tan{arcsin[cosπ/4]+arccos[sinπ/3]+arctan√3}
问题描述：
tan{arcsin[cosπ/4]+arccos[sinπ/3]+arctan√3}
答
解:
tan{arcsin[cosπ/4]+arccos[sinπ/3]+arctan√3}
=tan{arcsin[sinπ/4]+arccos[cosπ/6]+arctan√3}
=tan{π/4+π/6+π/3}
=tan3∏/4
=-1
注:
arcsin[sin∏/4]=∏/4
arccos[cosπ/6]=∏/6
也可把cos∏/4和sinπ/3算出,代入解.
多运用基本公式解题,把底子弄扎实点.
答
tan{arcsin[cosπ/4]+arccos[sinπ/3]+arctan√3}
=tan(arcsin*根号2/2+arccos*根号3/2+arctan根号3)
=tan(45+30+60)
=tan135
=-1
答
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.