<s>[INST] 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?
 [/INST] $${\sin x\over x}\approx1-{1\over6}x^2$$
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
$${\sin^2x\over x^2\sqrt{1-{\sin^2x\over3}}}\approx\left(1-{1\over3}x^2\right)\left(1+{1\over6}x^2\right)\approx1-{1\over6}x^2$$
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