Of course, $\cos(\pi x)$ is just $(-1)^x$ in disguise, at least on the integers. You're right that it's mostly convention, but it's a convention with a purpose behind it; in general, the solutions of any constant-coefficient (linear) recurrence relation $f(x) = a_1f(x-1) + a_2f(x-2) + \ldots + a_nf(x-n)$ can be represented as a sum of exponentials in $x$: $f(x) = c_1r_1^x+\ldots+c_nr_n^x$ , where $r_1, \ldots, r_n$ are the roots of the so-called characteristic polynomial of the recurrence, the polynomial $x^n-a_1x^{n-1}-\ldots-a_{n-1}x-a_n = 0$ (with a minor tweak when the polynomial has a multiple-root). In this case, the characteristic polynomial is $x^2-1$ with roots $\pm 1$, and Alpha writes its solution in this form; $c_11^x+c_2(-1)^x$, and of course the first term is just $c_1$. For more information on this sort of thing, check out the 'homogeneous linear' section of Wikipedia's recurrence relations page.