My Math Forum Inverse Evaluation

 Trigonometry Trigonometry Math Forum

 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 Display Modes Linear 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

 Contact - Home - Forums - Cryptocurrency Forum - Top