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

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

Mathematics
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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

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Other answers:

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?
you can start the period anywhere, as long as 1 full cycle takes place, however the period will always be 2pi/3
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
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
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
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
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
\[y=A*sin(p*(x-b))\] this is the form for sin graph where A is the amplitude,p determines the period and b is the shift \[y=3*sin(\frac{2\pi ~(x-\pi/12)}{2\pi/3} )\]
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
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?
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
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
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
so now I must plug in x values in for y?
What do you mean
http://i.imgur.com/CbqYn0Z.png
*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
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
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
\[Y=3* Sin(\frac {2\pi}{2pi/3} (X-\pi/12))\]
ahaha @imqwerty
sorry dan, but @imqwerty deserves that medal for that comment
imqwerty, I already did pay with the plan.
ok then (:
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.
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
u can get only this much with $50
ganeshie8, I put in a lot of effort thank you very much.
Maybe lets start over ?
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.
In which part are u having problem?
all of it.
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.
|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
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
Or I should put it as \[y = a*\sin(b(x-h \pi) ) +k \] similarly for cosine, just switch the trig function
I plug in just the amplitude and midpoint?
maybe it would help if you could tell exactly where you stand...have you understood what shifted sinusoidal function means?
maybe u shuld leave it (:

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