Answer: D
----------------------------------------------
$\cos^2\theta=\frac{\cos2\theta+1}{2}$ and $\cos^3\theta=\frac{\cos3\theta+3\cos\theta}{4}$.
Using these two identities we have:
$$\begin{align*}\cos^6\theta =&\left(\frac{\cos2\theta+1}{2}\right)^3 \\ =& \frac{\cos^3(2\theta)+1+3\cos^2(2\theta)+3\cos(2\theta)}{8}\\ =&\frac{\frac{\cos(6\theta)+3\cos(2\theta)}{4}+1+3\left(\frac{\cos(4\theta)+1}{2}\right)+3\cos(2\theta)}{8}\\ =& \text{some non-constant terms}+\frac{5}{16}\end{align*}$$
Hence, $\frac{5}{16}=\frac{10}{32}$.