Using Lagrange multipliers, and setting up the initial function:
$\Lambda(x, y, z, \lambda)=\sin(x)+\sin(y)+\sin(z)+\lambda(x+y+z-\pi)$
Setting the gradient $\nabla\Lambda=0$, we can write:
$\frac{\partial\Lambda}{\partial x}=\cos(x)+\lambda=0$ $\frac{\partial\Lambda}{\partial y}=\cos(y)+\lambda=0$ $\frac{\partial\Lambda}{\partial z}=\cos(z)+\lambda=0$ $\frac{\partial\Lambda}{\partial \lambda}=x+y+z-\pi=0$
The original equations give us $x=\arccos(-\lambda), y=\arccos(-\lambda),z=\arccos(-\lambda)$, substituting into the final partial derivative gives us:
$3\arccos(-\lambda)-\pi=0\implies-\lambda=\cos(\frac{\pi}{3})$
We can then use this to find the critical points, which are when $x=y=z=\frac{\pi}{3}$.
Plugging this value into our initial equation gives us the maximum value: $3\sin(\frac{\pi}{3})=\frac{3\sqrt{3}}{2}$.
In order to find the minimum, we note that any of the infinite number of $\cos^{-1}(-\lambda)$ are critical points, providing $x+y+x=\pi$. As we are looking to now minimize our solution, it makes sense to maximize one solution and minimize two others.
We can do this by observing that $\cos^{-1}(-\frac{1}{2})=2\pi-\frac{\pi}{3}$, and $\cos^{-1}(-\frac{1}{2})=-\frac{\pi}{3}$, plugging these values into our equation gives us:
$\sin(2\pi-\frac{\pi}{3})+2\sin(-\frac{\pi}{3})=-\frac{3\sqrt{3}}{2}$
Minimizing our objective function.