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:

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


CosineRuleCartesian.png

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:

CosineRule.png


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:

CosineRule2.png


This article is complete as far as it goes, but it could do with expansion.
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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:

CosineRule3.png

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.

CosineRule-Proof3-right.png
\(\ds a^2\) \(=\) \(\ds b^2 + c^2\) 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.

CosineRule-Proof3-acute.png

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.

CosineRule-Proof3-obtuse.png

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

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