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no

pfft, figures ... my period is incorrect

ok, i will wait

i dont understand

okay..... so what does that mean

it means you did something wrong ...

could you walk me through what i did wrong then.... I don't understand how to solve for Bx

yeah i understand that 2pi is the normal period of the sine function

your solution attributed this to: 2pi/k = 3pi instead of 2pi/k = 1

there was also some other mismathing going on

well... that didn't help

Some pictures will help, I think.
First, \(y = \sin x\)

is there a certain equation that I could use to find this?

multiply argument by a factor of \(\pi\)
closer, but still not enough

Multiplying argument by \(2\pi\) gets the right frequency: \(y = \sin 2\pi x\)

But comparing with problem graph, we still don't have the right amplitude, and our graph is inverted. Fix that by multiplying by -3, which scales and inverts:

The period business is just a simple proportion of the desired period and \(2\pi\)...

how is the period 1?

no i don't understand.

okay so there are 4 cycles

this isn't making any sense.....

|dw:1370965653907:dw|Here is a different sine curve.
What is the length of one cycle?

4?

Yes good :) Now look back at your graph, what is the length of one cycle?
That is your period.

Well, to get one cycle of \sin x, your x value has to go from 0 to \(2\pi\), right? If you need that to happen in more or less space along the x axis, you just have to multiply or divide to make that happen. Say the next problem asked you to make a sine function with period \(\pi/2\). That means that your new multiplier and the old multiplier (which is 1 by default, as in \(y = \sin 1x\)) have to satisfy a proportion:
\[\frac{1}{B} = \frac{T}{2\pi}\] where \(T\) is the new period, and \(B\) is the multiplier. If our period we want is \(2\pi\) then our value of \(B\) must be 1. That's just our starting point with \(y=\sin 1x\). If we want two cycles in the space normally occupied by 1, that means our period will be \(\pi\) instead of \(2\pi\) and \[\frac{1}{B} = \frac{T}{2\pi} = \frac{\pi}{2\pi} = \frac{1}{2}\] so \(B=2\). And if you graph \(y = \sin x\) and \(y = \sin 2x\) together you'll see that it does the right thing.

so [4=2\Pi/B\]

@whpalmer4 you have confused me. I don't know how to these..... sorry for wasting your time

What is the length of one cycle on the graph you were given?
Look at the x-axis.

1, because you told me

If no numbers were given, you would be unable to determine the period.

alright :/

ah, yes maybe :)

I know that was for me, years ago!

|dw:1370966450574:dw|
so what is the period of that?

Ah yes your labels are a little silly :) hehe

okay sure...... ugh!

|dw:1370966654243:dw|How bout this one instead :D

but i have trouble with the fraction ones....

Oh the fractions are the troublesome ones? ok my bad.

|dw:1370966770302:dw|

here are some tricky fractions for ya :U

2pi, pi/3?

yah pi/3 sounds right.
We started at x=0
ended at x=pi/3
after we did 2 loops.

6pi?

1/6 i mean

I think it's going to be \(\large B=6\)
Do those steps make sense?

|dw:1370967157840:dw|

|dw:1370967349882:dw|

ya looks good :)

okay, thank you!

So in our example, we would be able to write a sine function like this:
\(\large y=A\sin6x\)

Yes, good.