각 삼등분 넓이 극한

[출제의도] 삼각함수의 성질을 이용하여 삼각함수의 극한값을 구한다. $\displaystyle \mathrm{\angle} BPA = \frac{\pi}{2}$이므로 $\angle \mathrm{QBR}$$= \alpha$라 하면 $3 \alpha$$\displaystyle = \frac{\pi}{2} - \theta$ $\displaystyle \alpha = \frac{\pi}{6} - \frac{\theta}{3}$ $\displaystyle \mathrm{\overline{\mathrm{BP}}}$$= \sin \theta$이므로 $\displaystyle \overline{\mathrm{PQ}}$$= \sin \theta \tan 2 \alpha$ $\displaystyle \overline{\mathrm{PR}}$$= \sin \theta \tan \alpha$ 이다. 따라서 $S ( \theta )$$\displaystyle = \frac{1}{2} \times \overline{\mathrm{BP}} \times \overline{\mathrm{PQ}} - \frac{1}{2} \times \overline{\mathrm{BP}} \times \overline{\mathrm{PR}}$ $\displaystyle = \frac{1}{2} \sin ^{2} \theta \left( \tan 2 \alpha - \tan \alpha \right)$ $\displaystyle = \frac{1}{2} \sin ^{2} \theta \left\{ \tan \left( \frac{\pi}{3} - \frac{2}{3} \theta \right) - \tan \left( \frac{\pi}{6} - \frac{\theta}{3} \right) \right\}$ $\displaystyle \lim\limits _{\theta \rightarrow 0 +} {} \frac{\sin ^{2} \theta}{\theta ^{2}}$$\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {} \left( \frac{\sin \theta}{\theta} \right) ^{2}$$= 1$이므로 $\displaystyle \lim\limits _{\theta \rightarrow 0 +} {} \frac{S ( \theta )}{\theta ^{2}}$$\displaystyle = \frac{1}{2} \lim\limits _{\theta \rightarrow 0 +} {} \frac{\sin ^{2} \theta}{\theta ^{2}} \left\{ \tan \left( \frac{\pi}{3} - \frac{2}{3} \theta \right) - \tan \left( \frac{\pi}{6} - \frac{\theta}{3} \right) \right\}$ $\displaystyle = \frac{1}{2} \left( \tan \frac{\pi}{3} - \tan \frac{\pi}{6} \right)$ $\displaystyle = \frac{1}{2} \left( \sqrt{3} - \frac{\sqrt{3}}{3} \right)$ $\displaystyle = \frac{1}{2} \times \frac{2 \sqrt{3}}{3}$ $\displaystyle = \frac{\sqrt{3}}{3}$