- #1

özge

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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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- Thread starter özge
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- #1

özge

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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..

- #2

quantumdude

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Is it supposed to be this:

[tex]

y=cos(3x-1)

[/tex]

or this:

[tex]

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

[/tex]

?

- #3

arildno

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y=cos¶/3(x-1)?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..

Neither can I. Please type in the CORRECT formula.

- #4

Lisa!

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I think it's something else!Tom Mattson said:

Is it supposed to be this:

[tex]

y=cos(3x-1)

[/tex]

or this:

[tex]

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

[/tex]

?

- #5

özge

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y=cos pi/3(x-1)

now ok?

now ok?

- #6

arildno

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[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

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

- #8

quantumdude

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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

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(x-1) isn't near 3

- #10

quantumdude

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I don't know what that means.

- #11

arildno

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özge:

If Tom Mattson is right, here's how you could have written this properly:

y=cos(pi*(x-1)/3)

If Tom Mattson is right, here's how you could have written this properly:

y=cos(pi*(x-1)/3)

- #12

özge

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Tom Mattson said:

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

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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

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Do you know what the cosine function is??özge said:ok it is not that y is equal to cos multiplied by pi/3 multiplied by (x-1)

What is its argument?

- #15

özge

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was right I am sorry I am new to these things and i diddnt know how to write

- #16

quantumdude

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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

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ok, thanks

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