Докажите равенство cos160°-cos40°=(корень из 3)cos10°.

\[\cos{160{^\circ}} - \cos{40{^\circ}} = \sqrt{3}\cos{10{^\circ}}\text{\ \ \ \ }\]

\[- 2\sin\frac{200}{2} \cdot \sin\frac{120}{2} = \sqrt{3}\cos{10{^\circ}}\]

\[- 2\sin{100{^\circ}} \cdot \sin{60{^\circ}} = \sqrt{3}\cos{10{^\circ}}\]

\[- 2\sin{100{^\circ}} \cdot \frac{\sqrt{3}}{2} = \sqrt{3}\cos{10{^\circ}}\]

\[- \sqrt{3}\sin\left( \frac{\pi}{2} + 10{^\circ} \right) = \sqrt{3}\cos{10{^\circ}}\]

\[- \sqrt{3}\cos{10{^\circ}} \neq \ \sqrt{3}\cos{10{^\circ}}\]