삼각함수 수열 극한

$\displaystyle \sin \frac{\pi}{n} = \frac{a _{n}}{2 + a _{n}}$에서 $\displaystyle a _{n} = \frac{2 \sin \frac{\pi}{n}}{1 - \sin \frac{\pi}{n}}$$\displaystyle = \frac{2 \sin \frac{\pi}{n} \left( 1 + \sin \frac{\pi}{n} \right)}{\cos ^{2} \frac{\pi}{n}}$이고 $\displaystyle \sin \frac{\pi}{n} = \frac{b _{n}}{2 - b _{n}}$에서 $\displaystyle b _{n} = \frac{2 \sin \frac{\pi}{n}}{1 + \sin \frac{\pi}{n}}$$\displaystyle = \frac{2 \sin \frac{\pi}{n} \left( 1 - \sin \frac{\pi}{n} \right)}{\cos ^{2} \frac{\pi}{n}}$이므로 $\displaystyle a _{n} + b _{n} = \frac{4 \sin \frac{\pi}{n}}{\cos ^{2} \frac{\pi}{n}}$, $a _{n} - b _{n}$$\displaystyle = \frac{4 \sin ^{2} \frac{\pi}{n}}{\cos ^{2} \frac{\pi}{n}}$ $n \rightarrow \infty$일 때, $\displaystyle \frac{\pi}{n} \rightarrow + 0$이므로 $\displaystyle \frac{\pi}{n} = \theta$라 하면 $\displaystyle \frac{1}{\pi ^{3}} \lim\limits _{n\rightarrow \infty} n ^{3} ( a _{n} + b _{n} ) ( a _{n} - b _{n} )$ $\displaystyle = \lim\limits _{\theta \rightarrow 0} \frac{1}{\theta ^{3}} \cdot \frac{4 \sin \theta}{\cos ^{2} \theta} \cdot \frac{4 \sin ^{3} \theta}{\cos ^{2} \theta}$ $\displaystyle = \lim\limits _{\theta \rightarrow 0} 16 \cdot \left( \frac{\sin \theta}{\theta} \right) ^{3} \cdot \frac{1}{\cos ^{4} \theta}$ $= 16$