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A calculator is broken. The only keys that work are $\sin, \cos, \tan, \cot, \arcsin, \arccos$, and $\arctan$ buttons.

The original display is $0$.

In this problem, we will prove that given any positive rational number $q$, show that pressing some finite sequence of buttons will yield $q$.

Functions are always in radian form.

(a) Find and prove that there exists a sequence of buttons that will turn $\sqrt x$ into $\sqrt{x+1}$.

For this, I got lucky and tried out $\sec (\arctan(x))$ on my calculator and got $\sqrt{x^2+1}$, so I just used the reciprocal and got $\cos (\arctan (x)) = \frac{1}{\sqrt{x^2+1}}$. However, how I can actually prove this?

(b) Prove that there exists a sequence of buttons that will yield $\frac{3}{\sqrt{5}}$.

I know that to go from $x$ to $\frac{1}{x}$ it is $\cot (\arctan (x))$, and I will have to use part a) to get to part b).

How do I utilize this? I also know that $\sqrt{\frac{9}{5}}$ = $\frac{3}{\sqrt{5}}$, but after that, I'm stuck. Any hints?

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  • $\begingroup$ What do you know about triangles? $\endgroup$ – Euler....IS_ALIVE Jan 13 '15 at 6:38
  • $\begingroup$ This is an 1995 USAMO problem and has been asked a few times on Math.SE. However I cannot find the exact question in order to mark it as a duplicate. $\endgroup$ – Cyclohexanol. Jan 13 '15 at 6:43
  • $\begingroup$ @MathNoob: Yes, but the USAMO problem only asks to show that "given any positive rational number q, show that pressing some finite sequence of buttons will yield q". I don't believe that there are any similar questions here on Math.SE that have this specific a) and b). $\endgroup$ – Mathy Person Jan 13 '15 at 17:45
  • $\begingroup$ @Euler....IS_ALIVE: I do know that I have to make a right triangle with sides $1$ and $x$ and hypotenuse $\sqrt{x+1}$, but don't the functions $\sin, \cos, \tan,$, etc., but I don't get how we can plug in $\sqrt{x}$ into one of these functions as $\sqrt{x}$ is a side, not an angle. $\endgroup$ – Mathy Person Jan 13 '15 at 17:49
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(a) The relation you found could also be found on Wikipedia. Since you don't have a working $\sec$ button on your calculator, you haven't yet answered the question in a strict sense at this point. But you could use the $\cot(\arctan(x))=\frac1x$ you mentioned under (b) to turn your $\frac1{\sqrt{x^2+1}}$ into the desired $\sqrt{x^2+1}$.

(b) Repeated application of $(a)$ will giev you $0=\sqrt0,\sqrt1,\sqrt2,\dots,\sqrt9=3$ as the numerator. You could also use the same procedure to compute $\sqrt5$ or, perhaps even simpler, $\frac1{\sqrt5}$. So your main problem here is, how to multiply or divide without multiplication or division key, and without using the memory of the calculator to store either the numerator or the denominator. I'd suggest using continued fractions. I'll leave it at this level since you only asked for a hint.

If you want to check your findings, here is a possible sequence of intermediate result numbers which the calculator could display:

0.00000000000000
1.00000000000000
0.78539816339745
0.70710678118655
0.61547970867039
1.41421356237310
0.95531661812451
0.57735026918963
0.52359877559830
1.73205080756888
1.04719755119660
0.50000000000000
0.46364760900081
0.89442719099992
0.72972765622697
0.74535599249993
0.64052231267942
1.34164078649987
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For question (b): Just as MvG said above, we can get $\frac{1}{\sqrt{5}}$. Note that $\sec(\arcsin x)=\frac{1}{\sqrt{1-x^2}}$, so $\sqrt{\frac{5}{4}}$ may be obtained, which follows $\sqrt{\frac{4}{5}}$. Using $\sec(\arctan x)=\sqrt{x^2+1}$ can give the answer.

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(a) Since $\tan(\arctan(x))= x$, we can invert the equation to get $\frac{1}{\tan(\arctan(x))}=\boxed{\cot(\arctan(x))}$ (b) If you draw a triangle with legs 1 and $\sqrt{x}$, the hypotenuse would be $\sqrt{x+1}$. To have $\frac{1}{\sqrt{x+1}}$, you would do $\cos(\arctan(\sqrt{x}))$. To get $\sqrt{x+1}$ from $\sqrt{x}$, you would need to do $\frac{1}{\cos(\arctan(\sqrt{x}))}$. We know we can get the reciprocal from part (a). So, getting the reciprocal of $\frac{1}{\cos(\arctan(\sqrt{x}))}$, we get $\boxed{\cot(\arctan(\cos(\arctan(\sqrt{x})))}$ (c) Let's call the function we got a part (a) as $A$, and the function we got a part (b) as $B$. Then, starting from 0, we get: $$0 \overset{B}{\to} \sqrt 1 \overset{B}{\to} \sqrt 2 \overset{B}{\to} \sqrt 3 \overset{B}{\to} \sqrt 4 \overset{A}{\to} \sqrt \frac{1}{4} \overset{B}{\to} \sqrt \frac{5}{4} \overset{A}{\to} \sqrt \frac{4}{5} \overset{B}{\to} \sqrt \frac{9}{5} = \frac{3}{\sqrt 5}$$From, this we get the sequence of $\boxed{BBBBABAB}$, which is equal to... Something very long, which I do not need to type! Thus, I have proved that I can create the number $\frac{3}{\sqrt 5}$ from $0$!

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