题目

化简:sin(n∏-2/3∏)Xcos(n∏+4/3∏),n∈Z.

提问时间：2021-04-01

答案

原式=sin(nπ-2π/3)*cos(nπ+2π-2π/3)
=sin(nπ-2π/3)*cos(nπ-2π/3)
=1/2*sin[2(nπ-2π/3)]
=1/2*sin(2nπ-4π/3)
=1/2*sin(-4π/3)
=1/2*(-√3/2)
= -√3/4 .