I don't know how to define what a geometric proof is in general. However, I will define $\cos$ and $\sin$ by the differential equations
- $\cos(0) = 1$
- $\sin(0) = 0$
- $\forall x \in \mathbb{R}\cos'(x) = \sin(x)$
- $\forall x \in \mathbb{R}\sin'(x) = \cos(x)$
This definition ensures that the domains of both $\cos$ and $\sin$ are $\mathbb{R}$. I will also give some proofs of $\forall x \in \mathbb{R}\cos^2(x) + \sin^2(x) = 1$ that I call geometric proofs. After that, I will give a reason to suggest that since you can prove it geometrically, you can also prove it without any geometry. Then I will show that I can also prove the same statement another way without using any geometry at all.
Here's one proof. The distance formula is a binary function from $\mathbb{R}^2$ to $\mathbb{R}$ satisfying the following properties
- $\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R}d((x, y), (x + z, y + w)) = d((0, 0), (z, w))$
- $\forall x \in \mathbb{R}\forall y \in \mathbb{R}d((0, 0), (x, y)) \in \mathbb{R} - \mathbb{R}^-$
- $\forall x \in \mathbb{R} - \mathbb{R}^-d((0, 0), (x, 0)) = x$
- $\forall x \in \mathbb{R}\forall y \in \mathbb{R}d((0, 0), (x, -y)) = d((0, 0), (x, y))$
- $\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R}d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$
- The area of any square is the square of the length of its edges
- $\forall x \in \mathbb{R}d((0, 0), (\cos(x), \sin(x))) = 1$
where the area of any square in $\mathbb{R}^2$ is defined as the definite integral from $-\infty$ to $\infty$ of the the function that assigns to each real number $x$ the length of the intersection of the square and the set of all points in $\mathbb{R}^2$ whose first coordinate is $x$. Using only the first 5 assumptions, we can show that the distance formula is $d((x, y), (z, w)) = \sqrt{(z - x)^2 + (w - y)^2}$. Now using property 7, we can show that $\forall x \in \mathbb{R}1 = 1^2 = (d((0, 0), (\cos(x), \sin(x))))^2 = \sqrt{\cos^2(x) + \sin^2(x)}^2 = \cos^2(x) + \sin^2(x)$.
Here's another proof. Using the same 7 assumptions, take the sixth assumption. Take any square
Let $(x, y)$ be the displacement along any edge in either direction. We can see that the area of that square is $(x - y)^2 + 2xy = x^2 - 2xy + y^2 + 2xy = x^2 + y^2$. Using properties 1 and 6, we can show that the length of any of its edges in either direction is $\sqrt{x^2 + y^2}$. Also $\forall x \in \mathbb{R}\forall y \in \mathbb{R}$ where $x$ and $y$ are not both zero, there exists a square such that the displacement along one of its edges in one of the directions is $(x, y)$. This shows that $\forall x \in \mathbb{R}\forall y \in \mathbb{R}$ when $x$ and $y$ are not both zero, $d((0, 0), (x, y)) = \sqrt{x^2 + y^2}$. Using property 3, we can also show that $d((0, 0), (0, 0)) = 0 = \sqrt{0^2 + 0^2}$ so $\forall x \in \mathbb{R}\forall y \in \mathbb{R}d((0, 0), (x, y)) = \sqrt{x^2 + y^2}$. Using property 7, we can show that $\forall x \in \mathbb{R}1 = 1^2 = (d((0, 0), (\cos(x), \sin(x))))^2 = \sqrt{\cos^2(x) + \sin^2(x)}^2 = \cos^2(x) + \sin^2(x)$.
You may be thinking that since I showed using those 7 assumptions that $\forall x \in \mathbb{R}\cos^2(x) + \sin^2(x) = 1$, if we can show that $\forall x \in \mathbb{R}\cos^2(x) + \sin^2(x) = 1$, we can also give a nongeometric proof of $\forall x \in \mathbb{R}\cos^2(x) + \sin^2(x) = 1$. It makes sense to call something a nongeometric proof that $\forall x \in \mathbb{R}\cos^2(x) + \sin^2(x) = 1$ because that's a statement that gives no mention of the concept of distance but it probably does not make sense to call anything a nongeometric proof of the Pythagoren theorem. That's probably the reason the question Non-geometric Proof of Pythagorean Theorem got closed and because it already got closed and answered, I asked a fixed up version of it at Existence and uniqueness of function satisfying intuitive properties of distance in $\mathbb{R}^2$?.
Indeed, it can be shown that a function satisfying all 7 properties exists. We we define distance by the function $d((x, y), (z, w)) = \sqrt{x^2 + y^2}$, then it's easy to show that it satisfies the first 4 properties. It's not that hard to improvise the second proof of $\forall x \in \mathbb{R}\cos^2(x) + \sin^2(x) = 1$ into a proof without any assumptions of what properties the function $d((x, y), (z, w)) = \sqrt{x^2 + y^2}$ satisfies that that function satisfies the sixth property. It can also be shown to satisfy property 5 as follows. $\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R}d((0, 0), (xz - yw, xw + yz)) = \sqrt{(xz - yw)^2 + (xw + yz)^2} = \sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = \sqrt{x^2z^2 + x^2w^2 + y^2z^2 + w^2w^2} = \sqrt{(x^2 + y^2)(z^2 + w^2)} = \sqrt{x^2 + y^2}\sqrt{z^2 + w^2} = d((0, 0), (x, y))d((0, 0), (z, w))$. Finally, it can be proven to satisfy property 7 as follows. $\frac{d}{dx}(\cos^2(x) + \sin^2(x)) = \frac{d}{dx}\cos^2(x) + \frac{d}{dx}\sin^2(x) = 2\cos(x)(-\sin(x)) + 2\sin(x)\cos(x) = 0$. This shows that $\cos^2(x) + \sin^2(x)$ is constant. Also $\cos^2(0) + \sin^2(0) = 1$. Therefore, $\forall x \in \mathbb{R}d((0, 0), (\cos(x), \sin^2(x))) = \sqrt{\cos^2(x) + \sin^2(x)} = \sqrt{1} = 1$.
Since I showed that a binary function from $\mathbb{R}^2$ to $\mathbb{R}$ satisfying all 7 properties exists and gave a geometric proof of $\forall x \in \mathbb{R}\cos^2(x) + \sin^2(x) = 1$, a nongeometric proof of $\forall x \in \mathbb{R}\cos^2(x) + \sin^2(x) = 1$ probably also exists. Indeed, I actually used a nongeometric proof of $\forall x \in \mathbb{R}\cos^2(x) + \sin^2(x) = 1$ in the proof that a binary function from $\mathbb{R}^2$ to $\mathbb{R}$ satisfying all 7 properties exists.
Source: https://www.maa.org/press/periodicals/convergence/proportionality-in-similar-triangles-a-cross-cultural-comparison-the-student-module
