What spherical functions can the N-th order limited spherical harmonics represent? Background
It is known that the spherical harmonics $Y_n^m$ with order $n$ and degree $m$ (such that $n \ge 0, -n \le m \le n$) are functions on the sphere which form a complete, orthogonal infinite set in $L^2$ space on the sphere surface $S^2$.
Given any arbitrary square-integrable function $f(\theta, \phi) \in L^2(S^2)$, the spherical Fourier series with coefficients for order $n \le N$, denoted $S_N(f)$ converges absolutely in the 2-norm:
$$
\lim_{N \to \infty} \lVert f - S_N(f) \rVert_2 = 0
$$
For a finite, $N$-th order limited set of spherical harmonics, what functions can be exactly represented (in the norm convergence sense)?
Consider the spherical function:
$$
f(\theta, \phi) = \cos(2 \phi) \sin \theta
$$

Due to the identities:
$$
\cos \theta = \frac{e^{i \theta} + e^{-i \theta}}{2}, \quad \sin \theta = \frac{e^{i \theta} - e^{-i \theta}}{2i}
$$
it appears that $f(\theta, \phi)$ is a 2nd-order trigonometric polynomial, as one would conclude for a 1-dimensional function $g(x)$ with a traditional Fourier series (source):
$$
S_\infty (g) = \sum_{n=-\infty}^\infty c_n e^{i n x}
$$
which is equal to:
$$
c_0 + (c_1 e^{ix} + c_{-1} e^{-ix}) + \ldots + (c_n e^{inx} + c_{-n} e^{-inx}) + \ldots
$$
and we know that $e^{\pm i n \theta} = \cos (n \theta) \pm i \sin (n \theta)$. I think this expresses a relation between the trigonometric functions and the complex exponential, where order is the multiple of the dependent variable. I would assert that: the SHs are just this process in two dimensions. Therefore, the SHs of order 2 can exactly represent trig polynomials of order 2, regardless of whether it is in $\theta$ or $\phi$.
But I must be missing some crucial theory. Numerically, I can determine that there are SH coefficients that are non-zero beyond the 2nd-order:

Coefficients corresponding to order 2 and below are to the left of the red dashed line. (The ACN channel numbering is from Ambisonics to identify the specific SH coefficient according to $n^2 + n + m$.)
So the questions are: why can't the spherical harmonics of order $n \le 2$ represent $f(\theta, \phi)$? And furthermore, what exactly can they represent in terms of orders of polynomials? What math theory am I missing?
 A: Instead of polar, let us work in Cartesian coordinates $(x, y, z)\in\mathbb R^3$, with the constraint that $x^2+y^2+z^2=1$. You are representing functions on the sphere as $f=f(\theta, \phi)$, but we can equivalently represent them as functions of $(x, y, z)$ via the following change of coordinates formulas (physicist's convention):
$$\tag{1}
\begin{cases}
x=\sin\theta \cos \phi, \\
y=\sin\theta\sin\phi, \\
z=\cos \theta.
\end{cases}$$
The usefulness of this lies in the following.
Fact. The function $Y=Y(\theta, \phi)$ is a spherical harmonic of degree $n$ if and only if, letting
$$
Y(\theta, \phi)=H(x, y, z), $$
the function $H$ is a homogeneous polynomial of degree $n$ and moreover
$$
\frac{\partial^2 H}{\partial x^2} + \frac{\partial^2 H}{\partial y^2}+ \frac{\partial^2 H}{\partial z^2} =0,\qquad \forall (x, y, z)\in \mathbb R^3.$$
Notice that this last equation must hold for all $(x, y, z)\in\mathbb R^3$, not just on the sphere. In other words, $H$ must be a harmonic homogeneous polynomial.
Here there is a proof of this fact.
Examples of homogeneous harmonic polynomials $H=H(x, y, z)$ of various degrees are
$$
x,\quad y,\quad z,\quad x^2-z^2, $$
which, using (1), yield the following spherical harmonics in polar coordinates:
$$
\sin\theta  \cos\phi,\quad \sin\theta\sin\phi,\quad \cos\theta,\quad\sin^2\theta\cos^2\phi-\cos^2\theta.$$
We conclude from all this that, in particular, every linear combination of spherical harmonics of degree up to $N$ must be a harmonic polynomial in the Cartesian coordinates of degree up to $N$.

Now let us consider the example given above, namely
$$
f(\theta, \phi)=\cos(2\phi)\sin(\theta).$$
Using (1), we see that
$$
\begin{split}
f(\theta, \phi)&=(\cos^2\phi-\sin^2\phi)\sqrt{1-\cos^2\theta} \\
&=\frac{x^2-y^2}{x^2+y^2}\sqrt{1-z^2}\\
&=\frac{x^2-y^2}{\sqrt{1-z^2}}.
\end{split}$$
This is not a polynomial. Therefore, it is not a finite linear combination of spherical harmonics. It must possess nonzero spherical harmonics coefficients of arbitrarily high order, as observed numerically.
