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How can I prove this property

$2\sin^{-1}x = \begin{cases}\sin^{-1}(2x \sqrt{1-x^2})&\text{for }-\frac1{\sqrt2}<x<\frac1{\sqrt2} \\ \pi- \sin^{-1}(2x\sqrt{1-x^2})&\text{for }\frac1{\sqrt2}<x<1 \\ -\pi-\sin^{-1}(2x\sqrt{1-x^2})&\text{for }-1<x<-\frac1{\sqrt2}\end{cases}$

The first one is quite easy but how to proceed with other two

Arctic Char
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Saniya
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1 Answers1

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For the second case,

$\begin{align}\frac1{\sqrt2}<x<1 &\Rightarrow \frac\pi4<\sin^{-1}x<\frac\pi2 \\&\Rightarrow \frac\pi2<2\sin^{-1}x < \pi \\&\Rightarrow 0<\pi-2\sin^{-1}x <\frac{\pi}{2}\end{align}$.

Now let $\theta =\pi-2\sin^{-1}x $ and it lies in the range $(0,\frac\pi2)$. This is done to make the angle lie in the principal range $[-\frac\pi2,\frac\pi2]$ of the arcsine function.

$\begin{align}&\Rightarrow x = \sin\frac{\pi-\theta}{2 } = \cos\frac\theta2\\ &\Rightarrow 2x\sqrt{1-x^2} = 2\cos\frac\theta2\sin\frac\theta2 = \sin\theta \\ &\Rightarrow 2x\sqrt{1-x^2}= \sin(\pi-2\sin^{-1}x) \color{blue}{\text{ and the angle lies in the principal range of arcsine }}\\ &\Rightarrow\boxed{2\sin^{-1}x = \pi-\sin^{-1}(2x\sqrt{1-x^2})}\end{align}$

Similarly do with the third case, making the angle lie within the required range.

19aksh
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  • Can tan^-1x+tan-1y = tan^-1( x+y)/(1-xy) for xy<1 tan^-1x+tan-1y =pi+ tan^-1( x+y)/(1-xy) for xy>1 be proved in a similar manner – Saniya Nov 06 '20 at 13:54