Law of Cosines/Proof 1

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:

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

Let $\triangle ABC$ be embedded in a Cartesian coordinate system by identifying:

$C := \tuple {0, 0}$

$B := \tuple {a, 0}$

Thus by definition of sine and cosine:

$A = \tuple {b \cos C, b \sin C}$

$c = \sqrt {\paren {b \cos C - a}^2 + \paren {b \sin C - 0}^2}$

Hence:

$\ds c^2$

$=$

$\ds \paren {b \cos C - a}^2 + \paren {b \sin C - 0}^2$

squaring both sides of Distance Formula

$\ds $

$=$

$\ds b^2 \cos^2 C - 2 a b \cos C + a^2 + b^2 \sin^2 C$

$\ds $

$=$

$\ds a^2 + b^2 \paren {\sin^2 C + \cos^2 C} - 2 a b \cos C$

$\ds $

$=$

$\ds a^2 + b^2 - 2 a b \cos C$

$\blacksquare$

The Law of Cosines is also known as the Cosine Rule or Cosine Law.

It is known in France as Théorème d'Al-Kashi (Al-Kashi's Theorem) after Jamshīd al-Kāshī, who is believed to have first discovered it.