# Spherical law of cosines

In spherical trigonometry, the law of cosines (also called the cosine rule for sides[1]) is a theorem relating the sides and angles of spherical triangles, analogous to the ordinary law of cosines from plane trigonometry.

Spherical triangle solved by the law of cosines.

Given a unit sphere, a "spherical triangle" on the surface of the sphere is defined by the great circles connecting three points u, v, and w on the sphere (shown at right). If the lengths of these three sides are a (from u to v), b (from u to w), and c (from v to w), and the angle of the corner opposite c is C, then the (first) spherical law of cosines states:[2][1]

${\displaystyle \cos c=\cos a\cos b+\sin a\sin b\cos C\,.}$

Since this is a unit sphere, the lengths a, b, and c are simply equal to the angles (in radians) subtended by those sides from the center of the sphere. (For a non-unit sphere, the lengths are the subtended angles times the radius, and the formula still holds if a, b and c are reinterpreted as the subtended angles). As a special case, for C = π/2, then cos C = 0, and one obtains the spherical analogue of the Pythagorean theorem:

${\displaystyle \cos c=\cos a\cos b\,.}$

If the law of cosines is used to solve for c, the necessity of inverting the cosine magnifies rounding errors when c is small. In this case, the alternative formulation of the law of haversines is preferable.[3]

A variation on the law of cosines, the second spherical law of cosines,[4] (also called the cosine rule for angles[1]) states:

${\displaystyle \cos C=-\cos A\cos B+\sin A\sin B\cos c\,,}$

where A and B are the angles of the corners opposite to sides a and b, respectively. It can be obtained from consideration of a spherical triangle dual to the given one.

## ProofsEdit

### First proofEdit

A proof of the law of cosines can be constructed as follows.[2] Let u, v, and w denote the unit vectors from the center of the sphere to those corners of the triangle. Then, the lengths (angles) of the sides are given by the dot products:

${\displaystyle \cos a=\mathbf {u} \cdot \mathbf {v} }$
${\displaystyle \cos b=\mathbf {u} \cdot \mathbf {w} }$
${\displaystyle \cos c=\mathbf {v} \cdot \mathbf {w} }$

To get the angle C, we need the tangent vectors ta and tb at u along the directions of sides a and b, respectively. For example, the tangent vector ta is the unit vector perpendicular to u in the u-v plane, whose direction is given by the component of v perpendicular to u. This means:

${\displaystyle \mathbf {t} _{a}={\frac {\mathbf {v} -\mathbf {u} (\mathbf {u} \cdot \mathbf {v} )}{\left\|\mathbf {v} -\mathbf {u} (\mathbf {u} \cdot \mathbf {v} )\right\|}}={\frac {\mathbf {v} -\mathbf {u} \cos a}{\sin a}}}$

where for the denominator we have used the Pythagorean identity sin2 a = 1 − cos2 a and where ${\displaystyle \lVert \cdot \rVert }$  denotes the length of the vector in the denominator. Similarly,

${\displaystyle \mathbf {t} _{b}={\frac {\mathbf {w} -\mathbf {u} \cos b}{\sin b}}.}$

Then, the angle C is given by:

${\displaystyle \cos C=\mathbf {t} _{a}\cdot \mathbf {t} _{b}={\frac {\cos c-\cos a\cos b}{\sin a\sin b}}}$

from which the law of cosines immediately follows.

### Second proofEdit

To the diagram above, add a plane tangent to the sphere at u, and extend radii from the center of the sphere O through v and through w to meet the plane at points y and z. We then have two plane triangles with a side in common: the triangle containing u, y and z and the one containing O, y and z. Sides of the first triangle are tan a and tan b, with angle C between them; sides of the second triangle are sec a and sec b, with angle c between them. By the law of cosines for plane triangles (and remembering that sec2 of any angle is tan2 + 1),

{\displaystyle {\begin{aligned}\tan ^{2}a+\tan ^{2}b-2\tan a\tan b\cos C&=\sec ^{2}a+\sec ^{2}b-2\sec a\sec b\cos c\\[4pt]&=2+\tan ^{2}a+\tan ^{2}b-2\sec a\sec b\cos c.\end{aligned}}}

So

${\displaystyle -\tan a\tan b\cos C=1-\sec a\sec b\cos c}$

Multiply both sides by cos a cos b and rearrange.

### Third proofEdit

The angles and distances do not change if the sphere is rotated, so we can rotate the sphere so that ${\displaystyle \mathbf {u} }$  is at the north pole and ${\displaystyle \mathbf {v} }$  is somewhere on the prime meridian (longitude of 0). With this rotation, the spherical coordinates for ${\displaystyle \mathbf {v} }$  are ${\displaystyle (r,\theta ,\phi )=(1,a,0)}$  and the spherical coordinates for ${\displaystyle \mathbf {w} }$  are ${\displaystyle (r,\theta ,\phi )=(1,b,C)}$ . The Cartesian coordinates for ${\displaystyle \mathbf {v} }$  are ${\displaystyle (x,y,z)=(\sin a,0,\cos a)}$  and the Cartesian coordinates for ${\displaystyle \mathbf {w} }$  are ${\displaystyle (x,y,z)=(\sin b\cos C,\sin b\sin C,\cos b)}$ . The value of ${\displaystyle \cos c}$  is the dot product of the two Cartesian vectors, which is ${\displaystyle \sin a\sin b\cos C+\cos a\cos b}$ .

### Fourth proofEdit

The vectors ${\displaystyle \mathbf {u} \times \mathbf {v} }$  and ${\displaystyle \mathbf {u} \times \mathbf {w} }$  have lengths sin a and sin b respectively and the angle between them is C, so

${\displaystyle \sin a\sin b\cos C=(\mathbf {u} \times \mathbf {v} )\cdot (\mathbf {u} \times \mathbf {w} )=(\mathbf {u} \cdot \mathbf {u} )(\mathbf {v} \cdot \mathbf {w} )-(\mathbf {u} \cdot \mathbf {v} )(\mathbf {u} \cdot \mathbf {w} )=\cos c-\cos a\cos b\,,}$

using u · u = 1, u · v = cos a, u · w = cos b, v · w = cos c, cross products, and the Binet–Cauchy identity (p × q) · (r × s) = (p · r)(q · s) − (p · s)(q · r).

## RearrangementsEdit

The first and second spherical laws of cosines can be rearranged to put the sides (a, b, c) and angles (A, B, C) on opposite sides of the equations:

{\displaystyle {\begin{aligned}\cos C&={\frac {\cos c-\cos a\cos b}{\sin a\sin b}}\\\cos c&={\frac {\cos C+\cos A\cos B}{\sin A\sin B}}\end{aligned}}}

## Planar limit: small anglesEdit

For small spherical triangles, i.e. for small a, b, and c, the spherical law of cosines is approximately the same as the ordinary planar law of cosines,

${\displaystyle c^{2}\approx a^{2}+b^{2}-2ab\cos C\,.}$

To prove this, we will use the small-angle approximation obtained from the Maclaurin series for the cosine and sine functions:

${\displaystyle \cos a=1-{\frac {a^{2}}{2}}+O(a^{4}),\,\sin a=a+O(a^{3})}$

Substituting these expressions into the spherical law of cosines nets:

${\displaystyle 1-{\frac {c^{2}}{2}}+O(c^{4})=1-{\frac {a^{2}}{2}}-{\frac {b^{2}}{2}}+{\frac {a^{2}b^{2}}{4}}+O(a^{4})+O(b^{4})+\cos(C)(ab+O(a^{3}b)+O(ab^{3})+O(a^{3}b^{3}))}$

or after simplifying:

${\displaystyle c^{2}=a^{2}+b^{2}-2ab\cos C+O(c^{4})+O(a^{4})+O(b^{4})+O(a^{2}b^{2})+O(a^{3}b)+O(ab^{3})+O(a^{3}b^{3}).}$

Remembering the properties of big O notation, we can discard summands where the lowest exponent for a or b is greater than 1, so finally, the error in this approximation is:

${\displaystyle O(c^{4})+O(a^{3}b)+O(ab^{3}).\,\!}$