Law of Cosines
This article was Featured Proof .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 1
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 |
$\blacksquare$
Proof 2
Let $\triangle ABC$ be a triangle.
Case 1: $AC$ greater than $AB$
Using $AC$ as the radius, we construct a circle whose center is $A$.
Now we extend:
- $CB$ to $D$
- $AB$ to $F$
- $BA$ to $G$
- $CA$ to $E$.
$D$ is joined with $E$, thus:
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:
| \(\ds \paren {b + c} \paren {b - c}\) | \(=\) | \(\ds a \cdot BD\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {b^2 - c^2} a\) | \(=\) | \(\ds BD\) |
Next:
| \(\ds CD\) | \(=\) | \(\ds CB + BD\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds a + \frac {b^2 - c^2} a\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {a^2 + b^2 - c^2} a\) |
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
| \(\ds \cos C\) | \(=\) | \(\ds \frac {CD} {CE}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\paren {\dfrac {a^2 + b^2 - c^2} a} } {2 b}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {a^2 + b^2 - c^2} {2 a b}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds c^2\) | \(=\) | \(\ds a^2 + b^2 - 2 a b \cos C\) |
$\Box$
Case 2: $AC$ less than $AB$
When $AC$ is less than $AB$, the point $B$ lies outside the circle and so the diagram needs to be modified accordingly:
Draw a diagram for the case where $\angle ACB$ is a right angle and where it is a convex angle to show that the formula will be the same.
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Now we extend:
- $BA$ to $G$
- $CA$ to $E$.
Then we construct:
- $D$ as the point at which $CB$ intersects the circle
- $F$ as the point at which $AB$ intersects the circle.
Finally $D$ is joined to $E$.
Using the Secant Secant Theorem we have:
- $GB \cdot BF = CB \cdot BD$
$AF$ is a radius, so $AF = AC = b = GA$ and thus:
- $GB = GA + AB = b + c$
- $BF = AB - AF = b - c$
Thus:
| \(\ds \paren {b + c} \paren {b - c}\) | \(=\) | \(\ds CB \cdot BD\) | Secant Secant Theorem | |||||||||||
| \(\ds \paren {b + c} \paren {b - c}\) | \(=\) | \(\ds a \cdot BD\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {b^2 - c^2} a\) | \(=\) | \(\ds BD\) |
Next:
| \(\ds CD\) | \(=\) | \(\ds CB - BD\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds a - \frac {b^2 - c^2} a\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {a^2 - b^2 + c^2} a\) |
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
| \(\ds \cos C\) | \(=\) | \(\ds \frac {CD} {CE}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\paren {\dfrac {a^2 - b^2 + c^2} a} } {2 b}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {a^2 - b^2 + c^2} {2 a b}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds c^2\) | \(=\) | \(\ds a^2 + b^2 - 2 a b \cos C\) |
$\Box$
Case 3: $AC = AB$
When $AC = AB$ the points $B$, $D$ and $F$ coincide on the circumference of the circle:
We extend:
- $BA$ to $G$
- $CA$ to $E$
and immediately:
- $GB = CB$
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$\blacksquare$
Proof 3
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: $\cos C = \dfrac b a$ |
Hence the result.
$\blacksquare$
Acute Triangle
Let $\triangle ABC$ be an acute triangle.
Let $BD$ be dropped perpendicular to $AC$, and let us define $h = BD$, $e = CD$ and $f = AD$.
Then $\triangle CDB$ and $\triangle ADB$ are right triangles.
So we have both :
- $(1):\quad c^2 = h^2 + f^2 \quad$ Pythagoras's Theorem
- $(2):\quad a^2 = h^2 + e^2 \quad$ Pythagoras's Theorem
and also :
- $(3):\quad b^2 = (e + f)^2 = e^2 + f^2 + 2ef$
- $(4):\quad e = a \cos C \quad$ Definition:Cosine of Angle
We'll start with the first equation and use the rest of them to get the desired result :
| \(\ds c^2\) | \(=\) | \(\ds h^2 + f^2\) | $(1)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^2 - e^2 + f^2\) | $(2)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^2 - e^2 + f^2 + 2e^2 - 2e^2 + 2ef - 2ef\) | adding and subtracting $2e^2$ and $2ef$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^2 + (e^2 + f^2 + 2ef) - 2e(e + f)\) | rearanging | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^2 + b^2 - 2 a b \cos C\) | using $(3)$ to substitute both parentheses for $b^2$ and $b$ respectively, and $(4)$ to subst. e for $a \cos C$ |
$\blacksquare$
Obtuse Triangle
Let $\triangle ABC$ be an obtuse triangle.
Let $AC$ be extended and $BD$ be dropped perpendicular to $AC$, and let us define $h = BD$, $e = CD$ and $f = AD$.
Then $\triangle CDB$ and $\triangle ADB$ are right triangles.
So we have both :
- $(1):\quad c^2 = h^2 + f^2 \quad$ Pythagoras's Theorem
- $(2):\quad a^2 = h^2 + e^2 \quad$ Pythagoras's Theorem
and also :
- $(3):\quad e^2 = (b + f)^2 = b^2 + f^2 + 2bf$
- $(4):\quad e = a \cos C \quad$ Definition:Cosine of Angle
We'll start with the first equation and use the rest of them to get the desired result :
| \(\ds c^2\) | \(=\) | \(\ds h^2 + f^2\) | $(1)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^2 - e^2 + f^2\) | $(2)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^2 - b^2 - f^2 - 2bf + f^2\) | $(3)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^2 - b^2 - 2bf + 2b^2 - 2b^2\) | canceling out $f^2 - f^2$ and adding and subtracting $2b^2$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^2 + b^2 - 2b(b + f)\) | rearanging | |||||||||||
| \(\ds \) | \(=\) | \(\ds a^2 + b^2 - 2 a b \cos C\) | using $(4)$ to substitute $b + f = e$ with $a \cos C$ |
$\blacksquare$
Also known as
This result 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.
Also see
Historical Note
The Law of Cosines is believed to have been discovered by Jamshīd al-Kāshī.
Sources
- 1938: C.V. Durell and Alan Robson: Shorter Advanced Trigonometry ... (next): Chapter $\text I$: Properties of the Triangle
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: $5.93$
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): Entry: cosine rule (law of cosines): 1.
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Entry: cosine rule (law of cosines): 1.
- 2008: Ian Stewart: Taming the Infinite ... (previous) ... (next): Chapter $5$: Eternal Triangles: Logarithms
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Entry: triangle (iii)