Editing Law of sines (section)

=== Vector proof ===
Consider a unit sphere with three unit vectors {{math|'''OA'''}}, {{math|'''OB'''}} and {{math|'''OC'''}} drawn from the origin to the vertices of the triangle. Thus the angles {{math|''α''}}, {{math|''β''}}, and {{math|''γ''}} are the angles {{math|''a''}}, {{math|''b''}}, and {{math|''c''}}, respectively. The arc {{math|BC}} subtends an angle of magnitude {{math|''a''}} at the centre. Introduce a Cartesian basis with {{math|'''OA'''}} along the {{math|''z''}}-axis and {{math|'''OB'''}} in the {{math|''xz''}}-plane making an angle {{math|''c''}} with the {{math|''z''}}-axis. The vector {{math|'''OC'''}} projects to {{math|ON}} in the {{math|''xy''}}-plane and the angle between {{math|ON}} and the {{math|''x''}}-axis is {{math|''A''}}. Therefore, the three vectors have components:
<math display="block">\mathbf{OA} = \begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}, \quad
\mathbf{OB} = \begin{pmatrix}\sin c \\ 0 \\ \cos c\end{pmatrix}, \quad
\mathbf{OC} = \begin{pmatrix}\sin b\cos A \\ \sin b\sin A \\ \cos b\end{pmatrix}.</math>

The [[scalar triple product]], {{math|'''OA''' ⋅ ('''OB''' × '''OC''')}} is the volume of the [[parallelepiped]] formed by the position vectors of the vertices of the spherical triangle {{math|'''OA'''}}, {{math|'''OB'''}} and {{math|'''OC'''}}. This volume is invariant to the specific coordinate system used to represent {{math|'''OA'''}}, {{math|'''OB'''}} and {{math|'''OC'''}}. The value of the [[scalar triple product]] {{math|'''OA''' ⋅ ('''OB''' × '''OC''')}} is the {{math|3 × 3}} determinant with {{math|'''OA'''}}, {{math|'''OB'''}} and {{math|'''OC'''}} as its rows. With the {{math|''z''}}-axis along {{math|'''OA'''}} the square of this determinant is
<math display="block"> \begin{align}
\bigl(\mathbf{OA} \cdot (\mathbf{OB} \times \mathbf{OC})\bigr)^2
& = \left(\det \begin{pmatrix}\mathbf{OA} & \mathbf{OB} & \mathbf{OC}\end{pmatrix}\right)^2 \\[4pt]
& =  \begin{vmatrix}
0 & 0 & 1 \\   
\sin c & 0 & \cos c \\ 
\sin b \cos A & \sin b \sin A & \cos b 
\end{vmatrix} ^2
= \left(\sin b \sin c \sin A\right)^2.
\end{align}</math>
Repeating this calculation with the {{math|''z''}}-axis along {{math|'''OB'''}} gives {{math|(sin ''c'' sin ''a'' sin ''B'')<sup>2</sup>}}, while with the {{math|''z''}}-axis along {{math|'''OC'''}} it is {{math|(sin ''a'' sin ''b'' sin ''C'')<sup>2</sup>}}. Equating these expressions and dividing throughout by {{math|(sin ''a'' sin ''b'' sin ''c'')<sup>2</sup>}} gives
<math display="block">
\frac{\sin^2 A}{\sin^2 a}
= \frac{\sin^2 B}{\sin^2 b}
= \frac{\sin^2 C}{\sin^2 c}
= \frac{V^2}{\sin^2 (a) \sin^2 (b) \sin^2 (c)},
</math>
where {{mvar|V}} is the volume of the [[parallelepiped]] formed by the position vector of the vertices of the spherical triangle. Consequently, the result follows.

It is easy to see how for small spherical triangles, when the radius of the sphere is much greater than the sides of the triangle, this formula becomes the planar formula at the limit, since
<math display="block">\lim_{a \to 0} \frac{\sin a}{a} = 1</math>
and the same for {{math|sin ''b''}} and {{math|sin ''c''}}.

[[File:Sine law spherical small.svg|thumb|378x378px]]