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?
Best Answer
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.