Relation between $\sin(\cos(\alpha))$ and $\cos(\sin(\alpha))$ If $0\le\alpha\le\frac{\pi}{2}$, then which of the following is true?
A) $\sin(\cos(\alpha))<\cos(\sin(\alpha))$
B) $\sin(\cos(\alpha))\le \cos(\sin(\alpha))$ and equality holds for some $\alpha\in[0,\frac{\pi}{2}]$
C) $\sin(\cos(\alpha))>\cos(\sin(\alpha))$
D) $\sin(\cos(\alpha))\ge \cos(\sin(\alpha))$ and equality holds for some $\alpha\in[0,\frac{\pi}{2}]$
Testing for $\alpha=0$, I can say that the last two options will be incorrect. However which option among the first two will hold ?

Let $a=\cos{x}$, $b=\sin{x}$, and $a,b \in[0,1]$. We are now going to see which one ($\sin{a}$ or $\cos{b}$) is larger?
Noticg that $a$ and $b$ satisfy $a^2+b^2=1$, and if we regard the value $a$ and $b$ as a pair $(a,b)$ on the $(a,b)$-plane, it should be a circle with radius $1$ in the first quadrant.
If $\sin{a}=\cos{b}=\sin{(\frac{\pi}{2}-b)}, \forall a,b \in [0,1]$ holds, then $a=\frac{\pi}{2}-b$, (i.e. the line $a+b=\frac{\pi}{2}$), but $a+b=\frac{\pi}{2}$ does not touch the circle  $a^2+b^2=1$, so the equality does not hold.
Moreover, the region $a^2+b^2=1$ locates below the line $a+b=\frac{\pi}{2}$, that is, $a+b < \frac{\pi}{2}$. And, $\sin{\theta}$ is an increasing function for $\theta \in [0,1]$, we have $\sin{a} < \sin{(\frac{\pi}{2}-b)}=\cos{b}$. The answer is A).
