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 December 24th, 2010, 12:56 AM #1 Newbie   Joined: Dec 2010 Posts: 3 Thanks: 0 Help with Trigonometry Hi guys .... $\text{Find the smallest real number a such that for any }n\geq 3 \\ \\ \sum _{k=3}^n \frac{\sin \frac{\pi }{k(k+1)}}{\cos \frac{\pi }{k} \ \cos\frac{\pi }{k+1}}\ <\ a$
 January 1st, 2011, 01:51 AM #2 Senior Member   Joined: Nov 2010 Posts: 288 Thanks: 1 Re: Help with Trigonometry this expression is alway less or equal to squeroot(3) -1 when n=3 it is equal when n>3 it is always less than that
 January 1st, 2011, 04:30 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,942 Thanks: 2210 When n>3, the sum is obviously more than ?3 - 1, as more terms are summed (and all are positive). However, the sum seems never to exceed ?3.
January 1st, 2011, 09:28 PM   #4
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Re: Help with Trigonometry

Quote:
 Originally Posted by rayan Kalak Hi guys .... $\text{Find the smallest real number a such that for any }n\geq 3 \\ \\ \sum _{k=3}^n \frac{\sin \frac{\pi }{k(k+1)}}{\cos \frac{\pi }{k} \ \cos\frac{\pi }{k+1}}\ <\ a$
$\sin \frac{\pi}{k(k+1)}= \sin (\frac{\pi}{k}-\frac{\pi}{k+1})$

The summation can be written as

$\sum^{n}_{k=3}\frac{\sin \frac{\pi}{k}\cos \frac{\pi}{k+1}- \cos \frac{\pi}{k}\sin \frac{\pi}{k+1}}{\cos \frac{\pi}{k}\cos \frac{\pi}{k+1}}$

$=\sum^{n}_{k=3} \tan \frac{\pi}{k} -\tan \frac{\pi}{k+1}$

Using the method of differences, you are left with $\tan \frac{\pi}{3} - \tan \frac{\pi}{n}$. As n approaches infinity, tan (pi/n) approaches 0. Therefore, it can be deduced that the summation does not exceed sqrt(3) as mentioned by skipjack.

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