Without knowing how [tex]X[/tex] is distributed, or what its PMF happens to be, we can't completely answer the question. However, recall that
[tex]\cos(-kx)=\cos kx[/tex]
[tex]\sin(-kx)=-\sin kx[/tex]
If [tex]Y=\cos(\pi X)[/tex], then
[tex]\mathbb E[Y]=\displaystyle\sum_{k\in\mathbb Z}\cos(k\pi)\,\mathbb P(X=k)[/tex]
By the symmetry of [tex]X[/tex],
[tex]\mathbb E[Y]=\mathbb P(X=0)+2\displaystyle\sum_{k\ge1}\cos(k\pi)\,\mathbb P(X=k)[/tex]
Splitting on the even and odd [tex]k[/tex], and using the fact that [tex]\cos(k\pi)=(-1)^k[/tex] for [tex]k\in\mathbb Z[/tex], we can write this as
[tex]\mathbb E[Y]=\mathbb P(X=0)+2\displaystyle\sum_{\ell\ge1}\bigg(\mathbb P(X=2\ell)-\mathbb P(X=2\ell-1)\bigg)[/tex]
but without knowing how [tex]X[/tex] is distributed, we can't go any further as far as I'm aware.
For [tex]Y=\sin(\pi X)[/tex], we would have
[tex]\mathbb E[Y]=\mathbb E[\sin(\pi X)]=\displaystyle\sum_{k\in\mathbb Z}\sin(k\pi)\,\mathbb P(X=k)[/tex]
but [tex]\sin(k\pi)=0[/tex] for all integers [tex]k[/tex]. Thus [tex]\mathbb E[\sin(X\pi)]=0[/tex].
