三角比的關係
回答 (2)
2007-09-01 6:36 pm
✔ 最佳答案
( sin θ / 1 + cos θ ) + ( sin θ / 1 - cos θ ) =[sin θ(1-cos θ)+sin θ(1+cos θ)]/[(1+cos θ)(1-cos θ)]
=(sin θ-sin θ cos θ+sin θ+sin θ cos θ)/(1^2-cos^2 θ)
=2sin θ/sin ^2 θ
=2/sin θ//
2007-09-01 6:42 pm
( sin θ / 1 + cos θ ) + ( sin θ / 1 - cos θ )
= [sin θ(1-cos θ)] / [(1 + cos θ)(1- cos θ)] +[sin θ(1+cos θ)] / [(1 + cos θ)(1- cos θ)]
=[sin θ - sin θ cos θ + sin θ + sin θ cos θ]/[1-(cos θ)^2]
=2sin θ/(sin θ)^2
=2/sin θ
= [sin θ(1-cos θ)] / [(1 + cos θ)(1- cos θ)] +[sin θ(1+cos θ)] / [(1 + cos θ)(1- cos θ)]
=[sin θ - sin θ cos θ + sin θ + sin θ cos θ]/[1-(cos θ)^2]
=2sin θ/(sin θ)^2
=2/sin θ
參考: Math. Knowledge
收錄日期: 2021-04-30 20:41:36
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070901000051KK01210