Taking the sin of arccos When solving for the value of x in the equation
$$\sin^{-1}{(\sqrt{2x})}=\cos^{-1}(\sqrt{x})$$
one would take the sin of both sides of the equation cancelling out the arcsin leaving
$$\sqrt{2x}=\sin(\cos^{-1}(\sqrt{x}))$$
After researching online for relevant trigonometric identities I found that 
$$\sin(\cos^{-1}(\sqrt{x}))=\sqrt{1-x}$$
How do the trigonometric functions cancel out? 
Would it be the same for $\cos(\sin^{-1}(\sqrt{x}))$?
 A: study $\cos(\arcsin(\sqrt{x}))$:
suppose $\sin(\theta)=\sqrt{x}$, then $\theta=\arcsin(\sqrt{x})$, so
$$\cos(\arcsin(\sqrt{x}))=\cos(\theta)=\sqrt{1-\sin(\theta)^2}=\sqrt{1-x}$$
here we consider the case in the first quadrant. the $\sin(\arccos(\sqrt{x}))$ is obtained the same way.
A: The basic formula is
$$\arccos(a)=\arcsin(\sqrt{1-a^2})$$
now substitute $a=\sqrt{x}$ 
$$\arccos(\sqrt{x})=\arcsin(\sqrt{1-x})$$
and use $\sin(\arcsin(y))=y$ and get
$$\sin(\arccos(\sqrt{x}))=\sin(\arcsin(\sqrt{1-x}))=\sqrt{1-x}$$
The remaining task is to get the ranges for $x$ where these basics are valid, i.e. when is $(\sqrt{x})^2=x?\,$ etc.
A: Let $\displaystyle\cos^{-1}\sqrt x=\theta$
Using the definition of  Principal values, $\displaystyle0\le\theta\le\pi$
and $\displaystyle\cos\theta=\sqrt x\ge0\implies 0\le\theta\le\frac\pi2\implies\sin\theta=+\sqrt{1-x}\implies\theta=\sin^{-1}\sqrt{1-x}$
So, we have $\displaystyle\sin^{-1}\sqrt{2x}=\cos^{-1}\sqrt x=\theta=\sin^{-1}\sqrt{1-x}$
$\displaystyle\implies\sqrt{2x}=\sqrt{1-x}$
A: It's quite simple. Let's start from defining
$$y = \cos^{-1}(\sqrt{x})$$
Then
$$\cos y = \sqrt x$$
and at the same time
$$\sin^2y + \cos^2y = 1$$
so
$$\sin^2y = 1-\cos^2y = 1-(\sqrt x)^2 = 1-x$$
and finally
$$\sin y = \sqrt{\sin^2y} = \sqrt{1-x}$$
Of course we should carefully test for possible absolute values here when taking root of a square etc. but I meant to give a general outline rather than a detailed proof.
A: You can also construct a right triangle with a unit hypotenuse which specifically contains the angle $ \ \theta \ $ such that  
$$ \sin \ \theta \ = \ \sqrt{2x} \ \ , \ \ \cos \ \theta \ = \ \sqrt{x} \ \ . $$

Applying the Pythagorean Theorem will then give us an equation that permits us to solve directly for $ \ x \ $ . (The stated equation together with the domain of the square-root function require that this triangle lie in the first quadrant.)  
As for your second question, if we look at the complimentary angle in this particular triangle, $ \ \frac{\pi}{2} \ - \ \theta \ $ , we see that we will have the same result, $ \ \cos(\sin^{-1}(\sqrt{x})) \ = \ \sqrt{2x} \  $ . 
Upon solving for $ \ x \ $ , we do find that
$$ \sin(\cos^{-1}(\sqrt{x})) \ = \ \cos(\sin^{-1}(\sqrt{x})) \ = \ \sqrt{1-x} \ \  $$
is satisfied.
