I ran across exercise 2.8.4 in Oneill's Elementary Differential Geometry. It says "Given a frame field $E_1$ and $E_2$ on $R^2$ there is an angle function $\psi$ such that $E_1=\cos(\psi)U_1+\sin(\psi)U_2$, $E_2=-\sin(\psi)U_1+\cos(\psi)U2$
(where $U_1$, $U_2$, $U_3$ are the natural (rectangular) unit vectors)
express the connection form and dual 1-forms in terms of $\psi$ and the natural coordinates x,y."
For the connection forms I get $\omega_{12}=-\omega_{21}=d\psi$ with the other components zero. For the dual 1-forms I had difficulty. The definition of dual 1-forms states they are the 1-forms $\theta_i$ such that $\theta_i(v)=v\cdot E_i(p)$ for each tangent vector v. I have difficulty applying this definition. One reason is that it is independent of any coordinate system.
I noticed that the given frame field matches the frame field of cylindrical coordinates with the angle $\psi$. So that was my answer, the dual 1-forms of cylindrical coordinates: $\theta_1=dr$ where $r=\sqrt{x^2+y^2}$ and $\theta_2=rd\psi$.
So my question is, is my answer correct? Can I just pick a coordinate system and use it to compute the dual 1-forms?
Taking the dual forms of E to the natural frame field U preserves the coefficients, so each dual form is written with the coefficients from the attitude matrix