삼각함수 극한 도형

[출제의도] 수학내적 문제해결능력-삼각함수

$\displaystyle \angle PAB = \theta , \angle APB = \frac{\pi}{2} , {\overline{AB}} = 1$이므로 $\displaystyle {\overline{PB}} = \sin \theta$ $\angle BPQ = \theta$이므로 $\displaystyle {\overline{PQ}} = \sin \theta \cos \theta$ $\angle PQR = \theta$이므로 $\displaystyle {\overline{QR}} = \sin \theta \cos ^{2} \theta , {\overline{PR}} = \sin ^{2} \theta \cos \theta$ 두 삼각형 $PSB$와 $QSR$가 서로 닮음이므로 $\displaystyle {\overline{PS}} : {\overline{QS}} = {\overline{PB}} : {\overline{QR}} = \sin \theta : \sin \theta \cos ^{2} \theta = 1 : \cos ^{2} \theta$ 한편, 삼각형 $QPR$의 넓이를 $T ( \theta )$라 하면 $\displaystyle T ( \theta ) = \frac{1}{2} \times {\overline{PR}} \times {\overline{QR}}$ $\displaystyle = \frac{1}{2} \times \sin ^{2} \theta \cos \theta \times \sin \theta \cos ^{2} \theta$ $\displaystyle = \frac{\sin ^{3} \theta \cos ^{3} \theta}{2}$ $\displaystyle f ( \theta ) = T ( \theta ) \times \frac{{\overline{SQ}}}{{\overline{PQ}}} = T ( \theta ) \times \frac{{\overline{SQ}}}{{\overline{PS}} + {\overline{SQ}}}$ $\displaystyle = \frac{\sin ^{3} \theta \cos ^{3} \theta}{2} \times \frac{\cos ^{2} \theta}{1 + \cos ^{2} \theta} = \frac{\sin ^{3} \theta \cos ^{5} \theta}{2 ( 1 + \cos ^{2} \theta )}$ 따라서 $\displaystyle \lim\limits _{\theta \rightarrow 0 +} {} \frac{f ( \theta )}{\theta ^{3}} = \lim\limits _{\theta \rightarrow 0 +} {} \frac{\sin ^{3} \theta \cos ^{5} \theta}{2 \theta ^{3} ( 1 + \cos ^{2} \theta )} = 1 \times \frac{1}{2 \times 2} = \frac{1}{4}$