anonymous
  • anonymous
How to find the sin and cosine of a shifted sinusoidal function.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
dan815
  • dan815
looks like this is the sin graph shifted by pi/12 and the period is only pi/2 now
dan815
  • dan815
Let's figure out how to change the period

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dan815
  • dan815
sin(x) has what period initially
dan815
  • dan815
Hi, are you there?
anonymous
  • anonymous
i'm a little confused how you got pi/2
dan815
  • dan815
ah it's not pi/2
dan815
  • dan815
it's 2pi/3
dan815
  • dan815
http://i.imgur.com/KZfiTq6.png
anonymous
  • anonymous
what is 2pi/3? Sorry i'm lost
dan815
  • dan815
take a look at that picture
dan815
  • dan815
the length of this full cycle is called the period
anonymous
  • anonymous
where would the period start from?
dan815
  • dan815
you can start the period anywhere, as long as 1 full cycle takes place, however the period will always be 2pi/3
dan815
  • dan815
let me show you some examples
anonymous
  • anonymous
Ok thank you..
dan815
  • dan815
http://i.imgur.com/KnoNRXN.png
dan815
  • dan815
here i have picked 3 sets of pairs of points
anonymous
  • anonymous
Oh i see! Thank you :)
dan815
  • dan815
okay so lets start with the base graph y=sin(x), what period does this have initially
anonymous
  • anonymous
2pi?
dan815
  • dan815
right
dan815
  • dan815
y=sin(x) looks like this |dw:1444540802308:dw|
dan815
  • dan815
now lets see how we can possibly make this period only 2pi/3
dan815
  • 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
  • dan815
Tell me if that makes sense, this might be a little tricky when u first see it
anonymous
  • 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
  • dan815
right now can u figure out how to make this period 2pi/3 instead of 1
dan815
  • dan815
we saw how multiplying by 2pi in the brackets managed to speed up this period to 1 instead of 2pi
dan815
  • dan815
sin(ax) , what should this a be so that when x=2pi/3, we have a complete period
dan815
  • 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
  • anonymous
couldn't you just multiply it by 2pi?
dan815
  • dan815
2pi made the period go to 1
dan815
  • dan815
the answer is y=\[sin(\frac{2\pi ~x}{2\pi/3} )\]
dan815
  • dan815
see how when x=2pi/3 the stuff in the brackets turns to 2pi
dan815
  • dan815
now apply your shift and amplitude change
dan815
  • 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
  • anonymous
i'm still confused how we got 2pi from the 2pix/2pi/3?
dan815
  • dan815
when x=2pi/3
dan815
  • dan815
|dw:1444541639553:dw|
anonymous
  • anonymous
oh ok.. thank you
carlyleukhardt
  • carlyleukhardt
alright time to investigate sana xD
anonymous
  • anonymous
go ahead XD I have a bf
dan815
  • dan815
I think you need to understand better why y=sin(2pi x) in making the period 1 from 2pi previously
anonymous
  • 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
  • dan815
just looking to complete a whole period
dan815
  • 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
  • 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
  • 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
  • anonymous
so now I must plug in x values in for y?
dan815
  • dan815
What do you mean
dan815
  • dan815
http://i.imgur.com/CbqYn0Z.png
dan815
  • 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
  • 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
  • anonymous
is there a more simpler way to do this.. I feel like there is.
dan815
  • dan815
yeah you can memorize the formula
dan815
  • dan815
\[Y=A* Sin(\frac {2\pi}{L} (X-B))\] A=Amplitude L=Length B=Phase
nincompoop
  • nincompoop
grab a precal textbook
nincompoop
  • nincompoop
get the one by larson or stewart
anonymous
  • anonymous
my professor doesn't use the book, he expects us to "automatically know this"
anonymous
  • anonymous
and i've tried using the book
dan815
  • dan815
okay so can you plug into the formula?
dan815
  • dan815
You know the the length,the phase, the Amplitude
anonymous
  • anonymous
the amplitude would be 3?
dan815
  • dan815
yes
dan815
  • dan815
I already showed you what the answer should look like in the screenshot a couple comments ago
dan815
  • dan815
just ask one of these people if you are still confused, i have to go now
anonymous
  • anonymous
the 3sin(2pi(x-pi/12))/2pi/3)?
anonymous
  • anonymous
thanks for your time
carlyleukhardt
  • carlyleukhardt
i keep on getting notifications. oh nvm your done/
imqwerty
  • imqwerty
now u need to pay $50 for the help using paypal
dan815
  • dan815
\[Y=3* Sin(\frac {2\pi}{2pi/3} (X-\pi/12))\]
anonymous
  • anonymous
ahaha @imqwerty
anonymous
  • anonymous
sorry dan, but @imqwerty deserves that medal for that comment
anonymous
  • anonymous
imqwerty, I already did pay with the plan.
imqwerty
  • imqwerty
ok then (:
anonymous
  • 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
  • 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
  • imqwerty
u can get only this much with $50
anonymous
  • anonymous
ganeshie8, I put in a lot of effort thank you very much.
ganeshie8
  • ganeshie8
Maybe lets start over ?
anonymous
  • 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
  • imqwerty
In which part are u having problem?
anonymous
  • anonymous
all of it.
Astrophysics
  • 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
  • Astrophysics
|dw:1444544503214:dw| MID = red, MAX = pink, MIN = green to give you an idea of what I mean
anonymous
  • anonymous
then how would I find the equation.. i'm aware of the basics
Astrophysics
  • 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
  • Astrophysics
Or I should put it as \[y = a*\sin(b(x-h \pi) ) +k \] similarly for cosine, just switch the trig function
anonymous
  • anonymous
I plug in just the amplitude and midpoint?
baru
  • baru
maybe it would help if you could tell exactly where you stand...have you understood what shifted sinusoidal function means?
imqwerty
  • imqwerty
maybe u shuld leave it (:

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