http://img363.imageshack.us/img363/1020/20080522001yk9.jpg
Please go to the website above and check out the problem.
What I trying to do is to simplify them, and I have already spend half an hour but got nothing...
Do you know how to simplify them?
Thanks a lot~
Trig. Question (Simplifying)
2008-05-23 9:42 pm
回答 (2)
2008-05-23 9:55 pm
✔ 最佳答案
b. (sec^2 x -1) cot^2 x=[(1/cos^2 x) -1](1/tan^2 x)
=[(1-cos^2 x)/cos^2 x](cos^2 x/sin^2 x)
=[sin^2 x/cos^2 x](cos^2 x/sin^2 x)
=1
c. 2tan(pi/9)/(1-tan^2 (pi/9))
since tan2x = 2tanx/(1-tan^2 x)
so 2tan(pi/9)/(1-tan^2 (pi/9))
=tan(2pi/9)
d.(1-cos 2x)/sin2x
=(1- 1+2sin^2 x)/(2 sinx cosx)
=(2sin^2 x)/(2 sinx cosx)
=(sinx)/(cosx)
=tanx
參考: my math knowledge
2008-05-23 9:57 pm
(sec2 X - 1)cot2X
=(1+tan2X-1)cot2X
=tan2Xcot2X
=(tan2X)(1/tan2X)
=1
PS: 2 = square
=(1+tan2X-1)cot2X
=tan2Xcot2X
=(tan2X)(1/tan2X)
=1
PS: 2 = square
收錄日期: 2021-04-15 15:03:44
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080523000051KK00890