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$$\arcsin(2x^2-1)-2\arcsin x=-\dfrac{\pi}{2}$$

Thanks in advance

kimtahe6
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Stanson
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    The title isn't supposed to be the first line of your question. – Git Gud Oct 05 '13 at 09:22
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    For any $x$ which makes sense the following is true: $$\begin{align} \arcsin (2x^2-1)-2\arcsin (x)=-\dfrac \pi 2&\iff \arcsin (2x^2-1)=2\arcsin (x)-\dfrac \pi 2 \ &\implies \sin(\arcsin (2x^2-1))=\sin \left(2\arcsin (x)-\dfrac \pi 2\right)\ &\implies 2x^2-1 =\cos (2\arcsin(x))\ &\implies 2x^2-1=(\cos(\arcsin (x)))^2-(\sin (\arcsin(x)))^2\ &\implies 2x^2-1=\pm\sqrt{1-(\sin (\arcsin(x)))^2}-x^2\ &\implies 2x^2-1=\pm \sqrt{1-x^2} -x^2\ &,,,,,\vdots\end{align}$$ Warning: I didn't check to see if it's easy to do the rest. – Git Gud Oct 05 '13 at 09:32

1 Answers1

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Let $\arccos x=y\iff x=\cos y$ where $0\le y\le \pi$ which is the range of the principal values of $\arccos x$

$\implies 0\le2y\le2\pi$

as $\arccos x+\arcsin x=\frac\pi2$

$\displaystyle\arcsin(2x^2-1)-2\arcsin x$

$\displaystyle=\arcsin(2\cos^2y-1)-2\left(\frac\pi2-y\right)$

$\displaystyle=\arcsin(\cos2y)-\pi+2y$

$\displaystyle=\frac\pi2-\arccos(\cos2y)-\pi+2y$

Case $1:\arccos(\cos2y)=2y \iff 0\le 2y\le\pi\iff0\le y\le\frac\pi2$

Case $2:$ else we have $\frac\pi2<y\le\pi,$ consequently $\arccos(\cos2y)=2\pi-2y$

lab bhattacharjee
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