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looks like this is the sin graph shifted by pi/12 and the period is only pi/2 now

Let's figure out how to change the period

sin(x) has what period initially

Hi, are you there?

i'm a little confused how you got pi/2

ah it's not pi/2

it's 2pi/3

http://i.imgur.com/KZfiTq6.png

what is 2pi/3? Sorry i'm lost

take a look at that picture

the length of this full cycle is called the period

where would the period start from?

let me show you some examples

Ok thank you..

http://i.imgur.com/KnoNRXN.png

here i have picked 3 sets of pairs of points

Oh i see! Thank you :)

okay so lets start with the base graph
y=sin(x), what period does this have initially

2pi?

right

y=sin(x) looks like this |dw:1444540802308:dw|

now lets see how we can possibly make this period only 2pi/3

Tell me if that makes sense, this might be a little tricky when u first see it

I get that, it's just I'm a little confused how we use 2p/3 to help us find out the equation

right now can u figure out how to make this period 2pi/3 instead of 1

we saw how multiplying by 2pi in the brackets managed to speed up this period to 1 instead of 2pi

sin(ax) , what should this a be so that when x=2pi/3, we have a complete period

couldn't you just multiply it by 2pi?

2pi made the period go to 1

the answer is
y=\[sin(\frac{2\pi ~x}{2\pi/3} )\]

see how when x=2pi/3 the stuff in the brackets turns to 2pi

now apply your shift and amplitude change

i'm still confused how we got 2pi from the 2pix/2pi/3?

when x=2pi/3

|dw:1444541639553:dw|

oh ok.. thank you

alright time to investigate sana xD

go ahead XD I have a bf

I think you need to understand better why
y=sin(2pi x) in making the period 1 from 2pi previously

just looking to complete a whole period

think about it like this
y=sin(x), x has to be all the number from 0 to 2pi to complete a period

so now I must plug in x values in for y?

What do you mean

http://i.imgur.com/CbqYn0Z.png

is there a more simpler way to do this.. I feel like there is.

yeah you can memorize the formula

\[Y=A* Sin(\frac {2\pi}{L} (X-B))\]
A=Amplitude
L=Length
B=Phase

grab a precal textbook

get the one by larson or stewart

my professor doesn't use the book, he expects us to "automatically know this"

and i've tried using the book

okay so can you plug into the formula?

You know the the length,the phase, the Amplitude

the amplitude would be 3?

yes

I already showed you what the answer should look like in the screenshot a couple comments ago

just ask one of these people if you are still confused, i have to go now

the 3sin(2pi(x-pi/12))/2pi/3)?

thanks for your time

i keep on getting notifications. oh nvm your done/

now u need to pay $50 for the help using paypal

\[Y=3* Sin(\frac {2\pi}{2pi/3} (X-\pi/12))\]

imqwerty, I already did pay with the plan.

ok then (:

u can get only this much with $50

ganeshie8, I put in a lot of effort thank you very much.

Maybe lets start over ?

In which part are u having problem?

all of it.

|dw:1444544503214:dw| MID = red, MAX = pink, MIN = green to give you an idea of what I mean

then how would I find the equation.. i'm aware of the basics

I plug in just the amplitude and midpoint?

maybe u shuld leave it (: