A curious way of generating series expansion for $\cos x$ If we take the approximation $\sin x  \approx x$, then, using the trigonometric identity $1- \cos 2x = 2\sin^2 x$, and take $2\sin^2 x \approx 2x^2$, we get, after making the substitution $x \to x/2$, that
$$\cos x \approx 1 - \frac{x^2}{2}$$
Now, using the trigonometric identity
$$\cos(2x)=2\cos^2(x)-1$$
and using the last approximation  $\cos x \approx 1 - \frac{x^2}{2}$, then we get a new approximation for $\cos x$, namely
$$\cos(2x)\approx2\left(1-\frac{x^2}2\right)^2-1=1-2x^2+\frac{x^4}2$$
Then let $x\to\frac x2$ to get
$$\cos(x)\approx1-\frac{x^2}2+\frac{x^4}{32}$$
So, we repeat:
$$\cos(2x)=2\cos^2(x)-1\approx2\left(1-\frac{x^2}2+\frac{x^4}{32}\right)^2-1$$
and so on. This seems to generate a series expansion for $\cos x$, similar to Taylor's series, but with greater denominators. 
The question is: Does the iterative procedure described above generates better and better approximations to $\cos x$, that is, a Taylor-like series one, or this iterative procedure doesn't converge to $\cos x$ to arbitrary accuraty for real $x$?

You get weird coefficients for terms beyond $x^2$ because you re using an inaccurate input for the double angle formula.  If you have 
$\cos x = 1-(x^2/2)+O(x^4)$, 
you must accept 
$\cos 2x = 1-(2x^2)+(x^4/2)+O(x^4)$; 
and putting $x/2$ for $x$ then gives 
$\cos x = 1-(x^2/2)+(x^4/32)+O(x^4)$
You did not make the error term any smaller in order of magnitude than the quartic term you added, so you cannot defend the quartic term.
How to get the right quartic term?  Assume
$\cos x = 1-(x^2/2)+(ax^4)+O(x^6)$.
Apply the double angle formula; after simplifying you get
$\cos 2x = 1-(2x^2)+((4a+(1/2))x^4)+O(x^6)$,
where everything that's a multiple of $x^6$ or a higher power is lost in the "noise" of the error term. Put $x/2$ for $x$ to then get
$\cos x = 1-(x^2/2)+((a/4+(1/32))x^4)+O(x^6)$.
This matches your original assumption if $a=(a/4)+(1/32)$; thus, properly, $a=(1/24)$.
