$\cos (\cos x) > \sin (\sin x)$

We know that $|\sin(x)|<|x|$, which I assume is taken to be known, since you differentiated $\sin(x)$. Hence, it follows from this and Pythagorean identities that

$$\cos^2(\cos(x))=1-\sin^2(\cos(x))>1-\cos^2(x)=\sin^2(x)$$

$$\sin^2(\sin(x))<\sin^2(x)$$

Thus, it follows that

$$\cos^2(\cos(x))>\sin^2(x)>\sin^2(\sin(x))$$

And since $\cos(\cos(x))\in[\cos(1),1]\implies\cos(\cos(x))>0$ and $a^2>b^2\implies|a|>|b|\ge b$, it follows that

$$\cos^2(\cos(x))>\sin^2(\sin(x))\\\implies|\cos(\cos(x))|>|\sin(\sin(x))|\ge\sin(\sin(x))$$

We need to prove that $$\cos\cos{x}-\cos\left(\frac{\pi}{2}-\sin{x}\right)>0$$ or $$2\sin\frac{\frac{\pi}{2}-\sin{x}-\cos{x}}{2}\sin\frac{\frac{\pi}{2}-\sin{x}+\cos{x}}{2}>0,$$ which is true because By C-S $$\sin{x}\pm\cos{x}\leq\sqrt{(1^2+1^2)(\sin^2x+\cos^2x)}=\sqrt2<\frac{\pi}{2},$$ which gives $$0<\frac{\frac{\pi}{2}-\sin{x}\pm\cos{x}}{2}<\frac{\frac{\pi}{2}+\sqrt2}{2}<\pi$$