圖片參考：http://i707.photobucket.com/albums/ww74/stevieg90/01-73.jpg
更新1:
厲害!
Limit
2010-01-19 6:49 am
回答 (1)
2010-01-19 7:01 am
✔ 最佳答案
cos (θ/2)cos (θ/4)cos (θ/8) ... cos (θ/2n)= [cos (θ/2)cos (θ/4)cos (θ/8) ... cos (θ/2n)sin (θ/2n)]/sin (θ/2n)
= [(1/2n) sin θ]/sin (θ/2n)
lim (n → ∞) cos (θ/2)cos (θ/4)cos (θ/8) ... cos (θ/2n)
= lim (n → ∞) [(1/2n) sin θ]/sin (θ/2n)
= sin θ lim (n → ∞) (1/2n)/sin (θ/2n)
= (sin θ/θ) lim (n → ∞) (θ/2n)/sin (θ/2n)
= sin θ/θ
參考: Myself
收錄日期: 2021-04-22 00:46:29
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100118000051KK01868