How to simplify inverse trigonometric function

Question

How to simplify the following equation:

$$\sin(2\arccos(x))$$ I am thinking about:

$$\arccos(x) = t$$

Then we have:

$$\sin(2t) = 2\sin(t)\cos(t)$$

But then how to proceed?

Answer 1

Notice, let $\cos^{-1}x=\theta\iff \cos\theta=x$ $$\sin(2\theta)=2\sin\theta\cos \theta$$$$\sin(2\theta)=2\cos\theta\sqrt{1-\cos^2\theta}$$ setting the value of $\theta$, $$\sin(2\cos^{-1}x)=\color{red}{2x\sqrt{1-x^2}}$$ $\forall \ -1\le x\le 1$

Answer 2

So, you have reasoned that the expression is equivalent to:

$2\sin(\arccos(x))\cos(\arccos(x))$

Because $\arccos(x) ≡ \cos^{-1}(x)$, this is equivalent to

$2x\sin(\arccos(x))$

$\sin(\arccos(x))$ is in fact equivalent to $\sqrt{1+x^2}$ by identity. This can be shown by the Pythagorean Theorem.

The proof of the identity goes as follows: It is well known that $\sin^2(x)+\cos^2(x)=1$

Replacing $x$ with $\arccos(x)$, we have:

$\sin^2(\arccos(x))+ \cos^2(\arccos(x))=1$

Because $\cos$ and $\arccos$ are inverse functions, $\cos^2(\arccos(x))=x^2$

Therefore $\sin^2(\arccos(x))=1-x^2$, and

$\sin(\arccos(x))=\sqrt{1-x^2}$

Therefore the answer is $2x\sqrt{1-x^2}$.