simplification of $\cos\left( \frac{1}{2}\arcsin (x)\right)$ I know this identity :Simplifying $\cos(\arcsin x)$? but how I can simplify 

$$\cos\left( \frac{1}{2}\arcsin (x)\right)$$

if you have any idea. It could also be a sinus instead of a cosinus. When I do it I loop on something...
 A: Here is one way of visualizing it.

To find $\cos(\frac 12\arcsin x)$ we will need to use the Pythagorean theorem to find the length of the hypotenuse of that right triangle. Then use $\frac {\text {adjacent}}{\text{hypotenuse}}$
Alternatively, $\cos (\arcsin x) = \sqrt {1-x^2}$ and apply the half-angle identity.
A: using $$\cos { \frac { \alpha  }{ 2 }  } =\pm \sqrt { \frac { 1+\cos { \alpha  }  }{ 2 }  } $$
we get 
$$\\ \\ \\ \cos { \left( \frac { \arcsin { x }  }{ 2 }  \right)  } =\pm \sqrt { \frac { 1+\cos { \left( \arcsin { x }  \right)  }  }{ 2 }  } =\pm \sqrt { \frac { 1+\sqrt { 1-{ x }^{ 2 } }  }{ 2 }  } $$
A: Hint:
$$\pm\sqrt{1-\sin^2 (2a)} = \cos(2a) =\cos^2a-\sin^2a = 2\cos^2a -1  $$
Now, take $a=\frac{1}{2}\arcsin x$

A: Let $c=\cos\left(\frac12\arcsin x\right)$ and let $s=\sin\left(\frac12\arcsin x\right)$. Then  $2cs=\sin(\arcsin x)=x$ and $c^2+s^2=1$. Can you take it from here?
A: Note that $$\cos(2t)=2\cos^2(t)-1$$ now set $t=\arcsin(x)/2$ resulting:
\begin{align}
\cos(\arcsin(x))=2\cos^2(\arcsin(x)/2)-1
\end{align}
You can take it from here, right?
