1. Which is greater: $\frac{\sin 1^{\circ}}{\sin 2^{\circ}}$ or $\frac{\sin 3^{\circ}}{\sin 4^{\circ}}$?
Solution 1. Consider the difference of these numbers and transform it:

$\frac{\sin 1^{\circ}}{\sin 2^{\circ}}-\frac{\sin 3^{\circ}}{\sin 4^{\circ}}=\frac{\sin 1^{\circ} \cdot \sin 4^{\circ}-\sin 2^{\circ} \cdot \sin 3^{\circ}}{\sin 2^{\circ} \cdot \sin 4^{\circ}}=\frac{\left(\cos 3^{\circ}-\cos 5^{\circ}\right)-\left(\cos 1^{\circ}-\cos 5^{\circ}\right)}{2 \sin 2^{\circ} \cdot \sin 4^{\circ}}=$ $\frac{\cos 3^{\circ}-\cos 1^{\circ}}{2 \sin 2^{\circ} \cdot \sin 4^{\circ}}=\frac{-2 \sin 2^{\circ} \cdot \sin 1^{\circ}}{2 \sin 2^{\circ} \cdot \sin 4^{\circ}}=\frac{-\sin 1^{\circ}}{\sin 4^{\circ}}\cos 3^{\circ}$. Considering that these numbers are positive, we get that $\frac{\sin 1^{\circ}}{\sin 2^{\circ}}<\frac{\sin 3^{\circ}}{\sin 4^{\circ}}$

Answer: $\frac{\sin 3^{\circ}}{\sin 4^{\circ}}$