利用三角函数的单调性, 比较下列各组数的大小.

(1) $\sin \left(-\frac{\pi}{18}\right)$ 与 $\sin \left(-\frac{\pi}{10}\right)$;

(2)sin196与 $\cos 156^{\circ}$;

(3) $\cos \left(-\frac{23}{5} \pi\right)$ 与 $\cos \left(-\frac{17}{4} \pi\right)$.
(1) $\because-\frac{\pi}{2}<-\frac{\pi}{10}<-\frac{\pi}{18}<\frac{\pi}{2}$,

$\therefore \sin \left(-\frac{\pi}{18}\right)>\sin \left(-\frac{\pi}{10}\right)$.

(2) $\sin 196^{\circ}=\sin \left(180^{\circ}+16^{\circ}\right)=-\sin 16^{\circ}$,

$\cos 156^{\circ}=\cos \left(180^{\circ}-24^{\circ}\right)=-\cos 24^{\circ}=-\sin 66^{\circ}$,

$\because 0^{\circ}<16^{\circ}<66^{\circ}<90^{\circ}, \quad \therefore \sin 16^{\circ}<\sin 66^{\circ}$;

从而 $-\sin 16^{\circ}>-\sin 66^{\circ}$,

即 $\sin 196^{\circ}>\cos 156^{\circ}$.

(3) $\cos \left(-\frac{23}{5} \pi\right)=\cos \frac{23}{5} \pi$

$=\cos \left(4 \pi+\frac{3}{5} \pi\right)=\cos \frac{3}{5} \pi$,

$\cos \left(-\frac{17}{4} \pi\right)=\cos \frac{17}{4} \pi$

$=\cos \left(4 \pi+\frac{\pi}{4}\right)=\cos \frac{\pi}{4}$.

$\because 0<\frac{\pi}{4}<\frac{3}{5} \pi<\pi$, 且 $y=\cos x$ 在 $[0, \pi]$ 上是减函数,

$\therefore \cos _{5} \frac{3}{\pi} \pi<\cos ^{\frac{\pi}{4}}$,

即 $\cos \left(-\frac{23}{5} \pi\right)<\cos \left(-\frac{17}{4} \pi\right)$.