请数学高手解答 cosθ+cos2θ+······+cosnθ=?

显示全部楼层 · 2012-2-16 13:59:35

(cosx+cos2x+cos3x+...+cosnx)cosx=(cosxcosx+cosxcos2x+...+cosxcosnx)=(1/2)(cos2x+1)+(1/2)(cos3x+cosx)+(1/2)(cos4x+cos3x)+...+(1/2)(cosn+1)x+cosnx)=(1/2)(1+cos(n+1)x) +(1/2)(cosx+cos2x+...+cosnx)(cosx+cos2x+...+cosnx)(cosx-1/2)=(1/2)(1+cos(n+1)xcosx+cos2x+...+cosnx=(1+cos(n+1)x)/(2cosx-1)...

千问 · 2012-2-16 13:59:35

cosθ+cos2θ+······+cosnθ=n...