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## Differential equations proof

 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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 Quote by 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.
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.

 woud ε = be pi/2 - wt - theta?