pullback of 1-differential form

Given a $\phi_{0} \in \mathbb R$. Consider the function

$\displaystyle{\phi (t) = \phi_{0} + \int_0^t (ab'-a'b)(u) \text{du}}$

where $a^2(t) + b^2 (t)=1$

Prove that:

(i) $\phi(0)= \phi_{0}$

(ii) $\cos(\phi(t))=a(t)$ and $\sin(\phi(t))=b(t)$

Consider now the 1-differential form $\displaystyle{ \omega_0= \frac{x}{x^2 + y^2}\text{dy} -\frac{y}{x^2 + y^2}\text{dx}} \quad$ in $\mathbb R ^2$.

If $c:[0,1] \to \mathbb R ^2 -\{(0,0) \}$ is a $\text{C}^1$ class curve prove that: (iii)$c^{*}(\omega_0)={\text{d} \phi}$.

Here is what I did.

(i) Obviously holds.

(ii) We have that $\displaystyle{ (a\cos \phi +b\sin \phi)(t))^{'}=0 }$ using $\phi ' = ab'-ba'$ and $a^2+b^2=1$ and $aa'+bb'=0$.

So it is $(a\cos \phi +b\sin \phi)(t))= \text{const} = (a\cos \phi +b\sin \phi)(t))(0)= a(0)\cos \phi_0 + b(0) \sin \phi_0=1$

Why the last equality holds?

I need also some help on (iii).

Is there some connection between $\phi_0$, $x(0)$ and $y(0)$? (For you didn't specify $x(0)$ and $y(0)$). –  martini May 16 '12 at 11:51
What's $x(t)$ and $y(t)$ then? –  martini May 16 '12 at 11:57
It is from the definition of $\omega_0$. –  passenger May 16 '12 at 12:02
I mean in the beginning, before $\omega_0$ is defined, you use it to define $a$ and $b$. –  martini May 16 '12 at 12:03