The Pythagorean Theorem from the Half-Angle Formula for Cosine

I've been trying to rescue the Half-Angle Formulas from oblivion and give them the status they deserve by showing its many applications and equivalence with other fundamental theorems (i.e., the Law of Cosines or the Pythagorean Theorem). In this page the Pythagorean Theorem is presented as a mere corollary of the aforementioned formulas.

Let $\triangle{ABC}$ be a right-triangle with $BC=a$, $AC=b$, $AB=c$ and $\angle{ACB}=\gamma=90^\circ$. Also, let $s=\frac{a+b+c}{2}$. The proof is based on the formula

$$\cos^2{\frac{\gamma}{2}}=\frac{s(s-c)}{ab}=\frac{(a+b)^2-c^2}{4ab}\tag{1}$$

This is true for any triangle and can be proven independently of the Pythagorean theorem or the Pythagorean Identity. The proof of $(1)$ can be found here.

There has been controversy about whether it is possible to prove the Pythagorean Theorem using trigonometry. Elisha Loomis and many others believed and still believe that no trigonometric proof of the Pythagorean theorem is possible. This belief stems from the assumption that any such proof would rely on the most fundamental of trigonometric identities $\sin^2{\theta} + \cos^2{\theta} = 1$ which is nothing but a reformulation of the Pythagorean theorem proper. However, Jason Zimba showed it is possible to prove the Pythagoren Identity without a recourse to the Pythagorean theorem (see $[1]$). He then has used the Pythagorean Identity to prove the Pythagorean theorem. Another trigonometric proof is given by Luc Gheysens (see here).

Pythagorean Theorem. Given a right-triangle $\triangle{ABC}$ with $BC=a$, $AC=b$, $AB=c$ and $\angle{ACB}=\gamma=90^\circ$. Then

$$a^2+b^2=c^2$$

Proof. From the Double-Angle Formula for cosine we know

$$\cos{2\theta}=2\cos^2{\theta}-1\tag{2}$$

The proof of $(2)$ does not rely on the Pythagorean Theorem (see this video). Using the unit circle, the proof of $\cos{90^\circ}=0$ only depends on the definition of Cosine (see here). So applying $(2)$ we can calculate $\cos{45^\circ}=\frac{\sqrt{2}}{2}$ without risk of Petitio Principii. Hence

$$\cos^2{\frac{\gamma}{2}}=\frac{s(s-c)}{ab}=\frac{(a+b)^2-c^2}{4ab}=\frac{1}{2}$$

from which we have

$$\begin{align*}(a+b)^2-c^2&=2ab\\a^2+2ab+b^2-c^2&=2ab\\a^2+b^2-c^2&=0\end{align*}$$

I have to admit that this proof is not as satisfying as I would have hoped. Discussion of this proof along with geometric proofs (provided by Blue) of the Half-Angle Formulas can be seen here (pay no attention to the votedown, some intolerant people still believe in the impossibility of trigonometric proofs of the Pythagorean Theorem).

References

$[1]$ A. Bogomolny, More Trigonometric Proofs of the Pythagorean Theorem, from Cut-the-Knot.org