Deriving the Sine Formula for spherical trigonometry without using the Cosine Formula Is it possible to derive the sine formula for spherical triangle without the use of the cosine formula ? 
Every book on spherical trigonometry derives it from the cosine formula. 
Kindly provide any source to the proof without the use of cosine formula or any idea about how to proceed. 
 A: Let $O$ be the center of the sphere, say, with radius $1$, and consider the tetrahedron $OABC$. Sides $a$, $b$, $c$ of the spherical triangle are precisely the face angles $\angle BOC$, $\angle COA$, $\angle AOB$ at vertex $O$, and the spherical angles $A$, $B$, $C$ are the dihedral angles along $\overline{OA}$, $\overline{OB}$, $\overline{OC}$.
Drop perpendiculars from $C$ to points $P$ and $Q$ on $\overline{OA}$ and $\overline{OB}$; since $\overline{CP}$ and $\overline{CQ}$ are altitudes of respective faces $\triangle COA$ and $\triangle COB$, we have
$$\begin{align}
|CP| =|OC|\cdot \sin\angle COA = 1\cdot\sin b \\
|CQ|= |OC|\cdot \sin\angle BOC = 1\cdot \sin a 
\end{align}\tag{1}$$
Now, drop a perpendicular from $C$ to point R in face $AOB$ of the tetrahedron. We have that $\triangle CPR$ has a right angle at $R$; also, because the plane of $\triangle CPR$ is perpendicular to $\overline{OA}$, $\angle CPR$ matches the dihedral angle along $\overline{OA}$. Likewise, $\angle CQR$ matches the dihedral angle along $\overline{OB}$. Therefore,
$$|CP|\sin A = |CR| = |CQ|\sin B \quad\to\quad \sin b \sin A = \sin a\sin B \quad\to\quad \frac{\sin A}{\sin a}=\frac{\sin B}{\sin b}\tag{2}$$
The same kind of argument gives $\sin c \sin A = \sin a \sin C$, so that we may conclude
$$\frac{\sin A}{\sin a} = \frac{\sin B}{\sin b} = \frac{\sin C}{\sin c}\tag{$\star$}$$
A: In Triple product,
the cross product identity states, for any vectors $\,A,B,C,\,$
$$(A \times B)\times(A \times C) = (A \cdot(B \times C))A \tag{1}$$
where the factor of $\,A\,$ on the right side is the scalar triple product.
Three points $\, A,B,C \,$ on the unit sphere with
center at $\,O\,$ and $\, |A| = |B| = |C| = 1 \,$ form a spherical triangle
$\, ABC \,$ with opposite sides $\, a,b,c. \,$
From standard facts about the dot and cross product of two unit vectors,
 $$ \cos(c) =  A\cdot B, \quad
 \sin(c)^2 = |A\times B|^2. \tag{2}$$
The angle $\, C \,$ is determined by the intersection of the
planes given by $\, OAC \,$ and $\, OBC. \,$ That is, by their
normal vectors $\, A\times C\,$ and $\, B\times C.\,$
Using the cross product identity $(1)$ and $(2)$ we get
$$ \sin(C)^2 = \frac{|(A\times C)\times(B\times C)|^2}
{|A\times C|^2 \, |B\times C|^2}
 = \frac{\Delta^2 }{(\sin a \sin b)^2} \tag{3}$$
where $$ \Delta^2 := |C \cdot(A \times B)|^2 =
 (\sin a \sin b \sin C)^2. \tag{4}$$
 The law of sines follows from $(4)$ because
 $\, \Delta^2 \,$ is invariant under any permutation 
of its three operands.
A: Here is a direct proof, using nothing more than rotation matrices in $\mathbb{R}^3$ (you could also dress this up with unit quaternions if you like):-
Assume WLOG $a,b,c,A,B,C\in(0,\pi)$.  Put your spherical triangle ABC with $A$ on the positive $z$-axis, $B$ on the $xz$-plane (or equivalently, choose such coordinates axes).
Consider the following sequence of rotations:


*

*Rotate about $z$-axis by $\pi-A$ brings side $b$ to the $xz$-plane.

*Then rotate about $y$-axis by $b$ brings $C$ to the positive $z$-axis.

*Then rotate about $z$-axis by $\pi-C$ brings side $a$ to $xz$-plane

*Then rotate about $y$-axis by $a$ brings $B$ to the positive $z$-axis

*Then rotate about $z$-axis by $\pi-B$ brings side $c$ to the $xz$-plane

*Finally rotate about $y$-axis by $c$ brings $A$ to the positive $z$-axis.


The net effect is to leave the spherical triangle in its starting position, so it must be the identity on the entire sphere.  If  you write out in the matrix of
$$
R(c,\mathbf{e}_y)R(\pi-B,\mathbf{e}_z)
R(a,\mathbf{e}_y)R(\pi-C,\mathbf{e}_z)
R(b,\mathbf{e}_y)R(\pi-A,\mathbf{e}_z)=I
$$
(where $R(\theta,\mathbf{n})$ is the rotation by $\theta$ about $\mathbf{n}$) and you get as two of the components:
$$
\begin{align*}
\sin(\pi-B)\sin a-\sin(\pi-A)\sin b&=0\\
\sin(\pi-B)\sin c-\sin(\pi-C)\sin b&=0
\end{align*}
$$
So
$$
\frac{\sin(\pi-A)}{\sin a}=\frac{\sin(\pi-B)}{\sin b}=\frac{\sin(\pi-C)}{\sin c}
$$
or equivalently,
$$
\frac{\sin A}{\sin a}=\frac{\sin B}{\sin b}=\frac{\sin C}{\sin c}
$$
