원의 접선과 삼각함수 극한

그림에서 $\mathrm{\triangle} AOP$가 이등변삼각형이므로 $\mathrm{\angle} POQ = 2 \theta$ $\therefore \mathrm{P} ( \cos 2 \theta , \sin 2 \theta )$ 점 $\mathrm{P}$를 지나는 접선의 방정식은 $\cos 2 \theta x + \sin 2 \theta y = 1$ $\displaystyle \therefore \mathrm{Q} \left( \frac{1}{\cos 2 \theta} , 0 \right)$ $\displaystyle \therefore \mathrm{\overline{PQ}} - \overline{OQ}$$\displaystyle = \sqrt{\left( \cos 2 \theta - \frac{1}{\cos 2 \theta} \right) ^{2} + \sin ^{2} 2 \theta}$$\displaystyle - \frac{1}{\cos 2 \theta}$ $\displaystyle = \sqrt{1 - 2 + \frac{1}{\cos ^{2} 2 \theta}} - \frac{1}{\cos 2 \theta}$ $\displaystyle = \sqrt{\sec ^{2} 2 \theta - 1} - \frac{1}{\cos 2 \theta}$ $\displaystyle = \tan 2 \theta - \frac{1}{\cos 2 \theta}$$\displaystyle = \frac{\sin 2 \theta - 1}{\cos 2 \theta}$ 한편 $\displaystyle \theta - \frac{\pi}{4} = t$라 하면 $\displaystyle \theta = \frac{\pi}{4} + t$이므로 $\displaystyle \frac{\sin 2 \theta - 1}{\cos 2 \theta} = \frac{\sin \left( \frac{\pi}{2} + 2 t \right) - 1}{\cos \left( \frac{\pi}{2} + 2 t \right)}$$\displaystyle = \frac{\cos 2 t - 1}{- \sin 2 t}$$\displaystyle = \frac{- 2 \sin ^{2} t}{- \sin 2 t}$$\displaystyle = \frac{2 \sin ^{2} t}{\sin 2 t}$ $\displaystyle \therefore \lim\limits _{\theta \rightarrow \frac{\pi}{4} - 0} {} \frac{\mathrm{\overline{PQ} - \overline{OQ}}}{\theta - \frac{\pi}{4}}$$\displaystyle = \lim\limits _{t \rightarrow - 0} \frac{\frac{2 \sin ^{2} t}{\sin 2 t}}{t}$$\displaystyle = 2 \lim\limits _{t \rightarrow - 0} \frac{\sin ^{2} t}{t ^{2}} \cdot \frac{2 t}{\sin 2 t} \cdot \frac{1}{2}$$= 1$