Wavefunction of a particle on a polar potential Suppose you have the following hamiltonian:
$$
H=-\frac{\hbar^2}{2mR^2}\frac{d^2}{d\phi^2} \tag{1}
$$
which we can recognize as $\hat H=\frac{\hat L_z^2}{2mR^2}$. But working with (1) we see that:
$$
-\frac{\hbar^2}{2mR^2}\frac{d^2\psi}{d\phi^2}=E\psi
$$
yields:
$$
\psi(\phi)=A\sin(\omega \phi)+B\cos(\omega \phi)
$$
Applying boundary condition of $\psi(\phi)=\psi(\phi+2\pi)$ we get that $\omega=\frac{\sqrt{2mR^2E}}{\hbar}=n \to E_n=\frac{h^2n^2}{2mR^2}$. But what about the wave functions? I feel that this is one of those cases where we have well-defined parity and that we have $\psi(\phi)=\sin(n\phi)$ if $n$ is even/odd and $\psi(\phi)=\cos(n\phi)$ if $n$ is odd/even. But how to figure out which is which?
 A: Just a small addition to ZeroTheHero's answer, I think there's a more instructive reason to label the wavefunctions by $e^{in\phi}$ rather than sines and cosines. As you have correctly shown, in this problem (similar to the particle in the box), the energies are given by $$E_n = \frac{h^2 n^2}{2m R^2}.$$ However, what is different is that unlike the particle in the box, this problem has a degeneracy due to the boundary conditions: both $n$ and $-n$ produce different solutions that have the same energy. What distinguishes these solutions is the $z$ component of their angular momentum, $L_z$ (you could very crudely think of one of them as rotating to the ''right'' and the other to the ''left''). Thus, writing the wavefunctions as $e^{i n \phi}$ makes sure that these solutions are the simultaneous eigenfunctions of $H$, $L^2$, and $L_z$.
Of course, since any linear combination is also a solution, there's nothing wrong with using the sines and cosines, except that in this basis, the solutions will not have a definite value of $L_z$.
A: In principle one can use any normalized linear combination of $\cos(n\phi)$ and $\sin(n\phi)$ but 
the solutions for this special case are usually written in terms of exponentials:
$$
\psi_n(\phi)=\frac{1}{\sqrt{2\pi}}e^{in\phi}\, ,\qquad 
-\infty\le n \le \infty\, . \tag{1}
$$
Since $\psi_{\pm n}(\phi)$ are degenerate, an arbitrary linear combination
$$
A_n\psi_{n}(\phi) + B_n\psi_{-n}(\phi)
$$
is also a solution and an eigenfunction of $H$ with energy $E_n$, so one can take combinations of these two to convert to sine and cosine form, but (1) is enough as the functions are orthogonal over the angles:
$$
\langle n\vert m\rangle =\frac{1}{2\pi}\int_0^{2\pi} e^{i(m-n)\phi}=\delta_{nm}
$$
so that any function 
$$
f(\phi)=\sum_n c_n\psi_n(\phi)\, .
$$
Conversely, since $C_n(\phi)= \frac{1}{\sqrt{\pi}}\cos(n\phi)$ and $S_n(\phi)= \frac{1}{\sqrt{\pi}}\sin(n\phi)$ are also solutions, one can take combinations of those two to recover the exponentials.  The most general solution of the Schrödinger equation can then be expressed as
$$
\Psi_n(\phi,t)=e^{-iE_nt}\left( A_ne^{in\phi}+B_n e^{-in\phi} \right)= 
e^{-iE_nt}\left(a_n\cos(n\phi)+b_n \sin(n\phi)\right)
$$
with the various coefficients $A_n,B_n,a_n$ and $b_n$ determined by the initial condition, i.e. by $\Psi_n(\phi,0)$.
The exponential form is preferred because the exponentials are easier to manipulate than the trig functions under multiplication, derivative and integration.
