Since the cos(x) and sin(x) are both periodic with a period
of 2 pi, we can restrict the domain to (-pi,pi].

Since cos(sin(x)) and sin(cos(x)) are both even, we can
further restrict the domain to [0,pi].

Since cos(sin(x)) >= 0 and sin(cos(x)) < 0 in (pi/2,pi],
we can further restrict our domain to [0,pi/2].

     f(x) = sin(x) + cos(x) >= 0

    f'(x) = cos(x) - sin(x) = 0 ==> x = pi/4

    f"(x) = -[sin(x) + cos(x)] < 0 at x = pi/4

     f(0) = f(pi/2) = 1

Therefore, in [0,pi/2]

  sin(x) + cos(x) <= sin(pi/4) + cos(pi/4) = sqrt(2) < pi/2

Since the sin(x) is strictly increasing in [0,pi/2],

  sin(cos(x)) < sin(pi/2 - sin(x)) = cos(sin(x))

Therefore, 

  cos(sin(x)) > sin(cos(x))

for all real x.