# 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 $\cos(\sin^{-1}(\sqrt{x}))$?

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.

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.

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 $\displaystyle\sin^{-1}\sqrt{2x}=\cos^{-1}\sqrt x=\theta=\sin^{-1}\sqrt{1-x}$

$\displaystyle\implies\sqrt{2x}=\sqrt{1-x}$

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.

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.