# Urgent Help...Graphs of Sine and Cosine

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• Nov 15th 2007, 04:54 PM
aikenfan
Urgent Help...Graphs of Sine and Cosine
The Amplitude on 67 looks like it is 3...
How do I solve for the period...If I know that, then I can put it in the equation...

On 68, the amplitude is 2
And 70 also appears to be 2

http://i103.photobucket.com/albums/m...91919/11-1.jpg
• Nov 15th 2007, 05:10 PM
Jhevon
Quote:

Originally Posted by aikenfan
The Amplitude on 67 looks like it is 3...
How do I solve for the period...If I know that, then I can put it in the equation...

On 68, the amplitude is 2
And 70 also appears to be 2

http://i103.photobucket.com/albums/m...91919/11-1.jpg

look for when the graph begins to repeat itself. you will realize that for 67 the graph repeats itself twice between $0$ and $2 \pi$. use that to figure out the period. in 68, the graph has only one oscillation between $0$ and $2 \pi$, and for 70, we have one and a half periods between $0$ and $2 \pi$
• Nov 15th 2007, 05:21 PM
aikenfan
To figure out the period, I would do 2(pi)/b
But I do not understand where to get this from...
• Nov 15th 2007, 05:22 PM
Jhevon
Quote:

Originally Posted by aikenfan
To figure out the period, I would do 2(pi)/b
But I do not understand where to get this from...

you can't get it from the picture. just look at the graph and see how long it takes to repeat
• Nov 15th 2007, 06:07 PM
aikenfan
For 67, the five points are: (0,0) (pi/4, -4) (pi/2, 0 ) (3pi/4, 4) (pi, 0)
• Nov 15th 2007, 07:23 PM
mayflower29
Quote:

Originally Posted by Jhevon
look for when the graph begins to repeat itself. you will realize that for 67 the graph repeats itself twice between $0$ and $2 \pi$. use that to figure out the period. in 68, the graph has only one oscillation between $0$ and $2 \pi$, and for 70, we have one and a half periods between $0$ and $2 \pi$

Well, since you have the amplitude, to figure out the period, as someone already said, you figure out where it repeats. In 67, it looks like it stops at pi. I got this because it repeates twice between 0 and 2pi, so divide that in half, and you get pi for the period. However, when you put it into you equation you must use:

2pi/K = Period (K is the number in your equation)
2pi/K = pi
Therefore, k = 2, because the two's would cancel out leaving you with just pi. So your equation would be y = 3sin(pi)x . Hope that helps and makes the rest of them easier.
• Nov 15th 2007, 07:26 PM
Jhevon
Quote:

Originally Posted by mayflower29
Well, since you have the amplitude, to figure out the period, as someone already said, you figure out where it repeats. In 67, it looks like it stops at pi. I got this because it repeates twice between 0 and 2pi, so divide that in half, and you get pi for the period. However, when you put it into you equation you must use:

2pi/K = Period (K is the number in your equation)
2pi/K = pi
Therefore, pi = 2, because the two's would cancel out leaving you with just pi. So your equation would be y = 3sin(pi)x . Hope that helps and makes the rest of them easier.

you mean k = 2

i don't think the poster was having problems with writing the formula. it was actually identifying the period here that was the issue
• Nov 15th 2007, 07:31 PM
mayflower29
Quote:

Originally Posted by Jhevon
you mean k = 2

i don't think the poster was having problems with writing the formula. it was actually identifying the period here that was the issue

Yup, you're right, I meant K.
I didn't mean to quote you there, I meant to quote the original question.
And yes, it may have been the period, but I always forget or get a little confused once trying to figure out K, so I thought I would throw that in there. :)
• Nov 16th 2007, 03:57 AM
aikenfan
Wait, where did the 2 come from...
• Nov 16th 2007, 03:27 PM
mayflower29
Quote:

Originally Posted by aikenfan
Wait, where did the 2 come from...

The two came from:

In your graph, you can tell that the period is pi.
but you can't just put pi in your equation, because the "K" value in your equation is not the period, to get the period it is 2pi DIVIDED by K.

2pi/k = pi (you know pi from your graph)
What value of K gives you a period of pi?
2 does, because the twos cancel out and your left with pi as your period.
• Nov 18th 2007, 12:26 PM
aikenfan
This is what I have got for the others...Do they seem correct?
http://i103.photobucket.com/albums/m...1919/11002.jpg
• Nov 18th 2007, 01:58 PM
Jhevon
Quote:

Originally Posted by aikenfan
This is what I have got for the others...Do they seem correct?

they are incorrect
• Nov 18th 2007, 03:28 PM
aikenfan
Do you know what it is that I am doing wrong? I put the period in the equation right?
• Nov 18th 2007, 03:53 PM
Jhevon
Quote:

Originally Posted by aikenfan
Do you know what it is that I am doing wrong? I put the period in the equation right?

your amplitude is wrong as well as the period

recall, the general form is $y = a \sin k (x - b) + c$

where $|a|$ is the amplitude

$b$ is the phase shift

$c$ is the vertical shift

$\frac {2 \pi}{|k|}$ is the period, and

$\left[ b, b + \frac {2 \pi}{|k|}\right]$ is the appropriate graphing interval

thus, if your period is $4 \pi$, it means that,

$4 \pi = \frac {2 \pi}{k} \implies k = \frac 12$

also, your amplitude is 3 here. there is no vertical or phase shifts, thus, you equation for 68 is:

$y = 3 \sin \frac 12 x$

now try 70 again
• Nov 18th 2007, 04:08 PM
aikenfan
I'm sorry, I'm a bit confused...On 68, I thought that the Amplitude came from the graph...the line goes up to 2 (but the grid goes up to 3)
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