structural:scholar_edition, quality:corruption
Other proofs

A purely algebraic proof can be constructed from the spherical law of cosines. From the identity sin 2 A = 1 − cos 2 A {\displaystyle \sin ^{2}A=1-\cos ^{2}A} and the explicit expression for cos A {\displaystyle \cos A} from the spherical law of cosines sin 2 A = 1 − (cos a − cos b cos c sin b sin c) 2 = (1 − cos 2 b) (1 − cos 2 c) − (cos a − cos b cos c) 2 sin 2 b sin 2 c sin A sin a = 1 / 2 sin a sin b sin c. {\displaystyle {\begin{aligned}\sin ^{2}\!A&=1-\left({\frac {\cos a-\cos b\,\cos c}{\sin b\,\sin c}}\right)^{2}\\&={\frac {\left(1-\cos ^{2}\!b\right)\left(1-\cos ^{2}\!c\right)-\left(\cos a-\cos b\,\cos c\right)^{2}}{\sin ^{2}\!b\,\sin ^{2}\!c}}\\{\frac {\sin A}{\sin a}}&={\frac {\left^{1/2}}{\sin a\sin b\sin c}}.\end{aligned}}} Since the right hand side is invariant under a cyclic permutation of a, b, c {\displaystyle a,\;b,\;c} the spherical sine rule follows immediately.

The figure used in the Geometric proof above is used by and also provided in Banerjee (see Figure 3 in this paper) to derive the sine law using elementary linear algebra and projection matrices.