A calculator is broken so that the only keys that still work are the $\sin$, $\cos$, $\tan$, $\cot$, $\sin^{-1}$, $\cos^{-1}$, and $\tan^{-1}$ buttons. The display initially shows 0. (Assume that the calculator does real number calculations with infinite precision. All functions are in terms of radians.)

(a) Find, with proof, a sequence of buttons that will transform $x$ into $\frac{1}{x}$.

(b) Find, with proof, a sequence of buttons that will transform $\sqrt x$ into $\sqrt{x+1}$.

(c) Prove that there is a sequence of buttons that will produce $\frac{3}{\sqrt{5}}$.

This is a continuation of a closed problem a while ago. I have solved parts, a and b but c is a challenge.

(a) We know that $\tan(\arctan(x))= x$ so inversing the equation, you get $\frac{1}{\tan(\arctan(x))}=\boxed{\cot(\arctan(x))}$

(b) We can solve this using right-triangle trigonometry. With legs, 1 and $\sqrt{x}$, the hypotenuse would be $\sqrt{x+1}$. To have $\frac{1}{\sqrt{x+1}}$, you would write it as $\cos(\arctan(\sqrt{x}))$. To transform $\sqrt{x}$ ot $\sqrt{x+1}$, you would have $\frac{1}{\cos(\arctan(\sqrt{x}))}$. From part (a), we can find the reciprocal of anything. All we need is the reciprocal which is $\boxed{\cot(\arctan(\cos(\arctan(\sqrt{x})))}$

How can we solve c when the initial display is 0?


marked as duplicate by Joel Reyes Noche, JMoravitz, MJD, user223391, Jonas Meyer Jun 16 '15 at 3:54

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    $\begingroup$ All you have to do is get from $0$ to $\sqrt{\dfrac{9}{5}}$ using the operations $x \mapsto \dfrac{1}{x}$ and $\sqrt{x} \mapsto \sqrt{x+1}$. $\endgroup$ – JimmyK4542 Jun 16 '15 at 0:23
  • $\begingroup$ But how can you do that? $\endgroup$ – SAM Jun 16 '15 at 0:23
  • $\begingroup$ SAM, why on Earth did you reask the question instead of editing the older version to get it reopened? That's a no-no.by our norms. Mind you, this is a much improved version, and editing this to the original version would have worked. $\endgroup$ – Jyrki Lahtonen Jun 16 '15 at 5:23

The inverses of the operations $x \mapsto \dfrac{1}{x}$ and $\sqrt{x} \mapsto \sqrt{x+1}$ are $x \mapsto \dfrac{1}{x}$ and $\sqrt{x} \mapsto \sqrt{x-1}$.

Let $A$ denote the operation $x \mapsto \dfrac{1}{x}$ and $B$ denote the operation $\sqrt{x} \mapsto \sqrt{x-1}$.

Let's work backwards from $\dfrac{3}{\sqrt{5}} = \sqrt{\dfrac{9}{5}}$ by using $B$ if the current number is greater than or equal to $1$, and $A$ if the current number is less than $1$:

$\sqrt{\dfrac{9}{5}} \overset{B}{\to} \sqrt{\dfrac{4}{5}} \overset{A}{\to} \sqrt{\dfrac{5}{4}} \overset{B}{\to} \sqrt{\dfrac{1}{4}} \overset{A}{\to} \sqrt{4} \overset{B}{\to} \sqrt{3} \overset{B}{\to} \sqrt{2} \overset{B}{\to} \sqrt{1} \overset{B}{\to} \sqrt{0} = 0$.

Now, start with $0$ and apply the operations $B^{-1}$, $B^{-1}$, $B^{-1}$, $B^{-1}$, $A^{-1}$, $B^{-1}$, $A^{-1}$, $B^{-1}$ in that order to get $\dfrac{3}{\sqrt{5}}$


Use B four times to get $\sqrt{4}$. Then use A to get $\sqrt{\frac{1}{4}}$. Then use B to get $\sqrt{\frac{5}{4}}$. Then use A to get $\sqrt{\frac{4}{5}}$. Then use B to get $\sqrt{\frac{9}{5}}=\frac{3}{\sqrt{5}}$


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