Question

What are the absolute extrema of `f(x)= |sin(x) - cos(x)|` on the interval [-pi,pi]?

Answer 1

0 and `sqrt2`.

Explanation:

`0<=|sin theta|<=1`
#sin x - cos x = sin x -sin (pi/2-x) =2 cos ((x+pi/2-x)/2) sin ((x-(pi/2-x))/2) =- 2 cos (pi/4) sin (x-pi/4) = -sqrt2 sin (x-pi/4)#
so, #|sin x - cos x|=| -sqrt2 sin (x-pi/4)| =sqrt2|sin (x-pi/4)| <=sqrt2#.