Solid angle between vectors in n-dimensional space There is a formula of to calculate the angle between two normalized vectors:
$$\alpha=\arccos \frac {\vec{a} \cdot\ \vec{b}} {|\vec {a}||\vec {b}|}.$$
The formula of 3D solid angle between three normalized vectors is (borrowed from wikipedia):
$$\alpha = 2\arctan \frac{|\vec {a}\ \vec {b}\ \vec {c}|}{|\vec{a}| |\vec{b}| |\vec{c}| + (\vec{a} \cdot \ \vec{b})|\vec{c}| + (\vec{a} \cdot \vec {c})|\vec{b}| + (\vec{b} \cdot \vec{c})|\vec{a}|}.$$
How to figure out a formula of solid angle between $n$ normalized vectors in $n$-dimensional space?
UPDATE
OK, I've found out the following formula for the angle between two vectors in 2d space through the $\arctan$:
$$\alpha= 2\arctan  \frac {|\vec{a} \wedge \vec {b}|} {|\vec{a}| |\vec{b}| + \vec{a}\ \cdot\ \vec{b}}$$
How can I apply this $\arctan$ formula for the 4D space, for example?
For more detail, I explain my assumptions.
The solid angle of orthogonal basis in 4D space must be $\alpha = \frac {2 \cdot \pi ^ 2 \cdot R^3} {2 ^ 4} = \frac {\pi ^ 2 R^3} {8} $. Here we can see a $\pi^2$ factor. Does this mean that the 4D solid angle formula contains multiplication of two $\arctan$ there is one $\arctan$ in this formula? I think yes, but still have some difficulties with such formula inference.
 A: See the paper by Ribando, "Measuring Solid Angles Beyond Dimension Three", published in Discrete & Computational Geometry 2006. An electronic version may be found here: 
https://link.springer.com/content/pdf/10.1007%2Fs00454-006-1253-4.pdf
It seems there is no closed formula for solid angle in dimension > 3, but a multi-variable Taylor series is given in Theorem 2.2 there. Later in this paper, its radius of convergence is discussed.
A: I actually did a calculation for this a while back, so when I saw this question, I decided to search through my notebook for the answer.
Here’s what I came up with, assuming we’re finding the 4D solid angle of a pentachoron (tetrahedron based pyramid) bounded by vertices A, B, C, D, and the zero vector, this should be the solution (Assuming I got all the Latex right):
$$\frac{\cos^{(-1)}(\hat{A}\cdot \hat{B})\cos^{(-1)}(-\frac{(A\wedge B\wedge C)\cdot(A\wedge B\wedge D)}{|A\wedge B\wedge C||A\wedge B\wedge D|})}{2}-\frac{|A\wedge B\wedge \hat{C}\wedge \hat{D}|}{2}\int_{0}^{\infty}\frac{(A\wedge B)\cdot(B\wedge (\hat{C}+\hat{D}t))\cos^{(-1)}(\frac{\hat{B}\cdot (\hat{C}+\hat{D}t)}{|\hat{C}+\hat{D}t|})}{|B\wedge(\hat{C}+\hat{D}t)||A\wedge B\wedge (\hat{C}+\hat{D}t)|²}-\frac{(A\wedge B)\cdot(A\wedge (\hat{C}+\hat{D}t))\cos^{(-1)}(\frac{\hat{A}\cdot (\hat{C}+\hat{D}t)}{|\hat{C}+\hat{D}t|})}{|A\wedge(\hat{C}+\hat{D}t)||A\wedge B\wedge (\hat{C}+\hat{D}t)|²}dt$$
I had actually spent the longest time without ever finishing this problem, as I had realized that at this point, all I could do was integrate both entries by parts, with the arccosines being the terms that aren’t integrated.  Of course, if I did do that, we’d still have two integrals left, each of which would also be some inverse trig function times some ratio between vector operators, making it completely pointless, unless both resulting integrals end up being equal and opposite.  The only way too test the slim chance of this being to do some seriously difficult integration.  After coming here, based on the comments, I’d assume the answer is no: they don’t cancel out, and this is basically the simplest form I can get it in.
A: The 4-D solid angle shows up in calculations of the probability of certain election results when ranked-choice ballots are used, if one assumes voters think candidates so nearly equal that each voter selects one of the possible ballots at random.  The best reduction of these probabilities I had been able to make was to a double integral, which I evaluated numerically; a general formula, however complicated, without an integral, would be a great simplification, and might be of interest to election theorists other than me, so I encourage Math Machine to pursue one.  For the curious, the election results appear in my papers "The right way to read ranked-choice ballots: not instant runoff, but ranked pairs" available on the tab "The theoretical case from Charles Munger Jr." on the first page at https://www.betterchoices.vote/.  See paper C, Tables V, VI, and VII, and paper D, pp. 8--19 for my approach to these integrals.  At the very least, they would provide to Math Machine a scatter of results already known numerically to 9 digits, against which to test a general formula.
--Charles Munger, Jr.

*

*List item

A: Solid angles are a little more complicated than angles. From wolfram's mathworld:

The solid angle  subtended by a surface  is defined as the surface
  area  of a unit sphere covered by the surface's projection onto the
  sphere.

Did you notice the phrase 'surface area'? We're going to need some integrals! 
If the surface is called $S$, then the solid angle $\Omega$ is given by:
$$\Omega = \int_S \hat{\mathbf{n}}/\mathbf{r}^2 \cdot d \mathbf a $$
where $\hat{\mathbf{n}}$ is the unit vector from the origin, $d \mathbf{a}$ is the infinitesimal area of a path of a surface from $S$, and $r$ is the distance from the origin to the patch of surface.
See here for examples, and more info.
