Let $X$ be a complex manifold of dimension $n$. Thus, it's a real manifold of dimension $2n$.
Now cohomology is a topological concept so it should not depend upon the structure given on a topological space.
We know that $k^{th}$ Singular cohomology of $X$ is $0$ for $k \gt 2n$. We can also define a sheaf cohomology on that space using derived functor approach of Grothendieck. Then (by a result of Grothendieck) we know that k'th sheaf cohomology is $0$ for $k \gt n$.
Now, for constant sheaves [say R], the sheaf cohomology agrees with singular cohomology [with coefficient R]. Does this means that even the $k^{th}$ singular cohomology of $X$ vanishes for $k \gt n$?
[Edited] I now feel that the result which says that sheaf and singular agrees is actually this that $k^{th}$ sheaf cohomology [of a complex manifold and constant sheaf] will agree with $2k^{th}$ singular cohomology [of the underlying real manifold]. Is this correct?? I would still like others to comment.
Thanks.