삼각형 넓이 차 극한

[출제의도] 도형의 넓이를 삼각함수로 나타내고 극한값을 구할 수 있는가? 직각삼각형 $\mathrm{BMH}$에서 $\displaystyle {\mathrm{\overline{MB}}} = 1$ $\displaystyle \sin \theta = \frac{{\mathrm{\overline{MH}}}}{{\mathrm{\overline{MB}}}}$에서 $\displaystyle {\mathrm{\overline{MH}}} = {\mathrm{\overline{MB}}} \times \sin \theta = \sin \theta$ 삼각형 $\mathrm{DMC}$에서 $\displaystyle {\mathrm{\overline{MD}}} = {\mathrm{\overline{MH}}} = \sin \theta$, $\displaystyle {\mathrm{\overline{MC}}} = 1$, $\angle \mathrm{DMC} = \pi - \angle \mathrm{DMB}$$= \pi - \angle \mathrm{AMB}$$\displaystyle = \pi - \frac{\pi - \theta}{2}$$\displaystyle = \frac{\pi}{2} + \frac{\theta}{2}$ 이므로 $\displaystyle \triangle \mathrm{DMC} = \frac{1}{2} \times {\mathrm{\overline{MD}}} \times {\overline{MC}} \times \sin ( \angle DMC )$ $\displaystyle = \frac{1}{2} \times \sin \theta \times 1 \times \sin \left( \frac{\pi}{2} + \frac{\theta}{2} \right)$$\displaystyle = \frac{1}{2} \sin \theta \cos \frac{\theta}{2}$ 삼각형 $\mathrm{HMC}$에서 $\displaystyle {\mathrm{\overline{MH}}} = \sin \theta$, $\displaystyle {\mathrm{\overline{MC}}} = 1$, $\angle \mathrm{HMC} = \pi - \angle \mathrm{HMB}$$\displaystyle = \pi - \left( \frac{\pi}{2} - \theta \right)$$\displaystyle = \frac{\pi}{2} + \theta$ 이므로 $\displaystyle \triangle \mathrm{HMC} = \frac{1}{2} \times {\mathrm{\overline{MH}}} \times {\overline{MC}} \times \sin ( \angle HMC )$ $\displaystyle = \frac{1}{2} \times \sin \theta \times 1 \times \sin \left( \frac{\pi}{2} + \theta \right)$ $\displaystyle = \frac{1}{2} \sin \theta \cos \theta$ 이때 $f ( \theta ) - g ( \theta )$$= \triangle \mathrm{DMC} - \triangle \mathrm{HMC}$ $\displaystyle = \frac{1}{2} \sin \theta \cos \frac{\theta}{2} - \frac{1}{2} \sin \theta \cos \theta$ $\displaystyle = \frac{\sin \theta \left( \cos \frac{\theta}{2} - \cos \theta \right)}{2}$ 이므로 $\displaystyle \lim\limits _{\theta \rightarrow 0 +} {} \frac{f ( \theta ) - g ( \theta )}{\theta ^{3}}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {} \frac{\sin \theta \left( \cos \frac{\theta}{2} - \cos \theta \right)}{2 \theta ^{3}}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {} \frac{\sin \theta \left( \cos \frac{\theta}{2} - \cos \theta \right) \left( \cos \frac{\theta}{2} + \cos \theta \right)}{2 \theta ^{3} \left( \cos \frac{\theta}{2} + \cos \theta \right)}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {} \frac{\sin \theta \left( \cos ^{2} \frac{\theta}{2} - \cos ^{2} \theta \right)}{2 \theta ^{3} \left( \cos \frac{\theta}{2} + \cos \theta \right)}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0 +} {} \frac{\sin \theta \left( \sin ^{2} \theta - \sin ^{2} \frac{\theta}{2} \right)}{2 \theta ^{3} \left( \cos \frac{\theta}{2} + \cos \theta \right)}$ $\displaystyle = \frac{1}{2} \lim\limits _{\theta \rightarrow 0 +} {} \frac{\sin \theta}{\theta} \times \left\{ \lim\limits _{\theta \rightarrow 0 +} {} \left( \frac{\sin \theta}{\theta} \right) ^{2} \right.$ $\displaystyle - \frac{1}{4} \lim\limits _{\theta \rightarrow 0 +} {} \left( \frac{\sin \frac{\theta}{2}}{\frac{\theta}{2}} \right) ^{2} \}$ $\displaystyle \times \lim\limits _{\theta \rightarrow 0 +} {} \frac{1}{\cos \frac{\theta}{2} + \cos \theta}$ $\displaystyle = \frac{1}{2} \times 1 \times \left( 1 ^{2} - \frac{1}{4} \times 1 ^{2} \right) \times \frac{1}{1 + 1}$ $\displaystyle = \frac{3}{16}$ 따라서 $\displaystyle a = \frac{3}{16}$이므로 ∴ $\displaystyle 80 a = 80 \times \frac{3}{16} = 15$ <참고> 삼각형 HMC의 넓이는 직각삼각형 HBM의 넓이와 같다. $\displaystyle \triangle \mathrm{HMC} = \triangle \mathrm{HBM} = \frac{1}{2} \sin \theta \cos \theta$