tan²θ - sin²θ ≡sin²θtan²θ
詳細d 唔該
數學Prove identities
2009-04-14 4:06 am
回答 (3)
2009-04-14 4:40 am
✔ 最佳答案
L.H.S.=tanθ - sinθ圖片參考：http://g.imagehost.org/0267/ScreenHunter_02_Apr_13_20_38.gif
=sinθtanθ
=R.H.S.
2009-04-13 20:41:09 補充：
慢了,001最佳~!
2009-04-17 10:14 pm
sinθtanθ+ sinθ
≡ sinθ(tanθ+1)
≡ sinθsecθ
≡ sinθ/cosθ
≡ tanθ
so
tanθ - sinθ≡ sinθtanθ
2009-04-14 4:37 am
tanθ - sinθ
=(sinθ/cosθ)-(sinθ*cosθ/cosθ)
=(sinθ-sinθcosθ)/cosθ
=sinθ*(1-cosθ)/cosθ
=sinθ*sinθ/cosθ
=sinθ*tanθ
2009-04-13 20:38:50 補充：
哎...又係顯示唔到2次方
tan^2θ - sin^2θ
=(sin^2θ/cos^2θ)-(sin^2θ*cos^2θ/cos^2θ)
=(sin^2θ-sin^2θcos^2θ)/cos^2θ
=sin^2θ*(1-cos^2θ)/cos^2θ
=sin^2θ*sin^2θ/cos^2θ
=sin^2θ*tan^2θ
=(sinθ/cosθ)-(sinθ*cosθ/cosθ)
=(sinθ-sinθcosθ)/cosθ
=sinθ*(1-cosθ)/cosθ
=sinθ*sinθ/cosθ
=sinθ*tanθ
2009-04-13 20:38:50 補充：
哎...又係顯示唔到2次方
tan^2θ - sin^2θ
=(sin^2θ/cos^2θ)-(sin^2θ*cos^2θ/cos^2θ)
=(sin^2θ-sin^2θcos^2θ)/cos^2θ
=sin^2θ*(1-cos^2θ)/cos^2θ
=sin^2θ*sin^2θ/cos^2θ
=sin^2θ*tan^2θ
收錄日期: 2021-04-22 00:37:58
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090413000051KK01907