求和Sn=cosx+cos2x+cos3x+……+cosnx
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/07/19 14:26:37
求和Sn=cosx+cos2x+cos3x+……+cosnx
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
求和Sn=cosx+cos2x+cos3x+……+cosnx
化简:cosx+cos2x+cos3x+……cosnx=?
cosx+cos2x+cos3x+.+cosnx=
0.5+cosx+cos2x+cos3x…………cosnx,把这个式子化简成分子和分母的形式
COSX+COS2X+COS3X+COS4X+COS5X+COS6X+...+COSNX=1/2|{SIN(N+1/2)
求证:cosx+cos2x+…+cosnx=cosn+12x•sinn2xsinx2
怎么化解cosx+cos2x+.+cosnx
lim(cosx+cos2x+ cos3x+…+ cos nx-n)/(cosx-1) (x趋于0)
cosx*cos2x=0.5(cosx+cos3x)是如何推出的?
三角恒等变换证明cosx+cos2x+…+cosnx=[cos(n+1/2)·sinnx/2]/sinx/2怎么证明?
高中三角函数题:化简cosx+cos2x+...+cosnx
试求方程:cosx cos2x=cos3x cos4x