1. Simplify the following:
cosθ/1+sinθ - 1-sinθ/cosθ
2.
a) If sinθ = 1/2, find the value of sinθ+cosθ / 2sinθ+cosθ
b) If cosθ = 7/25, find the value of (sinθ-2cosθ / sinθ+3cosθ)tanθ
c) If tanθ = 1, find the value of 3cosθ-sinθ / 3sinθ+cosθ
F.3 Trigonometry
2009-02-25 3:17 am
回答 (3)
2009-02-25 6:57 am
2009-02-26 7:37 am
1)
cosθ/1+sinθ - 1-sinθ/cosθ
=[cos^2θ-(1-sinθ)(1+sinθ)] / [(1+sinθ)cosθ]
=[cos^2θ-(1-sin^2θ)] / [(1+sinθ)cosθ]
=0 / [(1+sinθ)cosθ]
=0
2a)
sinθ=1/2
cosθ=(√3) / 2
sinθ+cosθ / 2sinθ+cosθ
=[1/2 + (√3) / 2] / [2*1/2+√3) / 2]
=(1+√3) / (2+√3)
2b)
cosθ=7/25
sinθ=24/25
tanθ=24/7
(sinθ-2cosθ / sinθ+3cosθ)tanθ
={[24/25-2(7/25)] / [24/25+3(7/25)]}(24/7)
=16/21
2c)
tanθ=1
sinθ=1/2
cosθ=1/2
3cosθ-sinθ / 3sinθ+cosθ
=[3(1/2)-1/2] / [3(1/2)+(1/2)]
=1/2
2009-02-25 23:39:53 補充：
唔知點解打完就出晒D亂碼...
θ =θ
√ =√
cosθ/1+sinθ - 1-sinθ/cosθ
=[cos^2θ-(1-sinθ)(1+sinθ)] / [(1+sinθ)cosθ]
=[cos^2θ-(1-sin^2θ)] / [(1+sinθ)cosθ]
=0 / [(1+sinθ)cosθ]
=0
2a)
sinθ=1/2
cosθ=(√3) / 2
sinθ+cosθ / 2sinθ+cosθ
=[1/2 + (√3) / 2] / [2*1/2+√3) / 2]
=(1+√3) / (2+√3)
2b)
cosθ=7/25
sinθ=24/25
tanθ=24/7
(sinθ-2cosθ / sinθ+3cosθ)tanθ
={[24/25-2(7/25)] / [24/25+3(7/25)]}(24/7)
=16/21
2c)
tanθ=1
sinθ=1/2
cosθ=1/2
3cosθ-sinθ / 3sinθ+cosθ
=[3(1/2)-1/2] / [3(1/2)+(1/2)]
=1/2
2009-02-25 23:39:53 補充：
唔知點解打完就出晒D亂碼...
θ =θ
√ =√
參考: 我自己做咖... 希望可以幫到你啦:)
2009-02-25 4:01 am
1. Simplify the following:
圖片參考：http://g.imagehost.org/0460/ScreenHunter_01_Feb_23_19_44.gif
2009-02-24 20:09:44 補充：
(c)
http://g.imagehost.org/0369/ScreenHunter_02_Feb_23_19_53.gif
圖片參考：http://g.imagehost.org/0460/ScreenHunter_01_Feb_23_19_44.gif
2009-02-24 20:09:44 補充：
(c)
http://g.imagehost.org/0369/ScreenHunter_02_Feb_23_19_53.gif
收錄日期: 2021-04-23 20:40:18
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090224000051KK01420