6a) Find the points of intersection of these pairs of curves for -360<x<360, giving answers to 2 sig fig.
c) y=sec x, y=1 + cos x
d) y= tan x - sin^2 x, y= (cos x - sec x )^2
2.5 points for each... =P
SHow steps, 1000000000 thx.
^o^"懂"就是聰明Maths (傷心篇)
2010-09-22 2:05 am
回答 (1)
2010-09-26 8:25 pm
✔ 最佳答案
6a) by computer= =...6c) sec x = 1+cos x
1/cosx = 1+cos x
1=cos x+(cos x)^2
1-(cos x)^2=cos x
(sin x)^2=cos x
=> (tan x)(sin x)=0 => x=0
6d)
tan x- sin^2(x) = (cos x - 1/cosx)^2
tan x-sin^2(x) =(sin x*tan x)^2 (sin^2(x)/cos x=sin x*tan x)
tan x=sin^2(x)(tan^2(x)+1)
tan x=sin^2(x)(1/cos^2(x)) (tan^2(x)+1=sec^2(x))
tan x=(tan x)^2
=>tanx=0 or tanx=1
參考: me
收錄日期: 2021-04-23 20:43:02
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100921000051KK00901