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 ?
 A: 
peterwhy has proved what the plot shows.
A: So now the question is whether two curves
$$y=f(x)=\sin(\cos x)\\
y=g(x)=\cos(\sin x)\\
x \in\left[0,\frac\pi2\right]$$
touches each other, since both curves are differentiable in the given range and we are sure the curves does not cross instead, from the given choices. (Though I have no idea now how to prove that the curves do not cross)
Differentiate the two functions,
$$f'(x) = -\cos(\cos x)\cdot\sin x\\
g'(x) = -\sin(\sin x)\cdot\cos x$$
Equating the two,
$$\begin{align*}
\sin(\sin x)\cdot\cos x &= \cos(\cos x)\cdot\sin x\\
\sin(\sin x)&= \cos(\cos x)\cdot\tan x\\
\sin^2(\sin x)&= \cos^2(\cos x)\cdot\tan^2 x\\
1-\cos^2(\sin x)&= \left[1-\sin^2(\cos x)\right]\cdot\tan^2 x\\
\end{align*}$$
If there is/were indeed a touching point, $\cos(\sin x) = \sin(\cos x)$, so
$$\begin{align*}\tan^2 x &= 1\\
x &= \frac\pi4.
\end{align*}$$
But then
$$\cos\left(\sin \frac\pi4\right)=\cos\frac{\sqrt2}2\ne\sin\frac{\sqrt2}2=\sin\left(\cos\frac\pi4\right),$$
hence it contradicts. Equality does not hold in that range.
A: 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).
A: In $(0,\pi/2]$ the inequality
$$\sin x< x$$
holds, and $\cos$ is strictly decreasing. Therefore
$$\cos\sin x-\sin\cos x>\cos\sin x-\cos x> 0$$
For $x=0$ we have $\sin\cos 0=\sin 1<1=\cos\sin 0$. The correct option is A).
