If 2 sin^{-1} x = sin^{-1}(2x sqrt{1-x^2}),th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We know that $2 \sin^{-1} x = \sin^{-1}(2x \sqrt{1-x^2})$ holds true when $-\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}}$.Let $x = \sin 	heta$

Vedclass · Accepted Answer

$[\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]$

Vedclass · Answer

(C) We know that $2 \sin^{-1} x = \sin^{-1}(2x \sqrt{1-x^2})$ holds true when $-\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}}$.Let $x = \sin 	heta$,where $	heta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.Then $2 \sin^{-1}(\sin 	heta) = \sin^{-1}(2 \sin 	heta \sqrt{1-\sin^2 	heta}) = \sin^{-1}(2 \sin 	heta \cos 	heta) = \sin^{-1}(\sin 2	heta)$.For $2	heta = 2 \sin^{-1} x$ to be valid,$2	heta$ must lie in the principal value branch of $\sin^{-1}$,which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.So,$-\frac{\pi}{2} \le 2	heta \le \frac{\pi}{2} \implies -\frac{\pi}{4} \le 	heta \le \frac{\pi}{4}$.Taking sine on all sides,$\sin(-\frac{\pi}{4}) \le \sin 	heta \le \sin(\frac{\pi}{4})$,which gives $-\frac{1}{\sqrt{2}} \le x \le \frac{1}{\sqrt{2}}$.Thus,$x \in [-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}]$.

If $2 \sin^{-1} x = \sin^{-1}(2x \sqrt{1-x^2})$,then $x \in$ . . . . . . .

Similar Questions

If $\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}$,then the values of $x$ are

Considering only the principal values of the inverse trigonometric functions,the value of $\tan \left(\sin ^{-1}\left(\frac{3}{5}\right)-2 \cos ^{-1}\left(\frac{2}{\sqrt{5}}\right)\right)$ is

If $4\sin^{-1}x + \cos^{-1}x = \pi$,then $x$ is equal to

$\cot ^{-1}\left(2 \cdot 1^2\right)+\cot ^{-1}\left(2 \cdot 2^2\right)+\cot ^{-1}\left(2 \cdot 3^2\right)+\ldots \ldots \ldots \infty =$

Show that $\sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}x$ for $-\frac{1}{\sqrt{2}} \leq x \leq \frac{1}{\sqrt{2}}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module