Balti Tee 1991 · Ülesanne 10

Solution:

We use the equality

$$\sin 3^{\circ} = \sin (18^{\circ} - 15^{\circ}) = \sin 18^{\circ} \cos 15^{\circ} + \cos 18^{\circ} \sin 15^{\circ}$$

where

$$\sin 15^{\circ} = \sin \frac{30^{\circ}}{2} = \sqrt{\frac{1 - \cos 30^{\circ}}{2}} = \frac{\sqrt{6} - \sqrt{2}}{4}$$

and

$$\cos 15^{\circ} = \sqrt{1 - \sin^2 15^{\circ}} = \frac{\sqrt{6} + \sqrt{2}}{4}.$$

To calculate $\cos 18^{\circ}$ and $\sin 18^{\circ}$ note that $\cos (3 \cdot 18^{\circ}) = \sin (2 \cdot 18^{\circ})$. As $\cos 3x = \cos^3 x - 3 \cos x \sin^2 x = \cos x (1 - 4 \sin^2 x)$ and $\sin 2x = 2 \sin x \cos x$ we get $1 - 4 \sin^2 18^{\circ} = 2 \sin 18^{\circ}$. Solving this quadratic equation yields $\sin 18^{\circ} = \frac{\sqrt{5} - 1}{4}$ (we discard $\frac{-\sqrt{5} - 1}{4}$ which is negative) and $\cos 18^{\circ} = \frac{\sqrt{10 + 2\sqrt{5}}}{4}$.