证明1+cosA+cos2A+cos3A+……+cos(n-1)A=0 期中A=360/n

cosαinβ=[sin(α+β)-sin(α-β)]/2 nA=360°
1+cosA+cos2A+cos3A+……+cos(n-1)A
=sinA+cosAsinA+cos2AsinA+cos3AsinA+……+cos(n-1)AsinA
=sinA+[sin2A-sin0°+sin3A-sinA+sin4A-sin2A+……+sin(nA)-sin(n-2)A]/2
=sinA+[-sin0°-sinA+sin(n-1)A+sin(nA)]/2
=sinA+[-sinA-sinA]/2
=0
∵sinA≠0,∴1+cosA+cos2A+cos3A+……+cos(n-1)A=0