Problem 6. Compare the numbers $\frac{\cos 2014^{\circ}}{\cos 2015^{\circ}}$ and $\frac{\cos 2016^{\circ}}{\cos 2017^{\circ}}$.
Answer: The second expression is greater.

Solution. Note that $1800^{\circ}=5 \cdot 360^{\circ}$, so it is sufficient to compare the numbers $\frac{\cos 214^{\circ}}{\cos 215^{\circ}}$ and $\frac{\cos 216^{\circ}}{\cos 21^{\circ}}$. Note that the numbers $\cos 215^{\circ}$ and $\cos 217^{\circ}$ have the same sign, so we can multiply both numbers by the positive number $\cos 215^{\circ} \cdot \cos 217^{\circ}$.

It remains to compare the numbers

$$
\cos 214^{\circ} \cdot \cos 217^{\circ}=\frac{1}{2} \cdot\left(\cos \left(217^{\circ}-214^{\circ}\right)+\cos \left(217^{\circ}+214^{\circ}\right)\right)=\frac{1}{2}\left(\cos 3^{\circ}+\cos 431^{\circ}\right)
$$

and $\cos 216^{\circ} \cdot \cos 215^{\circ}=\frac{1}{2}\left(\cos 1^{\circ}+\cos 431^{\circ}\right)$. 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.