In this post, the Kac character formula for affine $\mathfrak{su}(2)_k$ $$\chi_{\ell}^{(k)}(\tau,z) = \frac{\Theta_{\ell+1,k+2}(\tau,z)-\Theta_{-\ell-1,k+2}(\tau,z)}{\Theta_{1,2}(\tau,z)-\Theta_{-1,2}(\tau,z)}$$ with $$\Theta_{\ell,k}(\tau,z)=\sum_{n\in \mathbb{Z}+\frac{\ell}{2k}} q^{kn^2}y^{kn}\, ,\qquad (q=e^{2\pi i\tau}\,,\:y=e^{2\pi iz})$$ is put in a nice form: $$\chi_{\ell}^{(k)}(\tau,z)=q^{m_{\ell}}\frac{\sum_{n\in\mathbb{Z}}\frac{\sin\left[(\ell+1+2n(k+2))\pi z\right]}{\sin(\pi z)}q^{n(\ell+1)+n^2(k+2)}}{\prod_{n>0}(1-q^n)(1-q^n y)(1-q^n y^{-1})}\, .$$
Some more details on this derivation would be much appreciated, or references where things are done a bit more explicitly!
Edit: since I've just answered my question, I'm sparing you my former waywardness :)
Well, it was actually pretty simple. One gets the sin(...) form this way: \begin{align*}\chi_{\ell}^{(k)}(\tau,z) &= q^{\frac{(\ell+1)^2}{4(k+2)}-\frac{1}{8}}\frac{\sum_{n\in\mathbb{Z}}q^{n^2(k+2)+n(\ell+1)}\left(y^{n(k+2) +\frac{\ell+1}{2}}-y^{-n(k+2)-\frac{\ell+1}{2}}\right)}{\sum_{n\in\mathbb{Z}}q^{2n^2+n}\left(y^{2n+\frac{1}{2}}-y^{-2n-\frac{1}{2}}\right)}\\&=q^{\frac{(\ell+1)^2}{4(k+2)}-\frac{1}{8}}\frac{\sum_{n\in\mathbb{Z}}q^{n^2(k+2)+n(\ell+1)}\sin\left(\pi z (2n(k+2) +\ell+1)\right)}{\sum_{n\in\mathbb{Z}}q^{2n^2+n}\sin\left(\pi z (4n+1)\right)}\end{align*} and the trick to put the denominator in a Jacobi triple product-form is given in Wakimoto's "Infinite-dimensional Lie algebras" section 3.6. The argument goes as follows: \begin{align*}\Theta_{1,2}(\tau,z)-\Theta_{-1,2}(\tau,z)&=\sum_{n\in\mathbb{Z}}\left(q^{2\left(n+\frac{1}{4}\right)^2}y^{2\left(n+\frac{1}{4}\right)}-q^{2\left(n-\frac{1}{4}\right)^2}y^{2\left(n-\frac{1}{4}\right)}\right)\\&= q^{\frac{1}{8}}y^{\frac{1}{2}}\sum_{n\in\mathbb{Z}}\left(q^{k\left(2n^2+n\right)}y^{2n}-q^{k\left(2n^2-n\right)}y^{2n-1}\right) \\&=q^{\frac{1}{8}}y^{\frac{1}{2}}\sum_{n\in\mathbb{Z}}\left(q^{\frac{\left(2n\right)^2+2n}{2}}y^{2n}-q^{\frac{\left(2n-1\right)^2+2n-1}{2}}y^{2n-1}\right) \\&=q^{\frac{1}{8}}y^{\frac{1}{2}}\sum_{n\in\mathbb{Z}}\left(-1\right)^n q^{\frac{n^2+n}{2}}y^{n} \\&=q^{\frac{1}{8}}y^{\frac{1}{2}}\prod_{m\geqslant 1}\left(1-q^m\right)\left(1-q^{m}\,y\right)\left(1-q^{m-1}\,y^{-1}\right)\,. \end{align*}
Of course one can use the same procedure on the denominator.