First proof
Let u, v, and w denote the unit vectors from the center of the sphere to those corners of the triangle. The angles and distances do not change if the coordinate system is rotated, so we can rotate the coordinate system so that is at the north pole and is somewhere on the prime meridian (longitude of 0). With this rotation, the spherical coordinates for are where θ is the angle measured from the north pole not from the equator, and the spherical coordinates for are The Cartesian coordinates for are and the Cartesian coordinates for are The value of is the dot product of the two Cartesian vectors, which is
Second proof
Let u, v, and w denote the unit vectors from the center of the sphere to those corners of the triangle. We have u · u = 1, v · w = cos c, u · v = cos a, and u · w = cos b. The vectors u × v and u × w have lengths sin a and sin b respectively and the angle between them is C, so
- sin a sin b cos C = (u × v) · (u × w) = (u · u)(v · w) − (u · v)(u · w) = cos c − cos a cos b,
using cross products, dot products, and the Binet–Cauchy identity (p × q) · (r × s) = (p · r)(q · s) − (p · s)(q · r).
Third proof
Let u, v, and w denote the unit vectors from the center of the sphere to those corners of the triangle. Consider the following rotational sequence where we first rotate the vector v to u by an angle followed by another rotation of vector u to w by an angle after which we rotate the vector w back to v by an angle The composition of these three rotations will form an identity transform.[clarification needed] That is, the composite rotation maps the point v to itself. These three rotational operations can be represented by quaternions:
where and are the unit vectors representing the axes of rotations, as defined by the right-hand rule, respectively. The composition of these three rotations is unity, Right multiplying both sides by conjugates we have where and This gives us the identity[5][6]
The quaternion product on the right-hand side of this identity is given by
Equating the scalar parts on both sides of the identity, we have
Here Since this identity is valid for any arc angles, suppressing the halves, we have
We can also recover the sine law by first noting that and then equating the vector parts on both sides of the identity as
The vector is orthogonal to both the vectors and and as such Taking dot product with respect to on both sides, and suppressing the halves, we have Now and so we have Dividing each side by we have
Since the right-hand side of the above expression is unchanged by cyclic permutation, we have