Show that $A=\{(r\cos\theta,r\sin\theta)\in\mathbb{R}^2|r\geq 0,\,0\leq \theta\leq 3\pi/4\}$ is diffeomorphic to $B=\{(r\cos\theta,r\sin\theta)\in\mathbb{R}^2|r\geq 0,\,0\leq \theta\leq \pi/2\}$.
Indeed, I have an idea of showing this but it seems a bit complicated, so I'd like to see whether we have a clearer way of doing it.
Besides, may I ask how one can say that $B=\{(r\cos\theta,r\sin\theta)\in\mathbb{R}^2|r\geq 0,\,0\leq \theta\leq \pi/2\}$ is non-diffeomorphic to $C=\{(r\cos\theta,r\sin\theta)\in\mathbb{R}^2|r\geq 0,\,0\leq \theta\leq 3\pi/2\}$?
Let $g$ be the map that takes $(r,\theta)\rightarrow (r,\beta\theta)$ (suitable $\beta$) in polar coordinates and on a suitable 'wedge'. It is $C^\infty$ outside of the origin. Let $h$ be a linear map, mapping (in cartesian coordinates) $(1;0)$ to $(1;0)$ and (suitable $\theta_0$): $$(\cos(\theta_0);\sin(\theta_0)) \mapsto (\cos(\beta\theta_0);\sin(\beta\theta_0))$$ Now let $\rho(r)$ be a $C^\infty$ function which is identical 1 for $0\leq r\leq \delta_0$ and identically zero for $r\geq \delta_1>\delta_0$. Then (here writing $h$ in polar coordinates): $$f(r,\theta) = \rho(r) h(r,\theta) + (1-\rho(r)) g(r,\theta)$$ will do the job in the first case (but not the second).
I think you are right that the two other sets are not diffeomorphic as a linear map may not take a wedge of angle $>\pi$ to one of angle $<\pi$ (and vice versa). Quite interesting btw.