Trig graph problem

  • Thread starter özge
  • Start date
  • #1
özge
7
0
Please Help,urgent

i have a trigonometry question and i really don't know how to solve i checked couple of books but still no hope here it is:

im supposed to draw the graph of y=cos¶/3(x-1) but i couldnt

so please help..
 

Answers and Replies

  • #2
quantumdude
Staff Emeritus
Science Advisor
Gold Member
5,575
23
Your equation is showing up funny on my browser.

Is it supposed to be this:

[tex]
y=cos(3x-1)
[/tex]

or this:

[tex]
y=cos(\frac{1}{3x-1})
[/tex]

?
 
  • #3
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
10,089
135
özge said:
i have a trigonometry question and i really don't know how to solve i checked couple of books but still no hope here it is:

im supposed to draw the graph of y=cos¶/3(x-1) but i couldnt

so please help..
y=cos¶/3(x-1)?
Neither can I. Please type in the CORRECT formula.
 
  • #4
Lisa!
Gold Member
639
97
Tom Mattson said:
Your equation is showing up funny on my browser.

Is it supposed to be this:

[tex]
y=cos(3x-1)
[/tex]

or this:

[tex]
y=cos(\frac{1}{3x-1})
[/tex]

?
I think it's something else!
 
  • #5
özge
7
0
y=cos pi/3(x-1)

now ok?
 
  • #6
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
10,089
135
Ok, so it is:
[tex]y=\cos(\frac{\pi}{3x-1})[/tex]

Hints:
1. Find the point of singularity, where y is not defined.
2. Determine the set of points were y is -1, 0, 1 respectively.
In particular, see what this implies for the function's behaviour very close to the singularity found in 1.
 
  • #7
özge
7
0
ok it is not that y is equal to cos multiplied by pi/3 multiplied by (x-1)
 
  • #8
quantumdude
Staff Emeritus
Science Advisor
Gold Member
5,575
23
Still not OK, and I suspect that it's because you aren't placing your parentheses properly.

What you have written is this:

[tex]y=cos(\frac{\pi}{3x-1})[/tex].

Now that's a perfectly legitimate function, but I strongly suspect that it is not the one you are supposed to graph, simply because it's too difficult for a trigonometry course. I am inclined to think that you really meant this:

[tex]y=cos[\frac{\pi}{3}(x-1)][/tex].

Am I right?
 
  • #9
özge
7
0
(x-1) isn't near 3
 
  • #10
quantumdude
Staff Emeritus
Science Advisor
Gold Member
5,575
23
I don't know what that means.
 
  • #11
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
10,089
135
özge:
If Tom Mattson is right, here's how you could have written this properly:
y=cos(pi*(x-1)/3)
 
  • #12
özge
7
0
Tom Mattson said:
Still not OK, and I suspect that it's because you aren't placing your parentheses properly.

What you have written is this:

[tex]y=cos(\frac{\pi}{3x-1})[/tex].

Now that's a perfectly legitimate function, but I strongly suspect that it is not the one you are supposed to graph, simply because it's too difficult for a trigonometry course. I am inclined to think that you really meant this:

[tex]y=cos[\frac{\pi}{3}(x-1)][/tex].

Am I right?

u are right
 
  • #13
quantumdude
Staff Emeritus
Science Advisor
Gold Member
5,575
23
özge said:
ok it is not that y is equal to cos multiplied by pi/3 multiplied by (x-1)

This can't be right. cos without an argument makes no sense.
 
  • #14
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
10,089
135
özge said:
ok it is not that y is equal to cos multiplied by pi/3 multiplied by (x-1)
Do you know what the cosine function is??
What is its argument?
 
  • #15
özge
7
0
what u wrote :[tex]y=cos[\frac{\pi}{3}(x-1)][/tex]
was right I am sorry I am new to these things and i diddnt know how to write
 
  • #16
quantumdude
Staff Emeritus
Science Advisor
Gold Member
5,575
23
özge said:
u are right

Alrighty then!

Here are a couple of tips to get you started.

*Examine the graph of [itex]y=cos(x)[/itex].

*Note that if you have the graph of some function [itex]y=f(x)[/itex] then you can obtain the graph of [itex]y=f(x+a)[/itex] by shifting the graph of [itex]y=f(x)[/itex] horizontally by [itex]a[/itex] units. You move to the right if [itex]a<0[/itex] and to the left if [itex]a>0[/itex].

*The period [itex]T[/itex] of a cosine function [itex]y=A\cos[k(x+a)][/itex] is [itex]T=\frac{2\pi}{k}[/itex].

Give it a shot and see how you do, OK?
 
  • #17
özge
7
0
ok, thanks
 

Suggested for: Trig graph problem

  • Last Post
Replies
7
Views
280
  • Last Post
Replies
25
Views
785
  • Last Post
Replies
8
Views
358
Replies
2
Views
311
  • Last Post
Replies
4
Views
318
  • Last Post
Replies
1
Views
335
Replies
2
Views
53
Replies
18
Views
797
Replies
4
Views
327
  • Last Post
Replies
18
Views
735
Top