sin² 변곡점 접선

$y = \sin ^{2} x$ $y' = 2 \sin x \cos x = \sin 2 x$ $\displaystyle y'' = 2 \cos 2 x = 0 \therefore x = \frac{\pi}{4} , \frac{3 \pi}{4}$ $\displaystyle x = \frac{\pi}{4}$일 때, 점$A$는 $\displaystyle ( \frac{\pi}{4} , \frac{1}{2} )$, $\displaystyle y' _{x = \frac{\pi}{4}} = 1$ $\displaystyle \therefore l _{1} : y = x - \frac{\pi}{4} + \frac{1}{2}$ $\cdots \text{①}$ $\displaystyle x = \frac{3 \pi}{4}$일 때, 점$B$는 $\displaystyle ( \frac{3 \pi}{4} , \frac{1}{2} )$, $\displaystyle y' _{x = \frac{3 \pi}{4}} = - 1$ $\displaystyle \therefore l _{2} : y = - x + \frac{3 \pi}{4} + \frac{1}{2}$ $\cdots \text{ ②}$ ①, ② 연립하면 $\displaystyle y = \frac{1}{2} + \frac{\pi}{4}$ $\displaystyle \therefore p = \frac{1}{2} , q = \frac{1}{4}$ $\therefore 40 ( p + q ) = 30$