Монгол Бодлогын Сан

Заавар: $\alpha=-\dfrac{\pi}{36}=-5^\circ$ ба $\cos x=\cos(-x)$. Иймд $$\dfrac{\sqrt3\cos3\alpha}{10(\cos9\alpha+\cos3\alpha)}=\dfrac{\sqrt3\cos15^\circ}{10(\cos45^\circ+\cos15^\circ)}$$ байна.

Бодолт: $\sin30^\circ=2\sin15^\circ\cos15^\circ$ тул $$\cos(30^\circ+15^\circ)=\cos30^\circ\cos15^\circ-\sin30^\circ\sin15^\circ=\cos15^\circ(\cos30^\circ-2\sin^215^\circ).$$ Мөн $-2\sin^215^\circ+1=\cos30^\circ$ тул $$\dfrac{\sqrt3\cos15^\circ}{10(\cos45^\circ+\cos15^\circ)}=\dfrac{\sqrt3\cos15^\circ}{10\cos15^\circ(\cos30^\circ-2\sin^215^\circ+1)}=$$ $$=\dfrac{\sqrt3}{10\cdot 2\cdot\cos 30^\circ}=\dfrac{\sqrt{3}}{10\cdot2\cdot\frac{\sqrt3}{2}}=\dfrac{1}{10}.$$