Finding the smooth inverse of a function Problem: The mapping $\phi: S^2 \longrightarrow S^2 $ by $$\phi(x,y,z)=(x\cos z+y\sin z,x\sin z-y \cos z,z)$$ is a diffeomorphism.  Where $S^2$ is a unit sphere in $\mathbb{R}^3$. 
I've already shown that $\phi$ is smooth and bijective. The only thing I can't find is $\phi^{-1}$ that is smooth.
Any help would be much appreciated!

Let $a=(x,y,z)\in S^2$. If we know that the derivative $d\phi_a: T_aS^2\to T_{\phi(a)}S^2$ is nonsingular, then we know that $\phi$ has a local inverse, defined and smooth in a neighborhood of $\phi(a)$.
Now, thinking for the moment of $\phi$ as a map on $\Bbb R^3$, we have
$$d\phi_a = \begin{bmatrix} \cos z & \sin z & -x\sin z + y\cos z \\
\sin z & -\cos z & x\cos z+\sin z \\
0 & 0 & 1\end{bmatrix}.$$
The determinant of $d\phi_a$ is $-1$ and hence $d\phi_a$ is invertible as a map $T_a \Bbb R^3\to T_{\phi(a)}\Bbb R^3$. It follows that the restriction of $d\phi_a$ to a map $T_a S^2\to T_{\phi(a)}S^2$ must be nonsingular (why?).
Therefore, it follows that $\phi$ has a local smooth inverse mapping a neighborhood of $\phi(a)\in S^2$ to $a\in S^2$. [If you haven't seen this application before, you can deduce it by using charts or parametrizations at $a$ and $\phi(a)$ and reducing to a question about the mapping of an open set in $\Bbb R^2$ to an open set in $\Bbb R^2$.]
