Suppose that $\bar{X}$ is the mean of $n$ independent standard normal random variables. I have seen it asserted that:
$$n (\cos(\bar{X}) - 1) \overset{\text{dist}}{\longrightarrow} -\frac{1}{2} \cdot \chi_1^2.$$
How do you go about isolating $\cos(\bar{X})$ so as to have an asymptotic distribution of it?
It is probably worth looking at this heuristically. Using the Maclaurin series for the cosine function, we have:
$$\cos x - 1 = - \frac{x^2}{2} + \mathcal{O}(x^4).$$
If we have $X_1,...,X_n \sim \text{IID N}(0,1)$ then we have $\bar{X} \sim \text{N}(0, 1/n)$, which means that $\sqrt{n} \bar{X} \sim \text{N}(0, 1)$. As $n \rightarrow \infty$ the higher order term will shrink rapidly to zero and so we have:
$$\begin{aligned} n(\cos(\bar{X}) - 1) &= -\frac{1}{2} \cdot n\bar{X}^2 + \mathcal{O}(n \bar{X}^4) \\[6pt] &= -\frac{1}{2} \cdot (\sqrt{n} \bar{X})^2 + \mathcal{O} \bigg( \frac{(\sqrt{n} \bar{X})^4}{n} \bigg) \\[6pt] &\rightarrow -\frac{1}{2} \cdot (\sqrt{n} \bar{X})^2 \\[6pt] &\sim -\frac{1}{2} \cdot \chi_1^2. \\[6pt] \end{aligned}$$
Thus, for large $n$, we have the approximate distribution:
$$\cos(\bar{X}) \overset{\text{Approx}}{\sim} 1 - \frac{1}{2n} \cdot \chi_1^2.$$
