Law of Cosines

Revision as of 15:42, 7 October 2007 by Temperal (talk | contribs) (Acute Triangle: Picture)
This is an AoPSWiki Word of the Week for Oct 4-Oct 10

The Law of Cosines is a theorem which relates the side-lengths and angles of a triangle. For a triangle with edges of length $a$, $b$ and $c$ opposite angles of measure $A$, $B$ and $C$, respectively, the Law of Cosines states:

$c^2 = a^2 + b^2 - 2ab\cos C$

$b^2 = a^2 + c^2 - 2ac\cos B$

$a^2 = b^2 + c^2 - 2bc\cos A$

In the case that one of the angles has measure $90^\circ$ (is a right angle), the corresponding statement reduces to the Pythagorean Theorem.

Proofs

Acute Triangle

picture+pic;
pair+A,B,C,D,E;
C=(30,70);
B=(0,0);
A=(100,0);
D=(30,0);
size(100);
draw(B--A--C--B);
draw(C--D);
label("A",A,(1,0));
dot(A);
label("B",B,(-1,-1));
dot(B);
label("C",C,(0,1));
dot(C);
draw(D--(30,4)--(34,4)--(34,0)--D);
label("f",(30,35),(1,0));
label("d",(15,0),(0,-1));
label("e",(50,0),(0,-1.5)); (Error making remote request. Unknown error_msg)

Let $a$, $b$, and $c$ be the side lengths, $C$ is the angle measure opposite side $c$, $f$ is the distance from angle $C$ to side $c$, and $d$ and $e$ are the lengths that $c$ is split into by $f$.

We use the Pythagorean theorem:

\[a^2+b^2-2f^2=d^2+e^2\]

We are trying to get $a^2+b^2-2f^2+2de$ on the LHS, because then the RHS would be $c^2$.

We use the addition rule for cosines and get:

\[\cos{C}=\dfrac{f}{a}*\dfrac{f}{b}-\dfrac{d}{a}*\dfrac{e}{b}=\dfrac{f^2-de}{ab}\]

We multiply by -2ab and get:

\[2de-2f^2=-2ab\cos{C}\]

Now remember our equation?

\[a^2+b^2-2f^2+2de=c^2\]

We replace the $-2f^2+2de$ by $-2ab\cos{C}$ and get:

\[c^2=a^2+b^2-2ab\cos{C}\]

We can use the same argument on the other sides.

Right Triangle

Since $C=90^{\circ}$, $\cos C=0$, so the expression reduces to the Pythagorean Theorem. You can find several proofs of the Pythagorean Theorem here

Obtuse Triangle

See also