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.
$\endgroup$2 Answers
$\begingroup$$\cos(2\arcsin x)$ will be $\ge0,$ if
$-\dfrac\pi2\le2\arcsin x\le\dfrac\pi2$
$\iff -\dfrac1{\sqrt2}\le x\le\dfrac1{\sqrt2}$
In that case $$|2x^2-1|=-(2x^2-1)$$
Check if $x>\dfrac1{\sqrt2}$
or $x<-\dfrac1{\sqrt2}$
$\endgroup$ 3 $\begingroup$At this point, we know$$ \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 $\cos(2\sin^{-1}x) = 1-2x^2$.
$\endgroup$ 7
