Tying to prove that the sequences $a_n$ and $b_n$ are bounded (to fill in the gap in user10676's approach), the following argument - using Baire category - turned up:
Let $\epsilon > 0$ be given. Define $F_N = \{x\in (c,d)\;|\;|f_n(x)|\le\epsilon \quad \forall n \ge N \}$. We have
$(c,d)=\bigcup_{N \in \mathbb N} F_N$ by pointwise convergence and furthermore the sets $F_N$ are closed by continuity of the functions $f_n$.
By the Baire category theorem there must be $N_0$ such that $F_{N_0}$ has nonempty interior. Choose two points $s,t$ such that $(s,t)\subset F_{N_0}$. Note that then in particular we have that $(s,t) \subset F_N$ for all $N>N_0$.
Supposing $N_0$ to be large enough (making it bigger will do no harm), the congruences \begin{eqnarray*} Nx &\equiv& 0 \pmod{2\pi}\\ Ny &\equiv& \frac\pi 2 \pmod{2\pi} \end{eqnarray*}
have solutions in $(s,t)$ for all $N>N_0$. But then for any $N>N_0$, choosing $x,y$ as above
\begin{eqnarray*} |a_N| &=& |a_N\cos(Nx) + b_N\sin(Nx)| \le \epsilon \\ |b_N| &=& |a_N\cos(Ny) + b_N\sin(Ny)| \le \epsilon \end{eqnarray*}
because $x,y\in (s,t)\subset F_{N_0}$.
This proves that the sequences $a_n$ and $b_n$ converge.