How Do You Graph y=cos(π/3)(x-1)?

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The discussion revolves around the correct interpretation of the equation y=cos(π/3)(x-1) for graphing purposes. Participants clarify that the intended formula is likely y=cos[π/3(x-1)], emphasizing the importance of proper parentheses in mathematical expressions. They discuss the implications of the function's behavior near singularities and provide hints for graphing, such as examining the basic cosine function and understanding horizontal shifts. The conversation highlights the necessity of clear communication in mathematical notation to avoid confusion. Overall, accurate representation of the equation is crucial for successfully graphing the function.
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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..
 
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Your equation is showing up funny on my browser.

Is it supposed to be this:

<br /> y=cos(3x-1)<br />

or this:

<br /> y=cos(\frac{1}{3x-1})<br />

?
 
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.
 
Tom Mattson said:
Your equation is showing up funny on my browser.

Is it supposed to be this:

<br /> y=cos(3x-1)<br />

or this:

<br /> y=cos(\frac{1}{3x-1})<br />

?
I think it's something else!
 
y=cos pi/3(x-1)

now ok?
 
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)
 
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) isn't near 3
 
  • #10
I don't know what that means.
 
  • #11
özge:
If Tom Mattson is right, here's how you could have written this properly:
y=cos(pi*(x-1)/3)
 
  • #12
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
 
  • #13
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
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
what u wrote :y=cos[\frac{\pi}{3}(x-1)]
was right I am sorry I am new to these things and i diddnt know how to write
 
  • #16
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&lt;0 and to the left if a&gt;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?
 
  • #17
ok, thanks
 
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