Is $\cos \frac{5\pi}{8}+i\sin\frac{5\pi}{8}$ in $U_8$, the multiplicative group of the $8$th roots of unity in $\mathbb{C}$?

I encountered this problem while reading Fraleigh, A First Course in Abstract Algebra, 4/e. (p.60 #19)

Let $U_8$ be the multiplicative group of the $8$th roots of unity in $\mathbb{C}$. The question asks to find the order of the cyclic subgroup of $U_8$ generated by $\cos \frac{5\pi}{8}+i\sin\frac{5\pi}{8}$.

My questions is: Is $\cos \frac{5\pi}{8}+i\sin\frac{5\pi}{8}$ in $U_8$ to begin with? According to my understanding of cyclic subgroups as defined in Fraleigh, a cyclic subgroup has to be generated from an element in the original group. (See Definition 1.10 and Theorem 1.5.) However, here, $\cos \frac{5\pi}{8}+i\sin\frac{5\pi}{8}$ raised to the eighth power is $-1$ instead of $1$. In this case, isn't "the subgroup of $U_8$ generated by $\cos \frac{5\pi}{8}+i\sin\frac{5\pi}{8}$" ill defined?

The suggested answers in the appendix of the book claims the answer to be $8$, which I am afraid is meaningless if the subgroup isn't defined in the first place.

-
Yup. You're right, it's a mistake in the book. Probably thought of $\frac{5\pi}{4} = \frac{5(2\pi)}{8}$. –  Daniel Fischer Aug 30 '14 at 0:19