三角恆等式10
回答 (1)
2011-03-04 2:08 am
✔ 最佳答案
sin(A-B) / (sinAsinB) + sin(B-C) / (sinBsinC) + sin(C-A) / (sinCsinA)= [sinC sin(A-B) + sinA sin(B-C) + sinB sin(C-A)] / (sinA sinB sinC)
= {(1/2)[cos(C - A+B) - cos(C + A-B] + (1/2)[cos(A -B+C) - cos(A + B-C)]
+ (1/2)[cos(B - C+A) - cos(B + C-A)] } / (sinA sinB sinC)
= 0
收錄日期: 2021-04-21 22:19:15
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110303000051KK00710