# Differential equations proof

1. Nov 8, 2012

### 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).

2. Nov 8, 2012

### 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 $B$.

3. Nov 8, 2012

### th3chemist

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

4. Nov 8, 2012

### 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?

5. Nov 8, 2012

### th3chemist

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

6. Nov 8, 2012

So.......?

7. Nov 8, 2012

### th3chemist

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

8. Nov 8, 2012

### LCKurtz

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

9. Nov 8, 2012

### th3chemist

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

10. Nov 8, 2012

### 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.

11. Nov 8, 2012

### 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.

12. Nov 8, 2012

### 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))

13. Nov 8, 2012

### 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.

14. Nov 8, 2012

### 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 :(.

15. Nov 8, 2012

### 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.

16. Nov 8, 2012

### 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?

17. Nov 8, 2012

### 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.

18. Nov 8, 2012

### LCKurtz

You have A and B in terms of r and theta. You want r and theta in terms of A and B.

19. Nov 8, 2012

### LCKurtz

You can get either a sine or cosine form. That shouldn't surprise you because anything that can be expressed as one can be expressed in terms of the other since $\sin\theta = \cos(\frac \pi 2 -\theta)$ and similarly for the cosine.

20. Nov 9, 2012

### th3chemist

woud ε = be pi/2 - wt - theta?