Law of Cosines

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:

\(\displaystyle c^2\)

\(=\)

\(\displaystyle \paren {b \cos C - a}^2 + \paren {b \sin C - 0}^2\)

squaring both sides of Distance Formula

\(\displaystyle \)

\(=\)

\(\displaystyle b^2 \cos^2 C - 2 a b \cos C + a^2 + b^2 \sin^2 C\)

Square of Difference

\(\displaystyle \)

\(=\)

\(\displaystyle a^2 + b^2 \paren {\sin^2 C + \cos^2 C} - 2 a b \cos C\)

Real Multiplication Distributes over Addition

\(\displaystyle \)

\(=\)

\(\displaystyle a^2 + b^2 - 2 a b \cos C\)

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:

\(\displaystyle \left({b + c}\right) \left({b - c}\right)\)

\(=\)

\(\displaystyle a \cdot BD\)

\(\displaystyle \implies \ \ \)

\(\displaystyle \frac {b^2 - c^2} a\)

\(=\)

\(\displaystyle BD\)

Next:

\(\displaystyle CD\)

\(=\)

\(\displaystyle CB + BD\)

\(\displaystyle \)

\(=\)

\(\displaystyle a + \frac {b^2 - c^2} a\)

\(\displaystyle \)

\(=\)

\(\displaystyle \frac {a^2 + b^2 - c^2} a\)

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

\(\displaystyle \cos C\)

\(=\)

\(\displaystyle \frac {CD} {CE}\)

\(\displaystyle \)

\(=\)

\(\displaystyle \frac {\left({\dfrac{a^2 + b^2 - c^2} a}\right)} {2 b}\)

\(\displaystyle \)

\(=\)

\(\displaystyle \frac {a^2 + b^2 - c^2} {2 a b}\)

\(\displaystyle \implies \ \ \)

\(\displaystyle c^2\)

\(=\)

\(\displaystyle a^2 + b^2 - 2 a b \cos C\)

$\Box$

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:

\(\displaystyle \left({b + c}\right) \left({b - c}\right)\)

\(=\)

\(\displaystyle CB \cdot BD\)

Secant Secant Theorem

\(\displaystyle \left({b + c}\right) \left({b - c}\right)\)

\(=\)

\(\displaystyle a \cdot BD\)

\(\displaystyle \implies \ \ \)

\(\displaystyle \frac {b^2 - c^2} a\)

\(=\)

\(\displaystyle BD\)

Next:

\(\displaystyle CD\)

\(=\)

\(\displaystyle CB - BD\)

\(\displaystyle \)

\(=\)

\(\displaystyle a - \frac {b^2 - c^2} a\)

\(\displaystyle \)

\(=\)

\(\displaystyle \frac {a^2 - b^2 + c^2} a\)

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

\(\displaystyle \cos C\)

\(=\)

\(\displaystyle \frac {CD} {CE}\)

\(\displaystyle \)

\(=\)

\(\displaystyle \frac {\left({\dfrac {a^2 - b^2 + c^2} a}\right)} {2 b}\)

\(\displaystyle \)

\(=\)

\(\displaystyle \frac {a^2 - b^2 + c^2} {2 a b}\)

\(\displaystyle \implies \ \ \)

\(\displaystyle c^2\)

\(=\)

\(\displaystyle a^2 + b^2 - 2 a b \cos C\)

$\Box$

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.

\(\displaystyle a^2\)

\(=\)

\(\displaystyle b^2 + c^2\)

Pythagoras's Theorem

\(\displaystyle c^2\)

\(=\)

\(\displaystyle a^2 - b^2\)

adding $-b^2$ to both sides and rearranging

\(\displaystyle \)

\(=\)

\(\displaystyle a^2 - 2 b^2 + b^2\)

adding $0 = b^2 - b^2$ to the right hand side

\(\displaystyle \)

\(=\)

\(\displaystyle a^2 - 2 a b \left({\frac b a}\right) + b^2\)

multiplying $2 b^2$ by $\dfrac a a$

\(\displaystyle \)

\(=\)

\(\displaystyle a^2 + b^2 - 2 a b \cos C\)

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 :

\(\displaystyle c^2\)

\(=\)

\(\displaystyle h^2 + f^2\)

$(1)$

\(\displaystyle \)

\(=\)

\(\displaystyle a^2 - e^2 + f^2\)

$(2)$

\(\displaystyle \)

\(=\)

\(\displaystyle a^2 - e^2 + f^2 + 2e^2 - 2e^2 + 2ef - 2ef\)

adding and subtracting $2e^2$ and $2ef$

\(\displaystyle \)

\(=\)

\(\displaystyle a^2 + (e^2 + f^2 + 2ef) - 2e(e + f)\)

rearanging

\(\displaystyle \)

\(=\)

\(\displaystyle a^2 + b^2 - 2 a b \cos C\)

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 :

\(\displaystyle c^2\)

\(=\)

\(\displaystyle h^2 + f^2\)

$(1)$

\(\displaystyle \)

\(=\)

\(\displaystyle a^2 - e^2 + f^2\)

$(2)$

\(\displaystyle \)

\(=\)

\(\displaystyle a^2 - b^2 - f^2 - 2bf + f^2\)

$(3)$

\(\displaystyle \)

\(=\)

\(\displaystyle a^2 - b^2 - 2bf + 2b^2 - 2b^2\)

canceling out $f^2 - f^2$ and adding and subtracting $2b^2$

\(\displaystyle \)

\(=\)

\(\displaystyle a^2 + b^2 - 2b(b + f)\)

rearanging

\(\displaystyle \)

\(=\)

\(\displaystyle a^2 + b^2 - 2 a b \cos C\)

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

• Law of Sines

Historical Note

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

Sources

• 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)