# Question About Small Angle Approximation

I'm currently going through "An Introduction to Mechanics" by Kleppner and Kolenkow. On page 38 of the text, he glosses over the small-angle approximation of $$sin(x)$$ and $$cos(x)$$.

Specifically, he uses the Taylor series of $$sin(x)$$ and $$cos(x)$$ to show that $$sin(x) \approx x$$ and $$cos(x) \approx 1 - \frac{1}{2}x^2$$.

He ends this section by remarking that "These expressions, which are sometimes called the small angle approximations, are valid up to terms of order $$x^3$$, denoted by $$O(x^3)$$".

I don't understand what being valid up to terms of order $$x^3$$ means. Could someone please explain?

Generally speaking, if a polynomial approximation $$p (x)$$ is accurate to $$O (x^n)$$, then

$$f (x) = p (x) + \varepsilon (x)$$

where $$\varepsilon (x)$$ approaches $$0$$ more rapidly than $$x^n$$; i.e., $$\lim_{x \to 0} \frac{\varepsilon (x)}{x^n} = 0$$.

For example, in the "small-angle" linear approximation for $$f (x) = \sin (x)$$, we have $$p (x) = x$$. Using Taylor expansion, we have an explicit expression for $$\varepsilon (x)$$, which is

$$\varepsilon (x) = -\frac{1}{3!} x^3 + \frac{1}{5!} x^5 - \dots$$

The linear approximation $$p$$ is accurate to $$O (x^2)$$, because

$$\lim_{x \to 0} \frac{\varepsilon (x)}{x^2} = \lim_{x \to 0} -\frac{1}{3!} x + \frac{1}{5!} x^3 - \dots = 0$$

But $$p$$ is not accurate to $$O (x^3)$$, since

$$\lim_{x \to 0} \frac{\varepsilon (x)}{x^3} = -\frac{1}{6} \ne 0$$

In a similar fashion, the quadratic-accurate polynomial approximation to $$f (x) = \cos x$$ is $$p (x) = 1 - \frac{1}{2} x^2$$. It turns out that this approximation is accurate to $$O (x^3)$$, since

$$\lim_{x \to 0} \frac{\varepsilon (x)}{x^3} = \lim_{x \to 0} \frac{\frac{1}{4!} x^4 - \frac{1}{6!} x^6 + \dots}{x^3} = \lim_{x \to 0} \frac{1}{4!} x - \frac{1}{6!} x^3 + \dots = 0$$