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March 27, 2015

March 27, 2015

Posted by **calculus** on Sunday, September 18, 2011 at 3:58pm.

- calculus -
**Steve**, Wednesday, September 21, 2011 at 4:17pmNot much calculus here, but here goes:

by definition

cos(cos^-1(x)) = x

cos^-1(cos(x)) = x

for x in suitable ranges. Now, for cos^-1(x) the function takes on principal values between 0 and π.

So, cos^-1(cos(15π/6)) = cos^-1(cos 5π/2) = cos^-1(0) = π/2.

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