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 July 11th, 2017, 11:50 PM #1 Member   Joined: Jul 2017 From: KOLKATA Posts: 49 Thanks: 3 Inverse Evaluation Simplify arccos(2x*sqrt(1 - x^2)) where 1/sqrt(2) < x < 1 July 12th, 2017, 12:27 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,931 Thanks: 2205 It simplifies to 2arcsin(x) - $\pi$/2. July 12th, 2017, 12:32 AM #3 Member   Joined: Jul 2017 From: KOLKATA Posts: 49 Thanks: 3 Thanks. Could you please write the steps? Last edited by skipjack; July 12th, 2017 at 12:55 AM. July 12th, 2017, 01:06 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,931 Thanks: 2205 Let x = cos(y), where 0 < y < $\pi$/4, then √(1 - x²) = sin(y) and arccos(2x√(1 - x²)) = arccos(2cosy)sin(y)) = $\pi$/2 - arcsin(sin(2y)) = $\pi$/2 - 2y = $\pi$/2 - 2arccos(x) = $\pi$/2 - 2($\pi$/2 - arcsin(x)) = 2arcsin(x) - $\pi$/2. July 12th, 2017, 03:35 AM #5 Member   Joined: Jul 2017 From: KOLKATA Posts: 49 Thanks: 3 Thanks Tags evaluation, inverse 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 Rejjy Calculus 11 December 25th, 2016 08:56 PM harley05 Calculus 3 May 9th, 2015 05:24 PM mathbalarka Calculus 5 February 18th, 2013 12:53 AM sj@vanbunningen.eu Calculus 5 October 20th, 2008 06:34 AM Soha Algebra 2 February 3rd, 2007 06:33 AM

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