Without loss of generality each of $x$, $y$, $z$ are $\le\pi/4$, since if for example $x\gt\pi/4$, then we can replace $x$ with $\pi/2-x$ which will not change the set $\{\cos(x),\sin(x)\}$.
So now $0\lt x \lt y \lt z \le \pi/4$, and we have the following ordering:
$\sin(x)\lt\sin(y)\lt\sin(z)\lt\cos(z)\lt\cos(y)\lt\cos(x)$
Hence the only possible pairings must be:
$A=\sin(x)+\cos(x)$
$B=\sin(y)+\cos(y)$
$C=\sin(z)+\cos(z)$
With $A=B=C$
(any other pairing will make one side heavier, term by term).
Now note $f(t)=\sin(t)+\cos(t)$ is a concave function since $\frac{d^2f(t)}{dt^2}=-\sin(t)-\cos(t)<0$ for $t \in (0,\pi/4)$.
So it follows that $B\gt \frac{A+C}{2}$.
So they can never be equal.
Edit: Fixed glitch spotted by Tom