Trigonometry of a tetrahedron

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.

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.

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.

Putting any of the four vertices in the role of O yields four such identities, but at most three of them are independent; if the "clockwise" sides of three of the four identities are multiplied and the product is inferred to be equal to the product of the "counterclockwise" sides of the same three identities, and then common factors are cancelled from both sides, the result is the fourth identity.

Three angles are the angles of some triangle if and only if their sum is 180° (π radians). What condition on 12 angles is necessary and sufficient for them to be the 12 angles of some tetrahedron? Clearly the sum of the angles of any side of the tetrahedron must be 180°. Since there are four such triangles, there are four such constraints on sums of angles, and the number of degrees of freedom is thereby reduced from 12 to 8. The four relations given by the sine law further reduce the number of degrees of freedom, from 8 down to not 4 but 5, since the fourth constraint is not independent of the first three. Thus the space of all shapes of tetrahedra is 5-dimensional.^[3]

See: Law of sines

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:

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

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

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.