Sue Liu, S6, Madras College St Andrews sent one of her super solutions in which she wrote $\sin(\cos x)$ as $\cos (\cos (x - (\pi/2)$ and then used the formula which gives the difference of two cosines as minus the product of two sines. Another triumph for Sue! After a lot more work with trig formulae Sue proves that $\cos(\sin x)$ is greater than $\sin(\cos x)$ for all $x$ and you can try this for yourselves.

There is another way of looking at this. You may like to sketch some graphs. First for $x$ between $0$ and $\pi/2$ the cosine function is decreasing and \[ 0 \leq \sin x \leq x \leq \pi/2 \] so it follows that \[1 = \cos 0 \geq \cos(\sin x) \geq \cos x \geq \cos(\pi/2) = 0 \ \ \ [1]\] Also, as for all $x$ in this interval, $ \cos x \geq 0$ and we also know that, for$ y \geq 0, \sin y \leq y$, we can put $y = cos x$ which gives\[ \sin (\cos x) \leq \cos x. \ \ \ [2]\] From [1] and [2] we see that, for $x$ between $0$ and $\pi/2$ \[ \cos(\sin x) \geq \cos x \geq \sin (\cos x). \] For $x$ between $ \pi/2 \mbox{ and } \pi$ it is even easier because in this interval $ \cos(\sin x) > 0\ \mbox{ and }\ \sin (\cos x) < 0.$ So far we have $\cos (\sin x) \geq \sin (\cos x)$ for x between 0 and $ \pi $.

For the interval $[-\pi, 0]$ put $y = - x$ then $y$ is in the interval $[-\pi, 0]$ and $x$ is in the interval $[0, \pi]$ so, using what we have already proved and the fact that sine is an odd function and cosine is an even function, we have

$$\begin{eqnarray} \cos (\sin y) = \cos (\sin - x) = \cos (- \sin x) \\ = \cos (\sin x) \\ \geq\sin (\cos x) \\ = \sin (\cos y). \end{eqnarray}$$

We have proved that $\cos (\sin x) \geq \sin (\cos x)$ for all x between $-\pi \mbox{ and } \pi$ and hence everywhere by periodicity.