# An elementary proof of the cosine law

The cosine law states that in triangle $ABC$ with side lengths $a, b,$ and $c$, the following equations are satisfied:

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

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

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

The discussion below shows how these equations were derived.

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Consider triangle $ABC$ with side lengths $a, b$, and $c$. Drop an altitude $h$ from $C$ and let $D$ be the intersection of $h$ and $AB$ as shown in the figure below.  If we let $x$ be the length of $AD$, then $BD = c - x$.

(1) Remembering the mnemonic SOHCAHTOA, in a triangle, the cosine of an angle is equal to length of the side adjacent to it divided by the length of the hypotenuse. Therefore, in triangle $ACD$,  $\cos A = \displaystyle\frac{x}{b}$. Simplifying, we have $x = b \cos A$.

(2) In triangle $BDC$, by the Pythagorean Theorem$a ^2 = h^2 + (c - x)^2$ which when simplified equals to $a ^2 = h^2 + c^2 - 2cx + x^2$.

(3) Also, in triangle $ACD$, by the Pythagorean theorem, $b^2 = x^2 + h^2$.

(4) Rearranging the equation in step (2), we have $a ^2 = c^2 - 2cx + (x^2 + h^2)$.  Now, substituting the equation in step (3) to the preceding equation and rearranging the terms, we have $a ^2 = b^2 + c^2 - 2cx$ .

(5) Substituting the equation in step (1) to the equation in step (4), we have $a ^2 = b^2 + c^2 - 2bc$ $\cos A$, which is equal to the first equation above *.

The formula for $b^2$ and $c^2$ follow from the proof above. This proves the cosine law.

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