Simplifying sin function and its inverse

Question

I am familiar with the simplifying $\sin(\arcsin(x))$ which equals $x$ as long as the domain of $\sin$ is $-1$ to $1$, however I cannot seem to simplify $\sin(2\arcsin(x/5))$, if we assume that $\sin$ is restricted to the domain of $-1$ to $1$, how would you go about solving this?

Answer 1

\begin{align}\sin\left(2\arcsin\left(\frac x5\right)\right)&=2\sin\left(\arcsin\left(\frac x5\right)\right)\cos\left(\arcsin\left(\frac x5\right)\right)\\&=2\times\frac x5\sqrt{1-\frac{x^2}{25}}\\&=\frac25x\sqrt{1-\frac{x^2}{25}}.\end{align}

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Note that $\cos(\arcsin x)=\sqrt{1-x^2}$, because $\cos(\arcsin x)\geqslant0$ and\begin{align}1&=\sin^2(\arcsin x)+\cos^2(\arcsin x)\\&=x^2+\cos^2(\arcsin x).\end{align}

Answer 2

Hint: $2\arcsin x=\arcsin (2x \sqrt{1-x^2})$

Answer 3

We use, that $\sin(2x)=2\sin(x)\cos(x)$, and $\sin(x)^2+\cos(x)^2=1$. We get:

$\sin(2\arcsin(x/5))=2\sin(\arcsin(x/5))\cos(\arcsin(x/5))=2\frac{x}{5}\cdot\cos(\arcsin(x/5))$

We have $\cos(x)=\sqrt{1-\sin(x)^2}$ and proceed:

$2\frac{x}{5}\cdot\sqrt{1-\sin(\arcsin(x/5))^2}=\frac{2x}{5}\sqrt{1-\frac{x^2}{25}}$