도형 넓이의 극한

[출제의도] 도형의 넓이를 삼각함수로 나타내고 극한값을 구할 수 있는가? 직각삼각형 $ABC$에서 $\displaystyle {\overline{AB}} = 1 , \angle CAB = \theta$이므로 $\displaystyle {\overline{AC}} = \sec \theta , {\overline{BC}} = \tan \theta$ 이때 직선 $CD$가 $\angle ACB$를 이등분하므로 $\displaystyle {\overline{AD}} : {\overline{BD}} = {\overline{AC}} : {\overline{BC}}$ 즉, $\displaystyle {\overline{AD}} = 1 \times \frac{\sec \theta}{\sec \theta + \tan \theta} = \frac{1}{1 + \sin \theta}$ 이므로 $\displaystyle S ( \theta ) = \frac{1}{2} \times \left( \frac{1}{1 + \sin \theta} \right) ^{2} \times \theta = \frac{1}{2} \times \frac{\theta}{\left( 1 + \sin \theta \right) ^{2}}$ 한편, $\displaystyle {\overline{CE}} = \sec \theta - \frac{1}{1 + \sin \theta}$이므로 $\displaystyle T ( \theta ) = \frac{1}{2} \times \tan \theta \times \left( \sec \theta - \frac{1}{1 + \sin \theta} \right) \times \sin \left( \frac{\pi}{2} - \theta \right)$ $\displaystyle = \frac{1}{2} \sin \theta \left( \sec \theta - \frac{1}{1 + \sin \theta} \right)$ 따라서 $\displaystyle \lim\limits _{\theta \rightarrow 0 +} {} \frac{\left\{ S ( \theta ) \right\} ^{2}}{T ( \theta )} = \lim\limits _{\theta \rightarrow 0 +} {} \frac{\left\{ \frac{1}{2} \times \frac{\theta}{\left( 1 + \sin \theta \right) ^{2}} \right\} ^{2}}{\frac{1}{2} \sin \theta \left( \sec \theta - \frac{1}{1 + \sin \theta} \right)}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {} \left\{ \frac{1}{2} \times \frac{\theta}{\sin \theta} \times \frac{\cos \theta}{\left( 1 + \sin \theta \right) ^{3}} \times \frac{\theta}{\sin \theta + 1 - \cos \theta} \right\}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {} \left\{ \frac{1}{2} \times \frac{\theta}{\sin \theta} \times \frac{\cos \theta}{\left( 1 + \sin \theta \right) ^{3}} \times \frac{1}{\frac{\sin \theta}{\theta} + \frac{1 - \cos \theta}{\theta}} \right\}$ $\displaystyle = \frac{1}{2} \times 1 \times \frac{1}{1} \times \frac{1}{1 + 0} = \frac{1}{2}$