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Differential equations proof 
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#1
Nov812, 09:50 AM

P: 66

1. The problem statement, all variables and given/known data
Show that Acos(ωt) + Bsin(ωt) can be written in the form r*sin(ωt  θ). Determine r and θ in terms of A and B. If Rcos(ωt  ε) = r*sin(ωt  θ), deermine the relationship among R, r, ε and θ. 2. Relevant equations Trig identities. 3. The attempt at a solution I'm really confused on this question. What I first tried to do is to use the sumdifference forumla on r*sin(ωt  θ) = r*sin(ωt)cos(θ)  r*cos(ωt)sin(θ). 


#2
Nov812, 11:57 AM

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PF Gold
P: 7,644

\sqrt{A^2+B^2}\left( \frac A {\sqrt{A^2+B^2}}\cos(\omega t)+ \frac B {\sqrt{A^2+B^2}} \sin(\omega t) \right)$$That quantity in the large parentheses looks like an addition formula. Think about a right triangle with legs ##A## and ##B##. 


#3
Nov812, 12:04 PM

P: 66

Is that from the Pythagorean theorem?
I'm still unsure how to set this problem up :( 


#4
Nov812, 12:09 PM

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PF Gold
P: 7,644

Differential equations proof



#5
Nov812, 12:14 PM

P: 66

It would be cos(theta) and sin(theta)



#7
Nov812, 12:41 PM

P: 66




#8
Nov812, 12:59 PM

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PF Gold
P: 7,644

Isn't ##\displaystyle{\frac A {\sqrt{A^2+B^2}}}## the cosine of some angle in your triangle?



#9
Nov812, 01:09 PM

P: 66




#10
Nov812, 01:36 PM

HW Helper
Thanks
PF Gold
P: 7,644




#11
Nov812, 01:46 PM

P: 66




#12
Nov812, 01:59 PM

P: 66

Here is something else I did. I got r*sin(wt)cos(theta)  r*cos(wt)sin(theta)
Then I set rewrote as r*cos(theta) + (rsin(theta)cos(wt) I can then get A = r*cos(theta) and B = (r*sin(theta)) 


#13
Nov812, 02:45 PM

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PF Gold
P: 7,644

A\cos(\omega t)+B\sin(\omega t) = \sqrt{A^2+B^2}\left( \frac A {\sqrt{A^2+B^2}}\cos(\omega t)+ \frac B {\sqrt{A^2+B^2}} \sin(\omega t) \right) = \sqrt{A^2+B^2}(\cos\theta\cos\omega t + \sin\theta \sin(\omega t) = \sqrt{A^2+B^2}\cos(\omega t  \theta)$$To get it to a sine function I think you have to work the last sentence in the original problem. 


#14
Nov812, 03:39 PM

P: 66




#15
Nov812, 06:03 PM

HW Helper
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PF Gold
P: 7,644




#16
Nov812, 06:34 PM

P: 66

Is A = r*cos(theta) and B = (r*sin(theta)) correct for the first part then? To get theta and r in terms of A and B?



#17
Nov812, 07:38 PM

P: 66

Hold on, I have a question. How can you assume that will give you cosine instead of sine? If I switch, A and B I would get sine.



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