May 25, 2024 · 7:58 am

Proof of the cosine rule

$a^2=b^2+c^2-2bccos(A)$

From the diagram above

$h^2=a^2-(b-x)^2\textnormal{ and }h^2=c^2-x^2$

$\therefore a^2-(b-x)^2=c^2-x^2$

$a^2-(b^2-2bx+x^2)=c^2-x^2$

$a^2-b^2+2bx-x^2-c^2+x^2=0$

(1) $\begin{equation*}a^2=b^2+c^2-2bx\end{equation*}$

From the diagram above, we can see

$cos A=\frac{x}{c}$

(2) $\begin{equation*}x=c cos A\end{equation*}$

Substitute equation 2 into equation 1

$a^2=b^2+c^2-2bc cosA$

It can also be handy to have the angle version

$cosA=\frac{b^2+c^2-a^2}{2bc}$

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