Form 3 Maths 幫幫手 - Q 3

2008-02-09 10:11 pm
點證明 sinθcosθ (1+tan^2 θ) = tanθ !!!???

更新1:

To §±²³¶ƒαβγδεζηθικλμνξ&om: L.H.S. 第三步果度,括號度少左由tanθ轉出黎ge sinθ喎...

回答 (2)

2008-02-09 10:17 pm
✔ 最佳答案
let theta be x

L.H.S = sinxcosx(1+tan^2x)
=sinxcosx[1+sin^2x/cos^2x]
=sinxcosx[(cos^2x+sin^2x)/cos^2x]
=sinxcosx(1/cos^2x) (cos^2x + sin^2x) = 1
=sinx/cosx
=tanx (sinx/cosx = tanx)
=R.H.S
so sinxcosx(1+tan^2x )= tanx is an identity.

2008-02-09 14:56:27 補充:
或許可向MSN問我[email protected]

2008-02-09 15:04:28 補充:
the details ans :http://hk.geocities.com/doortyui/maths1.jpg

2008-02-10 09:42:10 補充:
樓請注意. 人地先至係F.3如果佢唔係F.3我一早用左條式啦...
2008-02-09 11:46 pm
其實(1+tan^2θ) 是等於sec^2θ
而sec^2θ=1/(cos^2θ)
所以條問題
sinθcosθ (1+tan^2 θ) = tanθ
LHS:sinθ(cosθ)(1/cos^2θ)
=sinθ/cosθ
=tanθ
RHS=tanθ
so it's an identity

2008-02-09 15:47:51 補充:
有一條formula"1 tan^2θ=sec^2θ"所以先equal...唔知你學左味呢??"

2008-02-09 15:49:06 補充:
sorry..係"1 tan^2θ=sec^2θ"


收錄日期: 2021-04-22 23:31:48
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080209000051KK01064

檢視 Wayback Machine 備份