You start from 0 and have to try to get to any number. I've seen this before (of course I never remember solutions). All I can remember was a couple of hints we got, we had to prove a couple of things like you could get from x to something related to x (might have been double, might have been something else).. that eliminates needing to think about sin, cos and tan, and then use those two facts to try and get from 0 to any number. So try playing around to see if you can get to double a number or half or something like that.

I don't remember exactly how to do this problem. However, I do know that if you set up a table of sorts with the given functions, you will be able to find a way to generate all integers, and also be able to create the ratios of any two integers. If I have some time I may play with this problem and find the old solution.

Funny problem. The best solution would be to buy a new calculator ... Or : Let's define : f1 := sin(arcos(x)) = sqrt (1 - x^2); f2 := sin(arctan(x)) = x / sqrt (1 + x^2); f3 := cos(arctan(x)) = 1 / sqrt (1 + x^2); f4 := tan(arccos(x)) = sqrt (x^2 - 1) / x; f5 := tan(arcsin(x)) = x / sqrt (x^2 - 1); Then f4(f2(x)) = 1/x g(x) = f4(f2(f3(x))) = sqrt(1 + x^2) g(g(x) = sqrt (2 + x^2)); So : g^n(x) = sqrt (n + x^2); g^(n^2)(0) = n

Actually by this method, you get all the numbers of the form sqrt(m/n) , that is, it also includes any non-rational number whose square is rational.

My friend said me many years ago that it is Landau problem. And Landau solution was the same as yours, Gyan.

Myth - Is this the famous physicst Landau? (Liquid helium)?

As far as I know it is famous russian mathematician.

You can start by using the cosine of 0 to get 1, and then work your way around from there.

I believe the correct approach is induction on the numerator or denominator of $ q$. Edit: Maybe not. TripleM's approach seems viable.

t0rajir0u wrote: I believe the correct approach is induction on the numerator or denominator of $ q$. I think that actually, strong induction on the sum of the numerator and denominator of $ q$ is the most efficient strategy. See what you think of my wiki write-up : 1995 USAMO Problems/Problem 2.

We define the functions \begin{align*} f(x) &= \sin\arctan{x} = \frac{x}{\sqrt{1+x^2}},\\ g(x) &= \cos\arctan{x} = \frac{1}{\sqrt{1+x^2}},\\ h(x) &= \tan\arccos{x} = \frac{\sqrt{1-x^2}}{x}, \end{align*}where $f,g$ are defined for all positive $x$ and $h$ for positive $x<1$. We first prove by strong induction on $n$ that all numbers of the form $\sqrt{m/n}$ for $0<m\le n$ can be achieved. The base case $n=1$ is trivial since we start with $0$ and $\cos0=1$. For the inductive step, note that (keep in mind that $m\le n$ here) \[f\left(\sqrt{\frac{m}{n-m}}\right) = \frac{\sqrt{\frac{m}{n-m}}}{\sqrt{1+\frac{m}{n-m}}} = \sqrt{\frac{m}{n}},\quad g\left(\sqrt{\frac{n-m}{m}}\right) = \frac{1}{\sqrt{1+\frac{n-m}{m}}} = \sqrt{\frac{m}{n}}.\]Since either $m\ge n-m$ or $n-m\ge m$, at least one of \[\sqrt{\frac{m}{n-m}},\quad \sqrt{\frac{n-m}{m}}\]can be achieved by the inductive hypothesis, as desired. Now we consider the case $m>n$. From the induction, we have that $\sqrt{n/(m+n)}$ is achievable, so \[h\left(\sqrt{\frac{n}{m+n}}\right) = \frac{\sqrt{1-\frac{n}{m+n}}}{\sqrt{\frac{n}{m+n}}} = \sqrt{\frac{m}{n}}\]is as well, and we are done.

OK, this will be an annoying bump of an old topic, but since I'm a great fan of L. D. Landau... I have to fill in the details because Dau was mentioned here. Moisei Isakovich Kaganov talks about it in his article here (Quantum, march/april 1993). It was Landau's mental game - look at a license plate (back then, the plates had 4 numbers, in blocks by two) Image not found Replace the "-" with "=" and make the equality correct by using mathematical symbols on both sides and digits on each side. Y. Gandel has shown that the solution always exists since $\sec \arctan \sqrt{N}=\sqrt{N+1}$. That kills the spirit in the game, doesn't it? As Boris Gorobec notes here (Nauka i Zhisn, No. 1, 2000) that secant and cosecant are not in the Russian school curriculum anymore, and that regular students don't know what they are - hence, they couldn't use it in Landau's game, since if you use it as a cosine reciprocal - you need another 'one' for the numerator or exponent.

Since $\arccos\sin\theta = \frac{\pi}{2}-\theta$ and $\tan\left(\frac{\pi}{2}-\theta\right)=\frac{1}{\tan\theta}$ for $0<\theta <\frac{\pi}{2}$, we have for all $x>0$: \[\tan\arccos\sin\tan x = \tan\left(\frac{\pi}{2}-\arctan x\right)=\frac{1}{x} (*).\] Also for $x\ge 0$, we have \[\cos\arctan\sqrt{x}=\frac{1}{\sqrt{x+1}} (**).\] Hence, by $(*)$, \[\tan\arccos\sin\arctan\cos\arctan\sqrt{x}=\sqrt{x+1}.\] We induct on the denominator of $r$, to prove that $\sqrt{r}$, for all non-negative rational $r$, can be obtained using the operations \[\sqrt{x}\mapsto\sqrt{x+1}\text{and} x\mapsto\frac{1}{x}.\] If the denominator is $1$, we can obtain $\sqrt{r}$, for every non-negative integer $r$, by applying $\sqrt{x}\mapsto\sqrt{x+1}$ repeatedly. Now assume we can get $\sqrt{r}$ for all rational numbers $r$ with denominator up to $n$. In particular we can achieve \[\sqrt{\frac{n+1}{1}}, \sqrt{\frac{n+1}{2}}, \cdots , \sqrt{\frac{n+1}{n}}\]. Hence we can also get \[\sqrt{\frac{1}{n+1}}, \sqrt{\frac{2}{n+1}},\cdots , \sqrt{\frac{n}{n+1}},\] and $\sqrt{r}$, for any positive $r$ of denominator $n+1$ is obtainable by applying $\sqrt{x}\mapsto\sqrt{x+1}$. Hence for any positive rational $r$, we can obtain $\sqrt{r}$, and in particuar we can obtain $\sqrt{q^2}=q$. $\Box$