March 23, 2017

Post a New Question

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


  • calculus - , Wednesday, September 21, 2011 at 4:17pm

    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.

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question