# Sinusoidal Graph - sub intervals

1. Mar 16, 2011

### CrossFit415

1. The problem statement, all variables and given/known data

I can find the graphs amp and period. The only problem is finding the sub points or sub intervals. Say...

Y = 3 sin (4x)
Amp = 3
Period = 2pi/4 = pi/2

But.. don't know how to get the key points of the sub interval. The text book says I have to divide interval [0, pi/2] into four sub intervals Each of length pi/2 divided by 4. Then they got (0,0), (pi/8, 3), (pi/4, 0), (3pi/8, -3), (pi/2, 0) I don't understand how they got these. Thanks

2. Relevant equations

3. The attempt at a solution

2. Mar 16, 2011

### Mentallic

Draw the base graph y=sin(x) with $0\leq x\leq 2\pi$, It has a value of 0 at $x=0,\pi,2\pi$ and a value of 1 and -1 at $x=\pi/2, 3\pi/2$ respectively.
Basically, every sine graph of the form $y=Asin(Bx)$ will still have this same shape, but the amplitude (A) and period (B) will be different from the base graph y=sin(x).

What you should take away from this is that if the period of sin(x) is $2\pi$, then in between the two ends of the period 0 and $2\pi$ which is $\pi$, it will also be 0, and in between 0 and its half way mark which is $\pi$ we get the value of its amplitude (in this case 1), and between the half way mark and the end, $\pi$ and $2\pi$ we get the negative of its amplitude, -1.

3. Mar 16, 2011

### CrossFit415

Thanks, I know that but I don't know how to get the points in between the sin graph

I know if period = 2 pi then the middle point would me pi, but what if it has a different period. I don't know what to label on the graph on the middle part.

Last edited: Mar 16, 2011
4. Mar 16, 2011

### Mentallic

If the period is $2\pi$ then middle is half of that $$\frac{2\pi}{2}=\pi$$. If the period is some number x then the middle is x/2.

5. Mar 16, 2011

### CrossFit415

Ahhh I see now. Thank you my friend.

6. Mar 16, 2011

Good luck!