We have $C>0, h>0$ fixed constants. And $\cos \theta = 1-\frac{h^2}{2}C^2$, for $\theta \in [0,\pi]$. Using Taylor series expansion,
$$\theta = Ch + \frac{C^3}{24}h^3 + \frac{3C^5}{640}h^5 + o(h^5).$$ Now taking $n=1/h$, we have $$n\theta = C + \frac{C^3}{24}h^2 + \frac{3C^5}{640}h^4 + o(h^4).$$
Then, we get
$$\cos(n\theta) = \cos C - \frac{C^3}{24} \sin Ch^2 - \frac{27\sin C + 5C \cos C}{5760} C^5h^4 + o(h^4).$$
How do we get this last expansion for $\cos (n\theta)$? I would greatly appreciate any help.
We are given that $$ n\theta = C + D ,\quad D := \frac{C^3}{24}h^2 + \frac{3C^5}{640}h^4 + O(h^6). $$ Using the addition formula $\ \cos(x+y) = \cos x \cos y - \sin x \sin y \ $ we get the equation $$ \cos n\theta = \cos C \cos D - \sin C \sin D . $$ Now $\ \cos D = 1 - \frac{C^6}{1152}h^4 + O(h)^6 \ $ and $\ \sin D = \frac{C^3}{24}h^2 + \frac{3C^5}{640}h^4 + O(h)^6. \ $ Substituting we get $$ \cos n\theta = \cos C \Big( 1 - \frac{C^6}{1152}h^4 + O(h)^6 \Big) - \sin C \Big(\frac{C^3}{24}h^2 + \frac{3C^5}{640}h^4 + O(h)^6 \Big) $$ and your result follows.