<s>[INST] Simplifying $\cos(2\arcsin(x))$ using only pythagorean trigonometric identity I know that one can simplify $\cos(2\arcsin(x))$ using $\cos(a+b)=\cos(a)\cdot\cos(b)-\sin(a)\cdot\sin(b)$:
\begin{alignat}{1}
\cos(2\arcsin(x))&=\cos^2(\arcsin(x))-\sin^2(\arcsin(x))
\\&=1-2\sin^2(\arcsin(x))
\\&=1-2x^2
\end{alignat}

I tried to make this simplification using only $\sin^2(x)+\cos^2(x)=1$:
\begin{alignat}{1}
\cos^2(2\arcsin(x))&=1-\sin^2(2\arcsin(x))
\\ \left|\cos(2\arcsin(x))\right|&=\sqrt{1-\sin^2(2\arcsin(x))}
\\ &=\sqrt{1-\left(2x\sqrt{1-x^2}\right)^2}
\\ &=\sqrt{1-4x^2 |1-x^2|}
\\ &=\sqrt{1-4x^2(1-x^2)}
\\ &=\sqrt{1-4x^2+4x^4}
\\ &=\sqrt{\left(2x^2-1\right)^2}
\\ &=|2x^2-1|
\end{alignat}
And then I could not figure out how to proceed. So, how to get rid of $|\cdot|$ in:
$$\left|\cos(2\arcsin(x))\right|=|2x^2-1|$$
and get $1-2x^2$ ?

Note: this is a part of my attempt to solve the integral $\int x^2\cdot\sqrt{1-x^2}\,\,\mathrm dx$ by trigonometric substitution.
 [/INST] At this point, we know
$$
\cos(2\sin^{-1}x) = \pm(2x^2-1).
$$
Since both these functions are continuous, we can just check one value in each region where $2x^2-1 > 0$ and $2x^2-1 < 0$ to verify the sign to use in that region. For $|x|<1/\sqrt{2}$, check $x = 0$, and for $|x|>1/\sqrt{2}$, check $x = \pm 1$. You'll find they all work out to $\cos(2\sin^{-1}x) = 1-2x^2$.
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