\[ \sin\left(\frac{\arcsin(x)}{2}\right) = \sin\left(\arcsin(x) - \frac{\arcsin(x)}{2}\right) = \]
\[ \sin(\arcsin(x))\cos\left(\frac{\arcsin(x)}{2}\right) - \sin\left(\frac{\arcsin(x)}{2}\right)\cos(\arcsin(x)) = \]
\[ x\cos\left(\frac{\arcsin(x)}{2}\right) - \sin\left(\frac{\arcsin(x)}{2}\right)\sqrt{1-x^2} \Longrightarrow{} \]
\[ \sin\left(\frac{\arcsin(x)}{2}\right) + \sin\left(\frac{\arcsin(x)}{2}\right)\sqrt{1-x^2} = x\cos\left(\frac{\arcsin(x)}{2}\right) \Longrightarrow{} \]
\[ \sin\left(\frac{\arcsin(x)}{2}\right)(1 +\sqrt{1 - x^2}) = x\cos\left(\frac{\arcsin(x)}{2}\right) \Longrightarrow{} \]
\[ \sin\left(\frac{\arcsin(x)}{2}\right) = \frac{x\cos\left(\frac{\arcsin(x)}{2}\right)}{1 +\sqrt{1 - x^2}} \]
Usando que:
\[ \cos\left(\frac{\arcsin(x)}{2}\right) = \sqrt{\frac{1}{2}(1 + \sqrt{1-x^2}}) \]
Spoiler
\[ \cos(1/2 \arcsin(x)) = \sqrt{\cos^2(1/2 \arcsin(x))} = \sqrt{1-\sin^2(1/2 \arcsin(x))} \]
Usando que:
\[ \sin^2(x) = \frac{1 - \cos(2x)}{2} \]
Queda:
\[ \sqrt{1- \frac{1}{2}(1 - \cos(\arcsin(x)))} = \]
\[ \sqrt{1- \frac{1}{2}(1 - \sqrt{1-x^2})} = \]
\[ \sqrt{\frac{1}{2}(1 + \sqrt{1-x^2}}) \]
Queda:
\[ \sin\left(\frac{\arcsin(x)}{2}\right) = \frac{x\sqrt{\frac{1}{2}(1 + \sqrt{1-x^2}})}{1 +\sqrt{1 - x^2}} = \]
\[ \displaystyle\frac{x}{\sqrt{2(1 + \sqrt{1-x^2}})} \]