If sin ^{-1} x+sin ^{-1}(1-x)=cos ^{-1} x,the | vedclass

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

Question

(C) Given equation: $\sin ^{-1} x + \sin ^{-1}(1-x) = \cos ^{-1} x$ Since $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$,we have: $\sin ^{-1}(1-x) = \f

Vedclass · Accepted Answer

$\{0, \frac{1}{2}\}$

Vedclass · Answer

(C) Given equation: $\sin ^{-1} x + \sin ^{-1}(1-x) = \cos ^{-1} x$ Since $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$,we have: $\sin ^{-1}(1-x) = \frac{\pi}{2} - \sin ^{-1} x - \sin ^{-1} x = \frac{\pi}{2} - 2\sin ^{-1} x$ Taking $\sin$ on both sides: $1-x = \sin(\frac{\pi}{2} - 2\sin ^{-1} x) = \cos(2\sin ^{-1} x)$ Using the identity $\cos(2	heta) = 1 - 2\sin^2 	heta$,where $	heta = \sin ^{-1} x$: $1-x = 1 - 2(\sin(\sin ^{-1} x))^2$ $1-x = 1 - 2x^2$ $2x^2 - x = 0$ $x(2x - 1) = 0$ Thus,$x = 0$ or $x = \frac{1}{2}$. Checking the values: For $x=0$: $\sin ^{-1} 0 + \sin ^{-1} 1 = 0 + \frac{\pi}{2} = \frac{\pi}{2}$ and $\cos ^{-1} 0 = \frac{\pi}{2}$. (Valid) For $x=\frac{1}{2}$: $\sin ^{-1} \frac{1}{2} + \sin ^{-1} \frac{1}{2} = \frac{\pi}{6} + \frac{\pi}{6} = \frac{\pi}{3}$ and $\cos ^{-1} \frac{1}{2} = \frac{\pi}{3}$. (Valid) Therefore,$x \in \{0, \frac{1}{2}\}$.

If $\sin ^{-1} x+\sin ^{-1}(1-x)=\cos ^{-1} x$,then $x \in$

Similar Questions

How do you expect the metallic hydrides to be useful for hydrogen storage? Explain.

Select the correct arrangement of neural organization,according to the increasing degree of complexity.

The melting point of $Zn$ is lower as compared to those of the other elements of the $3d$ series because:

The amino acid which is not optically active is

In thermodynamics,which one of the following properties is not an intensive property?

Vedclass Test Series

Exam Paper Generator

Online Exam Module