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Topic: Trigonometry (Read 514 times) |
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Radiohead_man
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Trigonometry
« on: Dec 13th, 2004, 7:56am » |
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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.
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I'm a reasonable man
get off my case, get off my case.
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Barukh
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Re: Trigonometry
« Reply #1 on: Dec 13th, 2004, 9:16am » |
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on Dec 13th, 2004, 7:56am, Radiohead_man wrote:| cos(a); cos(2a); cos(4a); cos(8a); ... cos(2na), ... |
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Did you mean 2na ?
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John_Gaughan
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Re: Trigonometry
« Reply #2 on: Dec 13th, 2004, 12:58pm » |
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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 
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x == the_question
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Obob
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Re: Trigonometry
« Reply #3 on: Dec 13th, 2004, 2:13pm » |
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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.
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Barukh
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Re: Trigonometry
« Reply #4 on: Dec 13th, 2004, 11:08pm » |
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on Dec 13th, 2004, 2:13pm, Obob wrote:| Therefore there are no such a. |
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What about a = 2n+1[pi]/3 ?
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Obob
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Re: Trigonometry
« Reply #5 on: Dec 13th, 2004, 11:24pm » |
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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.
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Grimbal
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Re: Trigonometry
« Reply #6 on: Dec 14th, 2004, 3:59am » |
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on Dec 13th, 2004, 11:08pm, Barukh wrote:
What about a = 2n+1[pi]/3 ?
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Or (n[smiley=pm.gif]1/3)*2[pi]
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