I'm trying to show that:
$$(\nabla \times n)^2=(n \cdot \nabla \times n)^2+(n \times \nabla \times n)^2$$
I know that $n$ is a unit vector and that $-n$ is effectively the same as $n$ for the situation I'm dealing with. I've tried total expansion but nothing cancels out. Through this I've isolated the terms I need for the equality, but there are extra terms associated with each of those. Are there any formulas I can use for curl, cross product, etc? Or is there some trick I can use here?
Let $\nabla \times n=v$ and recall that $|n\cdot v|=|n||v|\cos\theta$, while $|n\times v|=|n||v|\sin\theta$ (where $\theta$ is the angle between $n$ and $v$). Then your equality immediately follows.