Solution
$$\cos18^{\circ\ }=\cos\left(180^{\circ\ }-162^{\circ\ }\right)=-\cos162^{\circ\ }.$$
$$\sin126^{\circ\ }=\sin\left(360^{\circ\ }-234^{\circ\ }\right)=-\sin234^{\circ\ }.$$
So,
$$\cos18^{\circ}+\cos162^{\circ}+\sin126^{\circ}+\sin234^{\circ}$$
$$=-\cos162^{\circ}+\cos162^{\circ}-\sin234^{\circ}+\sin234^{\circ}.$$
$$=0.$$
D is correct choice.
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