Let f(x) = sin(sin(sinx)), and g(x) = cos(cos(cosx)).
f'(x) = cos(sin(sinx))cos(sinx)cosx = 0 --> x = pi/2 and 3pi/2 only
Examine f(x) at both derivative and boundary points:
f(0) = 0, f(pi/2) = sin(sin1)), f(3pi/2)) = -sin(sin1)), f(2pi) = 0
So, x = pi/2 is absolute max, and x = 3pi/2 is absolute min.
g'(x) = -sin(cos(cosx))sin(cosx)cosx = 0 --> x = {0, pi/2 ,pi, 3pi/2, 2pi}
Examine g(x) at these points(they include boundaries, too):
g(0) = g(pi) = g(2pi) = cos(cos1)), g(pi/2) = g(3pi/2) = cos1
So, absolute min is x = 3pi/2, and absolute max is x = 0,pi,2pi. (cos(cos1) > cos1)
as x : 0 --> pi/2
f(0) < g(0) --> f(pi/2) = sin(sin1) > g(pi/2) = cos1 <-- There is a crossing here.
as x : pi/2 --> pi
f(pi/2) > g(pi/2) --> f(pi) = 0 < g(pi) <-- There is a crossing here, and g(x)'s deriative is never zero between pi/2 and pi.
After this point, f(x) crosses x = 0 line and does not return back since it's f'(x) = 0 point is at 3pi/2, where it is abs. min.
g(x) always positive as seen from its derivative and boundary points. So, there must be 2 roots.
