Formula for Determinant of Vectors given in spherical coordinates In 2D, one has an easy formula for the determinant of two vectors given in spherical coordinates, i.e.
$\begin{vmatrix} \cos(\phi_1) &\cos(\phi_2)\\ \sin(\phi_1) &\sin(\phi_2)\end{vmatrix}
=\sin(\phi_2-\phi_1)$.
I'm curious if there are similar formulae for higher dimensions, e.g. for 
$\begin{vmatrix}\sin(\theta_1)\sin(\phi_1)  &\sin(\theta_2)\sin(\phi_2) &\sin(\theta_3)\sin(\phi_3)\\ \sin(\theta_1)\cos(\phi_1)  &\sin(\theta_2)\cos(\phi_2) &\sin(\theta_3)\cos(\phi_3)\\ \cos(\theta_1)&\cos(\theta_2) &\cos(\theta_3)\end{vmatrix}=?$
 A: A rigid body rotation does not affect determinant in any dimension. A rigid body rotation is implemented as an orthogonal matrix with determinant +1 instead of -1. So you can reduce the $\mathbb{R}^{n}$ case to $\mathbb{R}^{n-1}$ by rotating until the first column of your matrix is along the first standard basis direction. Then the determinant is not changed by projecting all of the remaining vectors into the subspace which is orthogonal to the first element, because that's the same as subtracting a constant multiple of the new first column from all of the other columns. It is possible to hit a $0$ projection vector in this process, depending on the independence of the original unit vectors. Otherwise, projecting gives you a product of $n-1$ cosines that you can factor out, and the resulting matrix has a single $1$ in the corner and only $0$ terms in the same row and column as this $1$, and the $(n-1)\times (n-1)$ submatrix now contains unit vectors because of factoring out the cosine terms. This is a full reduction to the $\mathbb{R}^{n-1}$ case. Eventually you get down to the $\mathbb{R}^{2}$ case, or the algorithm stops because of linear dependence.
