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cosx+cos2x+cos3x+.+cosnx=
人气:301 ℃ 时间:2020-05-11 06:44:14
解答
cosx+cos2x+cos3x+.+cosnx
=sin(x/2)*[ cosx+cos2x+cos3x+.+cosnx] / sin(x/2) ( 将sin(x/2) 移入方括号里并化简)
= {sin[x(2n+1)/2] - sin(x/2) }/ [2sin(x/2)]
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