# 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:

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

$GB = CB$

$\blacksquare$

## 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.

## Historical Note

The Law of Cosines is believed to have been discovered by Jamshīd al-Kāshī.