If $x+\frac{1}{x}=2 \cos \theta$, then for $n \in N$, $x^{*}+\frac{1}{x^{n}}$ equals
(A) $2 \cos n$
(B) $2^{n} \cos \theta_{3}$
$(C) 20 \cos _{3} \theta_{3}$
(D) $2 \cos \theta$;
(E) $2^{n} \cos n \theta$
Let $n=2$, then $x^{2}+\frac{1}{x^{2}}=(x+\frac{1}{x})^{2}-2=2 \cos 2 \theta$, so the answer is $(A)$.