第7章 7.27

$y = \sin^2 x \cdot \cos x$

$y' = 2\sin x \cdot \cos x \cdot \cos x - \sin^2 x \cdot \sin x = 2\sin x \cdot \cos^2 x - (1-\cos^2 x)\sin x = 2\sin x \cdot \cos^2 x - \sin x + \cos^2 x \cdot \sin x$ $= 3\sin x \cdot \cos^2 x - \sin x$

$y' = 0$となるのは

$y' = 3\sin x \cdot \cos^2 x - \sin x = 0 \qquad \sin x(3\cos^2 x - 1) = 0 \qquad \sin x = 0 \qquad 3\cos^2 x - 1 = 0$

$x = 0 \qquad \cos x = \frac{1}{\sqrt{3}} \quad \text{のとき}$

$x = 0$のとき $y = \sin^2 0 \cdot \cos 0 = 0$

$\cos x = \dfrac{1}{\sqrt{3}}$のとき $y = (1-\cos^2 x)\cos x = \left(1 - \dfrac{1}{3}\right)\dfrac{1}{\sqrt{3}} = \dfrac{2}{3\sqrt{3}} = \dfrac{2\sqrt{3}}{9}$

$x = \dfrac{\pi}{2}$のとき $y = \sin^2 \dfrac{\pi}{2} \cdot \cos \dfrac{\pi}{2} = 0$

これから最大値は$\dfrac{2\sqrt{3}}{9}$、最小値は$0$となる