If π ≤ x ≤ 2π and y = cos–1 (cos x), find
y = cos-1 (cos x)
for x (π, 2π)
y = cos-1(cos x)
= cos-1(cos (π + (x - π)))
= cos-1(-cos (x-π))
= π – (x - π)
= 2π - x
[Since, cos(π+x) = - cos x and cos-1(-x) = π-x]
So,
For cos-1(cos x), x = n are the ‘sharp corners’ where slope changes from 1 to -1 or vice versa, i.e., the points where the curves are not differentiable.
So, for x [π,2π]
(Ans)