来自邱江涛的问题
tan{arcsin[cosπ/4]+arccos[sinπ/3]+arctan√3}
tan{arcsin[cosπ/4]+arccos[sinπ/3]+arctan√3}
1回答
2020-04-28 20:52
tan{arcsin[cosπ/4]+arccos[sinπ/3]+arctan√3}
tan{arcsin[cosπ/4]+arccos[sinπ/3]+arctan√3}
tan{arcsin[cosπ/4]+arccos[sinπ/3]+arctan√3}=
tan{arcsin(√2/2)+arccos(√3/2)+π/3}
=tan{π/4+π/6+π/3}
=tan{π/4+π/2}
=-tan(π/4)
=-1.