Small angle approximations - different answers

Question

I would like to approximate $$\frac{\cos^2{x}}{\sin(x) \tan(x)}$$ using the small angle approximations.

Throughout I will use $\sin(x) \approx x$, $\tan(x) \approx x$, $\cos(x) \approx 1 - \frac{x^2}{2}$.

Method 1: $$\frac{\cos^2{x}}{\sin(x) \tan(x)} \approx \frac{\left(1-\frac{x^2}{2}\right)^2}{x^2} = \frac{1}{4}(x^2 - x^{-2} - 1)$$

Method 2:

$$\frac{\cos^2{x}}{\sin(x) \tan(x)} = \frac{\cos^3{x}}{\sin^2(x)} \approx \frac{1}{x^2} \left(1 - \frac{x^2}{2}\right)^3$$

which when expanded out gives for example $-\frac{3}{2} \neq - \frac{1}{4}$ as the constant term.

Why the differences? Which is the better way to approximate this and why?

Answer 1

Because you are using less terms in your first expansion. What happens if you take $\tan(x) \approx \frac{x}{1 -x^2/2}$? It gives the second expansion. So the first one is a weaker approximation. In the first your are not keeping all terms of order $x^2$, but you do in the second. Note that the leading order terms for small $x$ are $x^{-2}$ which are the same.

Answer 2

Your function is singular in $x\to 0$. Hence, it does not make sense to approximate it by a Talyor expansion centered of around $x=0$.

The formal way in obtaining the Taylor series expansion would require us to calculate

$$f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n=f(x_0)+\sum_{n=1}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n$$

as $f(x=x_0)$ is singular the taylor expansion does not exist.