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I learned from Wolfram MathWorld about hypersine, as being a dimensional analog trig function for hypersolid angles. There it is being defined by

The hypersine ($n$-dimensional sine function) is a function of a vertex angle of an $n$-dimensional parallelotope or simplex. If the content of the parallelotope is $P$ and the contents of the $n$ facets of the parallelotope that meet at vertex $v_0$ are $P_k$, then the value of the $n$-dimensional sine of that vertex is $$\sin(v_0)=\frac {P^{n-1}}{\prod^n_{k=1}P_k}$$ […] The vertex simplex of a vertex of an $n$-dimensional parallelotope is the simplex that has as its vertices that vertex and the $n$ adjacent vertices of the parallelotope. Its content (and the content of all the other vertex simplices) is the content of the parallelotope divided by $n!$. If the content of an $n$-dimensional simplex is $S$, and the contents of the $n$ facets that meet at vertex $v_0$ are $S_1, S_2,..., S_n$, the simplex can be considered a vertex simplex of a parallelotope, and the facets also vertex simplices of the facets of the parallelotope with respect to the same vertex. Substituting in [the above] equation gives $$\sin(v_0)=\frac {(n!\ S)^{n-1}}{\prod^n_{k=1}(n-1)!\ S_k}$$

Confering the first defining equation to the area formula of a parallelogram, it becomes obvious that this definition is nothing but the usual sine function for $n=2$.

This article moreover continues on how to calculate the latter formula when knowing the respective dihedral angles $\alpha_{jk}$ between $S_j$ and $S_k$.

Applying this makes it easy to get the hypersine of the corner angle of any hypercube generally: $$\sin(v_0)=1$$

Even the hypersine of the corner angle of any regular simplex could be calculated to: $$\sin^2(v_0)=\frac{(n+1)^{n-1}}{n^n}=\frac1{n+1}\ \left(1+\frac1n\right)^n$$

But in order to calculate correspondingly the hypersine of the corner angle of the orthoplex one needs to extrapolate this function to non-simplicial corners as well. One clearly could easily subdivide the orthoplex corner symmetrically into $2^{n-1}$ equal (mostly right-angled) simplices, but then again one needs to know about addition theorems for this hypersine function.

Is there anything being known in this direction or could anything be derived for that purpose? - I would be very grateful to learn about it.

--- rk

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    $\begingroup$ That's not Wolfram Alpha, that is Wolfram Mathworld. $\endgroup$ Commented Jan 11, 2020 at 11:53

2 Answers 2

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This is the only reference found by MathSciNet:

Lerman, Gilad; Whitehouse, J. Tyler, On (d)-dimensional (d)-semimetrics and simplex-type inequalities for high-dimensional sine functions, J. Approx. Theory 156, No. 1, 52-81 (2009). ZBL1170.46025.

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From the paper mentioned in the other answer it looks to me, that addition theorems are kind out of near reach.

But at least we could come up with a result for the crosspolytopes (orthoplexes) none the less. This can be done by using the already mentioned dissection of the vertex corner angle into according subsimplices: Just consider $x3o3o...o3o4o$ (orthoplex) as being the vertex figure of $x4o3o3o...o3o4o$ (hypercubical honeycomb). That is, each looked for subsimplex is nothing but the vertex corner pyramid of an hypercube $-$ just that the vertex of consideration, i.e. which contributes to the vertex corner of the orthoplex, would be one of its base vertices.

Thus the contributing subsimplex corner angle could be derived as $$\sin_n^2(v_{0,\ subsimplex})=\left|\begin{array}{ccccc} 1 & 0 & … & 0 & -\frac1{\sqrt n}\\ 0 & 1 & … & 0 & -\frac1{\sqrt n}\\ … & … & … & … & …\\ 0 & 0 & … & 1 & -\frac1{\sqrt n}\\ -\frac1{\sqrt n} & -\frac1{\sqrt n} & … & -\frac1{\sqrt n} & 1 \end{array}\right| = \frac1n$$ And therefrom we clearly get at least $$\begin{array}{rcl} v_{0,\ orthoplex} & = &2^{n-1}\cdot v_{0,\ subsimplex}=\\ & = & 2^{n-1}\ \arcsin_n(\frac1{\sqrt n}) \end{array}$$ (Note that I attached the index $n$ to both the sine and the inverse sine functions in order to remind that we are dealing with hypersine instead of usual 2D sine only.)

--- rk

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