Editing Law of sines (section)

==Spherical law of sines==

The spherical law of sines deals with triangles on a sphere, whose sides are arcs of [[great circle]]s.

Suppose the radius of the sphere is&nbsp;1. Let {{math|''a''}}, {{math|''b''}}, and {{math|''c''}} be the lengths of the great-arcs that are the sides of the triangle. Because it is a unit sphere, {{math|''a''}}, {{math|''b''}}, and {{math|''c''}} are the angles at the center of the sphere subtended by those arcs, in radians. Let {{math|''A''}}, {{math|''B''}}, and {{math|''C''}} be the angles opposite those respective sides. These are [[dihedral angle]]s between the planes of the three great circles.

Then the spherical law of sines says:
<math display="block">\frac{\sin A}{\sin a} = \frac{\sin B}{\sin b} = \frac{\sin C}{\sin c}.</math>

[[File:Spherical trigonometry vectors.svg|thumb|right|200px]]

=== 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]]

=== Geometric proof ===
Consider a unit sphere with:
<math display="block">OA = OB = OC = 1</math>

Construct point <math>D</math> and point <math>E</math> such that <math>\angle ADO = \angle AEO = 90^\circ</math>

Construct point <math>A'</math> such that <math>\angle A'DO = \angle A'EO = 90^\circ</math>

It can therefore be seen that <math>\angle ADA' = B</math> and <math>\angle AEA' = C</math>

Notice that <math>A'</math> is the projection of <math>A</math> on plane <math>OBC</math>. Therefore <math>\angle AA'D = \angle AA'E = 90^\circ</math>

By basic trigonometry, we have:
<math display="block">\begin{align}
  AD &= \sin c \\
  AE &= \sin b
\end{align}</math>

But <math>AA' = AD \sin B = AE \sin C </math>

Combining them we have:
<math display="block">\begin{align}
\sin c \sin B &= \sin b \sin C \\
\Rightarrow \frac{\sin B}{\sin b} &=\frac{\sin C}{\sin c}
\end{align}</math>

By applying similar reasoning, we obtain the spherical law of sines:
<math display="block">\frac{\sin A}{\sin a} =\frac{\sin B}{\sin b} =\frac{\sin C}{\sin c} </math>

{{see also|Spherical trigonometry|Spherical law of cosines|Half-side formula}}

=== Other proofs ===
A purely algebraic proof can be constructed from the [[spherical law of cosines]]. From the identity <math>\sin^2 A = 1 - \cos^2 A</math> and the explicit expression for <math>\cos A</math> from the spherical law of cosines
<math display="block">\begin{align}
   \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}\\[8pt]
 \frac{\sin A}{\sin a}
 &= \frac{\left[1-\cos^2\!a-\cos^2\!b-\cos^2\!c + 2\cos a\cos b\cos c\right]^{1/2}}{\sin a\sin b\sin c}.
\end{align}</math>
Since the right hand side is invariant under a cyclic permutation of <math>a,\;b,\;c</math> the spherical sine rule follows immediately.

The figure used in the Geometric proof above is used by and also provided in Banerjee<ref name="banerjee">{{Citation | last = Banerjee | first = Sudipto | date = 2004 | title = Revisiting Spherical Trigonometry with Orthogonal Projectors | journal = The College Mathematics Journal | volume = 35 | issue = 5 | pages = 375–381 | publisher = Mathematical Association of America| doi = 10.1080/07468342.2004.11922099 | url = http://www.biostat.umn.edu/~sudiptob/ResearchPapers/banerjee.pdf | archive-url = https://web.archive.org/web/20041029141245id_/http://www.biostat.umn.edu/~sudiptob/ResearchPapers/banerjee.pdf | url-status = dead | archive-date = 2004-10-29 }}</ref> (see Figure 3 in this paper) to derive the sine law using elementary linear algebra and projection matrices.