Sine is not a sin cos cos is not a cause

We have $\cos(\sin{x})>\sin(\cos{x})$ for all $x$ , based on the following five observations.

1. $\cos(\sin{x})>\cos(x)>\sin(\cos{x})$ on the interval $(0,\pi/2)$ ,
2. $\cos(\sin{x})>0>\sin(\cos{x})$ on the interval $(\pi/2,\pi]$ ,
3. The inequality holds for $x=0$ and $x=\pi/2$ ,
4. Both functions have a period of $2\pi$ , and
5. Both functions are even.

One way to solve the problem is to realize that while $\sin(\cos x)$ may attain negative values for some x , $\cos(\sin x)$ will always be a positive number. Hence, the inference.

P.S. Evidently this is a pretty anomalous solution, but given the limited scope of options it makes solving pretty fast.

$\sin x+\cos x \leq \sqrt{2}$

And $\sqrt{2} < \dfrac{\pi}{2}$

So

$\sin x+\cos x \leq \sqrt{2} < \dfrac{\pi}{2}$

$\implies \sin x < \dfrac{\pi}{2} - \cos x$

$\implies \sin( \cos x) < \cos ( \sin x)$

If x = 0 then, = sin (cos 0) = sin(1) = 0.84

=cos(sin 0) = cos(0) = 1

0.84 < 1 Ans. cos (sin x)

This is valid if sinx or cosx is taken in degrees and not radians.

f(x)=sin(cos x) and g(x)=cos(sin x) so the images of f will variate between sin[0,1] which is [sin(0),sin(1)] = [0, 0.017] And the images of g will variate between cos[0,1] which is [0.999 ,1] therefore g(x)>f(x)