# 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.

## Proofs

### First proof

Let u, v, and w denote the unit vectors from the center of the sphere to those corners of the triangle. The angles and distances do not change if the coordinate system is rotated, so we can rotate the coordinate system 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 ${\displaystyle (r,\theta ,\phi )=(1,a,0)}$ , where θ is the angle measured from the north pole not from the equator, 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 ${\displaystyle \sin a\sin b\cos C+\cos a\cos b}$ .

### Second proof

Let u, v, and w denote the unit vectors from the center of the sphere to those corners of the triangle. We have u · u = 1, v · w = cos c, u · v = cos a, and u · w = cos b. The vectors u × v and u × w have lengths sin a and sin b respectively and the angle between them is C, so

sin a sin b cos C = (u × v) · (u × w) = (u · u)(v · w) − (u · v)(u · w) = cos c − cos a cos b,

using cross products, dot products, and the Binet–Cauchy identity (p × q) · (r × s) = (p · r)(q · s) − (p · s)(q · r).

### Third proof

Let u, v, and w denote the unit vectors from the center of the sphere to those corners of the triangle. Consider the following rotational sequence where we first rotate the vector v to u by an angle ${\displaystyle a}$ , followed another rotation of vector u to w by an angle ${\displaystyle b}$ , after which we rotate the vector w back to v by an angle ${\displaystyle c}$ . The composition of these three rotations will form an identity transform. That is, the composite rotation maps the point v to itself. These three rotational operations can be represented by quaternions:

{\displaystyle {\begin{aligned}q_{A}&=\cos {\frac {a}{2}}+\mathbf {A} \sin {\frac {a}{2}},\\q_{B}&=\cos {\frac {b}{2}}+\mathbf {B} \sin {\frac {b}{2}},\\q_{C}&=\cos {\frac {c}{2}}+\mathbf {C} \sin {\frac {c}{2}},\end{aligned}}}

where ${\displaystyle \mathbf {A} }$ , ${\displaystyle \mathbf {B} }$ , and ${\displaystyle \mathbf {C} }$  are the unit vectors representing the axes of rotations, as defined by the right-hand rule, respectively. The composition of these three rotations is unity, ${\displaystyle q_{C}q_{B}q_{A}=1.}$  Right multiplying both sides by conjugates ${\displaystyle q_{A}^{*}q_{B}^{*}}$ , we have ${\displaystyle q_{C}=q_{A}^{*}q_{B}^{*}}$ , where ${\displaystyle q_{A}^{*}=\cos {\frac {a}{2}}-\mathbf {A} \sin {\frac {a}{2}}}$  and ${\displaystyle q_{B}^{*}=\cos {\frac {b}{2}}-\mathbf {B} \sin {\frac {b}{2}}}$ . This gives us the identity[5][6]

${\displaystyle \cos {\frac {c}{2}}+\mathbf {C} \sin {\frac {c}{2}}=\left(\cos {\frac {a}{2}}-\mathbf {A} \sin {\frac {a}{2}}\right)\left(\cos {\frac {b}{2}}-\mathbf {B} \sin {\frac {b}{2}}\right).}$

The quaternion product on the right-hand side of this identity is given by

${\displaystyle \left(\cos {\frac {a}{2}}\cos {\frac {b}{2}}-\mathbf {A} \cdot \mathbf {B} \sin {\frac {a}{2}}\sin {\frac {b}{2}}\right)-\left(\mathbf {A} \sin {\frac {a}{2}}\cos {\frac {b}{2}}+\mathbf {B} \cos {\frac {a}{2}}\sin {\frac {b}{2}}-\mathbf {A} \times \mathbf {B} \sin {\frac {a}{2}}\sin {\frac {b}{2}}\right).}$

Equating the scalar parts on both sides of the identity, we have

${\displaystyle \cos {\frac {c}{2}}=\cos {\frac {a}{2}}\cos {\frac {b}{2}}-\mathbf {A} \cdot \mathbf {B} \sin {\frac {a}{2}}\sin {\frac {b}{2}}.}$

Here ${\displaystyle \mathbf {A} \cdot \mathbf {B} =\cos(\pi -C)=-\cos C}$ . Since this identity is valid for any angles, suppressing the halves, we have

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

We can also recover the sine law by first noting that ${\displaystyle \mathbf {A} \times \mathbf {B} =-\mathbf {u} \sin C}$  and then equating the vector parts on both sides of the identity as

${\displaystyle \mathbf {C} \sin {\frac {c}{2}}=-\left(\mathbf {A} \sin {\frac {a}{2}}\cos {\frac {b}{2}}+\mathbf {B} \cos {\frac {a}{2}}\sin {\frac {b}{2}}+\mathbf {u} \sin C\sin {\frac {a}{2}}\sin {\frac {b}{2}}\right).}$

The vector ${\displaystyle \mathbf {u} }$  is orthogonal to both the vectors ${\displaystyle \mathbf {A} }$  and ${\displaystyle \mathbf {B} }$ , and as such ${\displaystyle \mathbf {u} \cdot \mathbf {A} =\mathbf {u} \cdot \mathbf {B} =0}$ . Taking dot product with respect to ${\displaystyle \mathbf {u} }$  on both sides, and suppressing the halves, we have ${\displaystyle \mathbf {u} \cdot \mathbf {C} \sin c=-\sin C\sin a\sin b.}$  Now ${\displaystyle \mathbf {v} \times \mathbf {w} =-\mathbf {C} \sin c}$  and so we have ${\displaystyle \mathbf {u} \cdot (\mathbf {v} \times \mathbf {w} )=-\mathbf {u} \cdot \mathbf {C} \sin c=\sin C\sin a\sin b.}$  Dividing each side by ${\displaystyle \sin a\sin b\sin c}$ , we have

${\displaystyle {\frac {\sin C}{\sin c}}={\frac {\mathbf {u} \cdot (\mathbf {w} \times \mathbf {v} )}{\sin a\sin b\sin c}}.}$

Since the right-hand side of the above expression is unchanged by cyclic permutation, we have

${\displaystyle {\frac {\sin A}{\sin a}}={\frac {\sin B}{\sin b}}={\frac {\sin C}{\sin c}}.}$

## Rearrangements

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 angles

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:

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

Substituting these expressions into the spherical law of cosines nets:

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

or after simplifying:

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

The big O terms for a and b are dominated by O(a4) + O(b4) as a and b get small, so we can write this last expression as:

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