Law of Cosines

Theorem

Let $\triangle ABC$ be a triangle whose sides $a, b, c$ are such that $a$ is opposite $A$, $b$ is opposite $B$ and $c$ is opposite $C$.

Then $c^2 = a^2 + b^2 - 2ab \cos C$.

Proof 1

We can place this triangle onto a Cartesian coordinate system by plotting:

• $A = \left({b \cos C , b \sin C}\right)$
• $B = \left({a, 0}\right)$
• $C = \left({0, 0}\right)$

By the distance formula, we have:

$c = \sqrt{\left({b \cos C - a}\right)^2 + \left({b \sin C - 0}\right)^2}$

Now, we just work with this equation:

	\(\displaystyle \)	\(\displaystyle c^2\)	\(=\)	\(\displaystyle \left({b \cos C - a}\right)^2 + \left({b \sin C - 0}\right)^2\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle b^2 \cos ^2 C - 2ab\cos C + a^2 + b^2 \sin ^2 C\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a^2 + b^2 \left({\sin ^2 C + \cos ^2 C}\right) - 2 a b \cos C\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a^2 + b^2 - 2 a b \cos C\)	\(\displaystyle \)

$\blacksquare$

Proof 2

Let $\triangle ABC$ be a triangle.

Using $AC$ as the radius, we construct a circle.

Now we extend:

• $CB$ to $D$;
• $AB$ to $F$;
• $BA$ to $G$;
• $CA$ to $E$.

We join $D$ with $E$, and thus obtain this figure:

Using the Intersecting Chord Theorem we have $GB \cdot BF = CB \cdot BD$.

$AF$ is a radius, so $AF = AC = b = GA$ and thus:

• $GB = GA + AB = b + c$
• $BF = AF - AB = b - c$

Thus:

	\(\displaystyle \)	\(\displaystyle \left({b + c}\right) \left({b - c}\right)\)	\(=\)	\(\displaystyle a \cdot BD\)	\(\displaystyle \)
	\(\displaystyle \implies\)	\(\displaystyle \frac {b^2 - c^2} a\)	\(=\)	\(\displaystyle BD\)	\(\displaystyle \)

Next:

	\(\displaystyle \)	\(\displaystyle CD\)	\(=\)	\(\displaystyle CB + BD\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle a + \frac {b^2 - c^2} a\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {a^2 + b^2 - c^2} a\)	\(\displaystyle \)

As $CA$ is a radius, $CE$ is a diameter.

By Thales' Theorem, it follows that $\angle CDE$ is a right angle.

Then using the definition of cosine, we have

	\(\displaystyle \)	\(\displaystyle \cos C\)	\(=\)	\(\displaystyle \frac {CD} {CE}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {\left({\dfrac{a^2 + b^2 - c^2} a}\right)} {2b}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {a^2 + b^2 - c^2} {2 a b}\)	\(\displaystyle \)
	\(\displaystyle \implies\)	\(\displaystyle c^2\)	\(=\)	\(\displaystyle a^2 + b^2 - 2 a b \cos C\)	\(\displaystyle \)

$\blacksquare$

Sources

• Murray R. Spiegel: Mathematical Handbook of Formulas and Tables (1968): $5.93$