# Trigonometry/Law of Cosines

## Law of Cosines

The Pythagorean theorem is a special case of the more general theorem relating the lengths of sides in any triangle, the law of cosines:

$a^{2}+b^{2}-2ab\cos(\theta )=c^{2}$

where $\theta$  is the angle between sides $a$  and $b$  .

### Does the formula make sense?

This formula had better agree with the Pythagorean Theorem when $\theta =90^{\circ }$  .

So try it...

When $\theta =90^{\circ }$  , $\cos(\theta )=\cos(90^{\circ })=0$

The $-2ab\cos(\theta )=0$  and the formula reduces to the usual Pythagorean theorem.

## Permutations

For any triangle with angles $A,B,C$  and corresponding opposite side lengths $a,b,c$  , the Law of Cosines states that

$a^{2}=b^{2}+c^{2}-2bc\cdot \cos(A)$
$b^{2}=a^{2}+c^{2}-2ac\cdot \cos(B)$
$c^{2}=a^{2}+b^{2}-2ab\cdot \cos(C)$

### Proof

Dropping a perpendicular $OC$  from vertex $C$  to intersect $AB$  (or $AB$  extended) at $O$  splits this triangle into two right-angled triangles $AOC$  and $BOC$  , with altitude $h$  from side $c$  .

First we will find the lengths of the other two sides of triangle $AOC$  in terms of known quantities, using triangle $BOC$  .

$h=a\sin(B)$

Side $c$  is split into two segments, with total length $c$  .

${\overline {OB}}$  has length ${\overline {BC}}\cos(B)=a\cos(B)$
${\overline {AO}}={\overline {AB}}-{\overline {OB}}$  has length $c-a\cos(B)$

Now we can use the Pythagorean Theorem to find $b$  , since $b^{2}={\overline {AO}}^{2}+h^{2}$  .

 $b^{2}$ $={\bigl (}c-a\cos(B){\bigr )}^{2}+a^{2}\sin ^{2}(B)$ $=c^{2}-2ac\cos(B)+a^{2}\cos ^{2}(B)+a^{2}\sin ^{2}(B)$ $=a^{2}+c^{2}-2ac\cos(B)$ The corresponding expressions for $a$  and $c$  can be proved similarly.

The formula can be rearranged:

$\cos(C)={\frac {a^{2}+b^{2}-c^{2}}{2ab}}$

and similarly for $cos(A)$  and $cos(B)$  .

## Applications

This formula can be used to find the third side of a triangle if the other two sides and the angle between them are known. The rearranged formula can be used to find the angles of a triangle if all three sides are known. See Solving Triangles Given SAS.