Why can we use geometric proofs in algebra? For example, whenever I search for a proof of the Pythagorean theorem, I get a drawing of a geometric proof, yet we use the Pythagorean theorem to algebraically compute distance between points in an plane. Likewise sine cosine and tangent have geometric definitions, yet we determine their values not by drawing right triangles and measuring but by plugging angles into a function. 
I've read about how Cartesian Geometry combined Geometry and Algebra, but how can we be sure that the two are compatible? 
In other words, how do we know that we can just take a theorem like the Pythagorean Theorem, proved Geometrically, in the realm of rulers of compasses, and apply it to Algebra, in the realm of numbers? 
Geometry seems too empirical. Why is it that just because we draw something on paper and it roughly works out that we assume it's true? 
Thanks!
 A: 
Why can we use geometric proofs in algebra?
...but how can we be sure that the two are compatible

The reason is this: Euclidean geometry, formulated in full strength as in Hilbert's axioms including the completeness axiom, says that Euclidean geometry in two or three dimensions is exactly coordinate geometry over the field of real numbers. Algebra over the real numbers is, in a sense, interchangeable with this strong form of Euclidean geometry.
You could say that the basic operations in algebra (addition/subtraction/multiplication/divison) are encodings of geometric information. If you have a notion of "Euclidean length," then addition tells us how the lengths of two collinear line segments are related to the total length when placing them end to end. The area of a rectangle is based on multiplication of side-lengths. You can also treat area as an additive quantity (when the areas are disjoint, of course.)
Consider also that the Pythagorean theorem can be formulated purely in terms of real numbers, and purely in terms of geometry. They are statements of the same fact, just interpreted through slightly different languages.
Basically, the real numbers are uniquely suited to idealized measurement. Whether it be length or area, they are the "most complete" field suited for the job.
A: The "algebra" you mention is not modern, abstract algebra, but algebra as used in secondary school. Geometry is very intuitive, so using it at the secondary school level helps students to understand. That intuition seems to be built into our brains, since babies recognize faces and the like at very young ages.
However, in many higher-level classes, geometry is a result of analysis. For example, the sine and cosine functions are defined by infinite power series. The formula for the distance between two points is a definition, not derived from the Pythagorean theorem. A right angle is defined from the dot product of two vectors. The Pythagorean theorem then results pretty easily.
High school algebra is pretty much simplified analysis. So it is no wonder geometry and algebra are compatible: they both come from analysis, in their foundations if not in their presentation in secondary school.
All this does not answer the questions: why is the geometry we get from analysis such a good match for our intuition? And why is our mathematics such a good match for our universe? In my opinion, these matches are due to the fact that God is a mathematician. "Mathematics is the language with which God has written the universe" (Galileo).
A: Let me add to the previous answers. 

Geometry seems too empirical. Why is it that just because we draw
  something on paper and it roughly works out that we assume it's true?

We do not. We use axioms to prove theorems in any mathematical theory. Euclid was, as far as I know, the first one to try to build geometry on axiomatic system. This was a huge step, but his work wasn't completely satisfying because there were certain assumptions he made which could not be deduced from the axioms he has given (for example, he uses a fact that two circles of equal radii would intersect if they were close enough to construct equilateral triangle, which seems obvious, but couldn't be deduced from Euclid's axioms). It was Hilbert who was fist to give complete set of axioms of geometry as intended by Euclid and other Greek mathematicians, and that was not before 20th century. With set of axioms, you use deduction to produce theorems, such as Pythagoras theorem. Pictures/sketches are only used as a help to fully understand what is going on, and what do you actually need to prove what you want. If you carry out construction on paper, it is not a proof of the statement, rather you use abstract theory to prove certain construction will always produce what you want.
A: There is a mathematical definition of a  vector space, a vector space with inner product, a metric space. But there is not a definition (at least not one I know of) of what a geometry is. Yet there are at least three geometries which I hear every now and then, Euclidean Geometry, Hyperbolic Geometry and Paraboic geometry.
For the first one we have $5$ axioms (Due to Euclid) which tell us everything we can use to obtain a proof on euclidean geometry. It is possible to obtain a proof of Pythagoras theorem using these axioms.
On the other hand we have the plane $\mathbb R^2$ with the known metric and the 
 algebraic properties of $\mathbb R^2$ as a vector field. We can verify that the plane satisfies all of the axioms Euclid asks for, hence everything that can be proved using Euclid's axioms is also true in the plane. Although I'm not sure if every statement that can be expressed in "Euclid's language" that is true in the plane is necessarily provable from the $5$ axioms.
A: 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
