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## 'After Thought' printed from http://nrich.maths.org/

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.