반원 속 삼각형 넓이 극한

[출제의도] 도형의 성질을 활용하여 삼각함수의 극한 문제해결하기

삼각형 $\mathrm{QPB}$는 이등변삼각형이므로 $\displaystyle \mathrm{\angle} QBP = \frac{\pi}{2} - \frac{\theta}{2}$ 삼각형 $\mathrm{ABQ}$는 $\displaystyle \mathrm{\angle} Q = \frac{\pi}{2}$인 직각삼각형이므로 $\displaystyle \mathrm{\angle} QAB = \frac{\theta}{2}$ $\displaystyle \mathrm{\overline{QB}} = \mathrm{\overline{QP}} = 2 \sin \frac{\theta}{2}$ $S ( \theta )$$\displaystyle = \frac{1}{2} \times \mathrm{\overline{QB}} \mathit{\times} \mathrm{\overline{QP}} \mathit{\times} \sin \theta$ $\displaystyle \lim\limits _{\theta \rightarrow 0 +} {} \frac{S ( \theta )}{\theta ^{3}}$$\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {\frac{2 \sin ^{2} \frac{\theta}{2} \sin \theta}{\theta ^{3}}}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {\left\{ 2 \times \left( \frac{\sin \frac{\theta}{2}}{\frac{\theta}{2}} \right) ^{2} \times \frac{1}{4} \times \frac{\sin \theta}{\theta} \right\}}$ $\displaystyle = 2 \times 1 \times \frac{1}{4} \times 1 = \frac{1}{2}$