$\sin (\cos x)$ vs. $\cos (\sin x)$

$\sin x+\cos x\leq \sqrt 2 <\frac{π}{2}$ (since $3<π<4$ and $8<9\implies \sqrt 2 <1.5$ ), therefore $\cos x<\frac{π}{2}-\sin x\implies \boxed {\sin (\cos x) <\sin (\frac{π}{2}-\sin x) =\cos (\sin x)}$

Desmos

$B = \cos (\sin x) = \sin \left(\dfrac \pi 2 - \sin x \right)$ . Since $\sin x + \cos x = \sqrt 2 \sin \left(x + \dfrac \pi 4\right) \le \sqrt 2 < \dfrac \pi 2$ . Then $\dfrac \pi 2 - \sin x > \cos x$ . Therefore, $B = \sin \left(\dfrac \pi 2 - \sin x \right) > \sin (\cos x) = A$ or $\boxed B > A$ .

In the folllowing range we can,say that sinx<x......(1) ..so,cos(sinx)>cosx.....(2) Put...x=cosx in equation 1.. So,cosx>sin(cosx)......(3) Therefore, Cos(sinx)>sin(cosx)......... Please comment if you figure out some mistake