Quick way to expand $\cos^{-1}(\cos^2 x)$ up to $O(x^2)$ For a part of a question, I need to expand $\cos^{-1}(\cos^2 x)$ up to $O(x^2)$ about $x=0$. It took me quite a while to get an incorrect answer. What are some quick and efficient offline (i.e, no alpha) ways to get a good approximation?
EDIT: To elaborate a bit more, the first order term can be confidently stated to be $\sqrt{2}x$ as this expression is the length of a hypotenuse for a right spherical triangle, which must reduce to the flat triangle hypotenuse. No "reasoning" like this could work for second order term so I had to take the cumbersome way.

By symmetry, it is enough to consider the case $x \to 0^+$.
Here's a rigorous and not too complicated way to derive the asymptotic equality
$$
\arccos (\cos ^2 x) = \sqrt 2 x - \frac{{\sqrt 2 }}{{12}}x^3  + O(x^5 )\;\; {\rm as} \; x \to 0^+.
$$
From Wikipedia,
$$
\arccos x = 2\arctan \frac{{\sqrt {1 - x^2 } }}{{1 + x}},\;\; - 1 < x \le 1.
$$
Hence
$$
\arccos (\cos ^2 x) = 2\arctan \frac{{\sqrt {1 - \cos ^4 x} }}{{1 + \cos ^2 x}} = 2\arctan \frac{{\sin x}}{{\sqrt {1 + \cos ^2 x} }} = 2\arctan \frac{{\sin x}}{{\sqrt {2 - \sin ^2 x} }}.
$$
Next, from
$$
\frac{1}{{\sqrt {2 - x} }} = \frac{1}{{\sqrt 2 }} + \frac{1}{{4\sqrt 2 }}x + O(x^2 )
$$
we get
$$
\frac{{\sin x}}{{\sqrt {2 - \sin ^2 x} }} = \bigg(\frac{1}{{\sqrt 2 }} + \frac{1}{{4\sqrt 2 }}\sin ^2 x + O(x^4 )\bigg)\sin x.
$$
Then, from
$$
\sin x = x - \frac{{x^3 }}{6} + O(x^5 )
$$
we get
$$
\frac{{\sin x}}{{\sqrt {2 - \sin ^2 x} }} = \frac{1}{{\sqrt 2 }}\bigg(x - \frac{{x^3 }}{6}\bigg) + \frac{1}{{4\sqrt 2 }}x^3  + O(x^5 ) = \frac{1}{{\sqrt 2 }}x + \frac{1}{{12\sqrt 2 }}x^3  + O(x^5 ).
$$
Next, from 
$$
\arctan x = x - \frac{{x^3 }}{3} + O(x^5 )
$$
we get
$$
\arctan \frac{{\sin x}}{{\sqrt {2 - \sin ^2 x} }} = \frac{1}{{\sqrt 2 }}x + \frac{1}{{12\sqrt 2 }}x^3  - \frac{1}{{3\sqrt 2 ^3 }}x^3  + O(x^5 ) = \frac{1}{{\sqrt 2 }}x - \frac{1}{{12\sqrt 2 }}x^3  + O(x^5 ).
$$
Finally,
$$
\arccos (\cos ^2 x) = 2\arctan \frac{{\sin x}}{{\sqrt {2 - \sin ^2 x} }} = \sqrt 2 x - \frac{{\sqrt 2 }}{{12}}x^3  + O(x^5 ).
$$
