Law of Cosines/Proof 1
Jump to navigation
Jump to search
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:
- $c^2 = a^2 + b^2 - 2 a b \cos C$
Proof
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:
- $A = \tuple {b \cos C, b \sin C}$
By the Distance Formula:
- $c = \sqrt {\paren {b \cos C - a}^2 + \paren {b \sin C - 0}^2}$
Hence:
\(\ds c^2\) | \(=\) | \(\ds \paren {b \cos C - a}^2 + \paren {b \sin C - 0}^2\) | squaring both sides of Distance Formula | |||||||||||
\(\ds \) | \(=\) | \(\ds b^2 \cos^2 C - 2 a b \cos C + a^2 + b^2 \sin^2 C\) | Square of Difference | |||||||||||
\(\ds \) | \(=\) | \(\ds a^2 + b^2 \paren {\sin^2 C + \cos^2 C} - 2 a b \cos C\) | Real Multiplication Distributes over Addition | |||||||||||
\(\ds \) | \(=\) | \(\ds a^2 + b^2 - 2 a b \cos C\) | Sum of Squares of Sine and Cosine |
$\blacksquare$
Also known as
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.