MQ46 --- Series
回答 (1)
2012-09-16 9:35 am
✔ 最佳答案
cosx cos2x cos4x ... cos(2^n x)=sinx cosx cos2x cos4x ... cos(2^n x) / sinx
=(1/2) sin2x cos2x cos4x ... cos(2^n x) / sinx
=(1/2)^2 sin4x cos4x ... cos(2^n x) / sinx
=(1/2)^3 sin8x cos8x... cos(2^n x) / sinx
= ...
=(1/2)^(n+1) sin[2^(n+1) x] / sinx
收錄日期: 2021-04-13 18:59:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120915000051KK00673