Are trivial zeros of the zeta function important? Non-trivial zeros play an important (main) role in the distribution of prime numbers.
Are there theorems in which trivial zeros play an important (main) role?
 A: A conjecture of Quillen-Lichtenbaum
$$
\lim_{s \to n} (n-s)^{-\mu_n} \zeta_F (-s) = \pm \frac{\mid{K_{2n} (O_F)_{tor} }\mid}{\mid{K_{2n+1} (O_F)_{tor}}\mid} R_{F,n} * 2^{?}\\
$$
where $F$ is the number field like $\mathbb{Q}$ and $O_F$ is the integers therein generalizing $\mathbb{Z}$ in the case of Riemann zeta. $\mu_n$ is the multiplicity of the zero there. That makes sure you don't just get $0$. $K_\bullet (O_F)_{tor} $ means the torsion subgroup of algebraic K theory of that ring. $R_{F,n}$ is a so-called regulator and $2^?$ is for an unknown power of $2$. So you see the left hand side is leading information about when you have zeroes at negative integers.
This is proven in a bunch of cases via Voevodsky. In the $O_F=\mathbb{Z}$ case you just get the numerators and denominators of the Bernoulli numbers. That recovers the $-\frac{B_{n+1}}{n+1}$ above up to signs, powers of 2 and regulators.
So if you are interested in computing anything in the RHS, zooming in on the zeroes of the associated $\zeta_F$ are the most important.
A: Since you tagged this 'reference request', I'm going to answer with a theorem in my own paper "Euler, the symmetric group, and the Riemann zeta function." 
Let $\pi$ be a permutation in the symmetric group $S_n$.  An ascent is an occurrence of $\pi(j)<\pi(j+1)$ for $1\le j\le n-1$.  For
example, the permutation $(24513)$ has $3$ ascents.  The Eulerian number $\genfrac{\langle}{\rangle}{0pt}{1}{n}{k}$ is defined to be
the number of permutations in $S_n$ with exactly $k$ ascents.  (The Eulerian numbers are not to be confused with the
Euler numbers $E_n$.)   There is a surprising identity for alternating sums of Eulerian numbers: For integer $n\ge 1$ we have
$$
\zeta(-n)=\frac{ \sum_{k=1}^n  (-1)^{k} \genfrac{\langle}{\rangle}{0pt}{0}{n}{k}}{2^{n+1}(1-2^{n+1})}.
$$
Of course, $\zeta(-n)$ can be expressed in closed form in terms of the Bernoulli numbers by 
$$
\zeta(-n)=-\frac{B_{n+1}}{n+1},
$$
so the theorem is also an identity relating Eulerian numbers to Bernoulli numbers.  However, the proof is direct.  It's also not not deep; it consists of identities in Concrete Mathematics by Graham, Knuth and Patashnik, along with Abel summation.
