Using only trigonometric functions and a single instance of the number zero, derive a formula to calculate any positive integer n.

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# Trigonometry Formula to calculate positive integer n

### #1

Posted 19 November 2013 - 05:47 AM

### #2

Posted 20 November 2013 - 06:42 PM

### #3

Posted 03 December 2013 - 08:55 PM

Using only trigonometric functions and a single instance of the number zero, derive a formula to calculate any positive integer n.

Bumping this thread.

Does that include the argument the function takes?

If not, f(x) = 0 + tan(tan-1(x)) where x is assigned the desired integer n.

Or even simpler f(x) = 0 + x. No trig needed.

I know that I'm missing something.

Could you give an example as to form, one that is not the answer?.

*Vidi vici veni.*

### #4

Posted 06 December 2013 - 03:28 AM

this one breaks the op because it uses "floor command" and also sneaks in an implied (-1) but since you asked for a *bad example,* here you go:

(floor(arccos(-cos(0))))! using the main branch of the arccos [0,pi].

**Edited by BMAD, 06 December 2013 - 03:28 AM.**

### #5

Posted 06 December 2013 - 03:29 AM

Spoiler for Fun fun...

on the right track!

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