A simpler form of $\sin{(\frac{1}{2}\arcsin{x})}$ I've been wondering how to compute $$\sin{(\frac{1}{2}\arcsin{x})}, \quad|x|\le1,$$
I mean how to get rid of the sine and inverse sine function and acquire 
$$\frac{-\sqrt{1 - x}}{2} + \frac {\sqrt{1 + x}}{2},$$ which is the answer Wolfram Alpha provides.
My idea would be to substitute $\frac{1}{2}\arcsin{x}=t$, hence $x=\sin{2t}$ and my goal is to express $\sin{(\frac{1}{2}\arcsin{x})}=\sin{t}$ in terms of $x$.
$$x=\sin{2t}=2\sin{t}\cos{t}, \quad ?=\sin{t}$$
Substituting from $\sin^2{x}+\cos^2{x}=1$ takes me nowhere, could you help?
 A: By the half-angle formula,
$$|\sin(\theta/2)|=\sqrt{\frac{1-\cos(\theta)}2}$$
And by the Pythagorean theorem, assuming $\cos(\theta)\ge0$
$$\sqrt{\frac{1-\cos(\theta)}2}=\sqrt{\frac{1-\sqrt{1-\sin^2(\theta)}}2}$$
Now let $\sin(\theta)=x$ to finally get
$$|\sin(\arcsin(x)/2)|=\sqrt{\frac{1-\sqrt{1-x^2}}2}$$
This is a standard nested radical and decomposes into

$$\sin(\arcsin(x)/2)=\frac{\sqrt{1+x}-\sqrt{1-x}}2$$

Assuming $|x|\le1$.  (note the denesting process nicely came out to letting us drop the absolute value bars $\ddot\smile$)
A: Let $s=\sin\left(\frac12\arcsin x\right)$ and let $c=\cos\left(\frac12\arcsin x\right)$. Then $c^2+s^2=1$ and $2sc=\sin(\arcsin x)=x$. Solving the system$$\left\{\begin{array}{ll}c^2+s^2=1\\2sc=x,\end{array}\right.$$you get that $\displaystyle s=\frac{\sqrt{1-\sqrt{1-x^2}}}{\sqrt2}=\frac{\sqrt{1+x}-\sqrt{1-x}}2$.
A: Let $\arcsin x=y\implies-\dfrac\pi2\le y\le\dfrac\pi2,\sin y=x$
$\implies\cos\dfrac y2\ge0$ and $\cos y=+\sqrt{1-x^2}$
Now as $\cos y=1-2\sin^2\dfrac y2$
$\implies\sin\dfrac y2=\sqrt{\dfrac{1-\cos y}2}$ if $y\ge0$
else $\sin\dfrac y2=-\sqrt{\dfrac{1-\cos y}2}$
