Differential equations proof

by th3chemist
Tags: differential, equations, proof
 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 sum-difference forumla on r*sin(ωt - θ) = r*sin(ωt)cos(θ) - r*cos(ωt)sin(θ).
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 Quote by th3chemist 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 sum-difference forumla on r*sin(ωt - θ) = r*sin(ωt)cos(θ) - r*cos(ωt)sin(θ).
Start by writing your expression like this$$\sqrt{A^2+B^2}\left( \frac A {\sqrt{A^2+B^2}}\cos(\omega t)+ \frac B {\sqrt{A^2+B^2}} \sin(\omega t) \right)$$That quantity in the large parentheses looks like an addition formula. Think about a right triangle with legs ##A## and ##B##.
 P: 66 Is that from the Pythagorean theorem? I'm still unsure how to set this problem up :(
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Differential equations proof

 Quote by th3chemist Is that from the Pythagorean theorem? I'm still unsure how to set this problem up :(
Did you draw the triangle I suggested? Do those fractions look like sines and cosines of some angle? Can you make the expression in large parentheses look like an addition formula?
 P: 66 It would be cos(theta) and sin(theta)
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 Quote by th3chemist It would be cos(theta) and sin(theta)
So.......?
P: 66
 Quote by LCKurtz So.......?
If I use the sum-difference identity I would get cosine term not the sine :(
 HW Helper Thanks PF Gold P: 7,644 Isn't ##\displaystyle{\frac A {\sqrt{A^2+B^2}}}## the cosine of some angle in your triangle?
P: 66
 Quote by LCKurtz Isn't ##\displaystyle{\frac A {\sqrt{A^2+B^2}}}## the cosine of some angle in your triangle?
yeah so you get cos(θ)cos(ωt) + sin(θ)(sin(ωt))
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 Quote by th3chemist yeah so you get cos(θ)cos(ωt) + sin(θ)(sin(ωt))
This gives you the form ##r\cos(\omega t -\theta)## (don't forget there is that square root out front). Now getting it in a sine form is what I think last sentence in your problem is about.
P: 66
 Quote by LCKurtz This gives you the form ##r\cos(\omega t -\theta)## (don't forget there is that square root out front). Now getting it in a sine form is what I think last sentence in your problem is about.
But I did not show the first part. I'm quite sure though that if you have rcos(θ) and rsin(θ) they're constants.
 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))
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P: 7,644
 Quote by th3chemist But I did not show the first part.
Haven't we shown$$A\cos(\omega t)+B\sin(\omega t) = \sqrt{A^2+B^2}\left( \frac A {\sqrt{A^2+B^2}}\cos(\omega t)+ \frac B {\sqrt{A^2+B^2}} \sin(\omega t) \right) = \sqrt{A^2+B^2}(\cos\theta\cos\omega t + \sin\theta \sin(\omega t) = \sqrt{A^2+B^2}\cos(\omega t - \theta)$$To get it to a sine function I think you have to work the last sentence in the original problem.
P: 66
 Quote by LCKurtz Haven't we shown$$A\cos(\omega t)+B\sin(\omega t) = \sqrt{A^2+B^2}\left( \frac A {\sqrt{A^2+B^2}}\cos(\omega t)+ \frac B {\sqrt{A^2+B^2}} \sin(\omega t) \right) = \sqrt{A^2+B^2}(\cos\theta\cos\omega t + \sin\theta \sin(\omega t) = \sqrt{A^2+B^2}\cos(\omega t - \theta)$$To get it to a sine function I think you have to work the last sentence in the original problem.
What can I do though? Use the identity on both sides? The first line states to write it in the sin form. We did it in cos :(.
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 Quote by LCKurtz Haven't we shown$$A\cos(\omega t)+B\sin(\omega t) = \sqrt{A^2+B^2}\left( \frac A {\sqrt{A^2+B^2}}\cos(\omega t)+ \frac B {\sqrt{A^2+B^2}} \sin(\omega t) \right) = \sqrt{A^2+B^2}(\cos\theta\cos\omega t + \sin\theta \sin(\omega t) = \sqrt{A^2+B^2}\cos(\omega t - \theta)$$To get it to a sine function I think you have to work the last sentence in the original problem.
 Quote by th3chemist What can I do though? Use the identity on both sides? The first line states to write it in the sin form. We did it in cos :(.
Yes, we have the cosine. Did you read my previous post, copied above? That is part of your problem, to figure out how to change the cosine to a sine. I have helped you with half of your problem. Now you work on finishing it.
 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?
 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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 Quote by th3chemist 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?
You have A and B in terms of r and theta. You want r and theta in terms of A and B.

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