삼각형 넓이 극한

[출제의도] 삼각함수의 극한 이해하기

$\angle \mathrm{ABD}$$\displaystyle = \frac{\pi}{2}$이므로 선분 $\mathrm{AD}$는 원의 지름이다. $\angle \mathrm{ECD}$$\displaystyle = \frac{\pi}{2}$, $\angle \mathrm{DAB}$$= \angle \mathrm{CAD}$$\displaystyle = \frac{\theta}{2}$ $\displaystyle {\overline{\mathrm{AB}}}$$\displaystyle = 10 \cos \frac{\theta}{2}$, $\displaystyle {\overline{\mathrm{CD}}}$$\displaystyle = {\overline{\mathrm{BD}}}$$\displaystyle = 10 \sin \frac{\theta}{2}$ $\angle \mathrm{AEB}$$\displaystyle = \frac{\pi}{2} - \theta$이므로 $\angle \mathrm{CDE}$$= \theta$ $\displaystyle {\overline{\mathrm{CE}}}$$\displaystyle = 10 \sin \frac{\theta}{2} \tan \theta$ $f \left( \theta \right)$$\displaystyle = \frac{1}{2} \times \left( 10 \cos \frac{\theta}{2} \right) ^{2} \times \sin \theta$ $g \left( \theta \right)$$\displaystyle = \frac{1}{2} \times 10 \sin \frac{\theta}{2} \times 10 \sin \frac{\theta}{2} \tan \theta$ $\displaystyle \lim\limits _{\theta \rightarrow 0 +} {} \frac{g \left( \theta \right)}{\theta ^{2} \times f \left( \theta \right)}$$\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {} \frac{50 \sin ^{2} \frac{\theta}{2} \tan \theta}{\theta ^{2} \times 50 \cos ^{2} \frac{\theta}{2} \sin \theta}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {} \left\{ \frac{1}{\cos ^{2} \frac{\theta}{2}} \times \frac{\sin ^{2} \frac{\theta}{2} \tan \theta}{\theta ^{2} \times \sin \theta} \right\}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {} \left\{ 1 \times \frac{1}{4} \times \frac{\sin ^{2} \frac{\theta}{2}}{\left( \frac{\theta}{2} \right) ^{2}} \times \frac{\tan \theta}{\theta} \times \frac{\theta}{\sin \theta} \right\}$$\displaystyle = \frac{1}{4}$