I have a proof of the pythagorean theorem, where I, first trace the angles created by making two right triangles inside a big one. Following from that, the sides $a$ & $b$ are converted into hypotenuse, where the hypotenuse of the main triangle, $c$ is divided into two perpendicular lengths, $a$ $\sin \theta$ and $b$ $\cos \theta$.
Then, I use the basic axioms of sine and cosine for the angle α in the big right triangle, $\sin \theta$= $a/c$ and $\cos \theta$= $b/c$. Thus, the pythagorean theorem is easily extracted from the given axioms.
$$c=a\sin \theta+b\cos \theta$$
$$c=a(a/c)+b(b/c)$$
$$c^2=a^2+b^2$$
Is it a valid trigonometric proof of the pythagorean theorem? It does not presuppose $\sin^2 \theta + \cos^2\theta= 1$ in the proof, rather it uses the definitions of sine and cosine, and similarity of triangles ( which comes by the assumptions of trigonometry i.e. through angle- chasing, etc. ).
Yes, the proof is correct and it is not circular! What you do is essentially Einstein's proof, however, expressed using trigonometric definitions of "sine" and "cosine" as dimensionless ratios of two sides inside a right triangle.
In the following text, the symbol $:=$ means "equal by definition".
Only in flat space, we have equivalence between the trigonometric definition of "sine" as ratio of sides inside a right triangle
$\sin\alpha:=\frac{a}{c}$
and the analytic definition of the "sine" using Taylor series:
$\sin\alpha:=\alpha-\frac{\alpha^{3}}{3!}+\frac{\alpha^{5}}{5!}-\frac{\alpha^{7}}{7!}+\ldots$
In curved space, the two definitions produce different quantities, so they are not equivalent.
Why these two definitions are not equivalent?
It is because the geometric expression $\frac{a}{c}$ is curvature-dependent, whereas the algebraic expression $\alpha-\frac{\alpha^{3}}{3!}+\frac{\alpha^{5}}{5!}-\frac{\alpha^{7}}{7!}+\ldots$ is curvature-independent.
How to verify that these two definitions are not equivalent?
In the trigonometric definition, we have a function of two arguments that are dimensional and have units of "meters": the side length $a (m)$ and the hypotenuse $c (m)$:
$\sin\alpha:=\frac{a (m)}{c (m)}$
In contrast, in the analytic expression, we have a function of single argument $\alpha (1)$ that is a dimensionless quantity, i.e., pure number
$\sin\alpha:=\alpha (1) -\frac{\alpha(1)^{3}}{3!}+\frac{\alpha(1)^{5}}{5!}-\frac{\alpha(1)^{7}}{7!}+\ldots$
The above implies that only the expression in which "meters" appear is curvature-dependent. Thus, "trigonometric proofs" of the Pythagorean theorem are not "circular" as long as they use the trigonometric definition of "sine" as ratio of sides.
I have given a full explanation in Wikipedia "Trigonometric proof using Einstein's construction":
https://en.wikipedia.org/wiki/Pythagorean_theorem#Trigonometric_proof_using_Einstein's_construction