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July 12th, 2017, 12:50 AM   #1
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Inverse Evaluation

Simplify arccos(2x*sqrt(1 - x^2)) where 1/sqrt(2) < x < 1
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July 12th, 2017, 01:27 AM   #2
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It simplifies to 2arcsin(x) - $\pi$/2.
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July 12th, 2017, 01:32 AM   #3
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Thanks. Could you please write the steps?

Last edited by skipjack; July 12th, 2017 at 01:55 AM.
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July 12th, 2017, 02:06 AM   #4
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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.
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July 12th, 2017, 04:35 AM   #5
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Thanks
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