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化简:sin(n∏-2/3∏)Xcos(n∏+4/3∏),n∈Z.
人气:226 ℃ 时间:2020-05-10 00:59:44
解答
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 .
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