Let \(\theta = \cos^-1 \frac18\). Then \(\cos \theta = \frac18\).
We need \(\cos \frac\theta2\). Recall the half-angle formula:
\[
\cos \frac\theta2 = \sqrt \frac1 + \cos \theta2
\]
Since \(0 \leq \theta \leq \pi\), \(\cos \frac\theta2>0\), so we take the positive root.
Substitute \(\cos \theta = \frac18\):
\[
\cos \frac\theta2 = \sqrt \frac1 + \frac182
\]
Simplify:
\[
1 + \frac18 = \frac98
\]
So:
\[
\cos \frac\theta2 = \sqrt \frac\frac982 = \sqrt \frac916 = \frac34
\]
Thus, the value is \(\frac34\).