# Small-angle approximation of $\frac{\sin^2 x}{x^2 \sqrt{1-\frac{\sin^2 x}{3}}}$

I need to show the following:

$$\frac{\sin^2 x}{x^2 \sqrt{1-\frac{\sin^2 x}{3}}} \approx 1-\frac{x^2}{6}$$ when $$x$$ is small.

I think this problem is trickier than most other questions like it because in the original source there is comment saying "if you got $$1+\frac{x^2}{6}$$ [what I got] then think again!". My attempt was:

When $$x$$ is small, $$\sin x \approx x$$ so

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

Then using the binomial series approximation,

$$\left ( 1 - \frac{x^2}{3} \right )^{-\frac{1}{2}} \approx 1 - \frac{1}{2}\left ( -\frac{x^2}{3} \right ) + ... = 1 + \frac{x^2}{6}$$

...and so it looks like I've fallen into whatever trap the question set.

Where is my error?

Suppose for simplicity you had two polynomials $$P(z) = 1+ z + z^2 + 4z^4 + 7z^5$$ and $$Q(z) = 1 + 2z + 3z^2 + 4z^3$$, and I asked you to calculate the product $$P(z)Q(z)$$... but not the entire thing. Suppose I only want the terms up to quadratic term; i.e if we write \begin{align} P(z)Q(z) &= a_0 + a_1z + a_2z^2 + a_3z^3 + a_4z^4 + a_5z^5 +a_6z^6 + a_7z^7 +a_8z^8 \end{align} then I'm asking you to find the coefficients $$a_0,a_1,a_2$$ (but for now, let's just say for some reason I'm interested in what happens when $$|z|$$ is very small up to an accuracy of quadratic order, so I don't really care about the rest of the terms ). Well, we just multiply everything out: \begin{align} P(z)Q(z) &= (1+ z + z^2 + 4z^4 + 7z^5)(1 + 2z + 3z^2 + 4z^3) \\ &= 1 + (1\cdot 2z + z \cdot 1) + (1\cdot 3z^2 + z\cdot 2z + z^2 \cdot 1) \\ &+ \text{(terms involving z^3 or higher, which I don't care about for now)} \\ &= 1 + 3z + 6z^2 + O(z^3) \end{align} In other words, because in my final product, I'm only interested in calculating up to the quadratic term, I can simply truncate the polynomials $$P$$ and $$Q$$ to quadratic order, and then multiply them out (and then again only keep terms up to quadratic order): \begin{align} P(z)Q(z) &= (1 + z + z^2 + \cdots)(1 + 2z + 3z^2 + \cdots) \\ &= 1 + 3z + 6z^2 + O(z^3) \end{align}

Again, because I'm only interested up to quadratic order, there's no need for me to keep any terms beyond that for $$P(z)$$ and $$Q(z)$$, because if I approximate $$P(z) \approx 1+ z + z^2 + \color{red}{4z^4}$$ (i.e I keep the $$4^{th}$$ order term) and I multiply with $$Q(z) = 1+2z+3z^2 + 4z^3$$, then the red term multiplied with anything in $$Q(z)$$ will yield terms which are $$4^{th}$$ order or higher (which I don't care about).

But what you should not do is truncate $$P(z)$$ and $$Q(z)$$ up to linear order, and say that \begin{align} P(z)Q(z) & \approx (1+z)(1+2z) = 1 + 3z + 2z^2 \end{align} Because in this way, you're missing out other second order contributions (by multiplying constant term of $$P$$ with quadratic term of $$Q$$ and vice-versa).

This is how you know how many terms you need to use in your approximation. In your case, you want to approximate \begin{align} f(x) &= \dfrac{\sin^2x}{x^2\sqrt{1 - \frac{\sin^2x}{3}}} \end{align} up to $$2^{nd}$$ order. So, what do we do? We write things as a product first: \begin{align} f(x) &= \left(\dfrac{\sin x}{x}\right)\cdot\left(\dfrac{\sin x}{x}\right) \cdot \left(\dfrac{1}{\sqrt{1- \frac{\sin^2x}{3}}}\right)\tag{1} \end{align} Now, we have to expand each bracketed term up to atleast second order in $$x$$ and then multiply the result together. First: \begin{align} \dfrac{\sin x}{x} &= \dfrac{x - \dfrac{x^3}{6} + O(x^4)}{x} = 1 - \dfrac{x^2}{6} + O(x^3) \tag{2} \end{align} Next, we recall that \begin{align} \dfrac{1}{\sqrt{1-z}} &= 1+ \dfrac{z}{2} + \dfrac{3z^2}{8} + O(z^3) \end{align} Now, plug in $$z= \frac{\sin^2x}{3} = x + O(x^3)$$, to get \begin{align} \dfrac{1}{\sqrt{1-\frac{\sin^2x}{3}}} &= 1+ \dfrac{1}{2}\left(\dfrac{\sin^2x}{3}\right) + \dfrac{3}{8}\left(\frac{\sin^2x}{3}\right)^2 + O((\sin^2 x)^3) \\ &= 1 + \dfrac{1}{2}\left(\dfrac{x^2 + O(x^4)}{3}\right) + O(x^4) + O(x^6) \\ &= 1 + \dfrac{1}{6}x^2 + O(x^4)\tag{3}, \end{align} where in the second line, hopefully it's clear how I got the various terms: for example $$\sin x = x + O(x^3)$$, so $$\left(\frac{\sin^2x}{3}\right)^2 = \frac{1}{9}\sin^4x = \frac{1}{9} (x + O(x^3))^4 = O(x^4)$$. Therefore, the final answer is obtained by plugging $$(2)$$ and $$(3)$$ into $$(1)$$ : \begin{align} f(x) &= \left(1 - \dfrac{x^2}{6} + O(x^3)\right)^2 \cdot \left(1 + \dfrac{1}{6}x^2 + O(x^4)\right) \\ &= 1 - \dfrac{x^2}{6} + O(x^4) \end{align}

Long story short, if your end goal is to calculate up to second order, then at each stage of your algebra make sure you're keeping terms atleast up to $$x^2$$.

• +1 for a fantastic answer. While experienced users may not really want to read the long story, it's a boon for beginners. Jul 18 '20 at 0:49
• I wish I could choose 2 answers as accepted...this was very well explained, thanks :) Jul 18 '20 at 11:19

$${\sin x\over x}\approx1-{1\over6}x^2$$

so

$${\sin^2x\over x^2}\approx\left(1-{1\over6}x^2\right)^2\approx1-{1\over3}x^2$$

not just $$1$$. We get

$${\sin^2x\over x^2\sqrt{1-{\sin^2x\over3}}}\approx\left(1-{1\over3}x^2\right)\left(1+{1\over6}x^2\right)\approx1-{1\over6}x^2$$

• Is there any way to know when I should use more than one term to approximate sin(x) (like you did here)? Jul 17 '20 at 21:04
• @Nick_2440, it might be a good idea to carry asymptotics along explicitly in the form of $\sin x=x-{1\over6}x^3+O(x^5)$, etc. That way if you try $\sin x=x+O(x^3)$ you quickly see that ${\sin x\over x}=1+O(x^2)$ isn't good enough to give you an answer of the form $1-{1\over6}x^2+O(x^3)$. In fact it might be a good exercise to write out your own answer here doing that. Jul 17 '20 at 21:21

You can also note the the singularity at $$x=0$$ is removable and that $$f$$ is in fact at least 4 times differentiable. Taylor's expansion gives you the answer...

$$f(x)=f(0)+f'(0)x + \frac 12 f''(0) x^2 + O(x^3)$$

where

$$f(0)=\lim_{x\to 0}f(0) = 1, \quad f'(0) = \lim_{x\to 0}f'(x)=0, \quad f''(0)=\lim_{x\to 0}f''(x)= -\frac 13, \quad f'''(0)=0$$

yielding

$$f(x)=1-\frac 16 x^2 + O(x^4)\approx 1-\frac 16 x^2 (\textrm{for small } x ).$$

• Typo in last line. Jul 17 '20 at 21:47
• @spalein thanks. Jul 18 '20 at 10:35