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anonymous
 one year ago
How to find the sin and cosine of a shifted sinusoidal function.
anonymous
 one year ago
How to find the sin and cosine of a shifted sinusoidal function.

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dan815
 one year ago
Best ResponseYou've already chosen the best response.8looks like this is the sin graph shifted by pi/12 and the period is only pi/2 now

dan815
 one year ago
Best ResponseYou've already chosen the best response.8Let's figure out how to change the period

dan815
 one year ago
Best ResponseYou've already chosen the best response.8sin(x) has what period initially

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i'm a little confused how you got pi/2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what is 2pi/3? Sorry i'm lost

dan815
 one year ago
Best ResponseYou've already chosen the best response.8take a look at that picture

dan815
 one year ago
Best ResponseYou've already chosen the best response.8the length of this full cycle is called the period

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0where would the period start from?

dan815
 one year ago
Best ResponseYou've already chosen the best response.8you can start the period anywhere, as long as 1 full cycle takes place, however the period will always be 2pi/3

dan815
 one year ago
Best ResponseYou've already chosen the best response.8let me show you some examples

dan815
 one year ago
Best ResponseYou've already chosen the best response.8here i have picked 3 sets of pairs of points

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh i see! Thank you :)

dan815
 one year ago
Best ResponseYou've already chosen the best response.8okay so lets start with the base graph y=sin(x), what period does this have initially

dan815
 one year ago
Best ResponseYou've already chosen the best response.8y=sin(x) looks like this dw:1444540802308:dw

dan815
 one year ago
Best ResponseYou've already chosen the best response.8now lets see how we can possibly make this period only 2pi/3

dan815
 one year ago
Best ResponseYou've already chosen the best response.8currently 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
 one year ago
Best ResponseYou've already chosen the best response.8Tell me if that makes sense, this might be a little tricky when u first see it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I get that, it's just I'm a little confused how we use 2p/3 to help us find out the equation

dan815
 one year ago
Best ResponseYou've already chosen the best response.8right now can u figure out how to make this period 2pi/3 instead of 1

dan815
 one year ago
Best ResponseYou've already chosen the best response.8we saw how multiplying by 2pi in the brackets managed to speed up this period to 1 instead of 2pi

dan815
 one year ago
Best ResponseYou've already chosen the best response.8sin(ax) , what should this a be so that when x=2pi/3, we have a complete period

dan815
 one year ago
Best ResponseYou've already chosen the best response.8in 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
 one year ago
Best ResponseYou've already chosen the best response.0couldn't you just multiply it by 2pi?

dan815
 one year ago
Best ResponseYou've already chosen the best response.82pi made the period go to 1

dan815
 one year ago
Best ResponseYou've already chosen the best response.8the answer is y=\[sin(\frac{2\pi ~x}{2\pi/3} )\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.8see how when x=2pi/3 the stuff in the brackets turns to 2pi

dan815
 one year ago
Best ResponseYou've already chosen the best response.8now apply your shift and amplitude change

dan815
 one year ago
Best ResponseYou've already chosen the best response.8\[y=A*sin(p*(xb))\] 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
 one year ago
Best ResponseYou've already chosen the best response.0i'm still confused how we got 2pi from the 2pix/2pi/3?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0alright time to investigate sana xD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0go ahead XD I have a bf

dan815
 one year ago
Best ResponseYou've already chosen the best response.8I think you need to understand better why y=sin(2pi x) in making the period 1 from 2pi previously

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry 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
 one year ago
Best ResponseYou've already chosen the best response.8just looking to complete a whole period

dan815
 one year ago
Best ResponseYou've already chosen the best response.8think about it like this y=sin(x), x has to be all the number from 0 to 2pi to complete a period

dan815
 one year ago
Best ResponseYou've already chosen the best response.8when 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
 one year ago
Best ResponseYou've already chosen the best response.8the 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
 one year ago
Best ResponseYou've already chosen the best response.0so now I must plug in x values in for y?

dan815
 one year ago
Best ResponseYou've already chosen the best response.8*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
 one year ago
Best ResponseYou've already chosen the best response.8if 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
 one year ago
Best ResponseYou've already chosen the best response.0is there a more simpler way to do this.. I feel like there is.

dan815
 one year ago
Best ResponseYou've already chosen the best response.8yeah you can memorize the formula

dan815
 one year ago
Best ResponseYou've already chosen the best response.8\[Y=A* Sin(\frac {2\pi}{L} (XB))\] A=Amplitude L=Length B=Phase

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.2grab a precal textbook

nincompoop
 one year ago
Best ResponseYou've already chosen the best response.2get the one by larson or stewart

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0my professor doesn't use the book, he expects us to "automatically know this"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and i've tried using the book

dan815
 one year ago
Best ResponseYou've already chosen the best response.8okay so can you plug into the formula?

dan815
 one year ago
Best ResponseYou've already chosen the best response.8You know the the length,the phase, the Amplitude

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the amplitude would be 3?

dan815
 one year ago
Best ResponseYou've already chosen the best response.8I already showed you what the answer should look like in the screenshot a couple comments ago

dan815
 one year ago
Best ResponseYou've already chosen the best response.8just ask one of these people if you are still confused, i have to go now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 3sin(2pi(xpi/12))/2pi/3)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks for your time

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i keep on getting notifications. oh nvm your done/

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.5now u need to pay $50 for the help using paypal

dan815
 one year ago
Best ResponseYou've already chosen the best response.8\[Y=3* Sin(\frac {2\pi}{2pi/3} (X\pi/12))\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry dan, but @imqwerty deserves that medal for that comment

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0imqwerty, I already did pay with the plan.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0imqwerty, 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
 one year ago
Best ResponseYou've already chosen the best response.3just 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
 one year ago
Best ResponseYou've already chosen the best response.5u can get only this much with $50

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ganeshie8, I put in a lot of effort thank you very much.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Maybe lets start over ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.5In which part are u having problem?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4Well, 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
 one year ago
Best ResponseYou've already chosen the best response.4dw:1444544503214:dw MID = red, MAX = pink, MIN = green to give you an idea of what I mean

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then how would I find the equation.. i'm aware of the basics

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4Ok so your equation is of the form \[y=a*f(b(xh))+k\] so use what I've said above and see if you can figure it out

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.4Or I should put it as \[y = a*\sin(b(xh \pi) ) +k \] similarly for cosine, just switch the trig function

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I plug in just the amplitude and midpoint?

baru
 one year ago
Best ResponseYou've already chosen the best response.2maybe it would help if you could tell exactly where you stand...have you understood what shifted sinusoidal function means?

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.5maybe u shuld leave it (:
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