Spherical Law Of Cosines

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

Given a unit sphere, a "spherical triangle" on the surface of the sphere is defined by the great circles connecting three points , and on the sphere (shown at right). If the lengths of these three sides are (from to (from to ), and (from to ), and the angle of the corner opposite is , then the (first) spherical law of cosines states:Romuald Ireneus 'Scibor-Marchocki
Spherical trigonometry
''Elementary-Geometry Trigonometry'' web page (1997).W. Gellert, S. Gottwald, M. Hellwich, H. Kästner, and H. Küstner, ''The VNR Concise Encyclopedia of Mathematics'', 2nd ed., ch. 12 (Van Nostrand Reinhold: New York, 1989).

:$\cos c = \cos a \cos b + \sin a \sin b \cos C\,$

Since this is a unit sphere, the lengths , and 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 and are reinterpreted as the subtended angles). As a special case, for , then , and one obtains the spherical analogue of the Pythagorean theorem:

:$\cos c = \cos a \cos b\,$

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

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

:$\cos C = -\cos A \cos B + \sin A \sin B \cos c\,$

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

Proofs

First proof

Let , and 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 $\mathbf$ is at the north pole and $\mathbf$ is somewhere on the prime meridian (longitude of 0). With this rotation, the spherical coordinates for $\mathbf$ are $(r, \theta, \phi) = (1, a, 0)$, where ''θ'' is the angle measured from the north pole not from the equator, and the spherical coordinates for $\mathbf$ are $(r, \theta, \phi) = (1, b, C)$. The Cartesian coordinates for $\mathbf$ are $(x, y, z) = (\sin a, 0, \cos a)$ and the Cartesian coordinates for $\mathbf$ are $(x, y, z) = (\sin b \cos C, \sin b \sin C, \cos b)$. The value of $\cos c$ is the dot product of the two Cartesian vectors, which is $\sin a \sin b \cos C + \cos a \cos b$.

Second proof

Let , and denote the unit vectors from the center of the sphere to those corners of the triangle. We have , , , and . The vectors and have lengths and respectively and the angle between them is , so
:,
using cross products, dot products, and the Binet–Cauchy identity .

Third proof

Let , and 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 to by an angle $a$, followed another rotation of vector to by an angle $b$, after which we rotate the vector back to by an angle $c$. The composition of these three rotations will form an identity transform. That is, the composite rotation maps the point to itself. These three rotational operations can be represented by quaternions:

:$\begin q_A &= \cos \frac + \mathbf \sin \frac, \\ q_B &= \cos \frac + \mathbf \sin \frac, \\ q_C &= \cos \frac + \mathbf \sin \frac, \end$
where $\mathbf$, $\mathbf$, and $\mathbf$ 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, $q_C q_B q_A = 1.$ Right multiplying both sides by conjugates $q_A^* q_B^*$, we have $q_C = q_A^* q_B^*$, where $q_A^* = \cos \frac - \mathbf \sin \frac$ and $q_B^* = \cos \frac - \mathbf \sin \frac$. This gives us the identity

:$\cos \frac + \mathbf \sin \frac = \left(\cos \frac - \mathbf \sin \frac\right) \left( \cos \frac - \mathbf \sin \frac \right).$

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

:$\left(\cos \frac \cos \frac - \mathbf \cdot \mathbf \sin \frac \sin \frac \right) - \left(\mathbf \sin \frac \cos \frac + \mathbf \cos \frac \sin \frac - \mathbf \times \mathbf \sin \frac \sin \frac \right).$

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

:$\cos \frac = \cos \frac \cos \frac - \mathbf \cdot \mathbf \sin \frac \sin \frac.$

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

:$\cos c = \cos a \cos b + \cos C \sin a \sin b.$

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

:$\mathbf \sin \frac = -\left( \mathbf \sin \frac \cos \frac + \mathbf \cos \frac \sin \frac + \mathbf \sin C \sin \frac \sin \frac \right).$

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

:$\frac = \frac.$

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

:$\frac = \frac = \frac.$

Rearrangements

The first and second spherical laws of cosines can be rearranged to put the sides () and angles () on opposite sides of the equations:
:$\begin \cos C &= \frac \\ \\ \cos c &= \frac \\ \end$

Planar limit: small angles

For ''small'' spherical triangles, i.e. for small , and , the spherical law of cosines is approximately the same as the ordinary planar law of cosines,
: $c^2 \approx a^2 + b^2 - 2ab\cos C \,.$

To prove this, we will use the small-angle approximation obtained from the Maclaurin series for the cosine and sine functions:
: $\cos a = 1 - \frac + O\left(a^4\right),\, \sin a = a + O\left(a^3\right)$

Substituting these expressions into the spherical law of cosines nets:

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

or after simplifying:

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

The big O terms for and are dominated by as and {{math, ''b'' get small, so we can write this last expression as:

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

See also

* Half-side formula
* Hyperbolic law of cosines
* Solution of triangles
* Spherical law of sines

Notes

Spherical trigonometry
Articles containing proofs
Theorems in geometry

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