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Show that $$\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=-2 \pi+2 \cos ^{-1} x$$ if $-1 \leq x \leq-\frac{1}{\sqrt{2}}.$

I tried solving this question a lot but I’m unable to. My answer comes different.

$\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)$ Putting $x=\cos \theta$ $$ \begin{array}{l} =\sin ^{-1}\left(2 \cos \theta \sqrt{1-\cos ^{2} \theta}\right) \\ =\sin ^{-1}\left(2 \cos \theta \sqrt{\sin ^{2} \theta}\right) \quad\left(\because 1-\cos ^{2} \theta=\sin ^{2} \theta\right) \end{array} $$ $$ =\sin ^{-1}(2 \cos \theta \sin \theta) $$ $$ =\sin ^{-1}(2 \cos \theta \sin \theta) $$ $$ =-\sin ^{-1}(\sin 2\theta) $$ (Using $\sin 2 x=2 \sin x \cos x$)

$=2 \theta \quad$ As $x=\cos \theta$ $=2 \times \cos ^{-1} x$ $=2 \cos ^{-1} x$

EDIT: If any confusion with Q writing

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ACB
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S.M.T
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  • Use https://math.stackexchange.com/questions/672575/proof-for-the-formula-of-sum-of-arcsine-functions-arcsin-x-arcsin-y and https://math.stackexchange.com/questions/1116974/why-its-true-arcsinx-arccosx-frac-pi2 Not sure what is 10 ) ? – lab bhattacharjee Feb 17 '22 at 12:47
  • @labbhattacharjee That was just question number 10). – S.M.T Feb 17 '22 at 12:49
  • Should it be $\sin^{-1}(2\cos \theta \sin \theta) = \sin^{-1}(\sin2\theta)$? – peterwhy Feb 17 '22 at 12:57
  • @peterwhy How ? – S.M.T Feb 17 '22 at 13:28
  • $\sin^{-1}(\sin2\theta)$ = $sin^{-1}(2x*\sqrt{1-x^2}$ @peterwhy How are they equal ? It means this too right – S.M.T Feb 17 '22 at 13:32
  • @S.M.T $\sin 2\theta = 2\cos\theta\sin\theta = 2x\sqrt{1-x^2}$ is what you already know in your question (for that ranges of $x$ and $\theta$), so I omitted that part. – peterwhy Feb 17 '22 at 14:03

2 Answers2

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hint

Let $$f(x)=\arcsin(2x\sqrt{1-x^2})-2\arccos(x)$$

assuming $ f $ is differentiable at $ [-1,-\frac{1}{\sqrt{2}}] $,

prove that

$$f'(x)=0$$

So

$$f(x)=Cte=f(-1)=-2\pi$$

Other approach Put $$x=\cos(\frac{\theta}{2})$$

hamam_Abdallah
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Continuing from the substitution in the question, where OP put $x=\cos \theta$. Here I pick $\theta$ to be the particular value $\frac{3\pi}4\le \theta\le \pi$, so that $\theta$ is the principal value $\theta = \cos^{-1}x$.

Using the formula to find general solutions of inverse $\sin$, if given $\sin 2\theta$, $2\theta$ would be in the form

$$2\theta = n\pi + (-1)^n\sin^{-1}(\sin 2\theta)$$

For this case where $\frac{3\pi}2\le 2\theta\le 2\pi$, so $n=2$,

$$\begin{align*} 2\theta &= 2\pi + \sin^{-1}(\sin 2\theta)\\ \sin^{-1}(\sin 2\theta)&=-2\pi + 2\theta\\ \sin^{-1}\left(2x\sqrt{1-x^2}\right) &= -2\pi + 2\cos^{-1}x \end{align*}$$

The conversion from $\sin 2\theta = 2\cos\theta\sin\theta = 2x\sqrt{1-x^2}$ is what OP already knows.

($0 \le \sin \theta \le 1$ is important here, because that confirms that $\sqrt{\sin^2\theta} = \left|\sin\theta\right| = \sin\theta$.)

ACB
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peterwhy
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  • How do you prove the form of 2(theta) you have written in 1st statement ? Could you please explain it more.
    • – S.M.T Feb 17 '22 at 13:25
  • How is sin2(theta) = 2x*sqrt(1-$x^2)$ in the last equation u have written.
    • – S.M.T Feb 17 '22 at 13:25
  • @S.M.T 1) this is a general form that lists all the $t$ given $\sin t$; see https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Solutions_to_elementary_trigonometric_equations – peterwhy Feb 17 '22 at 13:47
  • @peterwhy K. About 2) part sir ? – S.M.T Feb 17 '22 at 13:48
  • @S.M.T 2) $\sin 2\theta = 2\cos\theta\sin\theta = 2x\sqrt{1-x^2}$ is what you already know in your question (for that ranges of $x$ and $\theta$), so I omitted that part. – peterwhy Feb 17 '22 at 14:03
  • Why did we use n=2 ? – S.M.T Feb 17 '22 at 14:51
  • @S.M.T Because $\frac{3\pi}2\le 2\theta\le 2\pi$, which is within $2\pi\pm \frac{\pi}2$, that range corresponds to $n=2$ in the general formula. – peterwhy Feb 17 '22 at 15:15
  • @peterwhy K , if the range corresponds. We can write them as equal to each other. Wow , I didn’t know it. – S.M.T Feb 17 '22 at 15:16
  • If I say sin2x=sinx cosx , then it means that their principal range is equal. – S.M.T Feb 17 '22 at 15:17
  • @S.M.T About the general formula $$2\theta = n\pi + (-1)^n\sin^{-1}(\sin 2\theta),$$ $n=0$ restricts to the range $-\frac\pi2\le 2\theta\le \frac\pi2$, $n=1$ gives $\frac\pi2\le 2\theta\le \frac{3\pi}2$, $n=2$ gives $\frac{3\pi}2\le 2\theta\le \frac{5\pi}2$, etc. – peterwhy Feb 17 '22 at 15:38