8. (17th "Hope Cup" Senior High School Grade 1 Training Question) If the function $f(x)$ is defined on $\mathbf{R}$ as an odd function and is increasing, for any $\theta \in \mathbf{R}$, the inequality $f(\cos 2 \theta-5)+f(2 m+4 \sin \theta)>0$ always holds, then the range of $m$ is
A. $m>5$
B. $m>2$
C. $24$
8. The given inequality can be transformed into
$$
f(\cos 2 \theta-5)>-f(2 m+4 \sin \theta) .
$$

Since $f(x)$ is an odd function on $\mathbf{R}$, we have
$$
f(\cos 2 \theta-5)>f(-2 m-4 \sin \theta),
$$

and since $f(x)$ is an increasing function on $\mathbf{R}$, we have
$$
\begin{array}{c}
\cos 2 \theta-5>-2 m-4 \sin \theta, \\
2 m>2 \sin ^{2} \theta-4 \sin \theta+4=2(\sin \theta-1)^{2}+2,
\end{array}
$$

When $\sin \theta=-1$, the maximum value of $2(\sin \theta-1)^{2}+2$ is 10, so $2 m>10$, which means $m>5$. Therefore, when $m>5$, the original inequality always holds. Choose A.