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数学
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化简:sin(npai-2/3pai)*cos(npai+4/3pai),n属于Z
人气:306 ℃ 时间:2020-09-20 23:14:45
解答
sin(nπ-2/3π)*cos(nπ+4/3π)
=sin(nπ-2/3π+2π)*cos(nπ+4/3π)
=sin(nπ+4/3π)*cos(nπ+4/3π)
=1/2*[2sin(nπ+4/3π)*cos(nπ+4/3π)]
=1/2*sin(2nπ+8/3π)
=1/2*sin(2/3π)
=1/2*(-√3/2)
=-√3/4
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