高一数学：tan[arcsin1+arctan(-√3/3)]+cos[arccos(-1/2)+arctan0]
问题描述：
高一数学：tan[arcsin1+arctan(-√3/3)]+cos[arccos(-1/2)+arctan0]
答
arcsin1=π/2
arctan(-√3/3)=-π/6
arcsin1+arctan(-√3/3)=π/3
arccos(-1/2)+arctan0=5π/6
所以tanπ/3+cos5π/6
=√3/3+√3/2=5√3/6