Putnam/Olympiad problem of the day - (Part A)

[http://math.harvard.edu/putnam(...) !!84u21X0KjW+ No.10694987]

A calculator is broken so that the only keys that still work are the [math] \, \sin, \; \cos, [/math] [math] \tan, \; \sin^{-1}, \; \cos^{-1}, \, [/math] and [math] \, \tan^{-1} \, [/math] buttons. The display initially shows 0. Given any positive rational number [math] \, q, \, [/math] show that pressing some finite sequence of buttons will yield [math] \, q [/math]. Assume that the calculator does real number calculations with infinite precision. All functions are in terms of radians.

[http://math.harvard.edu/putnam(...) !!84u21X0KjW+ No.10694989]

Previous Thread >>10623607

[http://math.harvard.edu/putnam(...) !!84u21X0KjW+ No.10695017]

Part B >>10695011

[Anonymous No.10695196]

Based

[Anonymous No.10695372]

>>10694987
If a is an angle that can be obtained then pi/2-a=sin^{-1}cosa is an angle that can be obtained.

If x is a positive lenght that can be obtined then
1/x=tan(pi/2-tan^{-1}) is a lenght that can be obtained.

If [math]\frac{1}{\sqrt{q}}[/math] is a lengh that can be obtained then [math]\frac{1}{\sqrt{q+1}}=sin tan^{-1}(\frac{1}{{\sqrt{q}})[/math]
is also.

So the set of rationals q such that [math]\sqrt{q}[/math] can be obtained is closed under the operations 1/q and q+1.

These two operations generate all positive rational numbers: from the q+1 operation you just have to obtain n/m for n<m, so the 1/q operation finishes by induction on m.

[Anonymous No.10695380]

>>10695372
[math]\frac{1}{\sqrt{q+1}}=sin(tan^{-1}(\frac{1}{\sqrt{q}}))[/math] is also. So the set of rationals q such that [math]\sqrt{q}[/math] can be obtained is closed onder the operations 1/q and q+1.
Tex preview showed it was fine. Trying the third line again.

[Anonymous No.10696591]

bump

[Anonymous No.10696670]

>>10694987
Thank you for reminding me I had to send 3 putnam problems to my students. Would have been embarrassing if I didn't after I told them I would.

[Anonymous No.10698085]

I like seeing these threads every day, thank op