Sin or Cos?

Since $\cos(\sin(x)) \gt \sin(\cos(x))$ for $x = 0, \frac{\pi}{4}, \frac{\pi}{2}$ it is reasonable to form the hypothesis that $\cos(\sin(x)) \gt \sin(\cos(x))$ for all real $x$ .

To test this hypothesis, it will suffice to show that the function

$f(x) = \cos(\sin(x)) - \sin(\cos(x))$

exceeds $0$ for all x. Now $f(x)$ is even and $2\pi$ -periodic, so we need only focus on the interval $[0,\pi]$ . We have already noted that $f(0) \gt 0$ . Also, since $\cos(\sin(x)) \gt 0$ and $\sin(\cos(x)) \lt 0$ on the subinterval $[\frac{\pi}{2}, \pi]$ we have that $f(x) \gt 0$ on this subinterval. Thus we can narrow our focus to the subinterval $I = (0, \frac{\pi}{2})$ .

Now on $I$ both $\sin(x)$ and $cos(x)$ lie on the interval $(0,1)$ . Since $\sin(z) \lt z$ for $0 \lt z \lt 1$ we have that $\sin(\cos(x)) \lt \cos(x)$ for $x$ in $I$ . Also, since $\sin(x) \lt x$ for $x$ in $I$ we have that $\cos(\sin(x)) \gt \cos(x)$ on $I$ as well. Thus for $x$ in $I$ we have that

$\sin(\cos(x)) \lt \cos(x) \lt \cos(\sin(x))$ .

We can thus finally conclude that $f(x) \gt 0$ for all real $x$ , i.e., that $\cos(\sin(x)) \gt \sin(\cos(x))$ for all real $x$ .

We know that for sin(x) and cos(x), the interval is [-1,1].

lets take interval [-1,1] as y.

now you can check sin(y) tends to 0, and cos(y) tends to 1.

one more point to note is sin(y) can be -ve and cos(y) is always +ve.

Hence you can easily compare that cos(sin x) always greater than sin(cos x).