$$S_1 = \sum_{k=1}^n (\sin \left[~ p(k) + \cos p(k)~ \right])$$
I wonder why this appears to give $$\frac{1}{n}S_2\sim 1/2 $$
Thanks for any insights or references.
Edited in light of comments--some code and a picture of $S_2.$
$S_2$= Table[Sum[Sin[ Prime[i] + Cos[ Prime[i]]], {i, 1, j}], {j, 1, 300}]]
This appears to be a pattern for all positive integers, not something special about primes. The average of $\sin(k+\cos k)$ over $1\le k\le n$ appears to converge to $\approx 0.44$. Here is a list plot of these averages for $ 200\le n\le 2000$:
Your sequence converges to $$J_1(1) = \frac{1}{2\pi} \int_0^{2\pi} \sin(t+\cos t) \, \mathrm{d} t \approx 0.440051. $$ The series expansion of the Bessel function gives an alternative expression for this constant: $$J_1(1) = \sum_{m=0}^\infty \frac{(-1)^m}{m!(m+1)!2^{2m+1}}.$$ This expression easily implies that $J_1(1)$ is irrational.
Consider first the simpler sequence $$ A_n = \frac{1}{n} \sum_{i=1}^n \sin(i+\cos i). $$ The function $f(t) = \sin(t+\cos t)$ is periodic with period $2\pi$. The equidistribution theorem implies that $i \pmod{2\pi}$ is asymptotically equidistributed in $[0,2\pi)$, and so $$ A_n \to \frac{1}{2\pi} \int_0^{2\pi} \sin(t+\cos t) \, \mathrm{d} t. $$
Next, your prime sequence $$ B_n = \frac{1}{n} \sum_{i=1}^n \sin(p_i+\cos p_i), $$ where $p_i$ is the $i$th prime. A deep theorem of Vinogradov implies that $p_i \pmod{2\pi}$ is asymptotically equidistributed in $[0,2\pi)$, and we get the same result.