Reach for the Summit - M-S2-A1

Note that $\cos (\sin x) = \sin \left(\frac \pi 2- \sin x \right)$ . Also note that $\frac \pi 2 - \sin x > \cos x$ for $0 \le x \le \frac \pi 2$ , because if $\frac \pi 2 - \sin x > \cos x$ , then $\frac \pi 2 > \sin x + \cos x = \sqrt 2 \sin \left(x + \frac \pi 4\right)$ , which is true, as $\frac \pi 2 > \sqrt 2 \ge \sin x + \cos x$ . Therefore $\cos (\sin x) = \sin \left(\frac \pi 2- \sin x \right) > \sin (\cos x)$ for $0 \le x \le \frac \pi 2$ .

For $\frac \pi 2 < x \le \pi$ , $\cos (\sin x) > 0$ , while $\sin (\cos x) < 0$ . Again $\cos (\sin x) > \sin (\cos x)$ .

Therefore $\boxed{\cos(\sin x) > \sin (\cos x)}$ for $0 \le x \le \pi$ .