11.1. $\quad$ Find the smallest period of the function $y=\cos ^{10} x+\sin ^{10} x$.
Answer: $\frac{\pi}{2}$. Solution. First, we check that $\frac{\pi}{2}$ is a period of the function. Indeed,

$$
y\left(x+\frac{\pi}{2}\right)=\left(\cos ^{2}\left(x+\frac{\pi}{2}\right)\right)^{5}+\left(\sin ^{2}\left(x+\frac{\pi}{2}\right)\right)^{5}=\sin ^{10} x+\cos ^{10} x
$$

To prove that this is the smallest period, we make the substitution $u=\sin ^{2} x$. Then $y=(1-u)^{5}+u^{5}$. Considering the function $y(u)$ on the interval [0,1] (this corresponds to the change of $x$ over the interval from 0 to $\frac{\pi}{2}$), we get that $y(u)$ decreases on the interval $\left[0, \frac{1}{2}\right]$ and increases on the interval $\left[\frac{1}{2}, 1\right]$ (this can be verified using the derivative: it is equal to $\left.-5(1-u)^{4}+5 u^{4}\right)$. From this, it follows that the function $y(x)=\cos ^{10} x+\sin ^{10} x$ cannot have a period smaller than $\frac{\pi}{2}$.