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

Question

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...

Answer 1

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.

Answer 2

using $$\cos { \frac { \alpha }{ 2 } } =\pm \sqrt { \frac { 1+\cos { \alpha } }{ 2 } } $$ we get $$\\ \\ \\ \cos { \left( \frac { \arcsin { x } }{ 2 } \right) } =\pm \sqrt { \frac { 1+\cos { \left( \arcsin { x } \right) } }{ 2 } } =\pm \sqrt { \frac { 1+\sqrt { 1-{ x }^{ 2 } } }{ 2 } } $$

Answer 3

Hint:

$$\pm\sqrt{1-\sin^2 (2a)} = \cos(2a) =\cos^2a-\sin^2a = 2\cos^2a -1 $$

Now, take $a=\frac{1}{2}\arcsin x$

Answer 4

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?

Answer 5

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?