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 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.

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