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 Trigonometry Trigonometry Math Forum

 August 12th, 2018, 08:54 PM #1 Senior Member   Joined: Jan 2012 Posts: 140 Thanks: 2 Series sum of inverse sin Hi, How can we evaluate $\displaystyle \sum_{i=1}^{2n}x_{i}$ given, $\displaystyle \sum_{i=1}^{2n}\sin^{-1} (x_{i})=n\pi$ Thx. August 12th, 2018, 09:25 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,553 Thanks: 1403 I see nothing that keeps us from keeping $x_i$ constant. $n \pi = \displaystyle \sum \limits_{i=1}^{2n}\sin^{-1}(x_i) = 2n \sin^{-1}(x)$ $\dfrac{\pi}{2} = \sin^{-1}(x)$ $x = 1$ and thus $\displaystyle \sum \limits_{i=1}^{2n}~x_i = 2n(1) = 2n$ We note that $x_i \leq 1$ and so $n\pi$ is the maximum value the 2nd sum can attain and it attains it only when $x_i = 1,~\forall i$ Thus the value of the first sum is unique. Thanks from happy21 August 13th, 2018, 08:32 AM #3 Senior Member   Joined: Jan 2012 Posts: 140 Thanks: 2 Thx. Tags inverse, series, sin, sum 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 capea Complex Analysis 0 March 5th, 2012 01:56 AM jaredbeach Algebra 1 November 17th, 2011 11:58 AM ZardoZ Real Analysis 10 September 21st, 2011 09:00 AM JOANA Abstract Algebra 0 March 18th, 2011 06:29 AM Barbarel Number Theory 9 April 11th, 2009 09:58 AM

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