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sin3A / sin2A = 2cosA - 1/(2cosA)
求solution,謝謝
trigonometrical identities 一問
2011-10-03 5:40 am
回答 (1)
2011-10-03 5:56 am
✔ 最佳答案
sin3A / sin2A= sin(A + 2A) / sin2A
= (sinA cos2A + sin2A cosA) / (2 sinA cosA)
= (sinA (2cos²A - 1) + 2 sinA cos²A) / (2 sinA cosA)
= (4 sinA cos²A - sinA) / (2 sinA cosA)
= 2 cosA - 1/(2cosA)
收錄日期: 2021-04-21 22:21:42
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20111002000051KK01147