Let $[\theta\_1,\theta\_2, \dots, \theta\_N]^\mathrm{T} \, \in \mathbb{R}^N$. The angles are not all identical (on the circle), i.e. $[\theta\_1,\theta\_2, \dots, \theta\_N] \not \equiv c [1,1,\dots, 1]^\mathrm{T}\,\, \mathrm{mod}\,\, 2\pi$. Define matrices $C$ and $S$ as:


\begin{align}
\begin{split}
[C]\_{jl}&= 1 \,\, \mathrm{if}\,\, j =l \\
[C]\_{jl}&= \cos(\theta\_j-\theta\_l)\,\, \mathrm{if} \,\, j \neq l
\end{split}
\end{align} 


and 


\begin{align}
\begin{split}
[S]\_{jl}&= 0 \,\, \mathrm{if}\,\, j =l \\
[S]\_{jl}&= \sin(\theta\_j-\theta\_l)\,\,\mathrm{if} \,\, j \neq l\,.
\end{split}
\end{align} 


Is it true that any $v$ that belongs to the nullspace of S also belongs to the nullspace of C? 


My repeated simulations in MATLAB with randomly generated $[\theta\_1,\theta\_2, \dots, \theta\_N]^\mathrm{T}$ seems to suggest that $C$ and $S$ are rank $2$ and any eigenvectors with $0$ eigenvalue for one is an eigenvector with $0$ eigenvalue for the other. 



Do these matrices have the same null space?
You must have made a mistake.


$E = -C + 2I + i S$ has entries $\exp(i (\theta\_j - \theta\_l)) = \exp(i\theta\_j) \exp(-i \theta\_l)$ and thus is the product $V V^\*$ where $V$ is the column vector with entries $\exp(i \theta\_j)$, so it has rank $1$.  

Then $S = (E - \overline{E})/(2i)$ has rank at most $2$ (it would be $0$ e.g. if all $\theta\_j$ are multiples of $\pi$). Similarly, $-C+2I$ has rank at most $2$, so $C$ has rank at least $N-2$ (that part at least you got right).
But "generically" $C$ will have rank $N$.


EDIT:
OK, now with the corrected question we have $E = C + iS$. Again $S = (E - \overline{E})/(2i)$ and now $C = (E + \overline{E})/(2i)$ have rank at most $2$. Moreover, $\ker(E) \cap \ker(\overline{E}) = (V^\*)^\perp \cap (\overline{V^\*})^\perp$ is contained in both $\ker(S)$ and $\ker(C)$.
Note that the following are equivalent:


1. $\theta\_j - \theta\_l \not \equiv 0 \mod \pi$ for some $j,l$.
2. $V$ and $\overline{V}$ are linearly independent.
3. $(V^\*)^\perp$ and $(\overline{V^\*})^\perp$ are distinct linear subspaces.
4. $\ker(E) \cap \ker(\overline{E})$ has dimension exactly $N-2$.


