# Trig graph problem

i have a trigonometry question and i really dont 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

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Tom Mattson
Staff Emeritus
Gold Member
Your equation is showing up funny on my browser.

Is it supposed to be this:

$$y=cos(3x-1)$$

or this:

$$y=cos(\frac{1}{3x-1})$$

?

arildno
Homework Helper
Gold Member
Dearly Missed
özge said:
i have a trigonometry question and i really dont 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

y=cos¶/3(x-1)???????????????????????????????????????????????????
Neither can I. Please type in the CORRECT formula.

Lisa!
Gold Member
Tom Mattson said:
Your equation is showing up funny on my browser.

Is it supposed to be this:

$$y=cos(3x-1)$$

or this:

$$y=cos(\frac{1}{3x-1})$$

?
I think it's something else!

y=cos pi/3(x-1)

now ok?

arildno
Homework Helper
Gold Member
Dearly Missed
Ok, so it is:
$$y=\cos(\frac{\pi}{3x-1})$$

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.

ok it is not that y is equal to cos multiplied by pi/3 multiplied by (x-1)

Tom Mattson
Staff Emeritus
Gold Member
Still not OK, and I suspect that it's because you aren't placing your parentheses properly.

What you have written is this:

$$y=cos(\frac{\pi}{3x-1})$$.

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:

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

Am I right?

(x-1) isnt near 3

Tom Mattson
Staff Emeritus
Gold Member
I don't know what that means.

arildno
Homework Helper
Gold Member
Dearly Missed
özge:
If Tom Mattson is right, here's how you could have written this properly:
y=cos(pi*(x-1)/3)

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:

$$y=cos(\frac{\pi}{3x-1})$$.

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:

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

Am I right?
u are right

Tom Mattson
Staff Emeritus
Gold Member
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.

arildno
Homework Helper
Gold Member
Dearly Missed
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?

what u wrote :$$y=cos[\frac{\pi}{3}(x-1)]$$
was right im sorry im new to these things and i diddnt know how to write

Tom Mattson
Staff Emeritus
Gold Member
özge said:
u are right
Alrighty then!

Here are a couple of tips to get you started.

*Examine the graph of $y=cos(x)$.

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

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

Give it a shot and see how you do, OK?

ok, thanks