Editing Heron's formula (section)

===Trigonometric proof using the law of cosines===
A modern proof, which uses [[algebra]] and is quite different from the one provided by Heron, follows.<ref>{{cite book
| author=Niven, Ivan
| title=Maxima and Minima Without Calculus
| url=https://archive.org/details/maximaminimawith0000nive
| url-access=registration
| publisher=The Mathematical Association of America
| year=1981
| pages=[https://archive.org/details/maximaminimawith0000nive/page/7 7–8]}}</ref>
Let {{tmath|a,}} {{tmath|b,}} {{tmath|c}} be the sides of the triangle and {{tmath|\alpha,}} {{tmath|\beta,}} {{tmath|\gamma}} the [[angle]]s opposite those sides.
Applying the [[law of cosines]] we get

<math display=block>\cos \gamma = \frac{a^2+b^2-c^2}{2ab}</math>

[[File:Triangle with notations 2 without points.svg|thumb|A triangle with sides {{mvar|a}}, {{mvar|b}} and {{mvar|c}}]]
From this proof, we get the algebraic statement that

<math display=block>\sin \gamma = \sqrt{1-\cos^2 \gamma} = \frac{\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2 }}{2ab}.</math>

The [[altitude (triangle)|altitude]] of the triangle on base {{tmath|a}} has length {{tmath|b\sin\gamma}}, and it follows

<math display=block>\begin{align}
A
&= \tfrac12 (\mbox{base}) (\mbox{altitude}) \\[6mu]
&= \tfrac12 ab\sin \gamma \\[6mu]
&= \frac{ab}{4ab}\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2} \\[6mu]
&= \tfrac14\sqrt{-a^4 - b^4 - c^4 + 2 a^2 b^2 + 2 a^2 c^2 + 2 b^2 c^2} \\[6mu]
&= \tfrac14\sqrt{(a + b + c)(- a + b + c)(a - b + c)(a + b - c)} \\[6mu]
&= \sqrt{
  \left(\frac{a + b + c}{2}\right)
  \left(\frac{- a + b + c}{2}\right)
  \left(\frac{a - b + c}{2}\right)
  \left(\frac{a + b - c}{2}\right)} \\[6mu]
&= \sqrt{s(s-a)(s-b)(s-c)}.
\end{align}</math>