calculus - Steve, Wednesday, September 21, 2011 at 4:17pm
Not much calculus here, but here goes:
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.