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


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evaluation, inverse 
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