반원 위 점과 코사인

$\displaystyle {\overline{\mathrm{OC}}} = {\overline{\mathrm{OD}}} = {\overline{\mathrm{OE}}} = 1$, $\displaystyle \angle \mathrm{COE} \mathit{=} \frac{\pi}{2}$이므로 $\displaystyle {\overline{\mathrm{CE}}} = \sqrt{2}$, $\angle \mathrm{OCD} \mathit{=} \angle \mathrm{ODC}$, $\angle \mathrm{ODE} \mathit{=} \angle \mathrm{OED}$ 사각형 $\mathrm{COED}$에서 $\displaystyle \angle \mathrm{OCD} \mathit{+} \angle \mathrm{CDE} \mathit{+} \angle \mathrm{OED} \mathit{=} 2 \pi - \frac{\pi}{2} = \frac{3}{2} \pi$, $\angle \mathrm{OCD} \mathit{+} \angle \mathrm{OED} \mathit{=} \angle \mathrm{CDE}$이므로 $\displaystyle 2 \angle \mathrm{CDE} \mathit{=} \frac{3}{2} \pi$, 즉 $\displaystyle \angle \mathrm{CDE} \mathit{=} \frac{3}{4} \pi$ 한편 $\displaystyle {\overline{\mathrm{CD}}} : {\overline{\mathrm{DE}}} = 1 : \sqrt{2}$이므로 $\displaystyle {\overline{\mathrm{CD}}} = a$, $\displaystyle {\overline{\mathrm{DE}}} = \sqrt{2} a \left( a > 0 \right)$이라 하자. 삼각형 $\mathrm{DCE}$에서 코사인법칙에 의하여 $\displaystyle {\overline{\mathrm{CE}}} ^{2}$$\displaystyle = a ^{2} + \left( \sqrt{2} a \right) ^{2} - 2 \times a \times \sqrt{2} a \times \cos \frac{3}{4} \pi$$= 5 a ^{2}$ 이고 $\displaystyle \overline{{\mathrm{CE}}} = \sqrt{5} a = \sqrt{2}$, 즉 $\displaystyle a = \frac{\sqrt{10}}{5}$ $\angle {\mathrm{OBE}} \mathit{=} \theta$라 하고 점 ${\mathrm{O}}$에서 선분 ${\mathrm{EB}}$에 내린 수선의 발을 ${\mathrm{H}}$라 하면 $\displaystyle {\overline{\mathrm{EB}}} = {\overline{\mathrm{DE}}} = \sqrt{2} a = \frac{2 \sqrt{5}}{5}$이므로 $\displaystyle \cos \theta = \frac{{\overline{\mathrm{BH}}}}{{\overline{\mathrm{OB}}}} = \frac{\frac{1}{2} {\overline{\mathrm{EB}}}}{{\overline{\mathrm{OB}}}} = \frac{\sqrt{5}}{5}$ [다른 풀이] $\displaystyle {\overline{\mathrm{OB}}} = {\overline{\mathrm{OD}}}$, $\displaystyle {\overline{\mathrm{EB}}} = {\overline{\mathrm{ED}}}$, $\displaystyle {\overline{\mathrm{OE}}}$는 공통이므로 삼각형 $\mathrm{OBE}$와 삼각형 $\mathrm{ODE}$는 합동이다. $\angle \mathrm{OBE} \mathit{=} \theta$라 하면 $\angle \mathrm{OEB} \mathit{=} \angle \mathrm{OED} \mathit{=} \angle \mathrm{ODE} \mathit{=} \theta$, $\angle \mathrm{EOB} \mathit{=} \angle \mathrm{DOE} \mathit{=} \pi - 2 \theta$, $\displaystyle \angle \mathrm{COD} \mathit{=} \frac{\pi}{2} - \angle \mathrm{DOE} \mathit{=} \frac{\pi}{2} - \left( \pi - 2 \theta \right) = 2 \theta - \frac{\pi}{2}$, $\displaystyle \angle \mathrm{ODC} \mathit{=} \frac{1}{2} \left( \pi - \angle \mathrm{COD} \right) = \frac{3}{4} \pi - \theta$, $\displaystyle \angle \mathrm{CDE} \mathit{=} \angle \mathrm{ODC} \mathit{+} \angle \mathrm{ODE} \mathit{=} \left( \frac{3}{4} \pi - \theta \right) + \theta = \frac{3}{4} \pi$ 한편 $\displaystyle {\overline{\mathrm{OC}}} = {\overline{\mathrm{OE}}} = 1$, $\displaystyle \angle \mathrm{COE} \mathit{=} \frac{\pi}{2}$이므로 $\displaystyle {\overline{\mathrm{CE}}} = \sqrt{2}$ $\displaystyle {\overline{\mathrm{CD}}} : {\overline{\mathrm{DE}}} = 1 : \sqrt{2}$이므로 $\displaystyle \overline{{\mathrm{CD}}} = a$, $\displaystyle {\overline{\mathrm{DE}}} = \sqrt{2} a \left( a > 0 \right)$이라 하자. 삼각형 ${\mathrm{DCE}}$에서 코사인법칙에 의하여 $\displaystyle {\overline{\mathrm{CE}}} ^{2} = a ^{2} + \left( \sqrt{2} a \right) ^{2} - 2 \times a \times \sqrt{2} a \times \cos \frac{3}{4} \pi = 5 a ^{2}$ 이고 $\displaystyle {\overline{\mathrm{CE}}} = \sqrt{5} a = \sqrt{2}$, $\displaystyle a = \frac{\sqrt{10}}{5}$, 즉 $\displaystyle \overline{{\mathrm{EB}}} = \overline{{\mathrm{DE}}} = \sqrt{2} a = \frac{2 \sqrt{5}}{5}$ 삼각형 ${\mathrm{OBE}}$에서 코사인법칙에 의하여 $\displaystyle \cos \theta = \frac{1 ^{2} + \left( \frac{2 \sqrt{5}}{5} \right) ^{2} - 1 ^{2}}{2 \times 1 \times \frac{2 \sqrt{5}}{5}} = \frac{\sqrt{5}}{5}$