cos@+cos2@+cos3@+....+cosn@化简后结果是 什么 啊？高手帮忙啊

cos@+cos2@+cos3@+....+cosn@
=(2sinacosa+2sinacos2a+...+2sinacosna)*1/2sina
=(sin2a+(sin3a-sina)+(sin4a-sin2a)+...+(sin(n+1)a-sin(n-1)a))*1/2sina
=(sin(n+1)a+sinna-sina)*1/2sina
=2sin(2n+1)a/2cosa/2*1/2sina-1/2
=sin(2n+1)a/2 / 2sina/2 -1/2

2sin(θ/2)×(cosθ＋cos2θ＋……＋cosnθ)
＝sin(3θ/2)－sin(θ/2)＋sin(5θ/2)－sin(3θ/2)＋……＋sin((2n+1)θ/2)－sin((2n-1)θ/2)
＝sin((2n+1)θ/2)－sin(θ/2)

所以，cosθ＋cos2θ＋……＋cosnθ＝[sin((2n+1)θ/2)－sin(θ/2)]/[2sin(θ/2)]