It's well known that: (https://en.wikipedia.org/wiki/Euler_numbers#In_terms_of_Stirling_numbers_of_the_second_kind) $$E_{n}=2^{2n-1}\sum_{\ell=1}^{n}\frac{(-1)^{\ell}S(n,\ell)}{\ell+1}\left(3\left(\frac{1}{4}\right)^{(\ell)}-\left(\frac{3}{4}\right)^{(\ell)}\right)$$
where $E_n$ are the Euler Numbers, $S(n,k)$ the Stirling Numbers of second kind, and $(x)^{(\ell)}$ denotes the rising factorial.
My question is: Is there some other less complicated way to express the Euler numbers in terms of the Stirling numbers? (For example using cosine)
Thanks.
I hope I can help here and motivate further discussion by proving the two identities from Wikipedia. Perhaps additional simplification is possible and we can help OP in this way. (Knuth notation for rising factorials.)
First identity
We seek to prove that with Euler numbers
$$E_{n} = 2^{2n-1} \sum_{\ell=1}^{n} \frac{(-1)^\ell}{\ell+1} {n\brace \ell} \left(3\left(\frac{1}{4}\right)^{\Large {\overline{\ell}}} - \left(\frac{3}{4}\right)^{\Large {\overline{\ell}}}\right).$$
With the usual generating functions we have for the RHS
$$2^{2n-1} n! [z^n] \sum_{\ell=1}^{n} \frac{(-1)^\ell}{\ell+1} (\exp(z)-1)^\ell \left(3 {-3/4+\ell\choose \ell} - {-1/4+\ell\choose \ell} \right) \\ = 2^{2n-1} n! [z^n] \sum_{\ell=1}^{n} \frac{1}{\ell+1} (\exp(z)-1)^\ell \left(3 {-1/4\choose \ell} - {-3/4\choose \ell} \right) \\ = 2^{2n-1} n! [z^n] \frac{1}{\exp(z)-1} \sum_{\ell\ge 0} \frac{1}{\ell+1} (\exp(z)-1)^{\ell+1} \\ \times [w^\ell] \left(3(1+w)^{-1/4} - (1+w)^{-3/4}\right).$$
We have extended the range both ways due to the coefficient extractor. Continuing,
$$2^{2n-1} n! [z^n] \frac{1}{\exp(z)-1} (4 \exp(z)^{3/4} - 4 \exp(z)^{1/4}) \\ = 2^{2n+1} n! [z^n] \frac{1}{\exp(z)-1} (\exp(z)^{3/4} - \exp(z)^{1/4}) \\ = 2 \times n! [z^n] \frac{1}{\exp(4z)-1} (\exp(3z) - \exp(z)) \\ = 2 \times n! [z^n] \frac{1}{\exp(2z)+1} \exp(z) = 2 \times n! [z^n] \frac{1}{\exp(z)+\exp(-z)} \\ = n! [z^n] \frac{1}{\cosh(z)}.$$
This is the claim.
Second identity
We seek to prove that with Euler numbers
$$E_{2n} = -4^{2n} \sum_{\ell=1}^{2n} \frac{(-1)^\ell}{\ell+1} {2n\brace \ell} \left(\frac{3}{4}\right)^{\Large {\overline{\ell}}}.$$
We again have using standard generating functions for the RHS
$$-4^{2n} (2n)! [z^{2n}] \sum_{\ell=1}^{2n} \frac{(-1)^\ell}{\ell+1} (\exp(z)-1)^\ell {-1/4+\ell\choose \ell} \\ = -4^{2n} (2n)! [z^{2n}] \sum_{\ell\ge 0} \frac{1}{\ell+1} (\exp(z)-1)^\ell {-3/4\choose \ell}$$
Here we have extended the range both ways due to the coefficient extractor. Continuing,
$$-4^{2n} (2n)! [z^{2n}] \frac{1}{\exp(z)-1} \sum_{\ell\ge 0} \frac{1}{\ell+1} (\exp(z)-1)^{\ell+1} {-3/4\choose \ell} \\ = -4^{2n} (2n)! [z^{2n}] \frac{1}{\exp(z)-1} \sum_{\ell\ge 0} \frac{1}{\ell+1} (\exp(z)-1)^{\ell+1} [w^\ell] (1+w)^{-3/4} \\ = -4^{2n+1} (2n)! [z^{2n}] \frac{1}{\exp(z)-1} (\exp(z)^{1/4}-1) \\ = -4 (2n)! [z^{2n}] \frac{1}{\exp(4z)-1} (\exp(z)-1).$$
Here we are extracting even coefficients so this is
$$-2 (2n)! [z^{2n}] \left[\frac{\exp(z)-1}{\exp(4z)-1} + \frac{\exp(-z)-1}{\exp(-4z)-1}\right] \\ = -2 (2n)! [z^{2n}] \left[\frac{\exp(z)-1}{\exp(4z)-1} + \frac{\exp(4z)-\exp(3z)}{\exp(4z)-1}\right] \\ = -2 (2n)! [z^{2n}] \frac{\exp(z)-1}{\exp(4z)-1} (1+ \exp(3z)).$$
Now as this is for $n\ge 1$ we should get a constant difference with the Euler number EGF, and indeed we have
$$\frac{2}{\exp(z)+\exp(-z)} = 2 \frac{\exp(z)}{\exp(2z)+1} = 2 \frac{\exp(z) (\exp(2z)-1)}{\exp(4z)-1} \\ = 2 \frac{\exp(z) (\exp(z)-1) (\exp(z)+1)}{\exp(4z)-1}$$
Substract to get
$$2 \frac{\exp(z)-1}{\exp(4z)-1} (\exp(2z)+\exp(z)+1+\exp(3z)) = 2.$$
This is the claim and we may conclude.