12．设$a = \sin 0.2$，$b = 0.16$，$c = \frac{1}{2}\ln\frac{3}{2}$，则（ ）
A．$a > c > b$ B．$b > a > c$ C．$c > b > a$ D．$c > a > b$
解 $a = \sin 0.2 < 0.2$，$b = 0.16 = \frac{4}{5} \times 0.2$，
设$f(x) = \sin x - \frac{4}{5}x\left( 0 < x < \frac{\pi}{2} \right)$，则$f'(x) = \cos x - \frac{4}{5}$，
$\therefore$当$\frac{4}{5} < \cos x < 1$时$f(x)$单增，当$0 < \cos x < \frac{4}{5}$时$f(x)$单减，
令$\cos x_0 = \frac{4}{5}$，则$\sin x_0 = \frac{3}{5}$，
$\therefore$当$0 < x < x_0$时$f(x)$单增，当$x_0 < x < \frac{\pi}{2}$时$f(x)$单减，
$\because \sin 0.2 < 0.2 < 0.6 < \sin x_0$，$\therefore 0.2 < x_0$，
$\because f(0) = 0$，$\therefore \sin 0.2 > \frac{4}{5} \times 0.2$，即$a > b$；
$c = \frac{1}{2}\ln\frac{3}{2} = \frac{1}{10}\ln\left( \frac{3}{2} \right)^5 > \frac{1}{10}\ln 7.59$，
$\because \text{e}^2 < 2.72 \times 2.72 = 7.3984$，$\therefore c > \frac{1}{10}\ln\text{e}^2 = 0.2$，
$\therefore c > a > b$．
另证$a > b$：设$g(x) = \sin x - x + \frac{1}{2}x^2\left( 0 < x < \frac{\pi}{2} \right)$，则$g'(x) = \cos x + x - 1$，
设$h(x) = \cos x + x - 1$，则$h'(x) = - \cos x + 1 \geq 0$，
$\because h(0) = 0$，$\therefore h(x) \geq 0$，$\therefore g(x)$单增，
$\because g(0) = 0$，$\therefore g(x) \geq 0$，
$\therefore \sin 0.2 - 0.2 + \frac{1}{2} \times 0.2^2 > 0$，$\therefore \sin 0.2 > 0.18$，$\therefore a > b$．