Question

化简:sin(n∏-2/3∏)Xcos(n∏+4/3∏),n∈Z.

Answer 1

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 。

Answer 2

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