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 $b^2=\stackrel{2}{\stackrel{‾}{AO}}+h^2$ .

$b^2$	$=(c-a\cos (B){)}^2+a^2{\sin }^2(B)$
	$=c^2-2ac\cos (B)+a^2{\cos }^2(B)+a^2{\sin }^2(B)$
	$=a^2+c^2-2ac\cos (B)$

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.