How to find the sin and cosine of a shifted sinusoidal function.

- anonymous

How to find the sin and cosine of a shifted sinusoidal function.

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- anonymous

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- dan815

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

- dan815

Let's figure out how to change the period

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## More answers

- dan815

sin(x) has what period initially

- dan815

Hi, are you there?

- anonymous

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

- dan815

ah it's not pi/2

- dan815

it's 2pi/3

- dan815

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

- anonymous

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

- dan815

take a look at that picture

- dan815

the length of this full cycle is called the period

- anonymous

where would the period start from?

- dan815

you can start the period anywhere, as long as 1 full cycle takes place, however the period will always be 2pi/3

- dan815

let me show you some examples

- anonymous

Ok thank you..

- dan815

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

- dan815

here i have picked 3 sets of pairs of points

- anonymous

Oh i see! Thank you :)

- dan815

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

- anonymous

2pi?

- dan815

right

- dan815

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

- dan815

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

- dan815

currently
y=sin(x), a period of 2pi
\[y=sin(2\pi x)\]
this graph would have a period of only 1, when you multiply by 2pi, because now, if you put in values for x, its being sped up by that 2pi factor, when u put in x=1 you can already see it will be sin(2pi*1) = sin(2*pi) and we know by then this would have completed a full cycle

- dan815

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

- anonymous

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

- dan815

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

- dan815

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

- dan815

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

- dan815

in other words
y=sin(ax)
a*2pi/3= 2pi -----> when you input x=2pi/3 you want to make sure this 'ax' in the brackets turns into 2pi so we know a full period was done

- anonymous

couldn't you just multiply it by 2pi?

- dan815

2pi made the period go to 1

- dan815

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

- dan815

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

- dan815

now apply your shift and amplitude change

- dan815

\[y=A*sin(p*(x-b))\] this is the form for sin graph where A is the amplitude,p determines the period and b is the shift
\[y=3*sin(\frac{2\pi ~(x-\pi/12)}{2\pi/3} )\]

- anonymous

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

- dan815

when x=2pi/3

- dan815

|dw:1444541639553:dw|

- anonymous

oh ok.. thank you

- carlyleukhardt

alright time to investigate sana xD

- anonymous

go ahead XD I have a bf

- dan815

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

- anonymous

sorry it's just so new for me it's hard to follow. So I understand how we got 2pi, are we following a formula or just looking for it to complete just a whole period?

- dan815

just looking to complete a whole period

- dan815

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

- dan815

when you multiply
y=sin(2pi * x), now x has to be all the number from x=0 to 1, because this will generate the whole set 0 to 2pi in the brackets anyway

- dan815

the point is you have to make sure you have interval lengths of 2pi showing up in the brackets for a sin function so that it completes a period

- anonymous

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

- dan815

What do you mean

- dan815

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

- dan815

*side note* when you multiply by p the period becomes 2pi/p, so you can figure out what p has to be to get the period you want

- dan815

if you set p=2pi/L
where L is the length of the period you read on the graph, thats a simple way to force your period to be L

- anonymous

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

- dan815

yeah you can memorize the formula

- dan815

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

- nincompoop

grab a precal textbook

- nincompoop

get the one by larson or stewart

- anonymous

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

- anonymous

and i've tried using the book

- dan815

okay so can you plug into the formula?

- dan815

You know the the length,the phase, the Amplitude

- anonymous

the amplitude would be 3?

- dan815

yes

- dan815

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

- dan815

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

- anonymous

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

- anonymous

thanks for your time

- carlyleukhardt

i keep on getting notifications. oh nvm your done/

- imqwerty

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

- dan815

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

- anonymous

ahaha @imqwerty

- anonymous

sorry dan, but @imqwerty deserves that medal for that comment

- anonymous

imqwerty, I already did pay with the plan.

- imqwerty

ok then (:

- anonymous

imqwerty, and it's sad to see that I'm paying so much money and not understanding these things. You guys should explain it step by step.

- ganeshie8

just saying.. paying money wont guarantee you quick understanding, it only guarantees you quick assistance, understanding the concept depends on how much effort you put and engage with the helper

- imqwerty

u can get only this much with $50

- anonymous

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

- ganeshie8

Maybe lets start over ?

- anonymous

I don't understand how to solve this problem. I know there is an equation that you follow. That's all that I got from it.

- imqwerty

In which part are u having problem?

- anonymous

all of it.

- Astrophysics

Well, I think the difficult part is, that there's many ways to do this, so you have to sort of find your own way...
A few steps that may help you:
1. Draw the Max and Min lines
2. Draw in the Mid line, i.) the mid line gives you the vertical displacement (k), ii.) the distance from the mid to the max gives you the amplitude (a)
3. Find the distance from crest to crest (or trough to trough) i.) This gives you the period (b), ii.) Period = 2pi/b
4. k, a, and b remain the same throughout the equation but note that a's sign can change
5. Determine the phase shift (h), which helps us determine whether you want sine or cosine i.) For cosine you need to start at a crest (MAX) or trough (MIN) ii.) For a sine you would have to start at the MID line
6. Determine whether your equation is positive or negative haha, there's many ways for this one
I think that's like a small tutorial on graphing and finding these equations, also know what amplitude, period, range, domain represent, if you want to know I can tell you I think this post is long enough though.

- Astrophysics

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

- anonymous

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

- Astrophysics

Ok so your equation is of the form \[y=a*f(b(x-h))+k\] so use what I've said above and see if you can figure it out

- Astrophysics

Or I should put it as \[y = a*\sin(b(x-h \pi) ) +k \] similarly for cosine, just switch the trig function

- anonymous

I plug in just the amplitude and midpoint?

- baru

maybe it would help if you could tell exactly where you stand...have you understood what shifted sinusoidal function means?

- imqwerty

maybe u shuld leave it (:

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