Law of Cosines
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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:
- $C := \left({0, 0}\right)$
- $B := \left({a, 0}\right)$
Thus by definition of sine and cosine:
- $A = \left({b \cos C , b \sin C}\right)$
By the Distance Formula:
- $c = \sqrt{\left({b \cos C - a}\right)^2 + \left({b \sin C - 0}\right)^2}$
Hence:
| \(\displaystyle c^2\) | \(=\) | \(\displaystyle \left({b \cos C - a}\right)^2 + \left({b \sin C - 0}\right)^2\) | $\quad$ squaring both sides of Distance Formula | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle b^2 \cos^2 C - 2 a b \cos C + a^2 + b^2 \sin^2 C\) | $\quad$ Square of Difference | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle a^2 + b^2 \left({\sin^2 C + \cos^2 C}\right) - 2 a b \cos C\) | $\quad$ Real Multiplication Distributes over Addition | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle a^2 + b^2 - 2 a b \cos C\) | $\quad$ Sum of Squares of Sine and Cosine | $\quad$ |
$\blacksquare$
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:
| \(\displaystyle \left({b + c}\right) \left({b - c}\right)\) | \(=\) | \(\displaystyle a \cdot BD\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \implies \ \ \) | \(\displaystyle \frac {b^2 - c^2} a\) | \(=\) | \(\displaystyle BD\) | $\quad$ | $\quad$ |
Next:
| \(\displaystyle CD\) | \(=\) | \(\displaystyle CB + BD\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle a + \frac {b^2 - c^2} a\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \frac {a^2 + b^2 - c^2} a\) | $\quad$ | $\quad$ |
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
| \(\displaystyle \cos C\) | \(=\) | \(\displaystyle \frac {CD} {CE}\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \frac {\left({\dfrac{a^2 + b^2 - c^2} a}\right)} {2 b}\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \frac {a^2 + b^2 - c^2} {2 a b}\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \implies \ \ \) | \(\displaystyle c^2\) | \(=\) | \(\displaystyle a^2 + b^2 - 2 a b \cos C\) | $\quad$ | $\quad$ |
$\Box$
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:
| \(\displaystyle \left({b + c}\right) \left({b - c}\right)\) | \(=\) | \(\displaystyle CB \cdot BD\) | $\quad$ Secant Secant Theorem | $\quad$ | |||||||||
| \(\displaystyle \left({b + c}\right) \left({b - c}\right)\) | \(=\) | \(\displaystyle a \cdot BD\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \implies \ \ \) | \(\displaystyle \frac {b^2 - c^2} a\) | \(=\) | \(\displaystyle BD\) | $\quad$ | $\quad$ |
Next:
| \(\displaystyle CD\) | \(=\) | \(\displaystyle CB - BD\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle a - \frac {b^2 - c^2} a\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \frac {a^2 - b^2 + c^2} a\) | $\quad$ | $\quad$ |
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
| \(\displaystyle \cos C\) | \(=\) | \(\displaystyle \frac {CD} {CE}\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \frac {\left({\dfrac {a^2 - b^2 + c^2} a}\right)} {2 b}\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \frac {a^2 - b^2 + c^2} {2 a b}\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \implies \ \ \) | \(\displaystyle c^2\) | \(=\) | \(\displaystyle a^2 + b^2 - 2 a b \cos C\) | $\quad$ | $\quad$ |
$\Box$
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:
- $GB = CB$
To discuss it in more detail, feel free to use the talk page.
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$\blacksquare$
Proof 3
Lemma: Right Triangle
Let $\triangle ABC$ be a right triangle such that $\angle A$ is right.
| \(\displaystyle a^2\) | \(=\) | \(\displaystyle b^2 + c^2\) | $\quad$ Pythagoras's Theorem | $\quad$ | |||||||||
| \(\displaystyle c^2\) | \(=\) | \(\displaystyle a^2 - b^2\) | $\quad$ adding $-b^2$ to both sides and rearranging | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle a^2 - 2 b^2 + b^2\) | $\quad$ adding $0 = b^2 - b^2$ to the right hand side | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle a^2 - 2 a b \left({\frac b a}\right) + b^2\) | $\quad$ multiplying $2 b^2$ by $\dfrac a a$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle a^2 + b^2 - 2 a b \cos C\) | $\quad$ Definition of Cosine: $\cos C = \dfrac b a$ | $\quad$ |
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 :
| \(\displaystyle c^2\) | \(=\) | \(\displaystyle h^2 + f^2\) | $\quad$ $(1)$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle a^2 - e^2 + f^2\) | $\quad$ $(2)$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle a^2 - e^2 + f^2 + 2e^2 - 2e^2 + 2ef - 2ef\) | $\quad$ adding and subtracting $2e^2$ and $2ef$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle a^2 + (e^2 + f^2 + 2ef) - 2e(e + f)\) | $\quad$ rearanging | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle a^2 + b^2 - 2 a b \cos C\) | $\quad$ using $(3)$ to substitute both parentheses for $b^2$ and $b$ respectively, and $(4)$ to subst. e for $a \cos C$ | $\quad$ |
$\blacksquare$
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 :
| \(\displaystyle c^2\) | \(=\) | \(\displaystyle h^2 + f^2\) | $\quad$ $(1)$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle a^2 - e^2 + f^2\) | $\quad$ $(2)$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle a^2 - b^2 - f^2 - 2bf + f^2\) | $\quad$ $(3)$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle a^2 - b^2 - 2bf + 2b^2 - 2b^2\) | $\quad$ canceling out $f^2 - f^2$ and adding and subtracting $2b^2$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle a^2 + b^2 - 2b(b + f)\) | $\quad$ rearanging | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle a^2 + b^2 - 2 a b \cos C\) | $\quad$ using $(4)$ to substitute $b + f = e$ with $a \cos C$ | $\quad$ |
$\blacksquare$
Historical Note
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.
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: $5.93$
- 2008: Ian Stewart: Taming the Infinite ... (previous) ... (next): Chapter $5$: Eternal Triangles
