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 Trigonometry Trigonometry Math Forum

 April 7th, 2019, 03:15 AM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 636 Thanks: 91 Prove inequality Prove inequality : $\displaystyle \underbrace{\sin \sin ... \sin }_{N-times} (N)\leq \sin \frac{1}{N}\;$ , $\displaystyle N\in \mathbb{N}$. To write it better in short-terms : $\displaystyle \sin_{N} (N)\leq\sin \frac{1}{N}$. April 7th, 2019, 01:36 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,552 Thanks: 1402 if this is a product chain it should be written as $\sin^N(N) \leq \sin\left(\dfrac 1 N \right),~N \in \mathbb{N}$ if it's a function composition chain I've seen it written as $\underset{\text{N times}}{\underbrace{\sin\circ \sin \circ \dots \sin(N)} }=\sin^{\circ N}(N) \leq \sin\left(\dfrac 1 N \right)$ which is it? Thanks from idontknow April 7th, 2019, 01:38 PM #3 Senior Member   Joined: Dec 2015 From: somewhere Posts: 636 Thanks: 91 The composition N-times of sin function. April 7th, 2019, 02:00 PM #4 Senior Member   Joined: Aug 2012 Posts: 2,386 Thanks: 746 Wolfram Alpha gives $\sin(\sin(2)) \approx 0.78907 \dots > \frac{1}{2}$. Of course, that's in radians. However, it's $< \frac{1}{2}$ for $2$ degrees. Just wanted to note that. OP must mean degrees. Thanks from topsquark and idontknow Last edited by skipjack; April 7th, 2019 at 03:21 PM. April 7th, 2019, 02:01 PM #5 Global Moderator   Joined: May 2007 Posts: 6,822 Thanks: 723 For $x \gt 0$ $\sin(x)\lt x$. Therefore $\sin(\sin(1/N))\lt \sin(1/N)$..Repeat N times to get what you want. Thanks from topsquark and idontknow Last edited by skipjack; April 7th, 2019 at 03:14 PM. April 7th, 2019, 02:18 PM #6 Senior Member   Joined: Dec 2015 From: somewhere Posts: 636 Thanks: 91 Tried all methods posted above but got no result . April 8th, 2019, 01:18 PM   #7
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 Originally Posted by idontknow Tried all methods posted above but got no result .
Did you understand my reply? April 8th, 2019, 01:19 PM   #8
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 Originally Posted by mathman Did you understand my reply?
No. April 9th, 2019, 12:11 PM   #9
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 Originally Posted by idontknow No. April 9th, 2019, 06:06 PM #10 Senior Member   Joined: Sep 2016 From: USA Posts: 645 Thanks: 407 Math Focus: Dynamical systems, analytic function theory, numerics Use induction and the fact that $\sin$ is entire and its Taylor expansion can be written as $\sin(N) = \sin(N-1) + \frac{1}{2} \cos(N-1) - \frac{1}{6}\sin(N-1) + \dotsc$. Thanks from idontknow Tags inequality, prove Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post rubis Elementary Math 0 October 11th, 2018 05:05 AM StillAlive Calculus 5 September 2nd, 2016 11:45 PM davedave Algebra 3 April 26th, 2014 05:19 PM Albert.Teng Algebra 2 April 9th, 2013 05:32 AM Albert.Teng Algebra 11 July 13th, 2012 05:42 AM

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