I already know about the relation between a circle and the square root of a certain number. It looks like this , where $b=\sqrt{x}$ and is actually very interesting. I wanted to check if I could take the image, and basically "write it down" in math form to come up with a, maybe useless, formula for the square root of $x$. The one I came up with is$$ \sqrt{x}=\frac{(x+1)\sin(\arccos\big(\frac{x1}{x+1}\big))}{2} $$ Now this formula is kinda long, and I want to know if, first of all, it is correct, and also if there is any possible way to simplify it without using a square root to achieve this, because that would be contradictory.

Well that's kind of why I want to know if it can be simplified or not. If it can't, well it's still nice to have but yeah it's not that effective – Kai Feb 07 '18 at 02:28

Also note that it is mainly for experimenting and for fun, I was just wondering – Kai Feb 07 '18 at 02:29

Take the square, use that $\sin^2=1\cos^2$ and you get $x$. – Feb 07 '18 at 02:30

But that would be against what I am trying to find: the square root of x – Kai Feb 07 '18 at 02:31

I knew that, but the thing is it requires a square root on itself. At this point I just think it can't really be done – Kai Feb 07 '18 at 02:33

There is no square root used. If you are going to prove that something is square root, the square root has to appear at the very end. That is what my proof does, and what appears at the end is the definition of the square root, not even its symbol. – Feb 07 '18 at 02:46

It's true but only because the RHS simplifies to $\sqrt x$. – DanielWainfleet Feb 07 '18 at 04:52
1 Answers
Too long for a comment.
I assume that you want to define the square root in general (not just $\sqrt x$ of a particular value of $x$), operated on the nonnegative reals.
If you want something that is algebraically more "elementary" than the square root operator, then what choices do you have left, besides the basic $+\times\div$?
If you would like to "think geometrically", then you must accept ideas such as: based on a circle, there are things that are geometrically very fundamental ("simpler" than square root) that appear algebraically less "simple".
Such fundamental operations can be (depending on how you frame your geometrical world view) sine, cosine, and their inverses.
Under this context, the formula you derived is correct and can be thought of as simple and NOT "kinda long".
At the same time, there's also the choice of restricting oneself to finitism or not. One can consider approximating a circle with regular polygons, and there can be algebraically tedious ways (yet geometrically intuitive) to approximate that vertical length $b$.
It takes infinite steps to really get to the square root. You have to decide whether this is an issue or not. Algebraically the whole real line has to be constructed with "infinite procedures" one kind or another.
Speaking or which, I don't know if the continued fraction view can be incorporated geometrically to a circle.
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