My chemistry textbook introduces the wave function as
$$\psi(x)= A \sin\left(\frac{2\pi x}{\lambda}\right)$$ Therefore, the Schrödinger Equation is:
$$\frac{d^2\psi(x)}{dx^2} = -~\left(\frac{2\pi}{\lambda}\right)^{2}\psi(x)$$
and by multiplying each side by $\frac{-h^2}{8\pi^2m}$ you get:
$$-~\frac{h^2}{8\pi^2m} \frac{d^2\psi(x)}{dx^2} = \frac{p^2}{2m}\psi(x) = (T)(\psi(x))$$ where $T$ is kinetic energy.
Now, I don't fully understand that last bit, and I am willing to see an explanation on it, but my main question is that, when the book transitions from discussing the Schrödinger Equation in the context of a "particle-in-a-box" to the context of a one-electron atom, it seemingly changes the wave function to
$$\psi_{(n,\ell,m)}(r,\theta,\phi) = R_{n,\ell}(r) Y_{\ell,m}(\theta,\phi)$$
where n = principle quantum number, $\ell$ = angular momentum number, m = magnetic quantum number, r = radius (distance between nucleus and electron), R = the "radial part" of the equation, Y = the "angular part" of the equation, and $\theta$ and $\phi$ are some angles that I still don't know how to get.
I guess my question is did the wave function $\psi$ change? Why is suddenly so different and why do r,$\theta$ and $\psi$ suddenly come into the picture? Is it because you are going from the one-dimensional, Cartesian "x" version of the wave function to the three-dimensional, polar coordinate version of it?