Approximating arcsin from above I am very new to function approximations, and I am interested in approximating arcsin with a function $f$, s.t. $f(x) \geq \arcsin(x)$ for all $x$. 
Taylor series would give me a function which is always at most $\arcsin(x)$, but I am wondering if there are maybe some trigonometric identities with which I can obtain a desired approximation. 
 A: For $0<x<\pi/2$ there is $\sin x<x<\tan x$, so for these $x$, an upper bound is
$$\arcsin(x)<\arcsin(\tan(x)),$$
while for $-\pi/2<x<0$ there is $\tan x<x<\sin x$, so for these $x$ an upper bound is
$$\arcsin(x)<x.$$
It looks like there is not a "simple" function which does the job on both sides of $0$ using the same formula. And also $\arcsin(\tan(x))$ is more complicated than the function you're trying to bound. Still, the taylor series for it may give a polynomial upper bound for $x>0$ [I haven't checked this].
EDIT: I just found some terms of the taylor series of $\arcsin(\tan(x))$. It starts out
$$x+(1/2)x^3+(3/8)x^5+(83/240)x^7+...$$
This will be greater than $\arcsin(x)$ if we use all the terms; if one wants the best upper bound from this, clearly one should not keep adding positive terms, but stop as soon as the series "beats" the series for $\arcsin(x)$. I haven't checked, but it looks like all we get from the taylor series for $\arcsin(\tan(x))$ is the simple statement
$$\arcsin(x)<x+(1/2)x^3,$$
which seems to hold for reasonably small positive $x$.
I guess what one would really want is a sequence of upper bounds that converged down to $\arcsin x$ for positive $x$. I don't know how to construct that.
