Solving $\arcsin\left(2x\sqrt{1-x^2}\right) = 2 \arcsin x$

If we have

$$\arcsin\left(2x\sqrt{1-x^2}\right) = 2 \arcsin x$$

we have to find the set of $x$ for which this is true.

I tried to solve it by putting $x = \sin a$ or $\cos a, but got no result. I am totally stuck how to do it. • First of all you should improve your title. The spelling looks weird. – callculus Apr 29 '16 at 18:19 • @callculus sorry now I have edited – user101522 Apr 29 '16 at 18:21 • @DougM I am not getting nothing like them – user101522 Apr 29 '16 at 18:34 5 Answers $$2\arcsin x =\begin{cases} \arcsin(2x\sqrt{1-x^2}) \;\;;2x^2\le 1 \\ \pi - \arcsin(2x\sqrt{1-x^2}) \;\;;2x^2 > 1, 0< x\le 1\\ -\pi - \arcsin(2x\sqrt{1-x^2}) \;\;;2x^2> 1, -1< x \le 0 \end{cases}$$ So the given equation clearly holds true for$2x^2\le1\iff-\dfrac1{\sqrt2}\le x\le\dfrac1{\sqrt2}$For the other two cases, we need $$\arcsin(2x\sqrt{1-x^2})=\pm\dfrac\pi2$$ $$\iff2x\sqrt{1-x^2}=\sin\left(\pm\dfrac\pi2\right)=\pm\sin\dfrac\pi2=\pm1$$ $$\iff(2x^2-1)^2=0$$ which has already been covered in the first case Note that if $$\arcsin(2x\sqrt{1-x^2})=2\arcsin(x) \tag 1$$ then taking the sine of both sides of$(1)yields \begin{align} 2x\sqrt{1-x^2}&=\sin(2\arcsin(x))\\\\ &=2\sin(\arcsin(x))\cos(\arcsin(x))\\\\ &=2x\sqrt{1-x^2} \end{align} However, taking the cosine of the left-hand side of(1)$yields $$\cos(\arcsin(2x\sqrt{1-x^2}))=|1-2x^2|$$ while taking the cosine of the right-hand side of$(1)$yields $$\cos(2\arcsin(x))=1-2x^2$$ Therefore, the equality $$\arcsin(2x\sqrt{1-x^2})=2\arcsin(x)$$ is valid for$|x|\le 1/\sqrt 2$. suppose x = sin a: arcsin(2sin a cos a) = 2 sin a arcsin(sin 2a) = 2a The range of arcsin is$[-\pi/2, \pi/2]$arcsin(sin 2a) = 2a is true when 2a is within that range. a in$[-\pi/4,\pi/4]$x in$[\sin (-\pi/4), \sin(\pi/4)]$• Doug,$\sqrt{1-\sin^2(a)}=|\cos(a)|$, not$\cos(a)$. – Mark Viola Apr 29 '16 at 18:49 This is essentially the same approach as in the answers of Doug M and Dr. MV, just written to separate the "if" from the "only if" in showing that$\arcsin(2x\sqrt{1-x^2})=2\arcsin x$if and only if$-1/\sqrt2\le x\le1/\sqrt2$. If$\arcsin(2x\sqrt{1-x^2})=2\arcsin x$, then the fact that range of the arcsine function is$[-\pi/2,\pi/2]$tells us $$-{\pi\over2}\le\arcsin(2x\sqrt{1-x^2})\le{\pi\over2}\implies-{\pi\over4}\le\arcsin x\le{\pi\over4}\implies-{1\over\sqrt2}\le x\le{1\over\sqrt2}$$ On the other hand, if$-1/\sqrt2\le x\le1/\sqrt2$, then$x$can be written as$\sin\theta$with$-\pi/4\le\theta\le\pi/4$. For such$\theta$'s, we have$\cos\theta=\sqrt{1-\sin^2\theta}$,$\arcsin(\sin\theta)=\theta$, and$\arcsin(\sin2\theta)=2\theta$, and it follows that $$\arcsin(2x\sqrt{1-x^2})=\arcsin(2\sin\theta\cos\theta)=\arcsin(\sin2\theta)=2\theta=2\arcsin(\sin\theta)=2\arcsin x$$ Thus the solution set is precisely the interval$-1/\sqrt2\le x\le1/\sqrt2$. • Barry, this is solid. +1 ... -Mark – Mark Viola Apr 29 '16 at 20:17 Consider the function $$f(x)=\arcsin(2x\sqrt{1-x^2})-2\arcsin x$$ which is defined in the domain where $$\begin{cases} |2x\sqrt{1-x^2}|\le 1\\[4px] |x|\le 1 \end{cases}$$ The first inequality becomes, writing$t=|x^2|$, $$4t-4t^2\le 1$$ that's satisfied for every$t$. So our function$f$is defined over$[-1,1]$. The derivative of$f$is $$f'(x)=\frac{1}{\sqrt{1-4x^2(1-x^2)}}\left(2\sqrt{1-x^2}-\frac{2x^2}{\sqrt{1-x^2}}\right)-\frac{2}{\sqrt{1-x^2}}$$ that can be simplified to $$f'(x)=\frac{2}{\sqrt{1-x^2}} \left(\frac{1-2x^2}{|1-2x^2|}-1\right)$$ The derivative exists in the set $$(-1,-1/\sqrt{2})\cup(-1/\sqrt{2},1/\sqrt{2})\cup(1/\sqrt{2},1)$$ and it can be simplified further as $$f'(x)=\begin{cases} -\dfrac{4}{\sqrt{1-x^2}} & \text{for 1/\sqrt{2}<|x|<1}\\[8px] 0 & \text{for |x|<1\sqrt{2}} \end{cases}$$ Thus the function is constant over$[-1/\sqrt{2},1/\sqrt{2}]$(also at the extremes, by continuity). Since$f(0)=0$, we have that the given identity is valid on this interval and nowhere else, because over$[-1,-1/\sqrt{2}]$and$[1/\sqrt{2},1]\$ the function is strictly decreasing. 