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Problem:

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A calculator is broken so that the only keys that still work are the sin, cos, tan, cot, asin, acos, and atan buttons. 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 1/x.

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

(c) The display initially shows 0. Prove that there is a sequence of buttons that will produce 3/sqrt(5).

 Jul 2, 2019
edited by Davis  Jul 16, 2019
edited by Davis  Jul 17, 2019
edited by Davis  Jul 17, 2019
 #1
avatar+8700 
+5

A calculator is broken so that the only keys that still work are the sin, cos, tan, cot, asin, acos, and atan buttons. 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 1/x.

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

(c) The display initially shows 0. Prove that there is a sequence of buttons that will produce 3/sqrt(5).

 

Provided x can be entered, and
if the calculator uses the reverse polish notation then:

 

a)  \(​ ​\frac{1}{x}\)         \(x \Rightarrow atan \Rightarrow cot\)       \(\frac{1}{x}=cot(atan (x)) \)      \(x \in \{x|x\in\mathbb {R}\}\) 

Works with DEG and RAD.

 

I pass.

crying  !

 Jul 3, 2019
edited by asinus  Jul 3, 2019
edited by asinus  Jul 3, 2019
edited by asinus  Jul 3, 2019
 #2
avatar+507 
+2

Thanks, asinus!

Davis  Jul 16, 2019
 #3
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+2

There's probably something quicker than this !

 

With thanks to asinis and hectictar, we have the two sequences

sequence [ 1 ] \(\displaystyle \cot(\arctan(x))=1/x,\)

and 

sequence [ 2 ] \(\displaystyle \cot(\arctan(\cos(\arctan(\sqrt{x}))))=\sqrt{x+1}.\)

 

Beginning with a zero on the display, selecting cos, will produce the number 1, which can be thought of the square root of 1.

So, then

sequence  [ 2 ] will get us sqrt(1 + 1 ) = sqrt( 2 ),

sequence  [ 2 ] will get us sqrt(2 + 1) = sqrt( 3 ),

sequence  [ 2 ] will get us sqrt(3 + 1) = sqrt( 4 ),

sequence  [ 1 ] will get us 1/sqrt(4) = sqrt(1/4),

sequence  [ 2 ] will get us sqrt(1/4 + 1) = sqrt(5/4),

sequence  [ 1 ] will get us 1/sqrt(5/4) = sqrt(4/5),

and finally,

sequence  [ 2 ] will get us sqrt(4/5 + 1) = sqrt( 9/5) = 3/sqrt(5).

 Jul 17, 2019
 #4
avatar+8829 
+5

Out of curiosity, I wanted to see a length comparison between your answer and my answer:

 

 

\(\cot(\arctan(\cos(\arctan(\cot(\arctan(\cot(\arctan(\cos(\arctan(\cot(\arctan(\cot(\arctan(\cos(\arctan(\cot(\arctan(\cos(\arctan(\cot(\arctan(\cos(\arctan(\cos(0)))))))))))))))))))))))))\)

 

25 total basic functions

 

Here is WolframAlpha's result: https://www.wolframalpha.com/input/?i=cot(arctan(. . .

 

 

versus

 

 

\(\cot(\arctan(\cos(\arctan(\cos(\arctan(\cos(\arcsin(\cos(\arctan(\cos(\arcsin(\cos(\arctan(\cos(\arctan(\cos(0)))))))))))))))))\)

 

17  total basic functions

 

Here is WolframAlpha's result: https://www.wolframalpha.com/input/?i=cot(arctan(. . .

.

smiley

 

.

hectictar  Jul 17, 2019
edited by hectictar  Jul 17, 2019
edited by hectictar  Jul 18, 2019
 #5
avatar+507 
+1

Thanks!

Davis  Jul 18, 2019

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