x）=cosx，
tan（ -x+3/2π）=tan(π/2-x)=cotx=cosx/sinx，
cos（ - π- x）=cos(π-x)=-cosx，
则f(x)=cos²x/-cosx=-cosx，
f（- 31/3 π）=-cos(-31/3π)=-cos(-π/3)=-cosπ/3=-1/2.

追问

是tan(-x＋π)啊