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?
Best Answer
(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: