sin A+sinB +sinC =4cosA/2cosB/2cosC/2
sin2A+sin2B-sin2C = 4cosAcosBsinC
cos2A+cos2B+cos2C+1 = -4cosAcosBcosC
三角恆等式17-19
2011-03-06 11:16 pm
回答 (1)
2011-03-08 5:57 am
收錄日期: 2021-04-23 23:23:41
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110306000051KK00746