为什么sinπ／n+sin2π／n.+sin(n-1)π／n=cotπ／2n?谢谢.

Question

为什么sinπ／n+sin2π／n.+sin(n-1)π／n=cotπ／2n?
谢谢.

Answer 1

2 sin [π／(2n)]·sin(π／n)= cos [π／n -π／(2n)]- cos [π／n +π／(2n)]= cos [π／(2n)]- cos [3π／(2n)]2 sin [π／(2n)]·sin(2π／n)= cos [2π／n -π／(2n)]- cos [2π／n+π／(2n)]= cos [3π／(2n)]- cos [5π／(2n)]2 sin [π／(2n)]·sin(3π／n)= cos [3π／n -π／(2n)]- cos [3π／n +π／(2n)]= cos [5π／(2n)]- cos [7π／(2n)]……2 sin [π／(2n)]·sin[(n-1)π／n]= cos [(n-1)π／n -π／(2n)]- cos [(n-1)π／n +π／(2n)]= cos [(2n-3）π／(2n)]- cos [(2n-1)π／(2n)]故：2 sin [π／(2n)] ·｛sin(π／n)+sin(2π／n)+.+sin[(n-1)π／n]｝= cos [π／(2n)]- cos [(2n-1)π／(2n)]= cos [π／(2n)]- cos [π-π／(2n)]=2 cos [π／(2n)]故：sin(π／n)+sin(2π／n)+.+sin[(n-1)π／n]= cos[π／(2n)]/ sin [π／(2n)]= cot [π／(2n)]