請各位幫小妹解答以下的問題。急！
請列式以方便小妹理解。
Simplify the following expressions:
1.3sinθ/cosθ
2.sin^3θ-sinθ+cos^2θsinθ
3.tan^2θ/1-cos^2θ
4.cosθsinθ(1/tan^2θ+1)
5.sinθ/cosθ+cosθ/sinθ
6.1/1-sinθ+1/1+sinθ
請大家盡量答吧，感激不盡。
中三問題 sin cos tan
2008-04-22 2:06 am
回答 (2)
2008-04-22 2:26 am
✔ 最佳答案
1:3tanθ
2:
=sinθ(sin^2θ+cos^2θ-1)
=sinθ(1-1)
=0
3:
=tan^2θ/sin^2θ
=1/cos^2θ
4:
=cosθsinθ[1/(sin^2θ/cos^2θ+1)]
=cosθsinθ[cos^2θ/(sin^2θ+cos^2θ)]
=cos^3θsinθ
5:
=(sin^2θ+cos^2θ)/sinθcosθ
=1/sinθcosθ
6:
=(1+sinθ+1-sinθ)/(1-sin^2θ)
=2/(1-sin^2θ)
=2/cos^2θ
2008-04-21 18:32:20 補充：
sorry 睇錯左第4條
solution應該係:
4:
=cosθsinθ(cos^2θ/sin^2θ +1)
=cosθsinθ[(cos^2θ+sin^2θ)/sin^2θ]
=cosθ/sinθ
=1/tanθ
參考: worked out myself,
2008-04-22 2:29 am
1. 3sinθ/cosθ
=3tanθ
2. sin3θ-sinθ+cos2θsinθ
=sin3θ-sinθ+(1-sin2θ)sinθ
=sin3θ-sinθ+sinθ-sin3θ
0
4. cosθsinθ(1/tan2θ+1)
=cosθsinθ(1/sin2θ)
=cosθ/sinθ
=1/tanθ
5.sinθ/cosθ+cosθ/sinθ
=(sin2θ+cos2θ)/sinθcosθ
=1/sinθcosθ
6. 1/(1-sinθ) +1 /(1+sinθ)
=[(1+sinθ)+(1-sinθ)]/[(1+sinθ)(1-sinθ)]
=2/cos2θ
=3tanθ
2. sin3θ-sinθ+cos2θsinθ
=sin3θ-sinθ+(1-sin2θ)sinθ
=sin3θ-sinθ+sinθ-sin3θ
0
4. cosθsinθ(1/tan2θ+1)
=cosθsinθ(1/sin2θ)
=cosθ/sinθ
=1/tanθ
5.sinθ/cosθ+cosθ/sinθ
=(sin2θ+cos2θ)/sinθcosθ
=1/sinθcosθ
6. 1/(1-sinθ) +1 /(1+sinθ)
=[(1+sinθ)+(1-sinθ)]/[(1+sinθ)(1-sinθ)]
=2/cos2θ
收錄日期: 2021-04-23 20:37:38
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080421000051KK01726