# 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 :)

• Taking square roots is usually easier than computing tangents Jul 18, 2016 at 13:43
• Do you know that: $\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+...$ ? Jul 18, 2016 at 13:49
• @EmilioNovati, this is kind of useless in the OP's case, since this expression will contain $\sqrt{x}$ Jul 18, 2016 at 13:50
• @EmilioNovati That will not help to detect the "nice" solutions. Jul 18, 2016 at 13:50
• Another one is $tan(22.5°)=\sqrt{2}-1$ Jul 18, 2016 at 13:50

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.

• Thank you for your answer :) Yes, I used the Taylor series of sin(x) and cos(x) to arrive at $\sqrt 3 = ...$ not very amazing, but I thought it was pretty cool ;) ... Anyways, back to the question... I started looking at this topic after seeing some infinite series for famous irrational numbers, such as $\pi$, $\phi$, $e$ and $\sqrt 2$, however, there were none for any other irrational numbers such as $\sqrt 5$, etc... But how else are we supposed to write irrational numbers? Can't be fractions... Can't write our infinitely many numbers.... So why not an infinite sum? Jul 21, 2016 at 15:09
• That's why I tried to look for square root 3 on the Internet, but no infinite sums were shown, leading me to try find one... And realizing $3 * tan(30) = \sqrt 3$, I started investigating trig functions... And Taylor series... If this question has no "answer" at this moment in time, it might take "new mathematics" to find an answer, or existing maths may be able to solve it... Either way, I feel that there should be a way to represent these irrational numbers. **Is there maybe a way to represent $\sqrt x$ as an infinite sum in terms of x? Jul 21, 2016 at 15:13

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)).

• I have, of course, used representations of $\arctan$ in radians, both because they're more pleasant, and because the EDIT is expressed in radians; to convert values to degrees, multiply by $180/\pi$. Jul 20, 2016 at 21:08
• Yes I also used radians to work out the formula for $\sqrt 3$. It is just sometimes easier to work with. With reference to your answer, is there absolutely no way to express $arctan(\sqrt x$) without using $\sqrt x$ (as shown in formula 2 of your post? Jul 21, 2016 at 5:11
• In the sense I think you're asking, the answer is "no": If there were a power series in $x$, then $\arctan\sqrt{x}$ would be analytic in a neighborhood of $0$, which it isn't. Again, if $a > 0$ is a number, it's possible to express $\sqrt{x}$ as a power series in $(x - a)$ over the interval $(0, 2a)$ (and therefore to express $\arctan \sqrt{x}$ as a power series in $(x - a)$ over some open neighborhood of $a$), but I don't see any particularly pleasant or illuminating way to do this. Jul 21, 2016 at 11:49