求证:cosx+cos2x+…+cosnx=cosn+12x•sinn2xsinx2.

问题描述:

求证:cosx+cos2x+…+cosnx=

cos
n+1
2
x•sin
n
2
x
sin
x
2

证明:∵2sinx2cosnx=sin(x2+nx)+sin(x2−nx).∴2sinx2(cosx+cos2x+…+cosnx)=(sin3x2−sinx2)+(sin5x2−3x2)+…+(sin1+2n2x−sin1−2n2x)=sin1+2n2x−sinx2=2cosn+12xsinn2x.∴cos+cos2x+…+cosnx=cosn+12x•...
答案解析:利用2sin

x
2
cosnx=sin(
x
2
+nx)
+sin(
x
2
−nx)
.及和差化积即可得出.
考试点:三角函数恒等式的证明.

知识点:本题考查了积化和差、和差化积,属于基础题.