Differential equations proof

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).

LCKurtz

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##.

th3chemist

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

LCKurtz

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?

th3chemist

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

LCKurtz

So.......?

th3chemist

If I use the sum-difference identity I would get cosine term not the sine :(

LCKurtz

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

th3chemist

yeah so you get cos(θ)cos(ωt) + sin(θ)(sin(ωt))

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.

th3chemist

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.

th3chemist

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))

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.

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 :(.

LCKurtz

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.

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?

th3chemist

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.