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        riddles >> putnam exam (pure math) >> Trigonometry 
        
(Message started by: Radiohead_man on Dec 13^th, 2004, 7:56am)

Title: Trigonometry
Post by Radiohead_man on Dec 13^th, 2004, 7:56am

Find all the values of a for which the series
cos(a); cos(2a); cos(4a); cos(8a); ... cos(2na), ...
consists of negative numbers only.

Title: Re: Trigonometry
Post by Barukh on Dec 13^th, 2004, 9:16am

on 12/13/04 at 07:56:48, Radiohead_man wrote:

cos(a); cos(2a); cos(4a); cos(8a); ... cos(2na), ...

Did you mean 2ⁿa ?

Title: Re: Trigonometry
Post by John_Gaughan on Dec 13^th, 2004, 12:58pm

Interesting. It seems that as n gets larger, a is restricted to larger and larger values. I am not sure if a converges on anything though. It appears not to, but my math skills only go up to Calculus III ;)

Title: Re: Trigonometry
Post by Obob on Dec 13^th, 2004, 2:13pm

The condition that cos a < 0 implies that [pi]<a<2[pi] (if a is a solution to the problem then so is 2[pi]k+a for any integer). Next, the condition that cos 2a < 0 implies that either [pi]/2<a<[pi] or 3[pi]/2<a<2[pi]. Taking into consideration the first restraint, we have 3[pi]/2<a<2[pi].

In general, the condition that cos 2^k a < 0 implies that a lies in one of the 2^k intervals of the form m[pi]/2^(k-1) +[pi]/2^k, for 0[le] m [le] (2^k)-1. Taking into consideration the restrictions imposed from the inequalities, cos a < 0,..., cos 2^(k-1) a <0, this clearly implies that 2[pi]-[pi]/2^k < a < 2[pi]. But if a is any number in [0,2[pi], then for some sufficiently large k, the inequality 2[pi]-[pi]/2^k < a < 2[pi] does not hold. Therefore there are no such a.

Title: Re: Trigonometry
Post by Barukh on Dec 13^th, 2004, 11:08pm

on 12/13/04 at 14:13:09, Obob wrote:

Therefore there are no such a.

What about a = 2ⁿ⁺¹[pi]/3 ? ;)

Title: Re: Trigonometry
Post by Obob on Dec 13^th, 2004, 11:24pm

Wow I'm a complete idiot. If you replace every instance of cos with sin in my above post, everything I said should be true.

Title: Re: Trigonometry
Post by Grimbal on Dec 14^th, 2004, 3:59am

on 12/13/04 at 23:08:37, Barukh wrote:

What about a = 2ⁿ⁺¹[pi]/3 ? ;)

Or (n[smiley=pm.gif]1/3)*2[pi]