Where does the following function take its extreme values?

$$
\sin \left(\cos ^{2} x\right)+\sin \left(\sin ^{2} x\right)
$$
Well-known trigonometric identities and the

$$
\sin \alpha+\sin \beta=2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}
$$

relationship can be used to transform our function into the form

$$
2 \sin \frac{1}{2} \cos \left(\frac{\cos 2 x}{2}\right)
$$

Here, $2 \sin \frac{1}{2}=0.9589$ is a constant, so our task is to find where the

$$
f(x)=\cos \left(\frac{\cos 2 x}{2}\right)=\cos \left(\frac{|\cos 2 x|}{2}\right)
$$

function takes its extreme values.

We know that the cosine function is strictly decreasing in the interval $[0, \pi]$. Furthermore,

$$
0 \leqq \frac{1}{2}|\cos 2 x| \leqq \frac{1}{2}<\pi
$$

thus, $f(x)$ takes its maximum where $\frac{1}{2}|\cos 2 x|$ is minimal, and its minimum where $\frac{1}{2}|\cos 2 x|$ is maximal.

The extreme values of the function in question and $\frac{1}{2}|\cos 2 x|$ are therefore taken at the same points. These latter are the values where $\cos 2 x=0$ or $|\cos 2 x|=1$. Based on these, the sought values are $x=k \pi / 4(k=0, \pm 1, \pm 2, \ldots)$. (I. S.)