What is greater: $\frac{\sin 1^{\circ}}{\sin 2^{\circ}}$ or $\frac{\sin 3^{\circ}}{\sin 4^{\circ}} ?$
The first method. Let's 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 $\frac{\sin 1^{\circ}}{\sin 2^{\circ}}0$. Therefore, on (0, π/2), the function $f(x)$ is increasing. These numbers are the values of this function at $x=\pi / 180$ and $x={ }^{3 \pi} / 180$ respectively, so the first number is less than the second.

## Answer

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