Problem 6. Compare the numbers $\frac{\cos 2016^{\circ}}{\cos 2017^{\circ}}$ and $\frac{\cos 2018^{\circ}}{\cos 2019^{\circ}}$.
Answer: The second expression is greater

Solution. Note that $1800^{\circ}=5 \cdot 360^{\circ}$, so it is sufficient to compare the numbers $\frac{\cos 216^{\circ}}{\cos 217^{\circ}}$ and $\frac{\cos 218^{\circ}}{\cos 219^{\circ}}$. Note that the numbers $\cos 217^{\circ}$ and $\cos 219^{\circ}$ have the same sign, so we can multiply both numbers by the positive number $\cos 217^{\circ} \cdot \cos 219^{\circ}$.

It remains to compare the numbers

$$
\cos 216^{\circ} \cdot \cos 219^{\circ}=\frac{1}{2} \cdot\left(\cos \left(219^{\circ}-216^{\circ}\right)+\cos \left(219^{\circ}+216^{\circ}\right)\right)=\frac{1}{2}\left(\cos 3^{\circ}+\cos 435^{\circ}\right)
$$

and $\cos 218^{\circ} \cdot \cos 217^{\circ}=\frac{1}{2}\left(\cos 1^{\circ}+\cos 435^{\circ}\right)$. Or, after transformations, $\cos 3^{\circ}$ and $\cos 1^{\circ}$. Since the cosine is decreasing on the interval $\left[0,90^{\circ}\right]$, we get $\cos 3^{\circ}<\cos 1^{\circ}$, i.e., the second expression was originally greater.