Answer
$\text{curl } \mathbf{F} = \cos(y - z)\mathbf{i} + \cos(z - x)\mathbf{j} + \cos(x - y)\mathbf{k}$
Work Step by Step
$F(x, y, z) = \sin(x - y)\mathbf{i} + \sin(y - z)\mathbf{j} + \sin(z - x)\mathbf{k}$
$\text{curl } \mathbf{F} = \nabla \times \mathbf{F} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
P & Q & R
\end{vmatrix}$
where $P = \sin(x - y)$, $Q = \sin(y - z)$, and $R = \sin(z - x)$.
$\text{curl } \mathbf{F} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
\sin(x - y) & \sin(y - z) & \sin(z - x)
\end{vmatrix}$
$\text{curl } \mathbf{F} = \left( \frac{\partial}{\partial y}(\sin(z - x)) - \frac{\partial}{\partial z}(\sin(y - z)) \right)\mathbf{i} - \left( \frac{\partial}{\partial x}(\sin(z - x)) - \frac{\partial}{\partial z}(\sin(x - y)) \right)\mathbf{j} + \left( \frac{\partial}{\partial x}(\sin(y - z)) - \frac{\partial}{\partial y}(\sin(x - y)) \right)\mathbf{k}$
$= \left( 0 - (-\cos(y - z)) \right)\mathbf{i} - \left( (-\cos(z - x)) - 0 \right)\mathbf{j} + \left( 0 - (-\cos(x - y)) \right)\mathbf{k}$
$= \left( \cos(y - z) \right)\mathbf{i} - \left( -\cos(z - x) \right)\mathbf{j} + \left( \cos(x - y) \right)\mathbf{k}$
$= \cos(y - z)\mathbf{i} + \cos(z - x)\mathbf{j} + \cos(x - y)\mathbf{k}$
