중점 y좌표 최대

[출제의도] 삼각함수의 덧셈정리를 활용하여 문제해결하기 점 $\mathrm{P}$의 좌표는 $\mathrm{\left( \cos \theta , \sin \theta \right)}$ 직선 $y = x$가 $x$축과 이루는 각의 크기는 $\displaystyle \frac{\pi}{4}$ $\displaystyle \angle \mathrm{POQ} = \theta - \frac{\pi}{4}$이므로 직각삼각형 $\mathrm{OQP}$에서 $\displaystyle \overline{\mathrm{OQ}} = \cos \left( \theta - \frac{\pi}{4} \right)$ $\displaystyle \mathrm{Q} \left( {\overline{\mathrm{OQ} \mathit{}}} \cos \frac{\pi}{4} , \overline{OQ} \sin \frac{\pi}{4} \right)$ $\displaystyle \mathrm{Q} \left( \cos \left( \theta - \frac{\pi}{4} \right) \cos \frac{\pi}{4} , \cos \left( \theta - \frac{\pi}{4} \right) \sin \frac{\pi}{4} \right)$이므로 점 $\mathrm{M}$의 $y$좌표는 $\displaystyle \frac{1}{2} \left\{ \sin \theta + \cos \left( \theta - \frac{\pi}{4} \right) \sin \frac{\pi}{4} \right\}$ $\displaystyle = \frac{1}{2} \left\{ \sin \theta + \left( \frac{\sqrt{2}}{2} \cos \theta + \frac{\sqrt{2}}{2} \sin \theta \right) \times \frac{\sqrt{2}}{2} \right\}$ $\displaystyle = \frac{1}{2} \left( \frac{3}{2} \sin \theta + \frac{1}{2} \cos \theta \right)$ $\displaystyle = \frac{1}{2} \times \frac{\sqrt{10}}{2} \sin \left( \theta + \alpha \right)$ $\displaystyle \left( \begin{array}{l} \begin{matrix} \\ \end{matrix} \end{array} \right.$단, $\displaystyle \sin \alpha = \frac{1}{\sqrt{10}}$, $\displaystyle \cos \alpha = \frac{3}{\sqrt{10}}$$\displaystyle \begin{matrix} \\ \end{matrix} )$ $\sin \left( \theta + \alpha \right) = 1$, 즉 $\displaystyle \theta = \frac{\pi}{2} - \alpha$일 때 점 $\mathrm{M}$의 $y$좌표는 최댓값을 갖는다. 따라서 $\displaystyle \tan \theta = \tan \left( \frac{\pi}{2} - \alpha \right) = \cot \alpha$ $\displaystyle = \frac{\cos \alpha}{\sin \alpha} = 3$