Монгол Бодлогын Сан

Заавар: $$\cos9\alpha+\cos3\alpha=2\cos\dfrac{9\alpha+3\alpha}{2}\,\cos\dfrac{9\alpha-3\alpha}{2}=2\cos6\alpha\cos3\alpha$$

Бодолт: $$\dfrac{\sqrt3\cos3\alpha}{10(\cos9\alpha+\cos3\alpha)}=\dfrac{\sqrt3\cos3\alpha}{10\cdot2\cos6\alpha\cos3\alpha}=\dfrac{\sqrt3}{20\cdot\cos6\alpha}$$ $$=\dfrac{\sqrt3}{20\cdot\cos\left(6\times\dfrac{\pi}{36}\right)}=\dfrac{\sqrt3}{20\cdot\cos\dfrac{\pi}{6}}=\dfrac{\sqrt3}{20\cdot\dfrac{\sqrt3}{2}}=\dfrac{1}{10}$$