comparing the smallest positive roots of $\cos(\sin x)=x$ and $\sin(\cos x)=x$ Given $x_1$ is a root of $$\sin(\cos x)=x$$ and $x_2$ is a root of
$$\cos(\sin x)=x$$ when $x_1,x_2 \in [0 \:\: \frac{\pi}{2}]$
Then which  is greater $x_1$ or $x_2$
we have first equation as $$\arcsin x=\cos x$$ and by graphs it has one solution $x_1 \lt 1$  and similarly from second equation
$$\arccos x=\sin x$$ and by graph  it also has $x_2 \lt 1$ as a solution. but any clue to decide which is greater 
 A: Let's argue that $\cos(\sin x)>\sin(\cos x)$ at least on $[0,\frac{\pi}{2}]$.
Suppose $x\in [0,\frac{\pi}{2}$].
First, we know $\sin x\le x$ so $\sin(\cos x)\le\cos x$ with equality only when $\cos x=0$, i.e., at $x=\frac{\pi}{2}$.
Second, since $\cos x$ is decreasing, and $\sin x<=x$, $\cos (\sin x)\ge\cos x$, with equality only then $\sin x=0$, i.e., at $x=0$.
Thus, $\cos(\sin x) > \sin(\cos x).$
Hence, $y=\cos(\sin x)$ intersects $y=x$ at a larger value of $x$ than $y=\sin(\cos x)$ does.
A: Hint $y=x $ is an increasing function. So the problem is determining whether $cos (sin (x))>sin (cos (x))$ or viceversa. Now we rearrange the given expressions as $sin (cos (x))=cos (\frac {\pi}{2}-cos (x)),cos (sin (x))=cos(cos (\frac {\pi}{2}-x))$. Now $\frac {\pi}{2}-cos (x)$ranges from $\frac {\pi}{2}-1$ to $\frac {\pi}{2}+1$. And cos ranges from $-1 to 1$ . Now as cos is a decreasing function from $0-\frac {\pi}{2}$ thus value of $cos (cos (0 to 1))>cos (\frac {\pi}{2}-1 to \frac {\pi}{2})$( ignoring the part from $\frac {\pi}{2}$ to$\frac {\pi}{2}+1$ thus we see that $cos (sinx)>sin (cosx) $ thus $x2>x1$
