arcsin[sin(19Π/12)]求值

题目

arcsin[sin(19Π/12)]求值
Π为3.1415926

答案

y=arcsin(x),定义域[-1,1] ,值域[-π/2,π/2]
arcsin[sin(19Π/12)]=arcsin[sin(2π - 5π/12)]=arcsin[sin(-5π/12)]= -5π/12