Classical trigonometric quantities

The following are trigonometric quantities generally associated to a general tetrahedron:

• The 6 edge lengths - associated to the six edges of the tetrahedron.
• The 12 face angles - there are three of them for each of the four faces of the tetrahedron.
• The 6 dihedral angles - associated to the six edges of the tetrahedron, since any two faces of the tetrahedron are connected by an edge.
• The 4 solid angles - associated to each point of the tetrahedron.

Let ${\displaystyle X={\overline {P_{1}P_{2}P_{3}P_{4}}}}$ be a general tetrahedron, where ${\displaystyle P_{1},P_{2},P_{3},P_{4}}$ are arbitrary points in three-dimensional space.

Furthermore, let ${\displaystyle e_{ij}}$ be the edge that joins ${\displaystyle P_{i}}$ and ${\displaystyle P_{j}}$ and let ${\displaystyle F_{i}}$ be the face of the tetrahedron opposite the point ${\displaystyle P_{i}}$; in other words:

• ${\displaystyle e_{ij}={\overline {P_{i}P_{j}}}}$
• ${\displaystyle F_{i}={\overline {P_{j}P_{k}P_{l}}}}$

where ${\displaystyle i,j,k,l\in \{1,2,3,4\}}$ and ${\displaystyle i\neq j\neq k\neq l}$.

Define the following quantities:

• ${\displaystyle d_{ij}}$ = the length of the edge ${\displaystyle e_{ij}}$
• ${\displaystyle \alpha _{i,j}}$ = the face angle at the point ${\displaystyle P_{i}}$ on the face ${\displaystyle F_{j}}$
• ${\displaystyle \theta _{ij}}$ = the dihedral angle between two faces adjacent to the edge ${\displaystyle e_{ij}}$
• ${\displaystyle \Omega _{i}}$ = the solid angle at the point ${\displaystyle P_{i}}$

Area and volume

Let ${\displaystyle \Delta _{i}}$ be the area of the face ${\displaystyle F_{i}}$. Such area may be calculated by Heron's formula (if all three edge lengths are known):

${\displaystyle \Delta _{i}={\sqrt {\frac {(d_{jk}+d_{jl}+d_{kl})(-d_{jk}+d_{jl}+d_{kl})(d_{jk}-d_{jl}+d_{kl})(d_{jk}+d_{jl}-d_{kl})}{16}}}}$

or by the following formula (if an angle and two corresponding edges are known):

${\displaystyle \Delta _{i}={\frac {1}{2}}d_{jk}d_{jl}\sin \alpha _{j,i}}$

Let ${\displaystyle h_{i}}$ be the altitude from the point ${\displaystyle P_{i}}$ to the face ${\displaystyle F_{i}}$. The volume ${\displaystyle V}$ of the tetrahedron ${\displaystyle X}$ is given by the following formula:

$${\displaystyle V={\frac {1}{3}}\Delta _{i}h_{i}}$$

It satisfies the following relation:^[2]

${\displaystyle 288V^{2}={\begin{vmatrix}2Q_{12}&Q_{12}+Q_{13}-Q_{23}&Q_{12}+Q_{14}-Q_{24}\\Q_{12}+Q_{13}-Q_{23}&2Q_{13}&Q_{13}+Q_{14}-Q_{34}\\Q_{12}+Q_{14}-Q_{24}&Q_{13}+Q_{14}-Q_{34}&2Q_{14}\end{vmatrix}}}$

where ${\displaystyle Q_{ij}=d_{ij}^{2}}$ are the quadrances (length squared) of the edges.

Affine triangle

Take the face ${\displaystyle F_{i}}$; the edges will have lengths ${\displaystyle d_{jk},d_{jl},d_{kl}}$ and the respective opposite angles are given by ${\displaystyle \alpha _{l,i},\alpha _{k,i},\alpha _{j,i}}$.

The usual laws for planar trigonometry of a triangle hold for this triangle.

Projective triangle

Consider the projective (spherical) triangle at the point ${\displaystyle P_{i}}$; the vertices of this projective triangle are the three lines that join ${\displaystyle P_{i}}$ with the other three vertices of the tetrahedron. The edges will have spherical lengths ${\displaystyle \alpha _{i,j},\alpha _{i,k},\alpha _{i,l}}$ and the respective opposite spherical angles are given by ${\displaystyle \theta _{ij},\theta _{ik},\theta _{il}}$.

The usual laws for spherical trigonometry hold for this projective triangle.

Alternating sines theorem

Take the tetrahedron ${\displaystyle X}$, and consider the point ${\displaystyle P_{i}}$ as an apex. The Alternating sines theorem is given by the following identity:

$${\displaystyle \sin(\alpha _{j,l})\sin(\alpha _{k,j})\sin(\alpha _{l,k})=\sin(\alpha _{j,k})\sin(\alpha _{k,l})\sin(\alpha _{l,j})}$$

One may view the two sides of this identity as corresponding to clockwise and counterclockwise orientations of the surface.

Law of sines for the tetrahedron

See: Law of sines

Law of cosines for the tetrahedron

The law of cosines for the tetrahedron^[4] relates the areas of each face of the tetrahedron and the dihedral angles about a point. It is given by the following identity:

${\displaystyle \Delta _{i}^{2}=\Delta _{j}^{2}+\Delta _{k}^{2}+\Delta _{l}^{2}-2(\Delta _{j}\Delta _{k}\cos \theta _{il}+\Delta _{j}\Delta _{l}\cos \theta _{ik}+\Delta _{k}\Delta _{l}\cos \theta _{ij})}$

Relationship between dihedral angles of tetrahedron

Take the general tetrahedron ${\displaystyle X}$ and project the faces ${\displaystyle F_{i},F_{j},F_{k}}$ onto the plane with the face ${\displaystyle F_{l}}$. Let ${\displaystyle c_{ij}=\cos \theta _{ij}}$.

Then the area of the face ${\displaystyle F_{l}}$ is given by the sum of the projected areas, as follows:

$${\displaystyle \Delta _{l}=\Delta _{i}c_{jk}+\Delta _{j}c_{ik}+\Delta _{k}c_{ij}}$$

By substitution of ${\displaystyle i,j,k,l\in \{1,2,3,4\}}$ with each of the four faces of the tetrahedron, one obtains the following homogeneous system of linear equations:

$${\displaystyle {\begin{cases}-\Delta _{1}+\Delta _{2}c_{34}+\Delta _{3}c_{24}+\Delta _{4}c_{23}=0\\\Delta _{1}c_{34}-\Delta _{2}+\Delta _{3}c_{14}+\Delta _{4}c_{13}=0\\\Delta _{1}c_{24}+\Delta _{2}c_{14}-\Delta _{3}+\Delta _{4}c_{12}=0\\\Delta _{1}c_{23}+\Delta _{2}c_{13}+\Delta _{3}c_{12}-\Delta _{4}=0\end{cases}}}$$

This homogeneous system will have solutions precisely when:

$${\displaystyle {\begin{vmatrix}-1&c_{34}&c_{24}&c_{23}\\c_{34}&-1&c_{14}&c_{13}\\c_{24}&c_{14}&-1&c_{12}\\c_{23}&c_{13}&c_{12}&-1\end{vmatrix}}=0}$$

By expanding this determinant, one obtains the relationship between the dihedral angles of the tetrahedron,^[1] as follows:

$${\displaystyle 1-\sum _{1\leq i<j\leq 4}c_{ij}^{2}+\sum _{j=2 \atop k\neq l\neq j}^{4}c_{1j}^{2}c_{kl}^{2}=2\left(\sum _{i=1 \atop j\neq k\neq l\neq i}^{4}c_{ij}c_{ik}c_{il}+\sum _{2\leq j<k\leq 4 \atop l\neq j,k}c_{1j}c_{1k}c_{jl}c_{kl}\right)}$$

Skew distances between edges of tetrahedron

Take the general tetrahedron ${\displaystyle X}$ and let ${\displaystyle P_{ij}}$ be the point on the edge ${\displaystyle e_{ij}}$ and ${\displaystyle P_{kl}}$ be the point on the edge ${\displaystyle e_{kl}}$ such that the line segment ${\displaystyle {\overline {P_{ij}P_{kl}}}}$ is perpendicular to both ${\displaystyle e_{ij}}$ & ${\displaystyle e_{kl}}$. Let ${\displaystyle R_{ij}}$ be the length of the line segment ${\displaystyle {\overline {P_{ij}P_{kl}}}}$.

To find ${\displaystyle R_{ij}}$:^[1]

First, construct a line through ${\displaystyle P_{k}}$ parallel to ${\displaystyle e_{il}}$ and another line through ${\displaystyle P_{i}}$ parallel to ${\displaystyle e_{kl}}$. Let ${\displaystyle O}$ be the intersection of these two lines. Join the points ${\displaystyle O}$ and ${\displaystyle P_{j}}$. By construction, ${\displaystyle {\overline {OP_{i}P_{l}P_{k}}}}$ is a parallelogram and thus ${\displaystyle {\overline {OP_{k}P_{i}}}}$ and ${\displaystyle {\overline {OP_{l}P_{i}}}}$ are congruent triangles. Thus, the tetrahedron ${\displaystyle X}$ and ${\displaystyle Y={\overline {OP_{i}P_{j}P_{k}}}}$ are equal in volume.

As a consequence, the quantity ${\displaystyle R_{ij}}$ is equal to the altitude from the point ${\displaystyle P_{k}}$ to the face ${\displaystyle {\overline {OP_{i}P_{j}}}}$ of the tetrahedron ${\displaystyle Y}$; this is shown by translation of the line segment ${\displaystyle {\overline {P_{ij}P_{kl}}}}$.

By the volume formula, the tetrahedron ${\displaystyle Y}$ satisfies the following relation:

$${\displaystyle 3V=R_{ij}\times \Delta ({\overline {OP_{i}P_{j}}})}$$

where ${\displaystyle \Delta ({\overline {OP_{i}P_{j}}})}$ is the area of the triangle ${\displaystyle {\overline {OP_{i}P_{j}}}}$. Since the length of the line segment ${\displaystyle {\overline {OP_{i}}}}$ is equal to ${\displaystyle d_{kl}}$ (as ${\displaystyle {\overline {OP_{i}P_{l}P_{k}}}}$ is a parallelogram):

$${\displaystyle \Delta ({\overline {OP_{i}P_{j}}})={\frac {1}{2}}d_{ij}d_{kl}\sin \lambda }$$

where ${\displaystyle \lambda =\angle OP_{i}P_{j}}$. Thus, the previous relation becomes:

$${\displaystyle 6V=R_{ij}d_{ij}d_{kl}\sin \lambda }$$

To obtain ${\displaystyle \sin \lambda }$, consider two spherical triangles:

1. Take the spherical triangle of the tetrahedron ${\displaystyle X}$ at the point ${\displaystyle P_{i}}$; it will have sides ${\displaystyle \alpha _{i,j},\alpha _{i,k},\alpha _{i,l}}$ and opposite angles ${\displaystyle \theta _{ij},\theta _{ik},\theta _{il}}$. By the spherical law of cosines:
   $${\displaystyle \cos \alpha _{i,k}=\cos \alpha _{i,j}\cos \alpha _{i,l}+\sin \alpha _{i,j}\sin \alpha _{i,l}\cos \theta _{ik}}$$

   
2. Take the spherical triangle of the tetrahedron ${\displaystyle X}$ at the point ${\displaystyle P_{i}}$. The sides are given by ${\displaystyle \alpha _{i,l},\alpha _{k,j},\lambda }$ and the only known opposite angle is that of ${\displaystyle \lambda }$, given by ${\displaystyle \pi -\theta _{ik}}$. By the spherical law of cosines:
   $${\displaystyle \cos \lambda =\cos \alpha _{i,l}\cos \alpha _{k,j}-\sin \alpha _{i,l}\sin \alpha _{k,j}\cos \theta _{ik}}$$

   

Combining the two equations gives the following result:

$${\displaystyle \cos \alpha _{i,k}\sin \alpha _{k,j}+\cos \lambda \sin \alpha _{i,j}=\cos \alpha _{i,l}\left(\cos \alpha _{i,j}\sin \alpha _{k,j}+\sin \alpha _{i,j}\cos \alpha _{k,j}\right)=\cos \alpha _{i,l}\sin \alpha _{l,j}}$$

Making ${\displaystyle \cos \lambda }$ the subject:

$${\displaystyle \cos \lambda =\cos \alpha _{i,l}{\frac {\sin \alpha _{l,j}}{\sin \alpha _{i,j}}}-\cos \alpha _{i,k}{\frac {\sin \alpha _{k,j}}{\sin \alpha _{i,j}}}}$$

Thus, using the cosine law and some basic trigonometry:

$${\displaystyle \cos \lambda ={\frac {d_{ij}^{2}+d_{ik}^{2}-d_{jk}^{2}}{2d_{ij}d_{ik}}}{\frac {d_{ik}}{d_{kl}}}-{\frac {d_{ij}^{2}+d_{il}^{2}-d_{jl}^{2}}{2d_{ij}d_{il}}}{\frac {d_{il}}{d_{kl}}}={\frac {d_{ik}^{2}+d_{jl}^{2}-d_{il}^{2}-d_{jk}^{2}}{2d_{ij}d_{kl}}}}$$

Thus:

$${\displaystyle \sin \lambda ={\sqrt {1-\left({\frac {d_{ik}^{2}+d_{jl}^{2}-d_{il}^{2}-d_{jk}^{2}}{2d_{ij}d_{kl}}}\right)^{2}}}={\frac {\sqrt {4d_{ij}^{2}d_{kl}^{2}-(d_{ik}^{2}+d_{jl}^{2}-d_{il}^{2}-d_{jk}^{2})^{2}}}{2d_{ij}d_{kl}}}}$$

So:

$${\displaystyle R_{ij}={\frac {12V}{\sqrt {4d_{ij}^{2}d_{kl}^{2}-(d_{ik}^{2}+d_{jl}^{2}-d_{il}^{2}-d_{jk}^{2})^{2}}}}}$$

${\displaystyle R_{ik}}$ and ${\displaystyle R_{il}}$ are obtained by permutation of the edge lengths.

Note that the denominator is a re-formulation of the Bretschneider-von Staudt formula, which evaluates the area of a general convex quadrilateral.