要證明
更新1:
三角恆等式
有條數五識??help me
回答 (1)
2007-02-08 3:41 am
✔ 最佳答案
(1-cosθ)(1+cosθ)/cos^2 (90 °-θ) = 1-cos^2 θ /cos^2 (90 ° - θ) [ Please remember that cos(90 °-θ)=sinθ ]
=sin^2 θ / sin^2 θ
=1
∵ L.H.S = R.H.S
∴(1-cos θ)(1+cos θ)/cos^2(90 ° - θ) ≡1
2007-02-07 19:44:06 補充：
加多點解釋 a, ^2 代表2次方b. (1-cosθ)(1 cosθ)= (1^2-cos^2θ)= 1-cos^2θc. cos^2 θ sin^2 θ=1
參考: My poor calculation
收錄日期: 2021-04-23 16:41:46
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070207000051KK03069