도형과 삼각함수의 극한

[출제의도] 도형과 관련된 삼각함수의 극한값을 구할 수 있는가?

$\angle \mathrm{OAP} = \angle \mathrm{OPA} = \theta$ $\angle \mathrm{BOP} = 2 \theta$ 따라서 $\displaystyle f \left( \theta \right) = \frac{1}{2} \sin 2 \theta$ 또한, $\displaystyle \overline{\mathrm{OA}} = 1$에서 $\displaystyle \overline{\mathrm{OS}} = \tan \theta$이므로 $\displaystyle \overline{\mathrm{CS}} = 1 - \tan \theta$ 이때, $\angle \mathrm{BOP} = \angle \mathrm{COQ} = 2 \theta$이고 삼각형 $\mathrm{OCQ}$는 이등변삼각형이므로 $\displaystyle \angle \mathrm{SCQ} = \frac{\pi}{2} - \theta$ 또한, $\displaystyle \angle \mathrm{CSR} = \theta + \frac{\pi}{2}$이므로 $\displaystyle \angle \mathrm{QRS} = \frac{\pi}{2}$ 따라서 점 $\mathrm{S}$에서 변 $\mathrm{CQ}$에 내린 수선의 발을 $\mathrm{H}$라 하면 $\angle \mathrm{CSH} = \theta$ 이므로 $\displaystyle \overline{\mathrm{SH}} = \overline{\mathrm{RQ}} = \left( 1 - \tan \theta \right) \cos \theta$ $\displaystyle \overline{\mathrm{CH}} = \left( 1 - \tan \theta \right) \sin \theta$ 이고 $\displaystyle \overline{\mathrm{CQ}} = \overline{\mathrm{BP}} = 2 \sin \theta$ $\displaystyle \overline{\mathrm{RS}}$$\displaystyle = \overline{\mathrm{QH}} = \overline{\mathrm{CQ}} - \overline{\mathrm{CH}}$ $= 2 \sin \theta - \left( \sin \theta - \sin \theta \tan \theta \right)$$= \sin \theta + \sin \theta \tan \theta$ 따라서 $g \left( \theta \right)$$\displaystyle = \frac{1}{2} \times \left( \overline{\mathrm{CQ}} + \overline{\mathrm{RS}} \right) \times \overline{\mathrm{QR}}$ $\displaystyle = \frac{1}{2} \times \left( 2 \sin \theta + \sin \theta + \sin \theta \tan \theta \right) \times \left( 1 - \tan \theta \right) \cos \theta$ $\displaystyle = \frac{1}{2} \times \left( 3 \sin \theta + \sin \theta \tan \theta \right) \left( 1 - \tan \theta \right) \cos \theta$ 이므로 $3 f \left( \theta \right) - 2 g \left( \theta \right)$$\displaystyle = \frac{3}{2} \sin 2 \theta - \left( 3 \sin \theta + \sin \theta \tan \theta \right) \left( 1 - \tan \theta \right) \cos \theta$ $= 3 \sin \theta \cos \theta - \sin \theta \cos \theta \left( 3 + \tan \theta \right) \left( 1 - \tan \theta \right)$ $= \sin \theta \cos \theta \tan \theta \left( \tan \theta + 2 \right)$ 따라서 $\displaystyle \lim\limits _{\theta \rightarrow 0 +} {} {\frac{3 f \left( \theta \right) - 2 g ( \theta )}{\theta ^{2}}}$$\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {} {\frac{\sin \theta \cos \theta \tan \theta \left( \tan \theta + 2 \right)}{\theta ^{2}}}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {} {\left\{ \frac{\sin \theta}{\theta} \times \frac{\tan \theta}{\theta} \times \cos \theta \times \left( \tan \theta + 2 \right) \right\}}$ $= 1 \times 1 \times 1 \times 2 = 2$