Law of Cosines/Proof 3
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
Lemma: Right Triangle
Let $\triangle ABC$ be a right triangle such that $\angle A$ is right.
| \(\ds a^2\) | \(=\) | \(\ds b^2 + c^2\) | Pythagoras's Theorem | |||||||||||
| \(\ds c^2\) | \(=\) | \(\ds a^2 - b^2\) | adding $-b^2$ to both sides and rearranging | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^2 - 2 b^2 + b^2\) | adding $0 = b^2 - b^2$ to the right hand side | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^2 - 2 a b \left({\frac b a}\right) + b^2\) | multiplying $2 b^2$ by $\dfrac a a$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^2 + b^2 - 2 a b \cos C\) | Definition of Cosine: $\cos C = \dfrac b a$ |
$\Box$
Acute Triangle
Let $\triangle ABC$ be an acute triangle.
Let $BD$ be dropped perpendicular to $AC$.
Let:
| \(\ds h\) | \(=\) | \(\ds BD\) | ||||||||||||
| \(\ds e\) | \(=\) | \(\ds CD\) | ||||||||||||
| \(\ds f\) | \(=\) | \(\ds AD\) |
We have that $\triangle CDB$ and $\triangle ADB$ are right triangles.
Hence:
| \(\text {(1)}: \quad\) | \(\ds c^2\) | \(=\) | \(\ds h^2 + f^2\) | Pythagoras's Theorem | ||||||||||
| \(\text {(2)}: \quad\) | \(\ds a^2\) | \(=\) | \(\ds h^2 + e^2\) | Pythagoras's Theorem | ||||||||||
| \(\text {(3)}: \quad\) | \(\ds b^2\) | \(=\) | \(\ds \paren {e + f}^2\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds e^2 + f^2 + 2ef\) | ||||||||||||
| \(\text {(4)}: \quad\) | \(\ds e\) | \(=\) | \(\ds a \cos C\) | Definition of Cosine of Angle |
Then:
| \(\ds c^2\) | \(=\) | \(\ds h^2 + f^2\) | from $(1)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^2 - e^2 + f^2\) | from $(2)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^2 - e^2 + f^2 + 2 e^2 - 2 e^2 + 2 e f - 2 e f\) | adding and subtracting $2 e^2$ and $2 e f$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^2 + \paren {e^2 + f^2 + 2 e f} - 2 e \paren {e + f}\) | rearanging | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^2 + b^2 - 2 a b \cos C\) | using $(3)$ to substitute for $b^2$ and $b$, and $(4)$ to substitute $e$ for $a \cos C$ |
$\Box$
Obtuse Triangle
Let $\triangle ABC$ be an obtuse triangle.
Let $AC$ be extended and $BD$ be dropped perpendicular to $AC$.
Let:
| \(\ds h\) | \(=\) | \(\ds BD\) | ||||||||||||
| \(\ds e\) | \(=\) | \(\ds CD\) | ||||||||||||
| \(\ds f\) | \(=\) | \(\ds AD\) |
We have that $\triangle CDB$ and $\triangle ADB$ are right triangles.
Hence:
| \(\text {(1)}: \quad\) | \(\ds c^2\) | \(=\) | \(\ds h^2 + f^2\) | Pythagoras's Theorem | ||||||||||
| \(\text {(2)}: \quad\) | \(\ds a^2\) | \(=\) | \(\ds h^2 + e^2\) | Pythagoras's Theorem | ||||||||||
| \(\text {(3)}: \quad\) | \(\ds e^2\) | \(=\) | \(\ds \paren {b + f}^2\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds b^2 + f^2 + 2 b f\) | ||||||||||||
| \(\text {(4)}: \quad\) | \(\ds e\) | \(=\) | \(\ds a \cos C\) | Definition of Cosine of Angle |
Then:
| \(\ds c^2\) | \(=\) | \(\ds h^2 + f^2\) | from $(1)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^2 - e^2 + f^2\) | from $(2)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^2 - b^2 - f^2 - 2 b f + f^2\) | substituting for $e^2$ from $(3)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^2 - b^2 - 2 b f + 2 b^2 - 2 b^2\) | simplifying and adding and subtracting $2 b^2$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^2 + b^2 - 2 b \paren {b + f}\) | rearranging | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^2 + b^2 - 2 a b \cos C\) | using $(4)$ to substitute $b + f = e$ with $a \cos C$ |
$\blacksquare$
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.
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text V$. Trigonometry: The cos formula and the sine formula