원 속 삼각형 넓이 미분

원의 중심을 $\mathrm{O}$, 선분 $\mathrm{AB}$와 선분 $\mathrm{PQ}$의 교점을 $\mathrm{H}$, $\displaystyle \mathrm{\overline{CP}}$$= x$라고 하면 $\displaystyle \mathrm{\overline{OC}} = 1$, $\displaystyle \mathrm{\overline{OP}} = 5$이므로 코사인법칙에 의하여 $25 = x ^{2} + 1 - 2 x \cos \theta$, $\left( x - \cos \theta \right) ^{2} = 24 + \cos ^{2} \theta$ $\displaystyle x = \cos \theta + \sqrt{24 + \cos ^{2} \theta}$($\because$ $x > 0$) $\displaystyle \mathrm{\overline{CP}} = \overline{CQ} = \mathit{x}$, $\angle \mathrm{PCQ} = 2 \theta$이므로 삼각형 $\mathrm{PCQ}$의 넓이 $S \left( \theta \right)$는 $S \left( \theta \right)$$\displaystyle = \frac{1}{2} \times \left( \cos \theta + \sqrt{24 + \cos ^{2} \theta} \right) ^{2} \sin 2 \theta$ $\displaystyle = \left( \cos \theta + \sqrt{24 + \cos ^{2} \theta} \right) ^{2} \frac{\sin 2 \theta}{2}$ $S' \left( \theta \right)$$\displaystyle = 2 \left( \cos \theta + \sqrt{24 + \cos ^{2} \theta} \right) \left( - \sin \theta - \frac{\sin \theta \cos \theta}{\sqrt{24 + \cos ^{2} \theta}} \right)$ $\displaystyle \times \frac{\sin 2 \theta}{2} + \left( \cos \theta + \sqrt{24 + \cos ^{2} \theta} \right) ^{2} \cos 2 \theta$ $\therefore$ $\displaystyle - 7 \times S' \left( \frac{\pi}{4} \right)$ $\displaystyle = - 7 \times \left( \frac{\sqrt{2}}{2} + \sqrt{24 + \frac{1}{2}} \right) \left( - \frac{\sqrt{2}}{2} - \frac{\frac{1}{2}}{\sqrt{24 + \frac{1}{2}}} \right)$ $\displaystyle = - 7 \times 4 \sqrt{2} \times \left( - \frac{4}{7} \sqrt{2} \right)$ $= 32$