求复变大神，把这个化成三角函数形式

Question

如题所述

Answer 1

原式 1-cos(n+1)θ-isin(n+1)θ
= --------------------------
1-cosθ-isinθ

2sin² (n+1)θ/2-2isin(n+1)θ/2cos(n+1)θ/2
=-------------------------------------------
2sin²θ/2-2isinθ/2cosθ/2

sin(n+1)θ/2[ sin(n+1)θ/2-icos(n+1)θ/2]
=-------------------------------------------
sinθ/2 [sinθ/2-icosθ/2]

sin(n+1)θ/2 [cos((n+1)θ/2-π/2)+isin((n+1)θ/2-π/2)]
=-----------------------------------------------------
sinθ/2 [cos(θ/2-π/2)+isin(θ/2-π/2)]

sin(n+1)θ/2[cos(nθ/2)+isin(nθ/2)]
=------------------------------------
sinθ/2