Решите, пожалуйста (;

Решите задачу:

$x=tg\frac{5\pi}{6}=tg(\pi -\frac{\pi}{6})=-tg\frac{\pi}{6}=-\frac{1}{\sqrt3}\approx -0,58\\\\y=cos\frac{2\pi}{5}=cos72^\circ \ \textgreater \ 0\; ,\; t.k.\; \; 72^\circ \in (0,90^\circ )\\\\z=tg(-\frac{\pi}{8}) =-tg\frac{\pi}{8}=- \frac{1-cos\frac{\pi}{4}}{sin\frac{\pi}{4}} =- \frac{1-\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}} =-\frac{2-\sqrt2}{\sqrt2}=\\\\=1-\sqrt2\approx -0,41\\\\y\ \textgreater \ z\ \textgreater \ x$