calculus
posted by calculus on .
cos^1(cos15pi/6)

Not 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.