The relationship between tan(x) and square roots Please note: I am working in DEGREES
I think the easiest way to illustrate my point is by showing some examples:


*

*$ \tan(0^\circ) = \sqrt 0 = 0$

*$ \tan(22.5^\circ) = \sqrt 2 -1$ 

*$ 3 \cdot \tan(30 ^\circ) =\sqrt 3$

*$ \tan(45 ^\circ) =\sqrt 1 = 1$

*$ \tan(60 ^\circ) =\sqrt 3$ 

*$ \tan(75 ^\circ) = 2 + \sqrt 3$ 


Ok, so there are some nice examples that have whole numbers, but then there are some that are less "pretty":


*

*$ \tan(54.73561... ^\circ) = \sqrt 2$ 

*$ \tan(65.90515... ^\circ) = \sqrt 5$ 
... and so on ...

These aren't very impressive, because once can easily generate these irrational numbers in order to get the square root of something by simply using the arctan(x) function. (arctan($\sqrt 2$) = 54.73561...

Let's look at this useless equation:
$ \tan( \arctan(\sqrt x)) = \sqrt x$ * SHOCKER *
So, I was wondering, is there possibly a way to represent this without using arctan($\sqrt x$) by the use of an infinite series - for example:
$ \tan(\sum_{n=0}^\infty $ something in terms of x ) = $\sqrt x$
I look forward to any answers or responses I may get :)
Kind regards
Joshua :)

EDIT: An interesting "discovery" (not really)
I'm putting "discovery" in inverted commas because I am fairly certain that I am not the first to find this, but I haven't been able to find it on the internet (that I have searched).
\begin{align}
\sqrt{3} & =  \frac{\sum_{n=0}^\infty \frac{(-1)^n (\frac{\pi}{3})^{2n+1}}{(2n+1)!}}{\sum_{n=0}^\infty \frac{(-1)^n (\frac{\pi}{3})^{2n}}{(2n)!}} \\
\end{align}
Now notice that $\sqrt3$ = infinite sum (THAT CONTAINS $\frac{\pi}{3}$) (The transcendental number pi divided by 3, which is the number we started with) - I do not mind having $\pi$, $e$ or $\phi$ in the infinite sum, as these are transcendental numbers. However, I would like to avoid non-transcendental numbers such as $\sqrt 2$, etc.
Let me simplify a bit more:
The bottom infinite sum converges to 0.5
\begin{align}
\sqrt{3} & =  \frac{\sum_{n=0}^\infty \frac{(-1)^n (\frac{\pi}{3})^{2n+1}}{(2n+1)!}}{0.5} \\
& = 2\sum_{n=0}^\infty \frac{(-1)^n (\frac{\pi}{3})^{2n+1}}{(2n+1)!}
\end{align}

Now all you observant people would notice that the equation can be written as $$\frac{\sqrt{3}}{2} = \sum_{n=0}^\infty \frac{(-1)^n (\frac{\pi}{3})^{2n+1}}{(2n+1)!} = sin(60^\circ)
$$
And
\begin{align}
\sqrt{3} & =  \frac{sin(60^\circ)}{cos(60^\circ)} = tan(60^\circ)\\
\end{align}

Maybe this helps someone work it out a little bit further :)
 A: Glad to see that you have made some original discoveries although these are pretty well known results in elementary calculus. It would be interesting to know as to how you arrived at them. Using differential and integral calculus it is easy to prove the following three formulas:
\begin{align}
\sin x &= x - \frac{x^{3}}{3!} + \cdots + (-1)^{n}\frac{x^{2n + 1}}{(2n + 1)!} + \cdots\tag{1}\\
\cos x &= 1 - \frac{x^{2}}{2!} + \cdots + (-1)^{n}\frac{x^{2n}}{(2n)!} + \cdots\tag{2}\\
\arctan x &= x - \frac{x^{3}}{3} + \cdots + (-1)^{n}\frac{x^{2n + 1}}{2n + 1} + \cdots\tag{3}
\end{align}
The formulas $(1), (2)$ are valid for all values of $x$ and here $x$ is the radian measure so that if you wish to calculate $\sin(60^{\circ})$ you need to write $$\sin (60^{\circ}) = \sin \left(\frac{\pi}{180}\cdot 60\right) = \sin\left(\frac{\pi}{3}\right)$$ and put $x = \pi/3$ in $(1)$ to get value of $\sin 60^{\circ}$. Thus using formulas $(1)$ and $(2)$ we have $$\frac{\sqrt{3}}{2} = \sin 60^{\circ} = \sum_{n = 0}^{\infty}(-1)^{n}\frac{(\pi/3)^{2n + 1}}{(2n + 1)!}, \frac{1}{2} = \cos 60^{\circ} = \sum_{n = 0}^{\infty}(-1)^{n}\frac{(\pi/3)^{2n}}{(2n)!}$$ And this leads to your interesting discovery.
On the other hand the formula $(3)$ holds only for $|x| \leq 1$ and the result we get is in radians and not degrees. Thus putting $x = 1$ we get $\arctan 1$ and the series in $(3)$ will give you the value $\pi/4$ which is radian equivalent of $45^{\circ}$.
Next note that the solution to $\tan f(x) = \sqrt{x}$ for $x \geq 0$ is given by $$f(x) = n\pi + \arctan{\sqrt{x}}$$ where $n$ is any integer. There is no other solution to this. If $\sqrt{x} \leq 1$ then it is possible to use $(3)$ to express $\arctan(\sqrt{x})$ as an infinite series. If $\sqrt{x} \geq 1$ then we can note that $$\arctan(\sqrt{x}) = \frac{\pi}{2} - \arctan\left(\frac{1}{\sqrt{x}}\right)$$ and since $1/\sqrt{x} < 1$ it is possible to apply formula $(3)$ to express $\arctan(1/\sqrt{x})$ as an infinite series. Thus for all values of $x \geq 0$ it is possible to express $\arctan(\sqrt{x})$ as an infinite series.
Elementary proofs of $(1), (2), (3)$ based on a combination of intuition and geometrical arguments were available with ancient Indian mathematicians (most notably Madhava) and these proofs, although not rigorous, are very much similar in spirit to the modern proofs based on differential and integral calculus.
A: If $k$ is an integer, and $f$ is a continuous, real-valued function defined on some real interval $I$ and satisfying
$$
(k - \tfrac{1}{2})\pi
  < f(x)
  < (k + \tfrac{1}{2})\pi
  \quad\text{for all $x$ in $I$,}
$$
and if $\tan f(x) = \sqrt{x}$ for all $x$ in $I$, then
$$
f(x) = k\pi + \arctan \sqrt{x}
  \quad\text{for all $x$ in $I$.}
$$
This doesn't leave any leeway for "something in terms of $x$", except to the extent that $\arctan\sqrt{x}$ can be represented by, e.g., the definite integral
$$
\arctan\sqrt{x} = \int_{0}^{\sqrt{x}} \frac{dt}{1 + t^{2}},
\tag{1}
$$
or—on the open interval $(0, 1)$—the power series (in $\sqrt{x}$)
$$
\arctan\sqrt{x}
  = \sum_{k=0}^{\infty} \frac{(-1)^{k} \sqrt{x}^{2k+1}}{2k + 1}
  = \sqrt{x}\sum_{k=0}^{\infty} \frac{(-1)^{k} x^{k}}{2k + 1},
\tag{2}
$$
or an infinite product, etc.
The series representation (2) is by no means unique. For example, the square root function is real-analytic on $(0, \infty)$ (the series representation centered at $a > 0$ has radius $a$), and $\arctan$ is real-analytic on the real line (the series centered at $c > 0$ has radius $\sqrt{c^{2} + 1}$, thanks to the complex singularities of the integrand in (1)).
