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求和Sn=cosx+cos2x+cos3x+……+cosnx
题目详情
求和Sn=cosx+cos2x+cos3x+……+cosnx
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答案和解析
2sin(x/2)[cosx+cos2x+cos3x+……+cosnx ]
=2sin(x/2)cosx+2sin(x/2)cos2x+2sin(x/2)cos3x+……+2sin(x/2)cosnx
=sin(3x/2)-sin(x/2)+sin(5x/2)-sin(3x/2)+sin(7x/2)-sin(5x/2)+……+sin(x/2+nx)-sin(nx-x/2)
=sin(x/2+nx)-sin(x/2)
所以 cosx+cos2x+cos3x+……+cosnx
=[sin(x/2+nx)-sin(x/2)]/[2sin(x/2)]
=sin(x/2+nx)/[2sin(x/2)]-1/2
=2sin(x/2)cosx+2sin(x/2)cos2x+2sin(x/2)cos3x+……+2sin(x/2)cosnx
=sin(3x/2)-sin(x/2)+sin(5x/2)-sin(3x/2)+sin(7x/2)-sin(5x/2)+……+sin(x/2+nx)-sin(nx-x/2)
=sin(x/2+nx)-sin(x/2)
所以 cosx+cos2x+cos3x+……+cosnx
=[sin(x/2+nx)-sin(x/2)]/[2sin(x/2)]
=sin(x/2+nx)/[2sin(x/2)]-1/2
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