Wikipedia:Sandbox

Proof

A proof of the law of cosines can be constructed as follows. 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:

$\cos(a)=\mathbf {u} \cdot \mathbf {v}$

$\cos(b)=\mathbf {u} \cdot \mathbf {w}$

$\cos(c)=\mathbf {v} \cdot \mathbf {w}$

To get the angle C, we need the tangent vectors t_a and t_b at u along the directions of sides a and b, respectively. For example, the tangent vector t_a 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:

$\mathbf {t} _{a}={\frac {\mathbf {v} -\mathbf {u} (\mathbf {u} \cdot \mathbf {v} )}{\left|\mathbf {v} -\mathbf {u} (\mathbf {u} \cdot \mathbf {v} )\right|}}={\frac {\mathbf {v} -\mathbf {u} (\mathbf {u} \cdot \mathbf {v} )}{(\mathbf {v} \cdot \mathbf {v} +\mathbf {u} \cdot \mathbf {u} (\mathbf {u} \cdot \mathbf {v} )^{2}-2(\mathbf {u} \cdot \mathbf {v} )cos(a))^{1/2}}}={\frac {\mathbf {v} -\mathbf {u} (\mathbf {u} \cdot \mathbf {v} )}{(1+cos^{2}(a)-2cos^{2}(a))^{1/2}}}={\frac {\mathbf {v} -\mathbf {u} \cos(a)}{\sin(a)}}$

where for the denominator we have used the Pythagorean identity sin²(a) = 1 − cos²(a). Similarly,

$\mathbf {t} _{b}={\frac {\mathbf {w} -\mathbf {u} \cos(b)}{\sin(b)}}$

Then, the angle C is given by:

$\cos(C)=\mathbf {t} _{a}\cdot \mathbf {t} _{b}$

$={\frac {\mathbf {v} \cdot \mathbf {w} -\mathbf {v} \cdot \mathbf {u} \cos(b)-\mathbf {u} \cos(a)\cdot \mathbf {w} +\mathbf {u} \cos(a)\cdot \mathbf {u} \cos(b)}{\sin(a)\sin(b)}}$

$={\frac {\cos(c)-\cos(b)\cos(a)-\cos(a)\cos(b)+\cos(a)\cos(b)}{\sin(a)\sin(b)}}$

$\cos(C)={\frac {\cos(c)-\cos(a)\cos(b)}{\sin(a)\sin(b)}}$

from which the law of cosines immediately follows.

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