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Mathematics Question on Properties of Inverse Trigonometric Functions

The formula 2 sin⁡−1x=sin⁡−1(2x1−x2)2\,\sin^{-1}x=\sin^{-1}(2x\sqrt{1-x^2}) holds for

A

x ∈ [0,1]x\,\in\,[0,1]

B

x ∈ [−12,12]x\,\in\,\left[-\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right]

C

x ∈ (−1,0)x\,\in\,(-1,0)

D

x ∈ [−32,32]x\,\in\,\left[-\frac{\sqrt3}{2},\frac{\sqrt3}{2}\right]

Answer

x ∈ [−12,12]x\,\in\,\left[-\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right]

Explanation

Solution

Let x=sin θx= sin\,\theta Then sin−1(2x1−x2)=sin−1(sin 2θ)sin^{-1}\left(2x\sqrt{1-x^{2}} \right) = sin^{-1} \left(sin \,2\theta\right) =2θ=2 sin−1x = 2\theta = 2\,sin^{-1} x if −π2≤2θ≤π2-\frac{\pi}{2} \le 2\theta \le \frac{\pi}{2} i.e., if −π4≤θ≤π4-\frac{\pi}{4} \le \theta \le \frac{\pi}{4} i.e., if sin(−π4)≤sin θ≤sinπ4sin\left(-\frac{\pi}{4}\right)\le sin \,\theta\le sin \frac{\pi}{4} i.e., if −12≤x≤12-\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}} ∴x∈[−12,12]\therefore x \in \left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]