Монгол Бодлогын Сан

Заавар: $\sin 22.5^{\circ}$, $\cos 22.5^{\circ} >0$ байна.

Бодолт: $$\sin^222.5=\dfrac{1-\cos(2\cdot 22.5^\circ)}{2}=\dfrac{1-\cos45^\circ}{2},$$ $$\cos^222.5=\dfrac{1+\cos(2\cdot22.5^\circ)}{2}=\dfrac{1+\cos45^\circ}{2}$$ $\sin 22.5^{\circ}$, $\cos 22.5^{\circ} >0$ тул $$\sin22.5^\circ=\sqrt{\dfrac{1-\sqrt{2}/2}{2}}=\dfrac{\sqrt{2-\sqrt{2}}}{2}$$ $$\cos22.5^\circ=\sqrt{\dfrac{1+\sqrt{2}/2}{2}}=\dfrac{\sqrt{2+\sqrt{2}}}{2}$$ ба $$\tg22.5^\circ=\dfrac{\dfrac{\sqrt{2-\sqrt{2}}}{2}}{\dfrac{\sqrt{2+\sqrt{2}}}{2}}=\dfrac{\sqrt{2-\sqrt{2}}}{\sqrt{2+\sqrt{2}}}=\sqrt{2}-1$$