Wikipedia states in this article about fundamental solutions that if $F\left( x \right) = \tfrac{1}{2} \left| x \right|$, then
$$\left( F \ast \sin \right)\left( x \right) := \int\limits_{-\infty}^{\infty} \frac{1}{2} \left| x - y \right| \cdot \sin\left( y \right)\, \operatorname{d}y = -\sin\left( x \right).$$
I can't prove a wiki statement about convolutions.
The best I can come up with is to prove that: $\left\langle \int_{-\infty}^{\infty} \tfrac{1}{2} \left| x - y \right| \sin\left( y \right)\, \operatorname{d}y,\, \phi \right\rangle = \left\langle -\sin\left( x \right),\, \phi \right\rangle$ where all $\phi$ are test functions.
But as far as I can tell the convolution: $\left( F \ast \sin \right)\left( x \right)$ is not possible even in the sense of distributions.