Let 
, 
, and 
be the lengths of the legs of a triangle
opposite angles 
, 
,
and 
. Then the law of cosines states
 |  |  |
(1)
|
 |  |  |
(2)
|
 |  |  |
(3)
|
Solving for the cosines yields the equivalent formulas
 |  |  |
(4)
|
 |  |  |
(5)
|
 |  |  |
(6)
|
This law can be derived in a number of ways. The definition of the dot product incorporates the law of cosines, so that the length of the vector
from 
to 
is given by
 |  |  |
(7)
|
 |  |  |
(8)
|
 |  |  |
(9)
|
where 
is the angle between 
and 
.
The formula can also be derived using a little geometry and simple algebra. From the above diagram,
 |  |  |
(10)
|
 |  |  |
(11)
|
 |  |  |
(12)
|
The law of cosines for the sides of a spherical triangle states that
 |  |  |
(13)
|
 |  |  |
(14)
|
 |  |  |
(15)
|
(Beyer 1987). The law of cosines for the angles of a spherical triangle states that
 |  |  |
(16)
|
 |  |  |
(17)
|
 |  |  |
(18)
|
(Beyer 1987).
For similar triangles, a generalized law of cosines is given by
 |
(19)
|
(Lee 1997). Furthermore, consider an arbitrary tetrahedron 
with triangles 
, 
, 
, and 
. Let the areas of these triangles be 
, 
, 
, and 
, respectively, and denote the dihedral
angle with respect to 
and 
for 
by 
. Then
 |
(20)
|
which gives the law of cosines in a tetrahedron,
 |
(21)
|
(Lee 1997). A corollary gives the nice identity
 |
(22)
|
